Introduction
Four-dimensional S U ( N ) SU(N) S U ( N ) R 2 × T 2 \mathbb{R}^{2} \times T^{2} R 2 × T 2 [1 , 2 , 3 , 4 , 5 , 6 ] (see also the preceding relevant works [7 , 8 , 9 , 10 , 11 ] for the similar story on R × T 3 \mathbb{R} \times T^{3} R × T 3 [12 , 13 , 14 ] (see also [15 , 16 , 17 , 18 , 19 ] ), and the analytic structure is not yet fully understood except for the Abelian-like solution for the twisted T 4 T^{4} T 4 [20 ] .
In previous works by the present authors [21 , 22 , 23 ] , we obtained hints toward an analytic understanding by considering the hierarchy of periods for T t w i s t e d 2 T^{2}_{\mathrm{twisted}} T twisted 2 ℓ 3 ≫ N ℓ 4 \ell_{3} \gg N\ell_{4} ℓ 3 ≫ N ℓ 4 N ℓ 4 N\ell_4 N ℓ 4 [24 , 25 , 26 , 27 , 28 , 29 ] gives the N N N [21 , 22 , 23 ] , we have found that the alternating infinite chain of BPS and KK monopole constituents separated by ℓ 3 \ell_3 ℓ 3 R 2 × T 2 \mathbb{R}^2 \times T^2 R 2 × T 2
An important open question is whether the monopole-center vortex chain actually comes out of the fully non-Abelian self-dual solution (see Ref. [30] for its numerical support), and it naturally motivates us to look back on how the KvBLLY caloron on R 3 × S 1 \mathbb{R}^3 \times S^1 R 3 × S 1 [31 ] for the periodic instanton. The analysis by K.-M. Lee, C.-h. Lu, and P. Yi [24 , 25 , 26 ] relies on its D-brane interpretation: Starting from N N N N N N N N N [32 ] but with the impurity to derive the caloron solution with nontrivial asymptotic holonomies.
The papers by Kraan and van Baal [27 , 28 , 29 ] instead use the ADHM construction more directly for constructing the periodic instanton: They started from the ADHM data for the charge-k k k R 4 \mathbb{R}^4 R 4 k → ∞ k \to\infty k → ∞ [33 , 34 ] .
In this work, we extend the Bruckmann-Kraan-van Baal method to construct the doubly-periodic instanton starting from the ADHM data. When taking the formal k → ∞ k \to\infty k → ∞ k k k R 4 \mathbb{R}^4 R 4 ∞ \infty ∞ Z 2 \mathbb{Z}^2 Z 2 T 2 T^2 T 2
Brief review for the fractional instanton on R 2 × ( T 2 ) p / N -twisted \mathbb{R}^{2} \times(T^{2})_{p/N\text{-twisted}} R 2 × ( T 2 ) p / N -twisted
In this section, we give a brief review on the self-dual fractional instanton on R 2 × T 2 \mathbb{R}^{2} \times T^{2} R 2 × T 2 [21 , 22 , 23 ] , and we shall see how the fractional instanton is realized as the self-dual monopole-center vortex chain.
Setup of the problem
We consider the self-dual Yang-Mills equation on R 2 × T 2 \mathbb{R}^{2} \times T^{2} R 2 × T 2 p / N p/N p / N gcd ( N , p ) = 1 \operatorname{gcd}(N,p)=1 gcd ( N , p ) = 1 x = ( x , x 3 , x 4 ) ∈ R 2 × T 2 x=(x,x_{3},x_{4}) \in\mathbb{R}^{2} \times T^{2} x = ( x , x 3 , x 4 ) ∈ R 2 × T 2
x 3 ∼ x 3 + ℓ 3 , x 4 ∼ x 4 + ℓ 4 . (1) x_{3} \sim x_{3}+\ell_{3}, \qquad x_{4} \sim x_{4}+\ell_{4}. \tag*{(1)} x 3 ∼ x 3 + ℓ 3 , x 4 ∼ x 4 + ℓ 4 . (1) Let us introduce the S U ( N ) SU(N) S U ( N ) C , S C,S C , S S C = ω C S SC=\omega CS S C = ω C S
( C ) n m = e − π i ( N − 1 ) N e 2 π i N n δ n m , ( S ) n m = e − π i ( N − 1 ) N δ n + 1 , m . (2) (C)_{nm}=e^{-\frac{\pi i(N-1)}{N}}e^{\frac{2\pi i}{N}n}\delta_{nm}, \qquad(S)_{nm}=e^{-\frac{\pi i(N-1)}{N}}\delta_{n+1,m}. \tag*{(2)} ( C ) nm = e − N π i ( N − 1 ) e N 2 π i n δ nm , ( S ) nm = e − N π i ( N − 1 ) δ n + 1 , m . (2) Here, the S U ( N ) SU(N) S U ( N ) n , m n,m n , m N N N U ( 1 ) U(1) U ( 1 ) e − π i ( N − 1 ) N e^{-\frac{\pi i(N-1)}{N}} e − N π i ( N − 1 ) C , S ∈ S U ( N ) C,S \in SU(N) C , S ∈ S U ( N ) N N N
Throughout this paper, the gauge field A = A μ d x μ A=A_{\mu}dx^{\mu} A = A μ d x μ D = d + i A D=d+iA D = d + i A F = − i D ∧ D = d A + i A ∧ A F=-iD \wedge D=dA+iA \wedge A F = − i D ∧ D = d A + i A ∧ A P exp ( − i ∮ A ) P\exp(-i\oint A) P exp ( − i ∮ A ) R 2 × T 2 \mathbb{R}^{2} \times T^{2} R 2 × T 2 p / N p/N p / N P a = P exp ( − i ∫ x x + ℓ a e ^ a A ) P_{a}=P\exp\left(-i\int_{x}^{x+\ell_{a}\hat{e}_{a}} A\right) P a = P exp ( − i ∫ x x + ℓ a e ^ a A )
P 3 = S − p , P 4 = C . (3) P_{3}=S^{-p}, \quad P_{4}=C. \tag*{(3)} P 3 = S − p , P 4 = C . (3) With S C = ω C S SC=\omega CS S C = ω C S P 3 P 4 = ω − p P 4 P 3 P_{3}P_{4}=\omega^{-p}P_{4}P_{3} P 3 P 4 = ω − p P 4 P 3 S U ( N ) SU(N) S U ( N ) p / N p/N p / N [1 Appendix A] ). It is then convenient to take the algebraic gauge, where the gauge field (and also any adjoint fields) obeys
A μ ( x + ℓ a e ^ a ) = P a A μ ( x ) P a − 1 ( a = 3 , 4 ) . (4) A_{\mu}(x+\ell_{a}\hat{e}_{a})=P_{a}A_{\mu}(x)P_{a}^{-1} \qquad(a=3,4). \tag*{(4)} A μ ( x + ℓ a e ^ a ) = P a A μ ( x ) P a − 1 ( a = 3 , 4 ) . (4) where e ^ a \hat{e}_{a} e ^ a a a a A μ = 3 , 4 → 0 A_{\mu=3,4} \to0 A μ = 3 , 4 → 0 ∣ x ∣ → ∞ |x| \to\infty ∣ x ∣ → ∞
m ( N k 3 + r 3 , N k 4 + r 4 ) 2 = ( 2 π N ) 2 { ( N k 3 + r 3 ) 2 ℓ 3 2 + ( N k 4 + r 4 ) 2 ℓ 4 2 } (5) m^{2}_{(Nk_{3}+r_{3},Nk_{4}+r_{4})}=\left(\frac{2\pi}{N}\right)^{2}\left\{\frac{(Nk_{3}+r_{3})^{2}}{\ell_{3}^{2}}+\frac{(Nk_{4}+r_{4})^{2}}{\ell_{4}^{2}}\right\} \tag*{(5)} m ( N k 3 + r 3 , N k 4 + r 4 ) 2 = ( N 2 π ) 2 { ℓ 3 2 ( N k 3 + r 3 ) 2 + ℓ 4 2 ( N k 4 + r 4 ) 2 } (5) with ( r 3 , r 4 ) ≠ ( 0 , 0 ) (r_{3},r_{4}) \ne(0,0) ( r 3 , r 4 ) = ( 0 , 0 ) S U ( N ) SU(N) S U ( N ) ∣ x ∣ → ∞ |x| \to\infty ∣ x ∣ → ∞ R 2 \mathbb{R}^{2} R 2
We then consider the possible minimal-action configuration as the topological soliton for this setup. Since the gauge field A μ A_\mu A μ ∣ x ∣ → ∞ \lvert x\rvert\to\infty ∣ x ∣ → ∞ R 2 \mathbb{R}^2 R 2 T 2 T^2 T 2
R 2 × T 2 ⏟ p / N -twist ⇒ T 2 ⏟ − q / N -twist × T 2 ⏟ p / N -twist . (6) \mathbb{R}^2 \times\underbrace{T^2}_{p/N\text{-twist}} \Rightarrow\underbrace{T^2}_{-q/N\text{-twist}} \times\underbrace{T^2}_{p/N\text{-twist}} . \tag*{(6)} R 2 × p / N -twist T 2 ⇒ − q / N -twist T 2 × p / N -twist T 2 . (6) For such a twisted boundary condition, the topological charge Q t o p = 1 8 π 2 ∫ tr ( F ∧ F ) Q_{\mathrm{top}} = \frac{1}{8\pi^2} \int\operatorname{tr}(F \wedge F) Q top = 8 π 2 1 ∫ tr ( F ∧ F ) [43 ]
Q t o p ∈ p q N + Z , (7) Q_{\mathrm{top}} \in\frac{pq}{N} + \mathbb{Z}, \tag*{(7)} Q top ∈ N pq + Z , (7) which has the 1 / N 1/N 1/ N S = 1 g 2 ∫ tr ( F ∧ ⋆ F ) S = \frac{1}{g^2} \int\operatorname{tr}(F \wedge\star F) S = g 2 1 ∫ tr ( F ∧ ⋆ F ) S Y M ≥ 8 π 2 g 2 ∣ Q t o p ∣ S_{\mathrm{YM}} \ge\frac{8\pi^2}{g^2} \lvert Q_{\mathrm{top}}\rvert S YM ≥ g 2 8 π 2 ∣ Q top ∣ q q q
q p ≡ ± 1 m o d N , (8) qp \equiv\pm1 \bmod N, \tag*{(8)} q p ≡ ± 1 mod N , (8) so that
Q t o p = ± 1 N , S Y M = 8 π 2 N g 2 . (9) Q_{\mathrm{top}} = \pm\frac{1}{N}, \qquad S_{\mathrm{YM}} = \frac{8\pi^2}{Ng^2}. \tag*{(9)} Q top = ± N 1 , S YM = N g 2 8 π 2 . (9) We note that such a choice of q q q gcd ( p , N ) = 1 \gcd(p,N)=1 g cd( p , N ) = 1 q q q W ( C ) W(C) W ( C ) C ⊂ R 2 C \subset\mathbb{R}^2 C ⊂ R 2 W ( C ) → e 2 π i N q W ( C ) W(C) \to e^{\frac{2\pi i}{N}q}W(C) W ( C ) → e N 2 π i q W ( C ) [44 , 45 , 46 , 47 , 48 , 49 , 50 , 51 , 52 , 53 , 54 ] , and the problem here is whether it is realized as the (anti-)self-dual Yang-Mills solution, F = ± ⋆ F F = \pm\star F F = ± ⋆ F R 2 × T 2 \mathbb{R}^2 \times T^2 R 2 × T 2
Monopole-center vortex chain for ℓ 3 ≫ N ℓ 4 \ell_3 \gg N\ell_4 ℓ 3 ≫ N ℓ 4
Solving the self-dual Yang-Mills equation, F = ± ⋆ F F = \pm\star F F = ± ⋆ F R 2 × T 2 \mathbb{R}^2 \times T^2 R 2 × T 2 [12 , 13 , 14 , 15 , 16 ] , and they confirmed that the self-duality bound is achieved by taking the continuum limit within a numerical error. Thus, there is little doubt on the presence of the self-dual center-vortex configuration itself, but we still would like to understand its analytic structures.
On the other hand, the self-dual solutions on R 3 × S 1 \mathbb{R}^3 \times S^1 R 3 × S 1 LLY caloron [24 , 25 , 26 , 27 , 28 , 29 ] , so it is natural to wonder if it helps the understanding for the self-dual center-vortex instanton. Following Refs. [21 , 22 , 23 ] , let us introduce the hierarchy for the two sizes of the torus:
ℓ 3 ≫ N ℓ 4 . (10) \ell_3 \gg N\ell_4. \tag*{(10)} ℓ 3 ≫ N ℓ 4 . (10) As we have chosen the gauge P 4 = C P_4 = C P 4 = C 2 π N ℓ 4 \frac{2\pi}{N\ell_4} N ℓ 4 2 π ≫ N ℓ 4 \gg N\ell_4 ≫ N ℓ 4 A μ A_\mu A μ [24 , 25 , 26 , 27 , 28 , 29 ] , whose the magnetic charges are specified by the Affine simple roots, α ⃗ n = ν ⃗ n − ν ⃗ n + 1 \vec{\alpha}_n = \vec{\nu}_n - \vec{\nu}_{n+1} α n = ν n − ν n + 1 n = 1 , 2 , … , N n = 1, 2, \ldots, N n = 1 , 2 , … , N ν ⃗ n \vec{\nu}_n ν n n = 1 , … , N n = 1, \ldots, N n = 1 , … , N S U ( N ) SU(N) S U ( N ) P 4 = C P_4 = C P 4 = C Q t o p = 1 / N Q_{\mathrm{top}} = 1/N Q top = 1/ N
As we now consider the case ℓ 3 ≫ N ℓ 4 \ell_3 \gg N\ell_4 ℓ 3 ≫ N ℓ 4 R 2 × ( S 1 ) ℓ 3 \mathbb{R}^2 \times(S^1)_{\ell_3} R 2 × ( S 1 ) ℓ 3 α ⃗ n \vec{\alpha}_n α n ( x , x 3 ) = ( 0 , 0 ) (x,x_3) = (0,0) ( x , x 3 ) = ( 0 , 0 )
∇ 2 A 4 A b e l ( x , x 3 ) = − 2 π α ⃗ n δ ( 2 ) ( x ) δ ( x 3 ) , (11) \nabla^2 A_4^{\mathrm{Abel}}(x,x_3) = -2\pi\vec{\alpha}_n\delta^{(2)}(x)\delta(x_3), \tag*{(11)} ∇ 2 A 4 Abel ( x , x 3 ) = − 2 π α n δ ( 2 ) ( x ) δ ( x 3 ) , (11) where A 4 A b e l = ( [ A 4 ] m m ) m = 1 , … , N A_4^{\mathrm{Abel}} = ([A_4]_{mm})_{m=1,\ldots,N} A 4 Abel = ([ A 4 ] mm ) m = 1 , … , N A μ A_\mu A μ A μ = 1 , 2 , 3 A_{\mu=1,2,3} A μ = 1 , 2 , 3 [21 , 22 , 23 ] for the full discussion). This equation can be solved by the mirror-image method with paying attention to the boundary condition (4) , A 4 ( x , x 3 + ℓ 3 ) = S − p A 4 ( x , x 3 ) S p A_4(x,x_3+\ell_3) = S^{-p}A_4(x,x_3)S^p A 4 ( x , x 3 + ℓ 3 ) = S − p A 4 ( x , x 3 ) S p [ A 4 ] m m ( x , x 3 + ℓ 3 ) = [ A 4 ] m − p , m − p ( x , x 3 ) [A_4]_{mm}(x,x_3+\ell_3) = [A_4]_{m-p,m-p}(x,x_3) [ A 4 ] mm ( x , x 3 + ℓ 3 ) = [ A 4 ] m − p , m − p ( x , x 3 ) α ⃗ n \vec{\alpha}_n α n x 3 = k ℓ 3 x_3=k\ell_3 x 3 = k ℓ 3 α ⃗ n + p k \vec{\alpha}_{n+pk} α n + p k x 3 ∈ R x_3 \in\mathbb{R} x 3 ∈ R
∇ 2 A 4 A b e l ( x , x 3 ) = − ∑ k 2 π α ⃗ n + p k δ ( 2 ) ( x ) δ ( x 3 − k ℓ 3 ) , (12) \nabla^2 A_4^{\mathrm{Abel}}(x,x_3) = -\sum_k 2\pi\vec{\alpha}_{n+pk}\delta^{(2)}(x)\delta(x_3-k\ell_3), \tag*{(12)} ∇ 2 A 4 Abel ( x , x 3 ) = − k ∑ 2 π α n + p k δ ( 2 ) ( x ) δ ( x 3 − k ℓ 3 ) , (12) which can be solved as (note that A 4 → 0 A_4 \to0 A 4 → 0 ∣ x ∣ → ∞ |\boldsymbol{x}| \to\infty ∣ x ∣ → ∞ 1
A 4 A b e l = 1 2 ∑ k ∈ Z w ⃗ n + p k ( q ) ( 1 ∣ ( x , x 3 − k ℓ 3 ) ∣ − 1 ∣ ( x , x 3 − ( k − 1 ) ℓ 3 ) ∣ ) = 2 N ℓ 3 ∑ m = 0 N − 1 w ⃗ n + p m ( q ) ∑ k ′ = 1 ∞ K 0 ( 2 π N ℓ 3 k ′ ∣ x ∣ ) × { cos 2 π k ′ ( x 3 − m ℓ 3 ) N ℓ 3 − cos 2 π k ′ ( x 3 − ( m − 1 ) ℓ 3 ) N ℓ 3 } , (13) A_4^{\mathrm{Abel}} = \frac{1}{2}\sum_{k\in\mathbb{Z}}\vec{w}_{n+pk}^{(q)}\left(\frac{1}{\left|\left(\boldsymbol{x},x_3-k\ell_3\right)\right|}-\frac{1}{\left|\left(\boldsymbol{x},x_3-(k-1)\ell_3\right)\right|}\right)
= \frac{2}{N\ell_3}\sum_{m=0}^{N-1}\vec{w}_{n+pm}^{(q)}\sum_{k'=1}^{\infty}K_0\left(\frac{2\pi}{N\ell_3}k'|\boldsymbol{x}|\right)
\times\left\{\cos\frac{2\pi k'(x_3-m\ell_3)}{N\ell_3}-\cos\frac{2\pi k'(x_3-(m-1)\ell_3)}{N\ell_3}\right\}, \tag*{(13)} A 4 Abel = 2 1 k ∈ Z ∑ w n + p k ( q ) ( ∣ ( x , x 3 − k ℓ 3 ) ∣ 1 − ∣ ( x , x 3 − ( k − 1 ) ℓ 3 ) ∣ 1 ) = N ℓ 3 2 m = 0 ∑ N − 1 w n + p m ( q ) k ′ = 1 ∑ ∞ K 0 ( N ℓ 3 2 π k ′ ∣ x ∣ ) × { cos N ℓ 3 2 π k ′ ( x 3 − m ℓ 3 ) − cos N ℓ 3 2 π k ′ ( x 3 − ( m − 1 ) ℓ 3 ) } , (13) where
w ⃗ n ( q ) = ν ⃗ n + ν ⃗ n + p + ⋯ + ν ⃗ n + ( q − 1 ) p . (14) \vec{w}_n^{(q)}=\vec{\nu}_n+\vec{\nu}_{n+p}+\cdots+\vec{\nu}_{n+(q-1)p}. \tag*{(14)} w n ( q ) = ν n + ν n + p + ⋯ + ν n + ( q − 1 ) p . (14) Here we take q p ≡ + 1 m o d N qp\equiv+1\ \mathrm{mod}\ N q p ≡ + 1 mod N w ⃗ n + p k ( q ) − w ⃗ n + p ( k + 1 ) ( q ) = ν ⃗ n + p k − ν ⃗ n + p k + 1 = α ⃗ n + p k \vec{w}_{n+pk}^{(q)}-\vec{w}_{n+p(k+1)}^{(q)}=\vec{\nu}_{n+pk}-\vec{\nu}_{n+pk+1}=\vec{\alpha}_{n+pk} w n + p k ( q ) − w n + p ( k + 1 ) ( q ) = ν n + p k − ν n + p k + 1 = α n + p k q p = − 1 qp=-1 q p = − 1 p → − p p\to-p p → − p O ( N ℓ 3 ) O(N\ell_3) O ( N ℓ 3 ) Figure 1 .
Figure 1. Monopole–center vortex chain along x 3 x_3 x 3 w ⃗ n + p k ( q ) \vec{w}_{n+pk}^{(q)} w n + p k ( q ) α ⃗ n + p ( k − 1 ) \vec{\alpha}_{n+p(k-1)} α n + p ( k − 1 ) α ⃗ n + p k \vec{\alpha}_{n+pk} α n + p k k ∈ Z k \in\mathbb{Z} k ∈ Z N ℓ 4 N\ell_4 N ℓ 4 ≪ ℓ 3 \ll\ell_3 ≪ ℓ 3 O ( N ℓ 3 ) O(N\ell_3) O ( N ℓ 3 ) e 2 π i q / N e^{2\pi iq/N} e 2 π i q / N
By cutting out the segment 0 ≤ x 3 < ℓ 3 0\leq x_3<\ell_3 0 ≤ x 3 < ℓ 3 S Y M = 8 π 2 N g 2 S_{\mathrm{YM}}=\frac{8\pi^2}{Ng^2} S YM = N g 2 8 π 2 Q t o p = 1 / N Q_{\mathrm{top}}=1/N Q top = 1/ N w ⃗ n ( q ) \vec{w}_n^{(q)} w n ( q ) q q q e 2 π i q / N e^{2\pi iq/N} e 2 π i q / N
According to Figure 1 , the monopole serves as the junction of the center vortex to change its color-magnetic direction. Even beyond the weakly coupled regime, such junctions in the center-vortex model play the important role for the confinement dynamics [55 , 56 , 57 , 58 , 59 , 60 , 61 ] : For example, a nonzero instanton density requires non-oriented center-vortex worldsheets, i.e., monopole junctions [62 , 63 , 64 ] . Moreover, the importance of the monopole junctions for the mass gap is recently uncovered in the numerical simulation of the lattice Z N \mathbb{Z}_N Z N [65 , 66 ] , and then it has been proven that the center-vortex proliferation without monopoles necessarily leads to a gapless phase, such as the Coulomb phase [67 ] .
Doubly-periodic instanton in the far-field limit
We now consider the torus with the extended period, x 3 ∼ x 3 + L 3 x_3\sim x_3+L_3 x 3 ∼ x 3 + L 3 x 4 ∼ x 4 + L 4 x_4\sim x_4+L_4 x 4 ∼ x 4 + L 4
L 3 = N ℓ 3 , L 4 = ℓ 4 , (15) L_3=N\ell_3,\qquad L_4=\ell_4, \tag*{(15)} L 3 = N ℓ 3 , L 4 = ℓ 4 , (15) where the ’t Hooft twist is untwisted: P 3 P 4 = P 4 P 3 P_3P_4=P_4P_3 P 3 P 4 = P 4 P 3 L 4 L_4 L 4 P 4 P_4 P 4 S U ( N ) → H i g g s U ( 1 ) N − 1 SU(N)\xrightarrow{\mathrm{Higgs}}U(1)^{N-1} S U ( N ) Higgs U ( 1 ) N − 1
Let us discuss the far-field behavior for the self-dual gauge field with Q t o p = 1 Q_{\mathrm{top}}=1 Q top = 1 [24 , 25 , 26 , 27 , 28 , 29 ] . We write the location of n n n ( y n , y 3 , n ) ∈ R 2 × S 1 (\boldsymbol{y}_n,y_{3,n})\in\mathbb{R}^2\times S^1 ( y n , y 3 , n ) ∈ R 2 × S 1 n = 1 , … , N n = 1, \ldots, N n = 1 , … , N
∇ 2 A 4 A b e l ( x , x 3 ) = − ∑ n = 1 N 2 π α ⃗ n δ ( 2 ) ( x − y n ) δ ( x 3 − y 3 , n ) (16) \nabla^2 A^{\mathrm{Abel}}_4(x,x_3)=-\sum_{n=1}^{N}2\pi\vec{\alpha}_n\delta^{(2)}(x-y_n)\delta(x_3-y_{3,n}) \tag*{(16)} ∇ 2 A 4 Abel ( x , x 3 ) = − n = 1 ∑ N 2 π α n δ ( 2 ) ( x − y n ) δ ( x 3 − y 3 , n ) (16) with x 3 ∈ R / ( L 3 Z ) x_3 \in\mathbb{R}/(L_3\mathbb{Z}) x 3 ∈ R / ( L 3 Z ) m m m
ν ⃗ m ⋅ A 4 A b e l ( x , x 3 ) = 1 2 ∑ k ∈ Z { 1 ∣ ( x − y m , x 3 − y 3 , m − k L 3 ) ∣ − 1 ∣ ( x − y m − 1 , x 3 − y 3 , m − 1 − k L 3 ) ∣ } = 1 L 3 ln ∣ x − y m − 1 ∣ ∣ x − y m ∣ + 2 L 3 ∑ k ′ = 1 ∞ [ K 0 ( 2 π k ′ ∣ x − y m ∣ L 3 ) cos 2 π k ′ ( x 3 − y 3 , m ) L 3 − K 0 ( 2 π k ′ ∣ x − y m − 1 ∣ L 3 ) cos 2 π k ′ ( x 3 − y 3 , m − 1 ) L 3 ] . (17) \begin{aligned}
\vec{\nu}_m\cdot A^{\mathrm{Abel}}_4(x,x_3)
&=\frac{1}{2}\sum_{k\in\mathbb{Z}}\left\{\frac{1}{\lvert(x-y_m,x_3-y_{3,m}-kL_3)\rvert}-\frac{1}{\lvert(x-y_{m-1},x_3-y_{3,m-1}-kL_3)\rvert}\right\}\\
&=\frac{1}{L_3}\ln\frac{\lvert x-y_{m-1}\rvert}{\lvert x-y_m\rvert}+\frac{2}{L_3}\sum_{k'=1}^{\infty}\left[K_0\left(\frac{2\pi k'\lvert x-y_m\rvert}{L_3}\right)\cos\frac{2\pi k'(x_3-y_{3,m})}{L_3}\right.\\
&\qquad\left.-K_0\left(\frac{2\pi k'\lvert x-y_{m-1}\rvert}{L_3}\right)\cos\frac{2\pi k'(x_3-y_{3,m-1})}{L_3}\right]. \tag*{(17)}
\end{aligned} ν m ⋅ A 4 Abel ( x , x 3 ) = 2 1 k ∈ Z ∑ { ∣( x − y m , x 3 − y 3 , m − k L 3 )∣ 1 − ∣( x − y m − 1 , x 3 − y 3 , m − 1 − k L 3 )∣ 1 } = L 3 1 ln ∣ x − y m ∣ ∣ x − y m − 1 ∣ + L 3 2 k ′ = 1 ∑ ∞ [ K 0 ( L 3 2 π k ′ ∣ x − y m ∣ ) cos L 3 2 π k ′ ( x 3 − y 3 , m ) − K 0 ( L 3 2 π k ′ ∣ x − y m − 1 ∣ ) cos L 3 2 π k ′ ( x 3 − y 3 , m − 1 ) ] . (17) Only when y m = y m − 1 y_m=y_{m-1} y m = y m − 1 exp ( − 2 π ∣ x ∣ / L 3 ) \exp(-2\pi\lvert x\rvert/L_3) exp ( − 2 π ∣ x ∣ / L 3 ) ∣ x ∣ → ∞ \lvert x\rvert\to\infty ∣ x ∣ → ∞ y m ≠ y m − 1 y_m\ne y_{m-1} y m = y m − 1 m m m O ( 1 / ∣ x ∣ 2 ) O(1/\lvert x\rvert^2) O ( 1/ ∣ x ∣ 2 )
ADHM construction for the doubly-periodic instanton
We have seen the hint that the self-dual fractional instanton on R 2 × T p / N -twisted 2 \mathbb{R}^2\times T^2_{p/N\text{-twisted}} R 2 × T p / N -twisted 2 [21 , 22 , 23 ] is totally limited to the far-field Abelian components. Can one really show that it appears as the far-field behavior of the fully non-Abelian self-dual configuration?
In this section, we apply the ADHM technique for the charge-1 instanton on R 2 × T untwisted 2 \mathbb{R}^2\times T^2_{\text{untwisted}} R 2 × T untwisted 2 N N N T ^ 2 \hat{T}^2 T ^ 2 [39 , 40 , 41 ] . By extending the work by Bruckmann and van Baal [33 , 34 ] on multi-caloron solutions to the current setup, we derive the alternating monopole-center vortex chain from the ADHM data explicitly in the next section. This section serves as the theoretical foundation for it: We discuss the regularization scheme associated with the 2d delta-function impurities in the Abelian Hitchin system, and justify the way to reconstruct the gauge field of the doubly-periodic instanton with using the Green’s function.
ADHM data for the doubly-periodic instanton
In this subsection, we explain how the ADHM technique [31 ] can be applied for constructing the doubly-periodic instanton. For the review on ADHM, see [68 ] .
Let us start from the ADHM data for k k k S U ( N ) SU(N) S U ( N ) R 4 ∋ ( x μ ) μ = 1 , … , 4 \mathbb{R}^4\ni(x_\mu)_{\mu=1,\ldots,4} R 4 ∋ ( x μ ) μ = 1 , … , 4 τ j = 1 , 2 , 3 \tau_{j=1,2,3} τ j = 1 , 2 , 3
σ μ = ( i τ j , 1 ) , σ ˉ μ = ( − i τ j , 1 ) , (18) \sigma_\mu=(i\tau_j,\mathbb{1}),\qquad\bar{\sigma}_\mu=(-i\tau_j,\mathbb{1}), \tag*{(18)} σ μ = ( i τ j , 1 ) , σ ˉ μ = ( − i τ j , 1 ) , (18) so that σ 4 = σ 0 = 1 2 \sigma_4=\sigma_0=\mathbb{1}_2 σ 4 = σ 0 = 1 2 σ 1 , 2 , 3 \sigma_{1,2,3} σ 1 , 2 , 3
We define B ( x ) : = B − x = σ μ ( B μ − x μ 1 k ) B(x) := B - x = \sigma_\mu(B_\mu- x_\mu1_k) B ( x ) := B − x = σ μ ( B μ − x μ 1 k ) ( N + 2 k ) × 2 k (N + 2k) \times2k ( N + 2 k ) × 2 k
Δ ( x ) : = ( λ B ( x ) ) . (19) \Delta(x) := \begin{pmatrix} \lambda\\ B(x) \end{pmatrix}. \tag*{(19)} Δ ( x ) := ( λ B ( x ) ) . (19) The self-duality equation on the ADHM side can be expressed as
Δ ( x ) † Δ ( x ) = λ † λ + B ( x ) † B ( x ) = ! σ 0 ⊗ f x − 1 , (20) \Delta(x)^\dagger\Delta(x) = \lambda^\dagger\lambda+ B(x)^\dagger B(x) \stackrel{!}{=} \sigma_0 \otimes f_x^{-1}, \tag*{(20)} Δ ( x ) † Δ ( x ) = λ † λ + B ( x ) † B ( x ) = ! σ 0 ⊗ f x − 1 , (20) where f x f_x f x k × k k \times k k × k (20) should be understood as three self-duality equations, 1 2 tr 2 ( σ j [ λ † λ + B ( x ) † B ( x ) ] ) = 0 \frac{1}{2}\operatorname{tr}_2(\sigma_j[\lambda^\dagger\lambda+ B(x)^\dagger B(x)]) = 0 2 1 tr 2 ( σ j [ λ † λ + B ( x ) † B ( x )]) = 0 j = 1 , 2 , 3 j = 1, 2, 3 j = 1 , 2 , 3 1 2 tr 2 ( σ 0 [ λ † λ + B ( x ) † B ( x ) ] ) \frac{1}{2}\operatorname{tr}_2(\sigma_0[\lambda^\dagger\lambda+ B(x)^\dagger B(x)]) 2 1 tr 2 ( σ 0 [ λ † λ + B ( x ) † B ( x )]) tr 2 \operatorname{tr}_2 tr 2
The gauge field A μ ( x ) A_\mu(x) A μ ( x ) N N N v v v
Δ ( x ) † v = 0 , (21) \Delta(x)^\dagger v = 0, \tag*{(21)} Δ ( x ) † v = 0 , (21) where v v v ( N + 2 k ) × N (N + 2k) \times N ( N + 2 k ) × N A μ A_\mu A μ
A μ = − i v † ∂ μ v . (22) A_\mu= -iv^\dagger\partial_\mu v. \tag*{(22)} A μ = − i v † ∂ μ v . (22) Using the Green function f x f_x f x A μ A_\mu A μ [33 ]
A μ ( x ) = − i 2 φ ( x ) 1 / 2 [ ∂ ν ( λ [ η ˉ μ ν ⊗ f x ] λ † ) ] φ ( x ) − 1 / 2 − i 2 [ φ ( x ) − 1 / 2 , ∂ μ φ ( x ) 1 / 2 ] , (23) A_\mu(x) = -\frac{i}{2}\varphi(x)^{1/2}[\partial_\nu(\lambda[\bar{\eta}_{\mu\nu} \otimes f_x]\lambda^\dagger)]\varphi(x)^{-1/2} - \frac{i}{2}[\varphi(x)^{-1/2},\partial_\mu\varphi(x)^{1/2}], \tag*{(23)} A μ ( x ) = − 2 i φ ( x ) 1/2 [ ∂ ν ( λ [ η ˉ μν ⊗ f x ] λ † )] φ ( x ) − 1/2 − 2 i [ φ ( x ) − 1/2 , ∂ μ φ ( x ) 1/2 ] , (23) where η ˉ μ ν = 1 2 σ ˉ [ μ σ ν ] \bar{\eta}_{\mu\nu} = \frac{1}{2}\bar{\sigma}_{[\mu}\sigma_{\nu]} η ˉ μν = 2 1 σ ˉ [ μ σ ν ]
φ ( x ) − 1 = [ 1 N + λ ( B ( x ) † B ( x ) ) − 1 λ † ] − 1 = 1 N − λ [ σ 0 ⊗ f x ] λ † . (24) \varphi(x)^{-1} = \left[1_N + \lambda(B(x)^\dagger B(x))^{-1}\lambda^\dagger\right]^{-1} = 1_N - \lambda[\sigma_0 \otimes f_x]\lambda^\dagger. \tag*{(24)} φ ( x ) − 1 = [ 1 N + λ ( B ( x ) † B ( x ) ) − 1 λ † ] − 1 = 1 N − λ [ σ 0 ⊗ f x ] λ † . (24) In our convention, A μ ( x ) A_\mu(x) A μ ( x )
Next, let us bring this ADHM method for the doubly-periodic instanton, charge-1 instanton on R 2 × T 2 \mathbb{R}^2 \times T^2 R 2 × T 2
x 3 ∼ x 3 + L 3 , x 4 ∼ x 4 + L 4 , (25) x_3 \sim x_3 + L_3,\qquad x_4 \sim x_4 + L_4, \tag*{(25)} x 3 ∼ x 3 + L 3 , x 4 ∼ x 4 + L 4 , (25) and we distinguish it from the ones ℓ 3 , 4 \ell_{3,4} ℓ 3 , 4 P 3 P_3 P 3 P 4 P_4 P 4
i F = d v † ∧ d v + ( v † d v ) ∧ ( v † d v ) = d v † ∧ ( 1 − v v † ) d v = ( d v † ) Δ ∧ f x Δ † d v = v † ( d Δ ) ∧ f x ( d Δ † ) v . \begin{aligned}
iF = dv^\dagger\wedge dv + (v^\dagger dv) \wedge(v^\dagger dv) = dv^\dagger\wedge(1 - vv^\dagger)dv \\
&= (dv^\dagger)\Delta\wedge f_x\Delta^\dagger dv = v^\dagger(d\Delta) \wedge f_x(d\Delta^\dagger)v.
\end{aligned} i F = d v † ∧ d v + ( v † d v ) ∧ ( v † d v ) = d v † ∧ ( 1 − v v † ) d v = ( d v † ) Δ ∧ f x Δ † d v = v † ( d Δ ) ∧ f x ( d Δ † ) v . Since d Δ ( x ) = − σ μ d x μ d\Delta(x) = -\sigma_\mu dx_\mu d Δ ( x ) = − σ μ d x μ
F μ ν = − 2 i v † η μ ν f x v , F_{\mu\nu} = -2iv^\dagger\eta_{\mu\nu}f_xv, F μν = − 2 i v † η μν f x v , with the self-dual ’t Hooft eta symbol, η μ ν = 1 2 σ [ μ σ ˉ ν ] \eta_{\mu\nu} = \frac{1}{2}\sigma_{[\mu}\bar{\sigma}_{\nu]} η μν = 2 1 σ [ μ σ ˉ ν ] P 3 P 4 = P 4 P 3 P_{3}P_{4}=P_{4}P_{3} P 3 P 4 = P 4 P 3 P 3 , 4 P_{3,4} P 3 , 4
P a = diag ( e i L a μ a , 1 , … , e i L a μ a , N ) = ∑ m = 1 N e i L a μ a , m Π m , (26) P_{a}=\operatorname{diag}\left(e^{iL_{a}\mu_{a,1}},\ldots,e^{iL_{a}\mu_{a,N}}\right)=\sum_{m=1}^{N}e^{iL_{a}\mu_{a,m}}\Pi_{m}, \tag*{(26)} P a = diag ( e i L a μ a , 1 , … , e i L a μ a , N ) = m = 1 ∑ N e i L a μ a , m Π m , (26) where Π m \Pi_{m} Π m m m m
A μ ( x + L a e ^ a ) = P a A μ ( x ) P a − 1 ( a = 3 , 4 ) . (27) A_{\mu}(x+L_{a}\hat{e}_{a})=P_{a}A_{\mu}(x)P_{a}^{-1}\qquad(a=3,4). \tag*{(27)} A μ ( x + L a e ^ a ) = P a A μ ( x ) P a − 1 ( a = 3 , 4 ) . (27) To study the fractional-instanton situation in Sec. 2, we can unfold the twisted torus by extending the periodicity for the 3rd direction; x 3 ∼ x 3 + N ℓ 3 x_{3}\sim x_{3}+N\ell_{3} x 3 ∼ x 3 + N ℓ 3 x 4 ∼ x 4 + ℓ 4 x_{4}\sim x_{4}+\ell_{4} x 4 ∼ x 4 + ℓ 4
L 3 = N ℓ 3 , L 4 = ℓ 4 , (28) L_{3}=N\ell_{3},\qquad L_{4}=\ell_{4}, \tag*{(28)} L 3 = N ℓ 3 , L 4 = ℓ 4 , (28) with specifying the holonomies as
P 3 = ( S − p ) N ∝ 1 N , P 4 = C . (29) P_{3}=(S^{-p})^{N}\propto\mathbb{1}_{N},\qquad P_{4}=C. \tag*{(29)} P 3 = ( S − p ) N ∝ 1 N , P 4 = C . (29) Center-symmetry transformation allows us to bring P 3 = 1 N P_{3}=\mathbb{1}_{N} P 3 = 1 N μ 3 , m = 0 \mu_{3,m}=0 μ 3 , m = 0 μ 4 , m = 2 π N L 4 m \mu_{4,m}=\frac{2\pi}{NL_{4}}m μ 4 , m = N L 4 2 π m m m m det P 4 = 1 \det P_{4}=1 det P 4 = 1 Q t o p = 1 Q_{\mathrm{top}}=1 Q top = 1 § 3.4 , this also requires a special choice of the remaining data ζ \zeta ζ
To impose the boundary condition (27) , we should send k → ∞ k\to\infty k → ∞ I = ( n 3 , n 4 ) ∈ Z × Z I=(n_{3},n_{4})\in\mathbb{Z}\times\mathbb{Z} I = ( n 3 , n 4 ) ∈ Z × Z [35 ] , which is the straightforward extension for the KvBLLY caloron [27 , 28 , 29 ] . Define the n 3 n_{3} n 3 n 4 n_{4} n 4 X ( n 3 , n 4 ) X_{(n_{3},n_{4})} X ( n 3 , n 4 )
( S 3 X ) ( n 3 , n 4 ) = X ( n 3 − 1 , n 4 ) , ( S 4 X ) ( n 3 , n 4 ) = X ( n 3 , n 4 − 1 ) . (30) (S_{3}X)_{(n_{3},n_{4})}=X_{(n_{3}-1,n_{4})},\qquad(S_{4}X)_{(n_{3},n_{4})}=X_{(n_{3},n_{4}-1)}. \tag*{(30)} ( S 3 X ) ( n 3 , n 4 ) = X ( n 3 − 1 , n 4 ) , ( S 4 X ) ( n 3 , n 4 ) = X ( n 3 , n 4 − 1 ) . (30) By requiring the “periodicity condition” for the ADHM data as
S a λ † P a − 1 = λ † , S a B S a − 1 = B − L a σ a ( a = 3 , 4 ) , (31) S_{a}\lambda^{\dagger}P_{a}^{-1}=\lambda^{\dagger},\qquad S_{a}BS_{a}^{-1}=B-L_{a}\sigma_{a}\qquad(a=3,4), \tag*{(31)} S a λ † P a − 1 = λ † , S a B S a − 1 = B − L a σ a ( a = 3 , 4 ) , (31) we can find that A μ ( x ) A_{\mu}(x) A μ ( x ) (27) . This is solved as
λ ( n 3 , n 4 ) = P 3 n 3 P 4 n 4 ζ , B I , J = ( n 4 L 4 σ 4 + n 3 L 3 σ 3 ) δ I , J + A ^ I − J , (32) \lambda_{(n_{3},n_{4})}=P_{3}^{n_{3}}P_{4}^{n_{4}}\zeta,\qquad B_{I,J}=(n_{4}L_{4}\sigma_{4}+n_{3}L_{3}\sigma_{3})\delta_{I,J}+\hat{A}_{I-J}, \tag*{(32)} λ ( n 3 , n 4 ) = P 3 n 3 P 4 n 4 ζ , B I , J = ( n 4 L 4 σ 4 + n 3 L 3 σ 3 ) δ I , J + A ^ I − J , (32) where ζ : = λ ( 0 , 0 ) \zeta:=\lambda_{(0,0)} ζ := λ ( 0 , 0 ) A ^ I − J : = B I − J , 0 \hat{A}_{I-J}:=B_{I-J,0} A ^ I − J := B I − J , 0 P 3 , 4 P_{3,4} P 3 , 4
Abelian Hitchin system with N N N
To obtain the ADHM constraint for the ADHM data ( ζ , A ^ I ) (\zeta,\hat{A}_{I}) ( ζ , A ^ I ) T ^ 2 ∋ ξ = ( ξ 3 , ξ 4 ) \hat{T}^{2}\ni\xi=(\xi_{3},\xi_{4}) T ^ 2 ∋ ξ = ( ξ 3 , ξ 4 ) 2 π / L 3 , 4 2\pi/L_{3,4} 2 π / L 3 , 4 X ^ ( ξ ) = ∑ n 3 , n 4 e i ( n 3 L 3 ξ 3 + n 4 L 4 ξ 4 ) X ( n 3 , n 4 ) \hat{X}(\xi)=\sum_{n_{3},n_{4}}e^{i(n_{3}L_{3}\xi_{3}+n_{4}L_{4}\xi_{4})}X_{(n_{3},n_{4})} X ^ ( ξ ) = ∑ n 3 , n 4 e i ( n 3 L 3 ξ 3 + n 4 L 4 ξ 4 ) X ( n 3 , n 4 ) I = ( n 3 , n 4 ) ∈ Z 2 I=(n_{3},n_{4})\in\mathbb{Z}^{2} I = ( n 3 , n 4 ) ∈ Z 2 [27 , 28 , 29 , 35 ] . The detailed computation of the Fourier transform is done in Appendix § 7 , and let us quote the result: B I J = ( n 4 L 4 σ 4 + n 3 L 3 σ 3 ) + A ^ I − J B_{IJ} = (n_4L_4\sigma_4 + n_3L_3\sigma_3) + \hat{A}_{I-J} B I J = ( n 4 L 4 σ 4 + n 3 L 3 σ 3 ) + A ^ I − J T ^ 2 \hat{T}^2 T ^ 2
B ^ ( ξ , ξ ′ ) = V ^ D δ ( 2 ) ( ξ − ξ ′ ) , (33) \hat{B}(\xi,\xi') = \hat{V}D\delta^{(2)}(\xi-\xi'), \tag*{(33)} B ^ ( ξ , ξ ′ ) = V ^ D δ ( 2 ) ( ξ − ξ ′ ) , (33) where V ^ = 4 π 2 / ( L 3 L 4 ) \hat{V}=4\pi^2/(L_3L_4) V ^ = 4 π 2 / ( L 3 L 4 ) T ^ 2 \hat{T}^2 T ^ 2
D = − i σ 4 ∂ ξ 4 − i σ 3 ∂ ξ 3 + A ^ ( ξ ) = : σ μ D μ . (34) D=-i\sigma_4\partial_{\xi_4}-i\sigma_3\partial_{\xi_3}+\hat{A}(\xi)=:\sigma_\mu D_\mu. \tag*{(34)} D = − i σ 4 ∂ ξ 4 − i σ 3 ∂ ξ 3 + A ^ ( ξ ) =: σ μ D μ . (34) Here, D μ D_\mu D μ λ \lambda λ P 3 , 4 P_{3,4} P 3 , 4
λ ^ † ( ξ ) = V ^ ∑ m = 1 N δ ( 2 ) ( ξ − μ m ) ζ † Π m , (35) \hat{\lambda}^{\dagger}(\xi)=\hat{V}\sum_{m=1}^{N}\delta^{(2)}(\xi-\mu_m)\zeta^{\dagger}\Pi_m, \tag*{(35)} λ ^ † ( ξ ) = V ^ m = 1 ∑ N δ ( 2 ) ( ξ − μ m ) ζ † Π m , (35) where
μ m = ( μ 3 , m , μ 4 , m ) ∈ T ^ 2 , m = 1 , … , N . (36) \mu_m=(\mu_{3,m},\mu_{4,m})\in\hat{T}^2,\qquad m=1,\ldots,N. \tag*{(36)} μ m = ( μ 3 , m , μ 4 , m ) ∈ T ^ 2 , m = 1 , … , N . (36) The ADHM constraint (20) , Δ † Δ = λ † λ + B ( x ) † B ( x ) = σ 0 ⊗ f x − 1 \Delta^{\dagger}\Delta=\lambda^{\dagger}\lambda+B(x)^{\dagger}B(x)=\sigma_0\otimes f_x^{-1} Δ † Δ = λ † λ + B ( x ) † B ( x ) = σ 0 ⊗ f x − 1 σ l \sigma_l σ l Δ † Δ \Delta^{\dagger}\Delta Δ † Δ B ( x ) ↔ D x : = σ μ ( D μ − x μ ) B(x)\leftrightarrow D_x:=\sigma_\mu(D_\mu-x_\mu) B ( x ) ↔ D x := σ μ ( D μ − x μ ) B ( x ) † B ( x ) B(x)^{\dagger}B(x) B ( x ) † B ( x )
D x † D x = σ ˉ μ σ ν ( D μ − x μ ) ( D ν − x ν ) = σ 0 ( D μ − x μ ) 2 + 1 2 η ˉ μ ν l σ l [ D μ , D ν ] . (37) \begin{aligned}
D_x^{\dagger}D_x&=\bar{\sigma}_\mu\sigma_\nu(D_\mu-x_\mu)(D_\nu-x_\nu)\\
&=\sigma_0(D_\mu-x_\mu)^2+\frac{1}{2}\bar{\eta}^{\,l}_{\mu\nu}\sigma_l[D_\mu,D_\nu]. \tag*{(37)}
\end{aligned} D x † D x = σ ˉ μ σ ν ( D μ − x μ ) ( D ν − x ν ) = σ 0 ( D μ − x μ ) 2 + 2 1 η ˉ μν l σ l [ D μ , D ν ] . (37) The other term λ † λ \lambda^{\dagger}\lambda λ † λ
λ ^ † λ ^ ( ξ ) = V ^ ∑ m = 1 N δ ( 2 ) ( ξ − μ m ) ζ † Π m ζ . (38) \hat{\lambda}^{\dagger}\hat{\lambda}(\xi)=\hat{V}\sum_{m=1}^{N}\delta^{(2)}(\xi-\mu_m)\zeta^{\dagger}\Pi_m\zeta. \tag*{(38)} λ ^ † λ ^ ( ξ ) = V ^ m = 1 ∑ N δ ( 2 ) ( ξ − μ m ) ζ † Π m ζ . (38) Following the convention of [33 ] , we parametrize the 2 × 2 2\times2 2 × 2 ζ † Π m ζ \zeta^{\dagger}\Pi_m\zeta ζ † Π m ζ
ζ † Π m ζ = : 1 2 π [ σ 0 ∣ ρ ⃗ m ∣ − τ ⃗ ⋅ ρ ⃗ m ] . (39) \zeta^{\dagger}\Pi_m\zeta=:\frac{1}{2\pi}\left[\sigma_0|\vec{\rho}_m|-\vec{\tau}\cdot\vec{\rho}_m\right]. \tag*{(39)} ζ † Π m ζ =: 2 π 1 [ σ 0 ∣ ρ m ∣ − τ ⋅ ρ m ] . (39) Since ζ † Π m ζ \zeta^{\dagger}\Pi_m\zeta ζ † Π m ζ σ 0 \sigma_0 σ 0 ρ ⃗ m \vec{\rho}_m ρ m ζ † Π m ζ = χ m χ m † \zeta^{\dagger}\Pi_m\zeta=\chi_m\chi_m^{\dagger} ζ † Π m ζ = χ m χ m † χ m † \chi_m^{\dagger} χ m † m m m ζ \zeta ζ ρ m j = − π χ m † τ j χ m \rho_m^j=-\pi\chi_m^{\dagger}\tau_j\chi_m ρ m j = − π χ m † τ j χ m ∣ ρ ⃗ m ∣ = π χ m † χ m |\vec{\rho}_m|=\pi\chi_m^{\dagger}\chi_m ∣ ρ m ∣ = π χ m † χ m
We can now obtain the Hitchin system with N N N (20) , tr [ σ l Δ † Δ ] = 0 \operatorname{tr}[\sigma_l\Delta^{\dagger}\Delta]=0 tr [ σ l Δ † Δ ] = 0
F ^ 4 l + 1 2 ϵ l j k F ^ j k = V ^ 2 π ∑ m ρ m l δ ( 2 ) ( ξ − μ m ) , (40) \hat{F}_{4l}+\frac{1}{2}\epsilon_{ljk}\hat{F}_{jk}=\frac{\hat{V}}{2\pi}\sum_m\rho_m^l\delta^{(2)}(\xi-\mu_m), \tag*{(40)} F ^ 4 l + 2 1 ϵ l j k F ^ j k = 2 π V ^ m ∑ ρ m l δ ( 2 ) ( ξ − μ m ) , (40) where (setting ∂ ξ 1 , 2 = 0 \partial_{\xi_{1,2}}=0 ∂ ξ 1 , 2 = 0
F ^ μ ν = i [ D μ , D ν ] = ∂ ξ μ A ^ ν − ∂ ξ ν A ^ μ + i [ A ^ μ , A ^ ν ] . (41) \hat{F}_{\mu\nu}=i[D_\mu,D_\nu]=\partial_{\xi_\mu}\hat{A}_\nu-\partial_{\xi_\nu}\hat{A}_\mu+i[\hat{A}_\mu,\hat{A}_\nu]. \tag*{(41)} F ^ μν = i [ D μ , D ν ] = ∂ ξ μ A ^ ν − ∂ ξ ν A ^ μ + i [ A ^ μ , A ^ ν ] . (41) Since we consider the case with Q t o p = 1 Q_{\mathrm{top}}=1 Q top = 1 T ^ 2 \widehat{T}^2 T 2 A ^ 3 d ξ 3 + A ^ 4 d ξ 4 \widehat{A}_3\mathrm{d}\xi_3+\widehat{A}_4\mathrm{d}\xi_4 A 3 d ξ 3 + A 4 d ξ 4 U ( 1 ) U(1) U ( 1 ) T ^ 2 \widehat{T}^2 T 2 H ^ = A ^ 1 + i A ^ 2 \widehat{H}=\widehat{A}_1+i\widehat{A}_2 H = A 1 + i A 2 (40) can be rewritten as
{ F ^ 34 = − 2 π ∑ m ρ m 3 L 3 L 4 δ ( 2 ) ( ξ − μ m ) , 1 2 ( ∂ ξ 3 − i ∂ ξ 4 ) H ^ = − i π ∑ m ρ m 1 + i ρ m 2 L 3 L 4 δ ( 2 ) ( ξ − μ m ) . (42) \left\{
\begin{aligned}
\widehat{F}_{34}&=-2\pi\sum_m\frac{\rho_m^3}{L_3L_4}\delta^{(2)}(\xi-\mu_m),\\
\frac{1}{2}(\partial_{\xi_3}-i\partial_{\xi_4})\widehat{H}&=-i\pi\sum_m\frac{\rho_m^1+i\rho_m^2}{L_3L_4}\delta^{(2)}(\xi-\mu_m).
\end{aligned}
\right. \tag*{(42)} ⎩ ⎨ ⎧ F 34 2 1 ( ∂ ξ 3 − i ∂ ξ 4 ) H = − 2 π m ∑ L 3 L 4 ρ m 3 δ ( 2 ) ( ξ − μ m ) , = − iπ m ∑ L 3 L 4 ρ m 1 + i ρ m 2 δ ( 2 ) ( ξ − μ m ) . (42) This is the Abelian Hitchin system that is dual to the S U ( N ) SU(N) S U ( N ) [35 ] (see also [42 ] ). By integrating these equations on T ^ 2 \widehat{T}^2 T 2 ∑ m ρ ⃗ m = 0 \sum_m\vec{\rho}_m=0 ∑ m ρ m = 0 tr [ τ ⃗ ζ † ζ ] = 0 \operatorname{tr}[\vec{\tau}\zeta^\dagger\zeta]=0 tr [ τ ζ † ζ ] = 0 [29 ] , because A ^ ( ξ ) = ∑ n e i n ⋅ ( L ξ ) A ^ n \widehat{A}(\xi)=\sum_n e^{in\cdot(L\xi)}\widehat{A}_n A ( ξ ) = ∑ n e in ⋅ ( L ξ ) A n κ m : = ρ m 3 / ( L 3 L 4 ) \kappa_m:=\rho_m^3/(L_3L_4) κ m := ρ m 3 / ( L 3 L 4 ) 0 < ∣ κ m ∣ < 1 0<|\kappa_m|<1 0 < ∣ κ m ∣ < 1 m m m 3
Formally, one can obtain the gauge field A μ A_\mu A μ f x f_x f x (23) . The Fourier transform of f x f_x f x f ^ x ( ξ , ξ ′ ) \widehat{f}_x(\xi,\xi') f x ( ξ , ξ ′ ) 4
[ ∑ μ = 1 4 ( D μ − x μ ) 2 + V ^ 2 π ∑ m = 1 N ∣ ρ ⃗ m ∣ δ ( 2 ) ( ξ − μ m ) ] f ^ x ( ξ , ξ ′ ) = V ^ δ ( 2 ) ( ξ − ξ ′ ) (43) \left[\sum_{\mu=1}^{4}(D_\mu-x_\mu)^2+\frac{\widehat{V}}{2\pi}\sum_{m=1}^{N}|\vec{\rho}_m|\delta^{(2)}(\xi-\mu_m)\right]\widehat{f}_x(\xi,\xi')=\widehat{V}\delta^{(2)}(\xi-\xi') \tag*{(43)} [ μ = 1 ∑ 4 ( D μ − x μ ) 2 + 2 π V m = 1 ∑ N ∣ ρ m ∣ δ ( 2 ) ( ξ − μ m ) ] f x ( ξ , ξ ′ ) = V δ ( 2 ) ( ξ − ξ ′ ) (43) which is obtained by the Fourier transform of (20) . With the following formulas,
φ − 1 = 1 N − ∑ m , m ′ = 1 N f ^ x ( μ m , μ m ′ ) Π m ζ ζ † Π m ′ (44) \varphi^{-1}=\mathbb{1}_N-\sum_{m,m'=1}^{N}\widehat{f}_x(\mu_m,\mu_{m'})\Pi_m\zeta\zeta^\dagger\Pi_{m'} \tag*{(44)} φ − 1 = 1 N − m , m ′ = 1 ∑ N f x ( μ m , μ m ′ ) Π m ζ ζ † Π m ′ (44) λ [ σ j ⊗ f x ] λ † = ∑ m , m ′ = 1 N f ^ x ( μ m , μ m ′ ) Π m ζ σ j ζ † Π m ′ (45) \lambda[\sigma_j\otimes f_x]\lambda^\dagger=\sum_{m,m'=1}^{N}\widehat{f}_x(\mu_m,\mu_{m'})\Pi_m\zeta\sigma_j\zeta^\dagger\Pi_{m'} \tag*{(45)} λ [ σ j ⊗ f x ] λ † = m , m ′ = 1 ∑ N f x ( μ m , μ m ′ ) Π m ζ σ j ζ † Π m ′ (45) Eq. (23) determines A μ ( x ) A_\mu(x) A μ ( x ) N × N N\times N N × N
[ f ^ x ( μ m , μ m ′ ) ] m , m ′ = 1 , … , N (46) \left[\widehat{f}_x(\mu_m,\mu_{m'})\right]_{m,m'=1,\ldots,N} \tag*{(46)} [ f x ( μ m , μ m ′ ) ] m , m ′ = 1 , … , N (46) whose entries are the values of the Green function between the impurity points. However, there is a subtlety related to the treatment of the 2d δ \delta δ
The finite reconstruction of the self-dual gauge field
For the 1d Nahm equation of calorons, the δ \delta δ (43) only produce jump conditions, and f x ( μ m , μ m ′ ) f_x(\mu_m,\mu_{m'}) f x ( μ m , μ m ′ ) [33 ] . On the two-dimensional dual torus this is no longer true, and (44) –(45) have to be read with care. The origin is the k → ∞ k \to\infty k → ∞ λ ( n 3 , n 4 ) = Λ 3 n 3 Λ 4 n 4 ζ \lambda_{(n_3,n_4)} = \Lambda_3^{n_3}\Lambda_4^{n_4}\zeta λ ( n 3 , n 4 ) = Λ 3 n 3 Λ 4 n 4 ζ λ \lambda λ Z 2 \mathbb{Z}^2 Z 2 f ^ x ( μ m , μ m ′ ) \hat{f}_x(\mu_m,\mu_{m'}) f ^ x ( μ m , μ m ′ ) φ − 1 \varphi^{-1} φ − 1 A μ A_\mu A μ (23) , (44) and (45) , and the result becomes finite.
Normalized Weyl zero modes for the Hitchin system
For the case of N = 2 N = 2 N = 2 [37 , 38 ] . Following Ford and Pawlowski [39 , 40 , 41 ] , let us write
κ m = ρ m 3 L 3 L 4 , α m = ρ m 1 + i ρ m 2 L 3 L 4 , (47) \kappa_m = \frac{\rho_m^3}{L_3L_4}, \qquad\alpha_m = \frac{\rho_m^1 + i\rho_m^2}{L_3L_4}, \tag*{(47)} κ m = L 3 L 4 ρ m 3 , α m = L 3 L 4 ρ m 1 + i ρ m 2 , (47) and we introduce γ m = ∣ ρ ⃗ m ∣ / L 3 L 4 = κ m 2 + ∣ α m ∣ 2 \gamma_m = |\vec{\rho}_m|/L_3L_4 = \sqrt{\kappa_m^2 + |\alpha_m|^2} γ m = ∣ ρ m ∣/ L 3 L 4 = κ m 2 + ∣ α m ∣ 2 [37 ] , Jardim has proven that the S U ( 2 ) SU(2) S U ( 2 ) α m ≠ 0 \alpha_m \ne0 α m = 0 5 dim ker D x † = N = 2 \dim\ker D_x^\dagger= N = 2 dim ker D x † = N = 2 S U ( 2 ) SU(2) S U ( 2 )
[ A μ ( x ) ] i j = − i ∫ T ^ 2 d 2 ξ ψ i † ( ξ ; x ) ∂ μ ψ j ( ξ ; x ) , (48) [A_\mu(x)]_{ij} = -i \int_{\widehat{T}^2} d^2\xi\,\psi_i^\dagger(\xi;x)\partial_\mu\psi_j(\xi;x), \tag*{(48)} [ A μ ( x ) ] ij = − i ∫ T 2 d 2 ξ ψ i † ( ξ ; x ) ∂ μ ψ j ( ξ ; x ) , (48) where ψ i ( ξ ; x ) \psi_i(\xi;x) ψ i ( ξ ; x ) L 2 L^2 L 2 D x † D_x^\dagger D x † [39 , 40 , 41 ] for α m = 0 \alpha_m = 0 α m = 0 x x x
We interpret and reconstruct the Weyl zero modes of D x † D_x^\dagger D x † [37 , 38 ] . For the ADHM case, the zero modes v v v Δ ( x ) † v = 0 \Delta(x)^\dagger v = 0 Δ ( x ) † v = 0 [68 ] by putting
v = ( − 1 N u ) φ − 1 / 2 , u = ( B ( x ) † ) − 1 λ † , (49) v = \begin{pmatrix} -1_N \\ u \end{pmatrix}\varphi^{-1/2}, \qquad u = \left(B(x)^\dagger\right)^{-1}\lambda^\dagger, \tag*{(49)} v = ( − 1 N u ) φ − 1/2 , u = ( B ( x ) † ) − 1 λ † , (49) where φ − 1 / 2 > 0 \varphi^{-1/2}>0 φ − 1/2 > 0 v † v = 1 N v^\dagger v=1_N v † v = 1 N § 4 , the ultraviolet divergence causes φ − 1 → 0 \varphi^{-1}\to0 φ − 1 → 0 v † v = 1 N v^\dagger v=1_N v † v = 1 N
v → ( 0 ψ ( ξ ; x ) ) (50) v\to\begin{pmatrix} \mathbf{0} \\ \psi(\xi;x) \end{pmatrix} \tag*{(50)} v → ( 0 ψ ( ξ ; x ) ) (50) with u φ − 1 / 2 → ψ ( ξ ; x ) = [ ψ i ( ξ ; x ) ] i = 1 , … , N u\varphi^{-1/2}\to\psi(\xi;x)=[\psi_i(\xi;x)]_{i=1,\ldots,N} u φ − 1/2 → ψ ( ξ ; x ) = [ ψ i ( ξ ; x ) ] i = 1 , … , N φ \varphi φ f ^ x ( μ m , μ m ′ ) \hat{f}_x(\mu_m,\mu_{m'}) f ^ x ( μ m , μ m ′ ) A μ ( x ) A_\mu(x) A μ ( x )
Point splitting and the renormalization prescription
Here, we give a concrete prescription to evaluate (44) and (45) , and to obtain the finite self-dual gauge field A μ A_\mu A μ (23) . At points away from the impurities, f ^ x ( ξ , ξ ′ ) \hat{f}_x(\xi,\xi') f ^ x ( ξ , ξ ′ ) (43) as the 2d Green’s function; the ultraviolet problem is only that the Green-function formula for A μ A_\mu A μ μ m \mu_m μ m (43) is scale invariant at leading order: with ϱ = ∣ ξ − μ m ∣ \varrho=|\xi-\mu_m| ϱ = ∣ ξ − μ m ∣ κ m \kappa_m κ m ∣ H ^ ∣ 2 ≃ ∣ α m ∣ 2 / ϱ 2 |\hat{H}|^2\simeq|\alpha_m|^2/\varrho^2 ∣ H ^ ∣ 2 ≃ ∣ α m ∣ 2 / ϱ 2 [ ( ℓ − κ m ) 2 + ∣ α m ∣ 2 ] / ϱ 2 [(\ell-\kappa_m)^2+|\alpha_m|^2]/\varrho^2 [( ℓ − κ m ) 2 + ∣ α m ∣ 2 ] / ϱ 2 ℓ \ell ℓ μ m \mu_m μ m s s s ℓ = 0 \ell=0 ℓ = 0 ϱ ± γ m \varrho^{\pm\gamma_m} ϱ ± γ m f ^ x \hat{f}_x f ^ x f ^ x ( ξ , ξ ′ ) ∝ ϱ γ m \hat{f}_x(\xi,\xi')\propto\varrho^{\gamma_m} f ^ x ( ξ , ξ ′ ) ∝ ϱ γ m ξ → μ m \xi\to\mu_m ξ → μ m ξ ′ \xi' ξ ′
As a specific prescription, we split the points and average over directions: With some ϵ > 0 \epsilon>0 ϵ > 0 ξ = ξ 3 + i ξ 4 \xi=\xi_3+i\xi_4 ξ = ξ 3 + i ξ 4
f ^ x ϵ ( μ m , μ m ′ ) : = ∫ 0 2 π d θ 2 π ∫ 0 2 π d θ ′ 2 π f ^ x ( μ m + ϵ e i θ , μ m ′ + ϵ e i θ ′ ) , (51) \hat{f}^{\epsilon}_x(\mu_m,\mu_{m'}) := \int_0^{2\pi}\frac{d\theta}{2\pi}\int_0^{2\pi}\frac{d\theta'}{2\pi}\,\hat{f}_x\left(\mu_m+\epsilon e^{i\theta},\mu_{m'}+\epsilon e^{i\theta'}\right), \tag*{(51)} f ^ x ϵ ( μ m , μ m ′ ) := ∫ 0 2 π 2 π d θ ∫ 0 2 π 2 π d θ ′ f ^ x ( μ m + ϵ e i θ , μ m ′ + ϵ e i θ ′ ) , (51) i.e. each point source is replaced by the uniform distribution on the circle of radius ϵ \epsilon ϵ φ ϵ \varphi_\epsilon φ ϵ M ϵ , μ ν : = λ [ η ˉ μ ν ⊗ f ϵ ] † M_{\epsilon,\mu\nu}:=\lambda[\bar{\eta}_{\mu\nu}\otimes f_\epsilon]^\dagger M ϵ , μν := λ [ η ˉ μν ⊗ f ϵ ] † (44) and (45) with f ^ x ( μ m , μ m ′ ) \hat{f}_x(\mu_m,\mu_{m'}) f ^ x ( μ m , μ m ′ ) f ^ x ϵ ( μ m , μ m ′ ) \hat{f}^{\epsilon}_x(\mu_m,\mu_{m'}) f ^ x ϵ ( μ m , μ m ′ ) ϵ → 0 \epsilon\to0 ϵ → 0
Claim 1 (Point-split renormalization and scaling ansatz). Let Z : = diag ( ϵ 2 γ 1 , … , ϵ 2 γ N ) Z:=\operatorname{diag}(\epsilon^{2\gamma_1},\ldots,\epsilon^{2\gamma_N}) Z := diag ( ϵ 2 γ 1 , … , ϵ 2 γ N ) ϵ \epsilon ϵ φ ϵ − 1 → 0 \varphi_\epsilon^{-1}\to0 φ ϵ − 1 → 0 γ m < 1 \gamma_m<1 γ m < 1 m m m
Φ ~ − 1 ( x ) : = lim ϵ → 0 Z − 1 / 2 φ ϵ − 1 ( x ) Z − 1 / 2 (52) \widetilde{\Phi}^{-1}(x):=\lim_{\epsilon\to0}Z^{-1/2}\varphi_\epsilon^{-1}(x)Z^{-1/2} \tag*{(52)} Φ − 1 ( x ) := ϵ → 0 lim Z − 1/2 φ ϵ − 1 ( x ) Z − 1/2 (52) where Φ ~ − 1 \widetilde{\Phi}^{-1} Φ − 1
[ Φ ~ − 1 ] m m ′ = − χ m † χ m ′ lim ϵ → 0 ϵ − γ m − γ m ′ f ^ x ϵ ( μ m , μ m ′ ) ( m ≠ m ′ ) , \left[\widetilde{\Phi}^{-1}\right]_{mm'}=-\chi_m^\dagger\chi_{m'}\lim_{\epsilon\to0}\epsilon^{-\gamma_m-\gamma_{m'}}\hat{f}^{\epsilon}_x(\mu_m,\mu_{m'})\qquad(m\ne m'), [ Φ − 1 ] m m ′ = − χ m † χ m ′ ϵ → 0 lim ϵ − γ m − γ m ′ f ^ x ϵ ( μ m , μ m ′ ) ( m = m ′ ) , [ Φ ~ − 1 ] m m = lim ϵ → 0 ϵ − 2 γ m ( 1 − ∣ ρ ⃗ m ∣ π f ^ x ϵ ( μ m , μ m ) ) . (53) \left[\widetilde{\Phi}^{-1}\right]_{mm}=\lim_{\epsilon\to0}\epsilon^{-2\gamma_m}\left(1-\frac{|\vec{\rho}_m|}{\pi}\hat{f}^{\epsilon}_x(\mu_m,\mu_m)\right). \tag*{(53)} [ Φ − 1 ] mm = ϵ → 0 lim ϵ − 2 γ m ( 1 − π ∣ ρ m ∣ f ^ x ϵ ( μ m , μ m ) ) . (53) and
M ~ μ ν ( x ) : = lim ϵ → 0 Z − 1 / 2 ( M ϵ , μ ν − M μ ν ( 0 ) ) Z − 1 / 2 (54) \widetilde{M}_{\mu\nu}(x) := \lim_{\epsilon\to0} Z^{-1/2}\left(M_{\epsilon,\mu\nu}-M_{\mu\nu}^{(0)}\right)Z^{-1/2} \tag*{(54)} M μν ( x ) := ϵ → 0 lim Z − 1/2 ( M ϵ , μν − M μν ( 0 ) ) Z − 1/2 (54) exist together with their x x x 6 M μ ν ( 0 ) = − i diag ( η ˉ μ ν l ρ 1 l ∣ ρ ⃗ 1 ∣ , … , η ˉ μ ν l ρ N l ∣ ρ ⃗ N ∣ ) M_{\mu\nu}^{(0)}=-i\operatorname{diag}\left(\frac{\bar{\eta}_{\mu\nu}^{l}\rho_{1}^{l}}{|\vec{\rho}_{1}|},\ldots,\frac{\bar{\eta}_{\mu\nu}^{l}\rho_{N}^{l}}{|\vec{\rho}_{N}|}\right) M μν ( 0 ) = − i diag ( ∣ ρ 1 ∣ η ˉ μν l ρ 1 l , … , ∣ ρ N ∣ η ˉ μν l ρ N l )
For α m = 0 \alpha_m=0 α m = 0 [40 ] . There, we can take H ^ = 0 \hat{H}=0 H ^ = 0 A ^ = ∗ d ϕ \hat{A}=*\mathrm{d}\phi A ^ = ∗ d ϕ A ^ 3 = − ∂ ξ 4 ϕ \hat{A}_{3}=-\partial_{\xi_{4}}\phi A ^ 3 = − ∂ ξ 4 ϕ A ^ 4 = ∂ ξ 3 ϕ \hat{A}_{4}=\partial_{\xi_{3}}\phi A ^ 4 = ∂ ξ 3 ϕ ϕ \phi ϕ ∇ 2 ϕ = − 2 π ∑ m κ m δ ( 2 ) ( ξ − μ m ) \nabla^{2}\phi=-2\pi\sum_m\kappa_m\delta^{(2)}(\xi-\mu_m) ∇ 2 ϕ = − 2 π ∑ m κ m δ ( 2 ) ( ξ − μ m ) ϕ ≃ − κ m log ∣ ξ − μ m ∣ \phi\simeq-\kappa_m\log|\xi-\mu_m| ϕ ≃ − κ m log ∣ ξ − μ m ∣ μ m \mu_m μ m μ m \mu_m μ m exp ( ± ϕ ( μ m + ϵ e i θ ) ) ∼ ϵ ∓ κ m \exp(\pm\phi(\mu_m+\epsilon e^{i\theta}))\sim\epsilon^{\mp\kappa_m} exp ( ± ϕ ( μ m + ϵ e i θ )) ∼ ϵ ∓ κ m [40 Sec. 6] ), which gives (53) with Z m = ϵ 2 ∣ κ m ∣ Z_m=\epsilon^{2|\kappa_m|} Z m = ϵ 2∣ κ m ∣ 0 < ∣ κ m ∣ < 1 0<|\kappa_m|<1 0 < ∣ κ m ∣ < 1 ⇔ γ m < 1 \Leftrightarrow\gamma_m<1 ⇔ γ m < 1 L 2 L^2 L 2 [40 ] . For general α m \alpha_m α m
Accepting Claim 1, the gauge field is obtained as follows. At finite ϵ \epsilon ϵ A μ ( ϵ ) A_{\mu}^{(\epsilon)} A μ ( ϵ ) (23) evaluated with ϕ ϵ \phi_\epsilon ϕ ϵ M ϵ , μ ν M_{\epsilon,\mu\nu} M ϵ , μν Φ ~ ϵ : = Z 1 / 2 ϕ ϵ Z 1 / 2 \widetilde{\Phi}_\epsilon:=Z^{1/2}\phi_\epsilon Z^{1/2} Φ ϵ := Z 1/2 ϕ ϵ Z 1/2 ∂ ν M ~ ϵ , μ ν : = Z − 1 / 2 ∂ ν M ϵ , μ ν Z − 1 / 2 \partial_\nu\widetilde{M}_{\epsilon,\mu\nu}:=Z^{-1/2}\partial_\nu M_{\epsilon,\mu\nu}Z^{-1/2} ∂ ν M ϵ , μν := Z − 1/2 ∂ ν M ϵ , μν Z − 1/2 A μ ( ϵ ) A_{\mu}^{(\epsilon)} A μ ( ϵ ) ϵ \epsilon ϵ ϵ \epsilon ϵ ϕ ϵ > 0 \phi_\epsilon>0 ϕ ϵ > 0 Φ ~ − 1 \widetilde{\Phi}^{-1} Φ − 1
g ϵ ( x ) : = ϕ ϵ − 1 / 2 Z − 1 / 2 Φ ~ ϵ 1 / 2 . (55) g_\epsilon(x):=\phi_\epsilon^{-1/2}Z^{-1/2}\widetilde{\Phi}_\epsilon^{1/2}. \tag*{(55)} g ϵ ( x ) := ϕ ϵ − 1/2 Z − 1/2 Φ ϵ 1/2 . (55) Since Z − 1 / 2 Φ ~ ϵ Z − 1 / 2 = ϕ ϵ Z^{-1/2}\widetilde{\Phi}_\epsilon Z^{-1/2}=\phi_\epsilon Z − 1/2 Φ ϵ Z − 1/2 = ϕ ϵ
g ϵ g ϵ † = ϕ ϵ − 1 / 2 Z − 1 / 2 Φ ~ ϵ Z − 1 / 2 ϕ ϵ − 1 / 2 = 1 N , det g ϵ = ( det Φ ~ ϵ ) 1 / 2 ( det ϕ ϵ det Z ) 1 / 2 = 1 , (56) g_\epsilon g_\epsilon^\dagger=\phi_\epsilon^{-1/2}Z^{-1/2}\widetilde{\Phi}_\epsilon Z^{-1/2}\phi_\epsilon^{-1/2}=1_N,\qquad\det g_\epsilon=\frac{(\det\widetilde{\Phi}_\epsilon)^{1/2}}{(\det\phi_\epsilon\det Z)^{1/2}}=1, \tag*{(56)} g ϵ g ϵ † = ϕ ϵ − 1/2 Z − 1/2 Φ ϵ Z − 1/2 ϕ ϵ − 1/2 = 1 N , det g ϵ = ( det ϕ ϵ det Z ) 1/2 ( det Φ ϵ ) 1/2 = 1 , (56) so g ϵ g_\epsilon g ϵ S U ( N ) SU(N) S U ( N ) A μ ( ϵ ) A_\mu^{(\epsilon)} A μ ( ϵ ) g ϵ g_\epsilon g ϵ
( A μ ( ϵ ) ) g ϵ = − i 2 Φ ~ ϵ 1 / 2 [ ∂ ν M ~ ϵ , μ ν ] Φ ~ ϵ 1 / 2 − i 2 [ Φ ~ ϵ − 1 / 2 , ∂ μ Φ ~ ϵ 1 / 2 ] . (57) \left(A_\mu^{(\epsilon)}\right)^{g_\epsilon}=-\frac{i}{2}\widetilde{\Phi}_\epsilon^{1/2}\left[\partial_\nu\widetilde{M}_{\epsilon,\mu\nu}\right]\widetilde{\Phi}_\epsilon^{1/2}-\frac{i}{2}\left[\widetilde{\Phi}_\epsilon^{-1/2},\partial_\mu\widetilde{\Phi}_\epsilon^{1/2}\right]. \tag*{(57)} ( A μ ( ϵ ) ) g ϵ = − 2 i Φ ϵ 1/2 [ ∂ ν M ϵ , μν ] Φ ϵ 1/2 − 2 i [ Φ ϵ − 1/2 , ∂ μ Φ ϵ 1/2 ] . (57) This identity holds at every ϵ \epsilon ϵ ϵ → 0 \epsilon\to0 ϵ → 0
A μ ′ = − i 2 Φ ~ 1 / 2 [ ∂ ν M ~ μ ν ] Φ ~ 1 / 2 − i 2 [ Φ ~ − 1 / 2 , ∂ μ Φ ~ 1 / 2 ] , (58) A'_\mu=-\frac{i}{2}\widetilde{\Phi}^{1/2}\left[\partial_\nu\widetilde{M}_{\mu\nu}\right]\widetilde{\Phi}^{1/2}-\frac{i}{2}\left[\widetilde{\Phi}^{-1/2},\partial_\mu\widetilde{\Phi}^{1/2}\right], \tag*{(58)} A μ ′ = − 2 i Φ 1/2 [ ∂ ν M μν ] Φ 1/2 − 2 i [ Φ − 1/2 , ∂ μ Φ 1/2 ] , (58) which is written in terms of manifestly finite quantities.
A μ ( ϵ ) A_\mu^{(\epsilon)} A μ ( ϵ ) (23) . Let L ϵ : = Z 1 / 2 ϕ ϵ 1 / 2 L_\epsilon:=Z^{1/2}\phi_\epsilon^{1/2} L ϵ := Z 1/2 ϕ ϵ 1/2 L ϵ ( L ϵ ) † = Φ ~ ϵ L_\epsilon(L_\epsilon)^\dagger=\widetilde{\Phi}_\epsilon L ϵ ( L ϵ ) † = Φ ϵ L ϵ L_\epsilon L ϵ L L L L L † = Φ ~ LL^\dagger=\widetilde{\Phi} L L † = Φ γ m \gamma_m γ m 7
g ϵ → g : = L − 1 Φ ~ 1 / 2 ∈ S U ( N ) , A μ ( ϵ ) → A μ : = ( A μ ′ ) g − 1 , (59) g_\epsilon\to g:=L^{-1}\widetilde{\Phi}^{1/2}\in SU(N),\qquad A_\mu^{(\epsilon)}\to A_\mu:=(A'_\mu)^{g^{-1}}, \tag*{(59)} g ϵ → g := L − 1 Φ 1/2 ∈ S U ( N ) , A μ ( ϵ ) → A μ := ( A μ ′ ) g − 1 , (59) and
A μ = − i 2 L † [ ∂ ν M ~ μ ν ] L − i 2 [ ( L − 1 ∂ μ L ) − ( L − 1 ∂ μ L ) † ] . (60) A_\mu=-\frac{i}{2}L^\dagger[\partial_\nu\widetilde{M}_{\mu\nu}]L-\frac{i}{2}\left[(L^{-1}\partial_\mu L)-(L^{-1}\partial_\mu L)^\dagger\right]. \tag*{(60)} A μ = − 2 i L † [ ∂ ν M μν ] L − 2 i [ ( L − 1 ∂ μ L ) − ( L − 1 ∂ μ L ) † ] . (60) In this sense, the formulas (23) , (44) and (45) hold with the values of f ^ x \hat{f}_x f ^ x (51) and ϵ → 0 \epsilon\to0 ϵ → 0
Self-duality and periodicity. Let us discuss if the gauge field A μ A_\mu A μ T 2 T^2 T 2 λ \lambda λ (27) . The point splitting only regularizes divergent quantities whose combination in A μ A_\mu A μ (51) , the terms odd in θ \theta θ θ ′ \theta' θ ′ x x x O ( ϵ 2 ) O(\epsilon^2) O ( ϵ 2 ) O ( ϵ 2 − 2 γ m ) O(\epsilon^{2-2\gamma_m}) O ( ϵ 2 − 2 γ m ) Z − 1 / 2 Z^{-1/2} Z − 1/2 ϵ \epsilon ϵ O ( ϵ 2 − 2 γ max ) O(\epsilon^{2-2\gamma_{\max}}) O ( ϵ 2 − 2 γ m a x )
∥ F ( ϵ ) − ⋆ F ( ϵ ) ∥ = O ( ϵ 2 − 2 max m γ m ) , \lVert F^{(\epsilon)}-\star F^{(\epsilon)}\rVert=O(\epsilon^{2-2\max_m\gamma_m}), ∥ F ( ϵ ) − ⋆ F ( ϵ ) ∥ = O ( ϵ 2 − 2 m a x m γ m ) , ∥ A μ ( ϵ ) ( x + L a e ^ a ) − P a A μ ( ϵ ) ( x ) P a − 1 ∥ = O ( ϵ 2 − 2 max m γ m ) , (61) \lVert A_\mu^{(\epsilon)}(x+L_a\hat{e}_a)-P_aA_\mu^{(\epsilon)}(x)P_a^{-1}\rVert=O(\epsilon^{2-2\max_m\gamma_m}), \tag*{(61)} ∥ A μ ( ϵ ) ( x + L a e ^ a ) − P a A μ ( ϵ ) ( x ) P a − 1 ∥ = O ( ϵ 2 − 2 m a x m γ m ) , (61) and this vanishes as ϵ → 0 \epsilon\to0 ϵ → 0 γ m < 1 \gamma_m<1 γ m < 1
Constructing fractional instanton via ’t Hooft-twist cut
Let us now construct the fractional instanton on R 2 × T p / N -twisted 2 \mathbb{R}^2\times T^2_{p/N\text{-twisted}} R 2 × T p / N -twisted 2 (4) , a self-dual field obeys A μ ( x + ℓ 3 e ^ 3 ) = S − p A μ ( x ) S p A_\mu(x+\ell_3\hat{e}_3)=S^{-p}A_\mu(x)S^p A μ ( x + ℓ 3 e ^ 3 ) = S − p A μ ( x ) S p A μ ( x + ℓ 4 e ^ 4 ) = C A μ ( x ) C − 1 A_\mu(x+\ell_4\hat{e}_4)=CA_\mu(x)C^{-1} A μ ( x + ℓ 4 e ^ 4 ) = C A μ ( x ) C − 1 N N N x 3 ∼ x 3 + N ℓ 3 x_3\sim x_3+N\ell_3 x 3 ∼ x 3 + N ℓ 3 (27) with L 3 = N ℓ 3 L_3=N\ell_3 L 3 = N ℓ 3 L 4 = ℓ 4 L_4=\ell_4 L 4 = ℓ 4 P 3 = ( S − p ) N ∝ 1 N P_3=(S^{-p})^N\propto1_N P 3 = ( S − p ) N ∝ 1 N P 4 = C P_4=C P 4 = C Q t o p = 1 / N Q_{\mathrm{top}}=1/N Q top = 1/ N Q t o p = 1 Q_{\mathrm{top}}=1 Q top = 1
A μ ( x + ℓ 3 e ^ 3 ) = S − p A μ ( x ) S p . (62) A_\mu(x+\ell_3\hat{e}_3)=S^{-p}A_\mu(x)S^p. \tag*{(62)} A μ ( x + ℓ 3 e ^ 3 ) = S − p A μ ( x ) S p . (62) Conversely, any such doubly-periodic instanton descends to the twisted torus, and its fundamental domain 0 ≤ x 3 < ℓ 3 0\leq x_3<\ell_3 0 ≤ x 3 < ℓ 3 Q t o p = 1 / N Q_{\mathrm{top}}=1/N Q top = 1/ N S Y M = 8 π 2 / ( N g 2 ) S_{\mathrm{YM}}=8\pi^2/(Ng^2) S YM = 8 π 2 / ( N g 2 )
Let us note that the construction of the Nahm transform on the twisted T 4 T^4 T 4 [72 ] (see also Refs. [73 , 74 ] ); the twist cut here is its counterpart for R 2 × T 2 \mathbb{R}^2\times T^2 R 2 × T 2
To translate (62) into a condition on the ADHM data ( ζ , A ^ ) (\zeta,\hat{A}) ( ζ , A ^ ) T ^ 2 \hat{T}^2 T ^ 2 (62) act on the ADHM data. The translation A μ ( x ) ↦ A μ ( x + ℓ 3 e ^ 3 ) A_\mu(x)\mapsto A_\mu(x+\ell_3\hat{e}_3) A μ ( x ) ↦ A μ ( x + ℓ 3 e ^ 3 ) B ↦ B − ℓ 3 σ 3 B\mapsto B-\ell_3\sigma_3 B ↦ B − ℓ 3 σ 3 B ( x ) = B − x B(x)=B-x B ( x ) = B − x
A ^ 3 ( ζ ) ↦ A ^ 3 ( ζ ) − ℓ 3 , (63) \hat{A}_3(\zeta)\mapsto\hat{A}_3(\zeta)-\ell_3, \tag*{(63)} A ^ 3 ( ζ ) ↦ A ^ 3 ( ζ ) − ℓ 3 , (63) with A ^ 4 \hat{A}_4 A ^ 4 H ^ \hat{H} H ^ ζ \zeta ζ ℓ 3 = L 3 / N \ell_3 = L_3/N ℓ 3 = L 3 / N U ( 1 ) U(1) U ( 1 ) ξ 3 \xi_3 ξ 3 ω = e 2 π i / N \omega= e^{2\pi i/N} ω = e 2 π i / N T ^ 2 \hat{T}^2 T ^ 2 A μ ↦ S − p A μ S p A_\mu\mapsto S^{-p}A_\mu S^p A μ ↦ S − p A μ S p λ ↦ S − p λ \lambda\mapsto S^{-p}\lambda λ ↦ S − p λ (32) with P 3 = 1 N P_3 = 1_N P 3 = 1 N S − p C S p = ω − p C S^{-p} C S^p = \omega^{-p}C S − p C S p = ω − p C ( S − p λ ) ( n 3 , n 4 ) = ω − p n 4 C n 4 ( S − p ζ ) (S^{-p}\lambda)_{(n_3,n_4)} = \omega^{-pn_4} C^{n_4}(S^{-p}\zeta) ( S − p λ ) ( n 3 , n 4 ) = ω − p n 4 C n 4 ( S − p ζ ) ω − p n 4 \omega^{-pn_4} ω − p n 4 U ( k ) U(k) U ( k ) diag ( ω − p n 4 ) \operatorname{diag}(\omega^{-pn_4}) diag ( ω − p n 4 ) A ^ ( n 3 , n 4 ) \hat{A}_{(n_3,n_4)} A ^ ( n 3 , n 4 ) ω − p n 4 = e − i n 4 L 4 ⋅ 2 π p / ( N L 4 ) \omega^{-pn_4} = e^{-in_4L_4\cdot2\pi p/(NL_4)} ω − p n 4 = e − i n 4 L 4 ⋅ 2 π p / ( N L 4 ) A ^ ( ξ ^ ) ↦ A ^ ( ξ ^ − 2 π p N L 4 e ^ 4 ) \hat{A}(\hat{\xi}) \mapsto\hat{A}\left(\hat{\xi} - \frac{2\pi p}{NL_4}\hat{e}_4\right) A ^ ( ξ ^ ) ↦ A ^ ( ξ ^ − N L 4 2 π p e ^ 4 ) S − p ζ S^{-p}\zeta S − p ζ ζ \zeta ζ p p p
( A ^ ( ξ ^ ) , H ^ ( ξ ^ ) ) ↦ ( A ^ ( ξ ^ − 2 π p N L 4 e ^ 4 ) , H ^ ( ξ ^ − 2 π p N L 4 e ^ 4 ) ) , ρ ⃗ m ↦ ρ ⃗ m − p . (64) \left(\hat{A}(\hat{\xi}),\hat{H}(\hat{\xi})\right) \mapsto\left(\hat{A}\left(\hat{\xi} - \frac{2\pi p}{NL_4}\hat{e}_4\right),\hat{H}\left(\hat{\xi} - \frac{2\pi p}{NL_4}\hat{e}_4\right)\right), \qquad\vec{\rho}_m \mapsto\vec{\rho}_{m-p}. \tag*{(64)} ( A ^ ( ξ ^ ) , H ^ ( ξ ^ ) ) ↦ ( A ^ ( ξ ^ − N L 4 2 π p e ^ 4 ) , H ^ ( ξ ^ − N L 4 2 π p e ^ 4 ) ) , ρ m ↦ ρ m − p . (64) This is the translation of the whole Hitchin configuration by p p p ξ 4 \xi_4 ξ 4 μ m − p + 2 π p N L 4 e ^ 4 = μ m \mu_{m-p} + \frac{2\pi p}{NL_4}\hat{e}_4 = \mu_m μ m − p + N L 4 2 π p e ^ 4 = μ m (62) are fractional operations on T ^ 2 \hat{T}^2 T ^ 2
( A ^ 3 − ℓ 3 , A ^ 4 , H ^ ; ρ ⃗ m ) ∣ ( ξ 3 , ξ 4 ) ∼ ( A ^ , H ^ ; ρ ⃗ m − p ) ∣ ( ξ 3 , ξ 4 − 2 π p / ( N L 4 ) ) . (65) \left.(\hat{A}_3-\ell_3,\hat{A}_4,\hat{H};\vec{\rho}_m)\right|_{(\xi_3,\xi_4)} \sim\left.(\hat{A},\hat{H};\vec{\rho}_{m-p})\right|_{(\xi_3,\xi_4-2\pi p/(NL_4))}. \tag*{(65)} ( A ^ 3 − ℓ 3 , A ^ 4 , H ^ ; ρ m ) ( ξ 3 , ξ 4 ) ∼ ( A ^ , H ^ ; ρ m − p ) ( ξ 3 , ξ 4 − 2 π p / ( N L 4 )) . (65) i.e. they should be identical up to the U ( 1 ) U(1) U ( 1 ) μ m \mu_m μ m
α m = 0 , and κ m ≡ q N ( m o d 1 ) . (66) \alpha_m = 0,\qquad\text{and}\qquad\kappa_m \equiv\frac{q}{N}\pmod1. \tag*{(66)} α m = 0 , and κ m ≡ N q ( mod 1 ) . (66) To see α m = 0 \alpha_m = 0 α m = 0 H ^ \hat{H} H ^ U ( 1 ) U(1) U ( 1 ) ξ 4 ↦ ξ 4 − 2 π p N L 4 \xi_4 \mapsto\xi_4 - \frac{2\pi p}{NL_4} ξ 4 ↦ ξ 4 − N L 4 2 π p α m − p = α m \alpha_{m-p} = \alpha_m α m − p = α m gcd ( p , N ) = 1 \gcd(p,N) = 1 g cd( p , N ) = 1 α m \alpha_m α m ∑ m α m = 0 \sum_m \alpha_m = 0 ∑ m α m = 0 α m = 0 \alpha_m = 0 α m = 0 H ^ \hat{H} H ^ (42) gives the constant H ^ = y ⊥ \hat{H} = y_\perp H ^ = y ⊥
Let us next derive the condition for κ m \kappa_m κ m T ^ 2 ∖ { μ m } \hat{T}^2 \setminus\{\mu_m\} T ^ 2 ∖ { μ m } μ m \mu_m μ m e 2 π i κ m e^{2\pi i\kappa_m} e 2 π i κ m ξ 4 ↦ ξ 4 − 2 π p N L 4 \xi_4 \mapsto\xi_4 - \frac{2\pi p}{NL_4} ξ 4 ↦ ξ 4 − N L 4 2 π p ( κ : = ) κ m − p ≡ κ m (\kappa:=)\kappa_{m-p} \equiv\kappa_m ( κ := ) κ m − p ≡ κ m ξ 3 \xi_3 ξ 3 ξ 4 \xi_4 ξ 4 e − 2 π i κ m e^{-2\pi i\kappa_m} e − 2 π i κ m ξ 4 \xi_4 ξ 4 μ 4 , m \mu_{4,m} μ 4 , m 2 π p N L 4 \frac{2\pi p}{NL_4} N L 4 2 π p p p p e 2 π i p κ e^{2\pi ip\kappa} e 2 π i p κ A ^ 3 ↦ A ^ 3 − ℓ 3 \hat{A}_3 \mapsto\hat{A}_3 - \ell_3 A ^ 3 ↦ A ^ 3 − ℓ 3 e 2 π i / N e^{2\pi i/N} e 2 π i / N e 2 π i p κ = e 2 π i / N e^{2\pi ip\kappa} = e^{2\pi i/N} e 2 π i p κ = e 2 π i / N ∑ m κ m = 0 \sum_m \kappa_m = 0 ∑ m κ m = 0 N κ ∈ Z N\kappa\in\mathbb{Z} N κ ∈ Z q q q e 2 π i κ = e 2 π i q p κ = e 2 π i q / N e^{2\pi i\kappa} = e^{2\pi iqp\kappa} = e^{2\pi iq/N} e 2 π iκ = e 2 π i q p κ = e 2 π i q / N κ ≡ q / N \kappa\equiv q/N κ ≡ q / N
The twist cut condition (66) fixes all the impurity data, and thus the fractional instantons are parametrized only by the flat part of the Hitchin data:
the constant Higgs field H ^ = y ⊥ ∈ C \hat{H} = y_\perp\in\mathbb{C} H ^ = y ⊥ ∈ C R 2 \mathbb{R}^2 R 2
the holonomy of A ^ \hat{A} A ^ ξ 3 \xi_3 ξ 3 x 3 x_3 x 3
the holonomy of A ^ \hat{A} A ^ ξ 4 \xi_4 ξ 4 x 4 x_4 x 4
This gives the 4 real moduli expected from the index theorem, 4 N Q t o p = 4 4NQ_{\mathrm{top}} = 4 4 N Q top = 4 § 5 . The twist cut data have α m = 0 \alpha_m = 0 α m = 0 A ^ = ⋆ d ϕ \hat{A} = \star d\phi A ^ = ⋆ d ϕ ∇ 2 ϕ = − 2 π ∑ m κ m δ ( 2 ) ( ξ − μ m ) \nabla^2\phi= -2\pi\sum_m \kappa_m\delta^{(2)}(\xi-\mu_m) ∇ 2 ϕ = − 2 π ∑ m κ m δ ( 2 ) ( ξ − μ m ) [40 ] , and γ m = ∣ κ m ∣ < 1 \gamma_m = |\kappa_m| < 1 γ m = ∣ κ m ∣ < 1
Let us now show that the data (66) indeed give a fractional instanton on R 2 × T p / N -twisted 2 \mathbb{R}^{2} \times T^{2}_{p/N\text{-twisted}} R 2 × T p / N -twisted 2 (66) holds, the two sides of (65) have the same constant Higgs field and the same holonomies on T ^ 2 ∖ { μ m } \widehat{T}^{2} \setminus\{\mu_m\} T 2 ∖ { μ m } U ( 1 ) U(1) U ( 1 ) e i θ ( ξ ) e^{i\theta(\xi)} e i θ ( ξ ) κ m − κ m − p \kappa_m-\kappa_{m-p} κ m − κ m − p μ m \mu_m μ m D x + ℓ 3 e ^ 3 ( ξ ) = e i θ ( ξ ) D x ( ξ − 2 π p N L 4 e ^ 4 ) e − i θ ( ξ ) D_{x+\ell_3\hat{e}_3}(\xi)=e^{i\theta(\xi)}D_x\left(\xi-\frac{2\pi p}{NL_4}\hat{e}_4\right)e^{-i\theta(\xi)} D x + ℓ 3 e ^ 3 ( ξ ) = e i θ ( ξ ) D x ( ξ − N L 4 2 π p e ^ 4 ) e − i θ ( ξ ) U ( k ) U(k) U ( k ) ψ ( ξ ) ↦ e i θ ( ξ ) ψ ( ξ − 2 π p N L 4 e ^ 4 ) \psi(\xi)\mapsto e^{i\theta(\xi)}\psi\left(\xi-\frac{2\pi p}{NL_4}\hat{e}_4\right) ψ ( ξ ) ↦ e i θ ( ξ ) ψ ( ξ − N L 4 2 π p e ^ 4 ) L 2 L^2 L 2 D x + D_x^+ D x + D x + ℓ 3 e ^ 3 + D_{x+\ell_3\hat{e}_3}^+ D x + ℓ 3 e ^ 3 + N N N 0 < ∣ κ m ∣ < 1 0<|\kappa_m|<1 0 < ∣ κ m ∣ < 1
ψ ( ξ ; x + ℓ 3 e ^ 3 ) = e i θ ( ξ ) ψ ( ξ − 2 π p N L 4 e ^ 4 ; x ) Ω 3 ( x ) − 1 (67) \psi(\xi;x+\ell_3\hat{e}_3)=e^{i\theta(\xi)}\psi\left(\xi-\frac{2\pi p}{NL_4}\hat{e}_4;x\right)\Omega_3(x)^{-1} \tag*{(67)} ψ ( ξ ; x + ℓ 3 e ^ 3 ) = e i θ ( ξ ) ψ ( ξ − N L 4 2 π p e ^ 4 ; x ) Ω 3 ( x ) − 1 (67) with some Ω 3 ( x ) ∈ S U ( N ) \Omega_3(x)\in SU(N) Ω 3 ( x ) ∈ S U ( N ) A μ A_\mu A μ ψ \psi ψ A μ ( x + ℓ 3 e ^ 3 ) = Ω 3 ( x ) A μ ( x ) Ω 3 − 1 ( x ) A_\mu(x+\ell_3\hat{e}_3)=\Omega_3(x)A_\mu(x)\Omega_3^{-1}(x) A μ ( x + ℓ 3 e ^ 3 ) = Ω 3 ( x ) A μ ( x ) Ω 3 − 1 ( x ) e i θ e^{i\theta} e i θ θ \theta θ x x x ξ \xi ξ T ^ 2 \widehat{T}^{2} T 2
The twist is also read off from the zero modes. The ADHM frame obeys ψ ( ξ ; x + ℓ 4 e ^ 4 ) = e i ℓ 4 ξ 4 ψ ( ξ ; x ) C − 1 \psi(\xi;x+\ell_4\hat{e}_4)=e^{i\ell_4\xi_4}\psi(\xi;x)C^{-1} ψ ( ξ ; x + ℓ 4 e ^ 4 ) = e i ℓ 4 ξ 4 ψ ( ξ ; x ) C − 1 (31) . Computing ψ ( ξ ; x + ℓ 3 e ^ 3 + ℓ 4 e ^ 4 ) \psi(\xi;x+\ell_3\hat{e}_3+\ell_4\hat{e}_4) ψ ( ξ ; x + ℓ 3 e ^ 3 + ℓ 4 e ^ 4 ) ξ 4 ↦ ξ 4 − 2 π p N L 4 \xi_4\mapsto\xi_4-\frac{2\pi p}{NL_4} ξ 4 ↦ ξ 4 − N L 4 2 π p e i ℓ 4 ξ 4 e^{i\ell_4\xi_4} e i ℓ 4 ξ 4 e − i ℓ 4 ⋅ 2 π p / ( N L 4 ) = ω − p e^{-i\ell_4\cdot2\pi p/(NL_4)}=\omega^{-p} e − i ℓ 4 ⋅ 2 π p / ( N L 4 ) = ω − p Ω 3 ( x + ℓ 4 e ^ 4 ) = ω − p C Ω 3 ( x ) C − 1 \Omega_3(x+\ell_4\hat{e}_4)=\omega^{-p}C\Omega_3(x)C^{-1} Ω 3 ( x + ℓ 4 e ^ 4 ) = ω − p C Ω 3 ( x ) C − 1 Ω 3 ( x ) \Omega_3(x) Ω 3 ( x ) Ω 4 = C \Omega_4=C Ω 4 = C Ω 3 ( x + ℓ 4 e ^ 4 ) Ω 4 = ω − p Ω 4 Ω 3 ( x ) \Omega_3(x+\ell_4\hat{e}_4)\Omega_4=\omega^{-p}\Omega_4\Omega_3(x) Ω 3 ( x + ℓ 4 e ^ 4 ) Ω 4 = ω − p Ω 4 Ω 3 ( x ) R 2 × T p / N -twisted 2 \mathbb{R}^{2}\times T^{2}_{p/N\text{-twisted}} R 2 × T p / N -twisted 2 Q t o p = 1 / N Q_{\mathrm{top}}=1/N Q top = 1/ N (4) with P 3 = S − p P_3=S^{-p} P 3 = S − p P 4 = C P_4=C P 4 = C (62) holds exactly.
In § 4 , we shall see that the data (66) put all the constituents at the common transverse position y ⊥ y_\perp y ⊥ L 3 κ m ≡ q ℓ 3 L_3\kappa_m\equiv q\ell_3 L 3 κ m ≡ q ℓ 3 x 3 x_3 x 3
Monopole-center vortex chain in the far-field limit
In this section, we derive the Abelian gauge potential in the far-field limit studied in Sec. 2. To understand the monopole-center vortex array of the doubly-periodic instanton, we need the ( x , x 3 ) (x,x_3) ( x , x 3 ) x 4 x_4 x 4 L 4 L_4 L 4 [33 , 34 ] : Their Nahm parameter z ∈ [ 0 , 1 ] z\in[0,1] z ∈ [ 0 , 1 ] ξ 4 ∈ [ 0 , 2 π / L 4 ] \xi_4\in[0,2\pi/L_4] ξ 4 ∈ [ 0 , 2 π / L 4 ] k × k k\times k k × k L 2 ( S 2 π / L 3 1 ) L^2(S^1_{2\pi/L_3}) L 2 ( S 2 π / L 3 1 ) ξ 3 \xi_3 ξ 3
Far-field limit, the strip temporal gauge, and Abelianization
Here, we explain the setup, where we perform the concrete calculation for constructing the self-dual gauge field A μ A_\mu A μ P 4 = C P_4=C P 4 = C μ 4 , m = 2 π m N L 4 \mu_{4,m}=\frac{2\pi m}{NL_4} μ 4 , m = N L 4 2 π m m = 1 , … , N m=1,\ldots,N m = 1 , … , N N L 4 ≪ L 3 NL_4\ll L_3 N L 4 ≪ L 3 8
μ 4 , N + 1 = μ 4 , 1 + 2 π L 4 \mu_{4,N+1}=\mu_{4,1}+\frac{2\pi}{L_4} μ 4 , N + 1 = μ 4 , 1 + L 4 2 π T ^ 2 \widehat{T}^{2} T 2 N N N Figure 2 ),
Figure 2. Dual torus T ^ 2 \widehat{T}^{2} T 2 N = 3 N=3 N = 3 μ m = ( μ 3 , m , μ 4 , m ) \mu_m=(\mu_{3,m},\mu_{4,m}) μ m = ( μ 3 , m , μ 4 , m ) ξ 4 = μ 4 , m \xi_4=\mu_{4,m} ξ 4 = μ 4 , m 2 π / ( N L 4 ) 2\pi/(NL_4) 2 π / ( N L 4 ) I m = { μ 4 , m < ξ 4 < μ 4 , m + 1 } \mathcal{I}_m=\{\mu_{4,m}<\xi_4<\mu_{4,m+1}\} I m = { μ 4 , m < ξ 4 < μ 4 , m + 1 } F ^ 34 = 0 \widehat{F}_{34}=0 F 34 = 0 A ^ 4 = 0 \widehat{A}_4=0 A 4 = 0 A ^ 3 \widehat{A}_3 A 3 H ^ \widehat{H} H ξ 3 \xi_3 ξ 3 y 3 , m y_{3,m} y 3 , m y ⊥ , m y_{\perp,m} y ⊥ , m Y 3 , m Y_{3,m} Y 3 , m y 3 , m + L 3 Z y_{3,m}+L_3\mathbb{Z} y 3 , m + L 3 Z
I m = { ( ξ 3 , ξ 4 ) ∈ T ^ 2 | μ 4 , m < ξ 4 < μ 4 , m + 1 } . (68) \mathcal{I}_m=\left\{(\xi_3,\xi_4)\in\widehat{T}^{2}\mathrel{}\middle|\mathrel{}\mu_{4,m}<\xi_4<\mu_{4,m+1}\right\}. \tag*{(68)} I m = { ( ξ 3 , ξ 4 ) ∈ T 2 μ 4 , m < ξ 4 < μ 4 , m + 1 } . (68) Let us discuss the behaviors of the U ( 1 ) U(1) U ( 1 ) A ^ a = 3 , 4 \widehat{A}_{a=3,4} A a = 3 , 4 H ^ \widehat{H} H I m \mathcal{I}_m I m (42) can be solved with using the theta functions to obtain A ^ a \widehat{A}_a A a H ^ \widehat{H} H
y ⊥ ( ξ 4 ) = y 1 ( ξ 4 ) + i y 2 ( ξ 4 ) : = L 3 2 π ∫ 0 2 π / L 3 d ξ 3 H ^ ( ξ 3 , ξ 4 ) , (69) y_{\perp}(\xi_4)=y_1(\xi_4)+\mathrm{i}y_2(\xi_4):=\frac{L_3}{2\pi}\int_0^{2\pi/L_3}\mathrm{d}\xi_3\widehat{H}(\xi_3,\xi_4), \tag*{(69)} y ⊥ ( ξ 4 ) = y 1 ( ξ 4 ) + i y 2 ( ξ 4 ) := 2 π L 3 ∫ 0 2 π / L 3 d ξ 3 H ( ξ 3 , ξ 4 ) , (69) and then the Hitchin equation (42) tells
∂ ξ 4 y ⊥ ( ξ 4 ) = L 3 ∑ m α m δ ( ξ 4 − μ 4 , m ) . (70) \partial_{\xi_4}y_{\perp}(\xi_4)=L_3\sum_m\alpha_m\delta(\xi_4-\mu_{4,m}). \tag*{(70)} ∂ ξ 4 y ⊥ ( ξ 4 ) = L 3 m ∑ α m δ ( ξ 4 − μ 4 , m ) . (70) Thus, inside each segment I m \mathcal{I}_m I m y ⊥ ( ξ 4 ) y_{\perp}(\xi_4) y ⊥ ( ξ 4 )
y ⊥ , m = y 1 , m + i y 2 , m : = y ⊥ ∣ I m . (71) y_{\perp,m}=y_{1,m}+\mathrm{i}y_{2,m}:=y_{\perp}|_{\mathcal{I}_m}. \tag*{(71)} y ⊥ , m = y 1 , m + i y 2 , m := y ⊥ ∣ I m . (71) They are constrained by α m \alpha_m α m y ⊥ , m − y ⊥ , m − 1 = L 3 α m y_{\perp,m}-y_{\perp,m-1}=L_3\alpha_m y ⊥ , m − y ⊥ , m − 1 = L 3 α m (42) for ξ 4 ≈ μ 4 , m \xi_4\approx\mu_{4,m} ξ 4 ≈ μ 4 , m I m − 1 ∪ I m \mathcal{I}_{m-1}\cup\mathcal{I}_m I m − 1 ∪ I m N L 4 ≪ L 3 NL_4\ll L_3 N L 4 ≪ L 3 H ^ → y ⊥ , m \widehat{H}\to y_{\perp,m} H → y ⊥ , m ξ 4 → + ∞ \xi_4\to+\infty ξ 4 → + ∞ H ^ → y ⊥ , m − 1 \widehat{H}\to y_{\perp,m-1} H → y ⊥ , m − 1 ξ 4 → − ∞ \xi_4\to-\infty ξ 4 → − ∞
H ^ ( ξ 3 , ξ 4 ) = y ⊥ , m + y ⊥ , m − 1 2 − i ( y ⊥ , m − y ⊥ , m − 1 ) 2 cot L 3 { ( ξ 3 − μ 3 , m ) − i ( ξ 4 − μ 4 , m ) } 2 . (72) \widehat{H}(\xi_3,\xi_4)=\frac{y_{\perp,m}+y_{\perp,m-1}}{2}-\frac{\mathrm{i}(y_{\perp,m}-y_{\perp,m-1})}{2}\cot\frac{L_3\{(\xi_3-\mu_{3,m})-\mathrm{i}(\xi_4-\mu_{4,m})\}}{2}. \tag*{(72)} H ( ξ 3 , ξ 4 ) = 2 y ⊥ , m + y ⊥ , m − 1 − 2 i ( y ⊥ , m − y ⊥ , m − 1 ) cot 2 L 3 {( ξ 3 − μ 3 , m ) − i ( ξ 4 − μ 4 , m )} . (72) Thus, H ^ \widehat{H} H y ⊥ , m y_{\perp,m} y ⊥ , m O ( e − L 3 ∣ ξ 4 − μ 4 , m ∣ ) O(e^{-L_3|\xi_4-\mu_{4,m}|}) O ( e − L 3 ∣ ξ 4 − μ 4 , m ∣ ) I m \mathcal{I}_m I m
For the U ( 1 ) U(1) U ( 1 ) A ^ 4 = 0 \widehat{A}_4=0 A 4 = 0 I m \mathcal{I}_m I m A ^ 3 \widehat{A}_3 A 3 ξ 3 \xi_3 ξ 3 I m \mathcal{I}_m I m F ^ 34 = 0 \widehat{F}_{34}=0 F 34 = 0
y 3 , m : = L 3 2 π ∮ 0 2 π / L 3 d ξ 3 A ^ 3 ∣ I m , (73) y_{3,m}:=\frac{L_3}{2\pi}\oint_{0}^{2\pi/L_3}d\xi_3\,\widehat{A}_3\big|_{\mathcal{I}_m}, \tag*{(73)} y 3 , m := 2 π L 3 ∮ 0 2 π / L 3 d ξ 3 A 3 I m , (73) and they are constrained by κ m \kappa_m κ m y 3 , m − y 3 , m − 1 = L 3 κ m y_{3,m}-y_{3,m-1}=L_3\kappa_m y 3 , m − y 3 , m − 1 = L 3 κ m
Y 3 , m : = D 3 ∣ I m = − i ∂ ξ 3 + A ^ 3 ∣ I m , (74) Y_{3,m}:=D_3\big|_{\mathcal{I}_m}=-i\partial_{\xi_3}+\widehat{A}_3\big|_{\mathcal{I}_m}, \tag*{(74)} Y 3 , m := D 3 I m = − i ∂ ξ 3 + A 3 I m , (74) and then its spectrum is given by
y 3 , m + L 3 Z . (75) y_{3,m}+L_3\mathbb{Z}. \tag*{(75)} y 3 , m + L 3 Z . (75) Combining the two, the strip I m \mathcal{I}_m I m ( y m : = ( y 1 , m , y 2 , m ) ) (\boldsymbol{y}_m:=(y_{1,m},y_{2,m})) ( y m := ( y 1 , m , y 2 , m ))
y ⃗ m , k : = ( y m , y 3 , m + k L 3 ) ∈ R 3 , \vec{y}_{m,k}:=(\boldsymbol{y}_m,y_{3,m}+kL_3)\in\mathbb{R}^3, y m , k := ( y m , y 3 , m + k L 3 ) ∈ R 3 , y ⃗ m , k − y ⃗ m − 1 , k = 1 L 4 ρ ⃗ m . (76) \vec{y}_{m,k}-\vec{y}_{m-1,k}=\frac{1}{L_4}\vec{\rho}_m. \tag*{(76)} y m , k − y m − 1 , k = L 4 1 ρ m . (76) The second relation combines the jumps of y ⊥ \boldsymbol{y}_\perp y ⊥ y 3 y_3 y 3 ∣ y ⃗ m , k − y ⃗ m − 1 , k ∣ = γ m L 3 |\vec{y}_{m,k}-\vec{y}_{m-1,k}|=\gamma_mL_3 ∣ y m , k − y m − 1 , k ∣ = γ m L 3 y ⃗ m , 0 = ( y m , y 3 , m ) ∈ R 2 × S L 3 1 \vec{y}_{m,0}=(\boldsymbol{y}_m,y_{3,m})\in\mathbb{R}^2\times S^1_{L_3} y m , 0 = ( y m , y 3 , m ) ∈ R 2 × S L 3 1 m m m [27 , 28 , 29 ] for the single-caloron case).
We define the far-field limit by
r min : = min m ∣ x ⊥ − y ⊥ , m ∣ ≫ N L 4 , (77) r_{\min}:=\min_m|\boldsymbol{x}_\perp-\boldsymbol{y}_{\perp,m}|\gg NL_4, \tag*{(77)} r m i n := m min ∣ x ⊥ − y ⊥ , m ∣ ≫ N L 4 , (77) where x ⊥ = x 1 + i x 2 \boldsymbol{x}_\perp=x_1+ix_2 x ⊥ = x 1 + i x 2 O ( e − 2 π r min / ( N L 4 ) ) O(e^{-2\pi r_{\min}/(NL_4)}) O ( e − 2 π r m i n / ( N L 4 ) ) f ^ x ( μ m , μ m + 1 ) \widehat{f}_{\boldsymbol{x}}(\mu_m,\mu_{m+1}) f x ( μ m , μ m + 1 ) ≫ 1 / L 3 \gg1/L_3 ≫ 1/ L 3 I m \mathcal{I}_m I m H ^ = y ⊥ , m \widehat{H}=\boldsymbol{y}_{\perp,m} H = y ⊥ , m
[ − ( ∂ ξ 4 − i x 4 ) 2 + ( Y 3 , m − x 3 ) 2 + ∣ y ⊥ , m − x ⊥ ∣ 2 ⏟ m a s s 2 ] f ^ x ( ξ , ξ ′ ) = V ^ δ ( 2 ) ( ξ − ξ ′ ) . (78) \left[-(\partial_{\xi_4}-ix_4)^2+(Y_{3,m}-x_3)^2+\underbrace{|\boldsymbol{y}_{\perp,m}-\boldsymbol{x}_\perp|^2}_{\mathrm{mass}^2}\right]\widehat{f}_{\boldsymbol{x}}(\xi,\xi')=\widehat{V}\delta^{(2)}(\xi-\xi'). \tag*{(78)} − ( ∂ ξ 4 − i x 4 ) 2 + ( Y 3 , m − x 3 ) 2 + mass 2 ∣ y ⊥ , m − x ⊥ ∣ 2 f x ( ξ , ξ ′ ) = V δ ( 2 ) ( ξ − ξ ′ ) . (78) Since m a s s 2 ≥ ∣ x ⊥ − y ⊥ , m ∣ 2 \mathrm{mass}^2\ge|\boldsymbol{x}_\perp-\boldsymbol{y}_{\perp,m}|^2 mass 2 ≥ ∣ x ⊥ − y ⊥ , m ∣ 2
∣ f ^ x ( ξ , ξ ′ ) ∣ ≲ exp ( − ∣ ξ 4 − ξ 4 ′ ∣ ∣ x ⊥ − y ⊥ , m ∣ ) . (79) |\widehat{f}_{\boldsymbol{x}}(\xi,\xi')|\lesssim\exp\left(-|\xi_4-\xi'_4|\,|\boldsymbol{x}_\perp-\boldsymbol{y}_{\perp,m}|\right). \tag*{(79)} ∣ f x ( ξ , ξ ′ ) ∣ ≲ exp ( − ∣ ξ 4 − ξ 4 ′ ∣ ∣ x ⊥ − y ⊥ , m ∣ ) . (79) The magnetic field and the δ \delta δ I m \mathcal{I}_m I m H ^ \widehat{H} H μ m \mu_m μ m H ^ ≃ − i α m / ( ξ ‾ − μ ‾ m ) \widehat{H}\simeq-i\alpha_m/(\overline{\xi}-\overline{\mu}_m) H ≃ − i α m / ( ξ − μ m ) ξ = ξ 3 + i ξ 4 \xi=\xi_3+i\xi_4 ξ = ξ 3 + i ξ 4 ξ → μ m \xi\to\mu_m ξ → μ m ξ ′ → μ m + 1 \xi'\to\mu_{m+1} ξ ′ → μ m + 1 δ \delta δ
f ^ x ( μ m , μ m + 1 ) ∼ Z m 1 / 2 Z m + 1 1 / 2 exp ( − 2 π N L 4 ∣ x ⊥ − y ⊥ , m ∣ ) . (80) \widehat{f}_{\boldsymbol{x}}(\mu_m,\mu_{m+1})\sim Z_m^{1/2}Z_{m+1}^{1/2}\exp\left(-\frac{2\pi}{NL_4}|\boldsymbol{x}_\perp-\boldsymbol{y}_{\perp,m}|\right). \tag*{(80)} f x ( μ m , μ m + 1 ) ∼ Z m 1/2 Z m + 1 1/2 exp ( − N L 4 2 π ∣ x ⊥ − y ⊥ , m ∣ ) . (80) For non-adjacent m m m m ′ m' m ′
[ Φ ~ ] m , m ′ , [ M ~ μ ν ] m m ′ = O ( e − 2 π r min / ( N L 4 ) ) (81) [\widetilde{\Phi}]_{m,m'},[\widetilde{M}_{\mu\nu}]_{mm'}=O\left(e^{-2\pi r_{\min}/(NL_{4})}\right) \tag*{(81)} [ Φ ] m , m ′ , [ M μν ] m m ′ = O ( e − 2 π r m i n / ( N L 4 ) ) (81) when m ≠ m ′ m\ne m' m = m ′ N × N N\times N N × N
One can also easily understand this Abelianization from the viewpoint of Weyl zero modes. Each of N N N ψ m \psi_m ψ m ξ 4 = μ 4 , m \xi_4=\mu_{4,m} ξ 4 = μ 4 , m ∣ ξ 4 − μ 4 , m ∣ ≳ 1 / ∣ x ⃗ ⊥ − y ⃗ ⊥ , m ∣ |\xi_4-\mu_{4,m}|\gtrsim1/|\vec{x}_{\perp}-\vec{y}_{\perp,m}| ∣ ξ 4 − μ 4 , m ∣ ≳ 1/∣ x ⊥ − y ⊥ , m ∣ ∫ d 2 ξ ψ m † ∂ μ ψ n \int d^2\xi\psi_m^\dagger\partial_\mu\psi_n ∫ d 2 ξ ψ m † ∂ μ ψ n m ≠ n m\ne n m = n O ( e − 2 π r min / ( N L 4 ) ) O(e^{-2\pi r_{\min}/(NL_4)}) O ( e − 2 π r m i n / ( N L 4 ) )
Abelianized gauge field and the monopole–vortex chain
Due to the Abelianization, we can focus on computing the diagonal components of A μ A_\mu A μ f ^ x ⃗ ( μ m , μ m ) \hat{f}_{\vec{x}}(\mu_m,\mu_m) f ^ x ( μ m , μ m ) [33 , 34 ] . Let us write the Nahm data in the strip temporal gauge as
R ⃗ ( ξ 4 ; x ⃗ ) : = ( x 1 − Re H ^ x 2 − Im H ^ x 3 − Y 3 ) . (82) \vec{R}(\xi_4;\vec{x}):=
\begin{pmatrix}
x_1-\operatorname{Re}\hat{H}\\
x_2-\operatorname{Im}\hat{H}\\
x_3-Y_3
\end{pmatrix}. \tag*{(82)} R ( ξ 4 ; x ) := x 1 − Re H ^ x 2 − Im H ^ x 3 − Y 3 . (82) Its components are understood as the operator L 2 ( S 2 π / L 3 1 ) → L 2 ( S 2 π / L 3 1 ) L^2(S^1_{2\pi/L_3})\to L^2(S^1_{2\pi/L_3}) L 2 ( S 2 π / L 3 1 ) → L 2 ( S 2 π / L 3 1 ) (42) becomes ∂ ξ 4 R ⃗ = i R ⃗ × R ⃗ \partial_{\xi_4}\vec{R}=i\vec{R}\times\vec{R} ∂ ξ 4 R = i R × R ξ 4 ≠ μ m \xi_4\ne\mu_m ξ 4 = μ m ∣ δ m ⟩ |\delta_m\rangle ∣ δ m ⟩ ⟨ ξ 3 ∣ δ m ⟩ = δ ( ξ 3 − μ 3 , m ) \langle\xi_3|\delta_m\rangle=\delta(\xi_3-\mu_{3,m}) ⟨ ξ 3 ∣ δ m ⟩ = δ ( ξ 3 − μ 3 , m ) I m − 1 \mathcal{I}_{m-1} I m − 1 I m \mathcal{I}_m I m ξ 4 = μ 4 , m \xi_4=\mu_{4,m} ξ 4 = μ 4 , m
Y 3 , m − Y 3 , m − 1 = 2 π κ m ∣ δ m ⟩ ⟨ δ m ∣ , Y_{3,m}-Y_{3,m-1}=2\pi\kappa_m|\delta_m\rangle\langle\delta_m|, Y 3 , m − Y 3 , m − 1 = 2 π κ m ∣ δ m ⟩ ⟨ δ m ∣ , H ^ ( ⋅ , μ 4 , m + 0 + ) − H ^ ( ⋅ , μ 4 , m − 0 + ) = 2 π α m ∣ δ m ⟩ ⟨ δ m ∣ , (83) \hat{H}(\cdot,\mu_{4,m}+0^+)-\hat{H}(\cdot,\mu_{4,m}-0^+)=2\pi\alpha_m|\delta_m\rangle\langle\delta_m|, \tag*{(83)} H ^ ( ⋅ , μ 4 , m + 0 + ) − H ^ ( ⋅ , μ 4 , m − 0 + ) = 2 π α m ∣ δ m ⟩ ⟨ δ m ∣ , (83) using the matrix notation. Let us also define, with the unit vector n ^ m : = ρ ⃗ m / ∣ ρ ⃗ m ∣ = ( Re α m , Im α m , κ m ) / γ m \hat{n}_m:=\vec{\rho}_m/|\vec{\rho}_m|=(\operatorname{Re}\alpha_m,\operatorname{Im}\alpha_m,\kappa_m)/\gamma_m n ^ m := ρ m /∣ ρ m ∣ = ( Re α m , Im α m , κ m ) / γ m
S m : = 2 π γ m ∣ δ m ⟩ ⟨ δ m ∣ = − n ^ m ⋅ [ R ⃗ ( μ 4 , m + 0 + ) − R ⃗ ( μ 4 , m − 0 + ) ] , (84) \begin{aligned}
S_m:=2\pi\gamma_m|\delta_m\rangle\langle\delta_m| \\
&=-\hat{n}_m\cdot\left[\vec{R}(\mu_{4,m}+0^+)-\vec{R}(\mu_{4,m}-0^+)\right], \tag*{(84)}
\end{aligned} S m := 2 π γ m ∣ δ m ⟩ ⟨ δ m ∣ = − n ^ m ⋅ [ R ( μ 4 , m + 0 + ) − R ( μ 4 , m − 0 + ) ] , (84) which describes the impurity contribution:
[ − ∂ ξ 4 2 + R ⃗ ( ξ 4 ; x ⃗ ) 2 + ∑ m δ ( ξ 4 − μ 4 , m ) S m ] f ^ x ⃗ ( ξ , ξ ′ ) = V ^ δ ( 2 ) ( ξ − ξ ′ ) . (85) \left[-\partial_{\xi_4}^{2}+\vec{R}(\xi_4;\vec{x})^{2}+\sum_m\delta(\xi_4-\mu_{4,m})S_m\right]\hat{f}_{\vec{x}}(\xi,\xi')=\hat{V}\delta^{(2)}(\xi-\xi'). \tag*{(85)} [ − ∂ ξ 4 2 + R ( ξ 4 ; x ) 2 + m ∑ δ ( ξ 4 − μ 4 , m ) S m ] f ^ x ( ξ , ξ ′ ) = V ^ δ ( 2 ) ( ξ − ξ ′ ) . (85) Here we have dropped x 4 x_4 x 4 e i x 4 ξ 4 e^{ix_4\xi_4} e i x 4 ξ 4 f ^ x ⃗ ( μ m , μ m ) \hat{f}_{\vec{x}}(\mu_m,\mu_m) f ^ x ( μ m , μ m ) S m S_m S m ∣ δ m ⟩ ⟨ δ m ∣ |\delta_m\rangle\langle\delta_m| ∣ δ m ⟩ ⟨ δ m ∣ U ( 1 ) U(1) U ( 1 ) I m \mathcal{I}_m I m R ⃗ ( ξ 4 , x ⃗ ) \vec{R}(\xi_4,\vec{x}) R ( ξ 4 , x )
R ⃗ m = ( x 1 − y 1 , m x 2 − y 2 , m x 3 − Y 3 , m ) (86) \vec{R}_m =
\begin{pmatrix}
x_1-y_{1,m}\\
x_2-y_{2,m}\\
x_3-Y_{3,m}
\end{pmatrix} \tag*{(86)} R m = x 1 − y 1 , m x 2 − y 2 , m x 3 − Y 3 , m (86) and R m : = ∣ R ⃗ m ∣ = R ⃗ m ⋅ R ⃗ m R_m := |\vec{R}_m| = \sqrt{\vec{R}_m\cdot\vec{R}_m} R m := ∣ R m ∣ = R m ⋅ R m I m \mathcal{I}_m I m
Let us write ⟨ ξ 3 ∣ ψ ( ξ 4 ) ⟩ : = f ^ x ( ξ , μ m ) \langle\xi_3|\psi(\xi_4)\rangle:= \hat{f}_x(\xi,\mu_m) ⟨ ξ 3 ∣ ψ ( ξ 4 )⟩ := f ^ x ( ξ , μ m ) ψ ( ξ 4 ) → 0 \psi(\xi_4)\to0 ψ ( ξ 4 ) → 0 ξ 4 − μ 4 , m → ± ∞ \xi_4-\mu_{4,m}\to\pm\infty ξ 4 − μ 4 , m → ± ∞ ξ 4 = μ 4 , m ± 1 \xi_4=\mu_{4,m\pm1} ξ 4 = μ 4 , m ± 1 [33 ] , we introduce the solutions R m ± ( ξ 4 ) R_m^\pm(\xi_4) R m ± ( ξ 4 ) I m \mathcal{I}_m I m
( R m ± ) 2 ± ∂ ξ 4 R m ± = R ⃗ ⋅ R ⃗ , ξ 4 ∈ I m , (87) \left(R_m^\pm\right)^2 \pm\partial_{\xi_4}R_m^\pm= \vec{R}\cdot\vec{R}, \qquad\xi_4\in\mathcal{I}_m, \tag*{(87)} ( R m ± ) 2 ± ∂ ξ 4 R m ± = R ⋅ R , ξ 4 ∈ I m , (87) and R m ± ≈ ∣ R ⃗ m ∣ R_m^\pm\approx|\vec{R}_m| R m ± ≈ ∣ R m ∣ ξ 4 \xi_4 ξ 4 I m \mathcal{I}_m I m ψ \psi ψ ξ 4 = μ 4 , m \xi_4=\mu_{4,m} ξ 4 = μ 4 , m
ψ ( ξ 4 ) = { P exp [ − ∫ μ 4 , m ξ 4 R m − ( z ) d z ] ψ ( μ 4 , m ) ( ξ 4 ∈ I m ) , P exp [ + ∫ μ 4 , m ξ 4 R m − 1 + ( z ) d z ] ψ ( μ 4 , m ) ( ξ 4 ∈ I m − 1 ) . (88) \psi(\xi_4)=
\begin{cases}
\mathcal{P}\exp\left[-\int_{\mu_{4,m}}^{\xi_4}R_m^-(z)\,\mathrm{d}z\right]\psi(\mu_{4,m}) & (\xi_4\in\mathcal{I}_m),\\
\mathcal{P}\exp\left[+\int_{\mu_{4,m}}^{\xi_4}R_{m-1}^+(z)\,\mathrm{d}z\right]\psi(\mu_{4,m}) & (\xi_4\in\mathcal{I}_{m-1}).
\end{cases} \tag*{(88)} ψ ( ξ 4 ) = ⎩ ⎨ ⎧ P exp [ − ∫ μ 4 , m ξ 4 R m − ( z ) d z ] ψ ( μ 4 , m ) P exp [ + ∫ μ 4 , m ξ 4 R m − 1 + ( z ) d z ] ψ ( μ 4 , m ) ( ξ 4 ∈ I m ) , ( ξ 4 ∈ I m − 1 ) . (88) This shows ∂ ξ 4 ψ = − R m − ψ \partial_{\xi_4}\psi=-R_m^-\psi ∂ ξ 4 ψ = − R m − ψ ξ 4 ∈ I m \xi_4\in\mathcal{I}_m ξ 4 ∈ I m ∂ ξ 4 ψ = R m − 1 + ψ \partial_{\xi_4}\psi=R_{m-1}^+\psi ∂ ξ 4 ψ = R m − 1 + ψ ξ 4 ∈ I m − 1 \xi_4\in\mathcal{I}_{m-1} ξ 4 ∈ I m − 1 ξ 4 = μ 4 , m \xi_4=\mu_{4,m} ξ 4 = μ 4 , m
Σ m ∣ ψ ⟩ = V ^ ∣ δ m ⟩ , Σ m : = R m − ( μ 4 , m ) + R m − 1 + ( μ 4 , m ) + S m , (89) \Sigma_m|\psi\rangle=\hat{V}|\delta_m\rangle,\qquad\Sigma_m:=R_m^-(\mu_{4,m})+R_{m-1}^+(\mu_{4,m})+S_m, \tag*{(89)} Σ m ∣ ψ ⟩ = V ^ ∣ δ m ⟩ , Σ m := R m − ( μ 4 , m ) + R m − 1 + ( μ 4 , m ) + S m , (89) and we find the far-field expression for f ^ x ( μ m , μ m ) \hat{f}_x(\mu_m,\mu_m) f ^ x ( μ m , μ m ) [33 Sec.4.1] ,
f ^ x ( μ m , μ m ) = ⟨ δ m ∣ ψ ⟩ = V ^ ⟨ δ m ∣ Σ m − 1 ∣ δ m ⟩ . (90) \hat{f}_x(\mu_m,\mu_m)=\langle\delta_m|\psi\rangle=\hat{V}\langle\delta_m|\Sigma_m^{-1}|\delta_m\rangle. \tag*{(90)} f ^ x ( μ m , μ m ) = ⟨ δ m ∣ ψ ⟩ = V ^ ⟨ δ m ∣ Σ m − 1 ∣ δ m ⟩ . (90) Using the matrix determinant lemma, we can rewrite it as
[ φ − 1 ] m m = 1 − ∣ ρ ⃗ m ∣ π f ^ x ( μ m , μ m ) = det [ Σ m − 2 S m ] det Σ m = det [ R m − ( μ 4 , m ) + R m − 1 + ( μ 4 , m ) − S m ] det [ R m − ( μ 4 , m ) + R m − 1 + ( μ 4 , m ) + S m ] . (91) \begin{aligned}
[\varphi^{-1}]_{mm}=1-\frac{|\vec{\rho}_m|}{\pi}\hat{f}_x(\mu_m,\mu_m)=\frac{\det[\Sigma_m-2S_m]}{\det\Sigma_m} \\
&=\frac{\det[R_m^-(\mu_{4,m})+R_{m-1}^+(\mu_{4,m})-S_m]}{\det[R_m^-(\mu_{4,m})+R_{m-1}^+(\mu_{4,m})+S_m]}. \tag*{(91)}
\end{aligned} [ φ − 1 ] mm = 1 − π ∣ ρ m ∣ f ^ x ( μ m , μ m ) = det Σ m det [ Σ m − 2 S m ] = det [ R m − ( μ 4 , m ) + R m − 1 + ( μ 4 , m ) + S m ] det [ R m − ( μ 4 , m ) + R m − 1 + ( μ 4 , m ) − S m ] . (91) Here, it is crucial that S m ∝ ∣ δ m ⟩ ⟨ δ m ∣ S_m\propto|\delta_m\rangle\langle\delta_m| S m ∝ ∣ δ m ⟩ ⟨ δ m ∣ 9 § 3.3.2 , φ \varphi φ A μ A_\mu A μ
Evaluation of the determinant. For α m = 0 \alpha_m=0 α m = 0 R m − = R m R_m^- = R_m R m − = R m R m − 1 + = R m − 1 R_{m-1}^+ = R_{m-1} R m − 1 + = R m − 1 (91) can be evaluated as for the axially symmetric multi-caloron [33 Eqs. (88)–(90)] . For α m ≠ 0 \alpha_m \ne0 α m = 0 H ^ \hat{H} H ^ O ( 1 / L 3 ) O(1/L_3) O ( 1/ L 3 ) R ± R^\pm R ± ξ 4 = μ 4 , m \xi_4=\mu_{4,m} ξ 4 = μ 4 , m 10
Proposition 1 (Determinant formula). For general α m \alpha_m α m
det [ R m − + R m − 1 + − S m ] det [ R m − + R m − 1 + + S m ] ∣ ξ 4 = μ 4 , m = ∏ k ∈ Z ∣ x ⃗ − y ⃗ m , k ∣ + n ^ m ⋅ ( x ⃗ − y ⃗ m , k ) ∣ x ⃗ − y ⃗ m − 1 , k ∣ + n ^ m ⋅ ( x ⃗ − y ⃗ m − 1 , k ) . (92) \left.\frac{\det[R_m^-+R_{m-1}^+-S_m]}{\det[R_m^-+R_{m-1}^++S_m]}\right|_{\xi_4=\mu_{4,m}}=\prod_{k\in\mathbb{Z}}\frac{|\vec{x}-\vec{y}_{m,k}|+\hat{n}_m\cdot(\vec{x}-\vec{y}_{m,k})}{|\vec{x}-\vec{y}_{m-1,k}|+\hat{n}_m\cdot(\vec{x}-\vec{y}_{m-1,k})}. \tag*{(92)} det [ R m − + R m − 1 + + S m ] det [ R m − + R m − 1 + − S m ] ξ 4 = μ 4 , m = k ∈ Z ∏ ∣ x − y m − 1 , k ∣ + n ^ m ⋅ ( x − y m − 1 , k ) ∣ x − y m , k ∣ + n ^ m ⋅ ( x − y m , k ) . (92) Proof. Near ξ 4 = μ 4 , m \xi_4=\mu_{4,m} ξ 4 = μ 4 , m i D x iD_x i D x ∂ ξ 4 + τ ⃗ ⋅ R ⃗ \partial_{\xi_4}+\vec{\tau}\cdot\vec{R} ∂ ξ 4 + τ ⋅ R (34) , and the ADHM equation becomes
( ∂ ξ 4 + τ ⃗ ⋅ R ⃗ ) † ( ∂ ξ 4 + τ ⃗ ⋅ R ⃗ ) = ( − ∂ ξ 4 2 + R ⃗ ⋅ R ⃗ ) ⊗ 1 2 ( ξ 4 ≠ μ 4 , m ) , (93) (\partial_{\xi_4}+\vec{\tau}\cdot\vec{R})^\dagger(\partial_{\xi_4}+\vec{\tau}\cdot\vec{R})=(-\partial_{\xi_4}^2+\vec{R}\cdot\vec{R})\otimes\mathbb{1}_2\qquad(\xi_4\ne\mu_{4,m}), \tag*{(93)} ( ∂ ξ 4 + τ ⋅ R ) † ( ∂ ξ 4 + τ ⋅ R ) = ( − ∂ ξ 4 2 + R ⋅ R ) ⊗ 1 2 ( ξ 4 = μ 4 , m ) , (93) which follows from ∂ ξ 4 R ⃗ = i R ⃗ × R ⃗ \partial_{\xi_4}\vec{R}=i\vec{R}\times\vec{R} ∂ ξ 4 R = i R × R (83) and (84) , R ⃗ \vec{R} R − n ^ m S m -\hat{n}_mS_m − n ^ m S m ξ 4 = μ 4 , m \xi_4=\mu_{4,m} ξ 4 = μ 4 , m n ^ m \hat{n}_m n ^ m n ^ m ⋅ τ ⃗ \hat{n}_m\cdot\vec{\tau} n ^ m ⋅ τ
τ ⃗ ⋅ R ⃗ = ( a b † b − a ) , a : = n ^ m ⋅ R ⃗ , b : = ( e ^ 1 + i e ^ 2 ) ⋅ R ⃗ , \vec{\tau}\cdot\vec{R}=\begin{pmatrix} a & b^\dagger\\ b & -a \end{pmatrix},\qquad a:=\hat{n}_m\cdot\vec{R},\quad b:=(\hat{e}_1+i\hat{e}_2)\cdot\vec{R}, τ ⋅ R = ( a b b † − a ) , a := n ^ m ⋅ R , b := ( e ^ 1 + i e ^ 2 ) ⋅ R , where ( e ^ 1 , e ^ 2 , n ^ m ) (\hat{e}_1,\hat{e}_2,\hat{n}_m) ( e ^ 1 , e ^ 2 , n ^ m ) b b b μ 4 , m \mu_{4,m} μ 4 , m
a ( μ 4 , m − 0 + ) − a ( μ 4 , m + 0 + ) = S m . a(\mu_{4,m}-0^+)-a(\mu_{4,m}+0^+)=S_m. a ( μ 4 , m − 0 + ) − a ( μ 4 , m + 0 + ) = S m . Let Ψ + \Psi_+ Ψ + ( ∂ ξ 4 + τ ⃗ ⋅ R ⃗ ) Ψ + = 0 (\partial_{\xi_4}+\vec{\tau}\cdot\vec{R})\Psi_+=0 ( ∂ ξ 4 + τ ⋅ R ) Ψ + = 0 I m \mathcal{I}_m I m (93) , both spin components decay, so ∂ ξ 4 Ψ = − R m − Ψ \partial_{\xi_4}\Psi=-R_m^-\Psi ∂ ξ 4 Ψ = − R m − Ψ ( τ ⃗ ⋅ R ⃗ − R m − ) Ψ + = 0 (\vec{\tau}\cdot\vec{R}-R_m^-)\Psi_+=0 ( τ ⋅ R − R m − ) Ψ + = 0 Ψ + = ( ψ + Γ + ψ + ) \Psi_+=\begin{pmatrix}\psi_+\\ \Gamma_+\psi_+\end{pmatrix} Ψ + = ( ψ + Γ + ψ + )
R m − − a = b † Γ + , R m − + a = b Γ + − 1 . R_m^- - a=b^\dagger\Gamma_+,\qquad R_m^-+a=b\Gamma_+^{-1}. R m − − a = b † Γ + , R m − + a = b Γ + − 1 . Likewise, the solutions Ψ − \Psi_- Ψ − I m − 1 \mathcal{I}_{m-1} I m − 1 ( τ ⃗ ⋅ R ⃗ + R m − 1 + ) Ψ − = 0 (\vec{\tau}\cdot\vec{R}+R_{m-1}^+)\Psi_-=0 ( τ ⋅ R + R m − 1 + ) Ψ − = 0 Ψ − = ( ψ − Γ − ψ − ) \Psi_-=\begin{pmatrix}\psi_-\\ \Gamma_-\psi_-\end{pmatrix} Ψ − = ( ψ − Γ − ψ − ) R m − 1 + + a = − b † Γ − R_{m-1}^++a=-b^\dagger\Gamma_- R m − 1 + + a = − b † Γ − R m − 1 + − a = − b Γ − − 1 R_{m-1}^+-a=-b\Gamma_-^{-1} R m − 1 + − a = − b Γ − − 1 ξ 4 = μ 4 , m \xi_4=\mu_{4,m} ξ 4 = μ 4 , m a ( μ 4 , m + 0 + ) a(\mu_{4,m}+0^+) a ( μ 4 , m + 0 + ) a ( μ 4 , m − 0 + ) a(\mu_{4,m}-0^+) a ( μ 4 , m − 0 + )
R m − + R m − 1 + + S m = b † ( Γ + − Γ − ) , R m − + R m − 1 + − S m = − b Γ + − 1 ( Γ + − Γ − ) Γ − − 1 . (94) R_m^-+R_{m-1}^++S_m=b^\dagger(\Gamma_+-\Gamma_-),\qquad R_m^-+R_{m-1}^+-S_m=-b\Gamma_+^{-1}(\Gamma_+-\Gamma_-)\Gamma_-^{-1}. \tag*{(94)} R m − + R m − 1 + + S m = b † ( Γ + − Γ − ) , R m − + R m − 1 + − S m = − b Γ + − 1 ( Γ + − Γ − ) Γ − − 1 . (94) The non-commuting factor Γ + − Γ − \Gamma_+-\Gamma_- Γ + − Γ −
det [ R m − + R m − 1 + − S m ] det [ R m − + R m − 1 + + S m ] ∣ ξ 4 = μ 4 , m = 1 ∣ det Γ + det Γ − ∣ ∣ ξ 4 = μ 4 , m . (95) \left.\frac{\det[R_m^-+R_{m-1}^+-S_m]}{\det[R_m^-+R_{m-1}^++S_m]}\right|_{\xi_4=\mu_{4,m}}=\left.\frac{1}{|\det\Gamma_+\det\Gamma_-|}\right|_{\xi_4=\mu_{4,m}}. \tag*{(95)} det [ R m − + R m − 1 + + S m ] det [ R m − + R m − 1 + − S m ] ξ 4 = μ 4 , m = ∣ det Γ + det Γ − ∣ 1 ξ 4 = μ 4 , m . (95) Both spin components of a decaying solution in I m \mathcal{I}_m I m P exp ( − ∫ R m d ξ 4 ) \mathcal{P}\exp(-\int R_m\,\mathrm{d}\xi_4) P exp ( − ∫ R m d ξ 4 ) Γ + ( ξ 4 ) \Gamma_+(\xi_4) Γ + ( ξ 4 ) I m \mathcal{I}_m I m ∂ ξ 4 Γ + = − [ R m , Γ + ] \partial_{\xi_4}\Gamma_+ = -[R_m,\Gamma_+] ∂ ξ 4 Γ + = − [ R m , Γ + ] det Γ + \det\Gamma_+ det Γ + ξ 4 \xi_4 ξ 4 Γ − \Gamma_- Γ − I m − 1 \mathcal{I}_{m-1} I m − 1 R ⃗ = R ⃗ j \vec{R}=\vec{R}_j R = R j Y 3 , j Y_{3,j} Y 3 , j R ∓ = R j R^\mp=R_j R ∓ = R j
Γ + = ( R m + n ^ m ⋅ R ⃗ m ) − 1 b m , Γ − = − ( b m − 1 † ) − 1 ( R m − 1 + n ^ m ⋅ R ⃗ m − 1 ) , (96) \Gamma_+ = (R_m+\hat{n}_m\cdot\vec{R}_m)^{-1}b_m,\qquad\Gamma_- = -(b_{m-1}^{\dagger})^{-1}(R_{m-1}+\hat{n}_m\cdot\vec{R}_{m-1}), \tag*{(96)} Γ + = ( R m + n ^ m ⋅ R m ) − 1 b m , Γ − = − ( b m − 1 † ) − 1 ( R m − 1 + n ^ m ⋅ R m − 1 ) , (96) where b j b_j b j b b b I j \mathcal{I}_j I j 11 y ⃗ m , k − y ⃗ m − 1 , k \vec{y}_{m,k}-\vec{y}_{m-1,k} y m , k − y m − 1 , k n ^ m \hat{n}_m n ^ m (76) , b m b_m b m b m − 1 b_{m-1} b m − 1 ∣ det b m ∣ = ∣ det b m − 1 ∣ |\det b_m|=|\det b_{m-1}| ∣ det b m ∣ = ∣ det b m − 1 ∣ det ( R m + n ^ m ⋅ R ⃗ m ) / det ( R m − 1 + n ^ m ⋅ R ⃗ m − 1 ) \det(R_m+\hat{n}_m\cdot\vec{R}_m)/\det(R_{m-1}+\hat{n}_m\cdot\vec{R}_{m-1}) det ( R m + n ^ m ⋅ R m ) / det ( R m − 1 + n ^ m ⋅ R m − 1 ) Y 3 , j Y_{3,j} Y 3 , j y 3 , j + k L 3 y_{3,j}+kL_3 y 3 , j + k L 3 R j + n ^ m ⋅ R ⃗ j = ∣ x ⃗ − y ⃗ j , k ∣ + n ^ m ⋅ ( x ⃗ − y ⃗ j , k ) R_j+\hat{n}_m\cdot\vec{R}_j=|\vec{x}-\vec{y}_{j,k}|+\hat{n}_m\cdot(\vec{x}-\vec{y}_{j,k}) R j + n ^ m ⋅ R j = ∣ x − y j , k ∣ + n ^ m ⋅ ( x − y j , k ) (92) .
Abelian gauge field and monopole chain
We now have all the ingredients to evaluate the gauge field A μ A_\mu A μ y ⃗ m − 1 , k ( s ) : = y ⃗ m − 1 , k + s n ^ m \vec{y}_{m-1,k}(s):=\vec{y}_{m-1,k}+s\hat{n}_m y m − 1 , k ( s ) := y m − 1 , k + s n ^ m d d s log [ ∣ x ⃗ − y ⃗ m − 1 , k ( s ) ∣ + n ^ m ⋅ ( x ⃗ − y ⃗ m − 1 , k ( s ) ) ] = − 1 / ∣ x ⃗ − y ⃗ m − 1 , k ( s ) ∣ \frac{\mathrm{d}}{\mathrm{d}s}\log\left[|\vec{x}-\vec{y}_{m-1,k}(s)|+\hat{n}_m\cdot(\vec{x}-\vec{y}_{m-1,k}(s))\right]=-1/|\vec{x}-\vec{y}_{m-1,k}(s)| d s d log [ ∣ x − y m − 1 , k ( s ) ∣ + n ^ m ⋅ ( x − y m − 1 , k ( s )) ] = − 1/∣ x − y m − 1 , k ( s ) ∣ y ⃗ m − 1 , k ( γ m L 3 ) = y ⃗ m , k \vec{y}_{m-1,k}(\gamma_mL_3)=\vec{y}_{m,k} y m − 1 , k ( γ m L 3 ) = y m , k (76) . Thus (91) and (92) give
− log [ φ − 1 ] m m = ∑ k ∫ 0 γ m L 3 d s ∣ x ⃗ − y ⃗ m − 1 , k ( s ) ∣ , (97) -\log[\varphi^{-1}]_{mm}=\sum_k\int_0^{\gamma_mL_3}\frac{\mathrm{d}s}{|\vec{x}-\vec{y}_{m-1,k}(s)|}, \tag*{(97)} − log [ φ − 1 ] mm = k ∑ ∫ 0 γ m L 3 ∣ x − y m − 1 , k ( s ) ∣ d s , (97) which is the Coulomb potential of a periodic array of uniformly charged rods. The k k k ≃ γ m / ∣ k ∣ \simeq\gamma_m/|k| ≃ γ m /∣ k ∣ ∣ k ∣ |k| ∣ k ∣ ∣ k ∣ ≤ K |k|\le K ∣ k ∣ ≤ K 2 γ m log K 2\gamma_m\log K 2 γ m log K x x x − log Z m -\log Z_m − log Z m K ∼ 1 / ( ϵ L 3 ) K\sim1/(\epsilon L_3) K ∼ 1/ ( ϵ L 3 ) μ m \mu_m μ m x x x Φ ~ m \widetilde{\Phi}_m Φ m Φ ~ \widetilde{\Phi} Φ
log Φ ~ m ( x ) = lim K → ∞ [ ∑ ∣ k ∣ ≤ K ∫ 0 γ m L 3 d s ∣ x ⃗ − y ⃗ m − 1 , k ( s ) ∣ − 2 γ m log K ] + c o n s t . (98) \log\widetilde{\Phi}_m(x)=\lim_{K\to\infty}\left[\sum_{|k|\le K}\int_0^{\gamma_mL_3}\frac{\mathrm{d}s}{|\vec{x}-\vec{y}_{m-1,k}(s)|}-2\gamma_m\log K\right]+\mathrm{const.} \tag*{(98)} log Φ m ( x ) = K → ∞ lim ∣ k ∣ ≤ K ∑ ∫ 0 γ m L 3 ∣ x − y m − 1 , k ( s ) ∣ d s − 2 γ m log K + const. (98) up to O ( e − 2 π r min / ( N L 4 ) ) O(e^{-2\pi r_{\min}/(NL_4)}) O ( e − 2 π r m i n / ( N L 4 ) ) (98) enter the gauge field below, and they are absolutely convergent sums.
Up to O ( e − 2 π r min / ( N L 4 ) ) O(e^{-2\pi r_{\min}/(NL_4)}) O ( e − 2 π r m i n / ( N L 4 ) ) Φ ~ \widetilde{\Phi} Φ M ~ μ ν \widetilde{M}_{\mu\nu} M μν (58) vanishes, and L = Φ ~ 1 / 2 L=\widetilde{\Phi}^{1/2} L = Φ 1/2 (54) , [ M ~ μ ν ] m m = i η ˉ μ ν l n ^ m l Φ ~ m − 1 [\widetilde{M}_{\mu\nu}]_{mm}=i\bar{\eta}^{l}_{\mu\nu}\hat{n}^{l}_m\widetilde{\Phi}^{-1}_m [ M μν ] mm = i η ˉ μν l n ^ m l Φ m − 1 (58) gives
[ A μ ] m m = − 1 2 η ˉ μ ν l n ^ m l ∂ ν log Φ ~ m . (99) [A_\mu]_{mm}=-\frac{1}{2}\bar{\eta}^{l}_{\mu\nu}\hat{n}^{l}_m\,\partial_\nu\log\widetilde{\Phi}_m. \tag*{(99)} [ A μ ] mm = − 2 1 η ˉ μν l n ^ m l ∂ ν log Φ m . (99) With η ˉ 4 j l = δ l j \bar{\eta}^{l}_{4j}=\delta_{lj} η ˉ 4 j l = δ l j η ˉ j k l = ϵ l j k \bar{\eta}^{l}_{jk}=\epsilon_{ljk} η ˉ j k l = ϵ l j k ∇ = ( ∂ 1 , ∂ 2 , ∂ 3 ) \nabla=(\partial_1,\partial_2,\partial_3) ∇ = ( ∂ 1 , ∂ 2 , ∂ 3 )
[ A 4 ] m m = − 1 2 n ^ m ⋅ ∇ log Φ ~ m , [A_4]_{mm}=-\frac{1}{2}\hat{n}_m\cdot\nabla\log\widetilde{\Phi}_m, [ A 4 ] mm = − 2 1 n ^ m ⋅ ∇ log Φ m , ( [ A 1 ] m m , [ A 2 ] m m , [ A 3 ] m m ) = 1 2 n ^ m × ∇ log Φ ~ m . (100) ([A_1]_{mm},[A_2]_{mm},[A_3]_{mm})=\frac{1}{2}\hat{n}_m\times\nabla\log\widetilde{\Phi}_m. \tag*{(100)} ([ A 1 ] mm , [ A 2 ] mm , [ A 3 ] mm ) = 2 1 n ^ m × ∇ log Φ m . (100) Since log Φ ~ m \log\widetilde{\Phi}_m log Φ m
∇ × A ⃗ m m = − 1 2 ( n ^ m ⋅ ∇ ) ∇ log Φ ~ m = ∇ [ A 4 ] m m , (101) \nabla\times\vec{A}_{mm}=-\frac{1}{2}(\hat{n}_m\cdot\nabla)\nabla\log\widetilde{\Phi}_m=\nabla[A_4]_{mm}, \tag*{(101)} ∇ × A mm = − 2 1 ( n ^ m ⋅ ∇ ) ∇ log Φ m = ∇ [ A 4 ] mm , (101) and thus (100) is an Abelian self-dual field. There is no x 4 x_4 x 4
Shifting x ⃗ \vec{x} x n ^ m \hat{n}_m n ^ m n ^ m ⋅ ∇ log Φ ~ m = ∑ k ( ∣ x ⃗ − y ⃗ m − 1 , k ∣ − 1 − ∣ x ⃗ − y ⃗ m , k ∣ − 1 ) \hat{n}_m\cdot\nabla\log\widetilde{\Phi}_m=\sum_k\left(|\vec{x}-\vec{y}_{m-1,k}|^{-1}-|\vec{x}-\vec{y}_{m,k}|^{-1}\right) n ^ m ⋅ ∇ log Φ m = ∑ k ( ∣ x − y m − 1 , k ∣ − 1 − ∣ x − y m , k ∣ − 1 ) μ m \mu_m μ m m m m ∑ m ρ ⃗ m = 0 \sum_m\vec{\rho}_m=0 ∑ m ρ m = 0 y ⃗ 0 , k = y ⃗ N , k \vec{y}_{0,k}=\vec{y}_{N,k} y 0 , k = y N , k
∇ 2 A 4 A b e l = − 2 π ∑ k ∈ Z ∑ m = 1 N α ⃗ m δ ( 3 ) ( x ⃗ − y ⃗ m , k ) , (102) \nabla^2 A_4^{\mathrm{Abel}}=-2\pi\sum_{k\in\mathbb{Z}}\sum_{m=1}^{N}\vec{\alpha}_m\,\delta^{(3)}(\vec{x}-\vec{y}_{m,k}), \tag*{(102)} ∇ 2 A 4 Abel = − 2 π k ∈ Z ∑ m = 1 ∑ N α m δ ( 3 ) ( x − y m , k ) , (102) which is nothing but (16) . Thus, in the far field, the doubly-periodic instanton is a periodic chain of the N N N α ⃗ m \vec{\alpha}_m α m ( y m , y 3 , m ) (y_m,y_{3,m}) ( y m , y 3 , m ) L 3 L_3 L 3 y ⃗ m − y ⃗ m − 1 = L 3 ( Re α m , Im α m ) \vec{y}_m-\vec{y}_{m-1}=L_3(\operatorname{Re}\alpha_m,\operatorname{Im}\alpha_m) y m − y m − 1 = L 3 ( Re α m , Im α m ) y 3 , m − y 3 , m − 1 = L 3 κ m y_{3,m}-y_{3,m-1}=L_3\kappa_m y 3 , m − y 3 , m − 1 = L 3 κ m
Wilson loop on R 2 \mathbb{R}^2 R 2 Lastly, we compute the Wilson loop along a large circle C R = { ∣ x ⃗ ∣ = R } C_R=\{|\vec{x}|=R\} C R = { ∣ x ∣ = R } R 2 \mathbb{R}^2 R 2 ( x 3 , x 4 ) (x_3,x_4) ( x 3 , x 4 ) O ( e − 2 π R / ( N L 4 ) ) O(e^{-2\pi R/(NL_4)}) O ( e − 2 π R / ( N L 4 ) ) (100) . For ∣ x ⃗ ∣ = R ≫ L 3 , ∣ y ⃗ m ∣ |\vec{x}|=R\gg L_3,|\vec{y}_m| ∣ x ∣ = R ≫ L 3 , ∣ y m ∣ (98) as
log Φ ~ m ≃ lim K → ∞ [ γ m L 3 ∑ k = − K K 1 ∣ x ⃗ ∣ 2 + ( k L 3 ) 2 − 2 γ m log K + c o n s t . ] ≃ lim K → ∞ [ γ m ∫ − K L 3 K L 3 d z ∣ x ⃗ ∣ 2 + z 2 − 2 γ m log K + c o n s t . ] = − 2 γ m log ∣ x ⃗ ∣ + c o n s t . + O ( ∣ x ⃗ ∣ − 1 ) , (103) \begin{aligned}
\log\widetilde{\Phi}_m &\simeq\lim_{K\to\infty}\left[\gamma_mL_3\sum_{k=-K}^{K}\frac{1}{\sqrt{|\vec{x}|^2+(kL_3)^2}}-2\gamma_m\log K+\mathrm{const.}\right]\\
&\simeq\lim_{K\to\infty}\left[\gamma_m\int_{-KL_3}^{KL_3}\frac{\mathrm{d}z}{\sqrt{|\vec{x}|^2+z^2}}-2\gamma_m\log K+\mathrm{const.}\right]\\
&=-2\gamma_m\log|\vec{x}|+\mathrm{const.}+O(|\vec{x}|^{-1}), \tag*{(103)}
\end{aligned} log Φ m ≃ K → ∞ lim [ γ m L 3 k = − K ∑ K ∣ x ∣ 2 + ( k L 3 ) 2 1 − 2 γ m log K + const. ] ≃ K → ∞ lim [ γ m ∫ − K L 3 K L 3 ∣ x ∣ 2 + z 2 d z − 2 γ m log K + const. ] = − 2 γ m log ∣ x ∣ + const. + O ( ∣ x ∣ − 1 ) , (103) where the x 3 x_3 x 3 e − 2 π ∣ x ⃗ ∣ / L 3 e^{-2\pi|\vec{x}|/L_3} e − 2 π ∣ x ∣/ L 3 (100) , only the x 3 x_3 x 3 n ^ m \hat{n}_m n ^ m γ m n ^ m 3 = κ m \gamma_m\hat{n}_m^3=\kappa_m γ m n ^ m 3 = κ m
( [ A 1 ] m m , [ A 2 ] m m ) = κ m ( x 2 , − x 1 ) ∣ x ⃗ ∣ 2 + O ( ∣ x ⃗ ∣ − 2 ) , \left([A_1]_{mm},[A_2]_{mm}\right)=\kappa_m\frac{(x_2,-x_1)}{|\vec{x}|^2}+O(|\vec{x}|^{-2}), ( [ A 1 ] mm , [ A 2 ] mm ) = κ m ∣ x ∣ 2 ( x 2 , − x 1 ) + O ( ∣ x ∣ − 2 ) , ∮ C R [ A ] m m = − 2 π κ m + O ( R − 1 ) . (104) \oint_{C_R}[A]_{mm}=-2\pi\kappa_m+O(R^{-1}). \tag*{(104)} ∮ C R [ A ] mm = − 2 π κ m + O ( R − 1 ) . (104) The Wilson loop at C R → ∞ C_R\to\infty C R → ∞
W ( C ∞ ) = lim R → ∞ P exp ( − i ∮ C R A ) = diag ( e 2 π i κ 1 , … , e 2 π i κ N ) . (105) W(C_\infty)=\lim_{R\to\infty}\mathcal{P}\exp\left(-i\oint_{C_R}A\right)=\operatorname{diag}\left(e^{2\pi i\kappa_1},\ldots,e^{2\pi i\kappa_N}\right). \tag*{(105)} W ( C ∞ ) = R → ∞ lim P exp ( − i ∮ C R A ) = diag ( e 2 π i κ 1 , … , e 2 π i κ N ) . (105) For generic impurity data, W ( C ∞ ) W(C_\infty) W ( C ∞ ) S U ( N ) SU(N) S U ( N ) 2 π κ m 2\pi\kappa_m 2 π κ m m m m R 2 \mathbb{R}^2 R 2 κ m ≡ q / N mod 1 \kappa_m\equiv q/N\ \operatorname{mod}1 κ m ≡ q / N mod 1 (105) becomes the center element,
W ( C ∞ ) = e 2 π i q / N 1 N . (106) W(C_\infty)=e^{2\pi iq/N}\mathbb{1}_N. \tag*{(106)} W ( C ∞ ) = e 2 π i q / N 1 N . (106) This reproduces the Wilson loop winding around the center vortex in Sec. 2, where q q q R 2 \mathbb{R}^{2} R 2 e − 2 π R / ( N L 4 ) e^{-2\pi R/(NL_{4})} e − 2 π R / ( N L 4 ) (105) is exact in the limit R → ∞ R \to\infty R → ∞
Summary and Discussion
Summary. In this paper, we have studied the S U ( N ) SU(N) S U ( N ) R 2 × T 2 \mathbb{R}^{2} \times T^{2} R 2 × T 2 [21 , 22 , 23 ] . Following Kraan and van Baal [27 , 28 , 29 ] (see also [35 , 42 ] ), we took the formal k → ∞ k \to\infty k → ∞ k k k R 4 \mathbb{R}^{4} R 4 Z 2 \mathbb{Z}^{2} Z 2 T 2 T^{2} T 2 T ^ 2 \widehat{T}^{2} T 2 N N N (42) . The positions μ m \mu_m μ m P 3 , 4 P_{3,4} P 3 , 4 κ m \kappa_m κ m α m \alpha_m α m (23) for A μ A_\mu A μ T ^ 2 \widehat{T}^{2} T 2 Φ ~ \widetilde{\Phi} Φ M ~ μ ν \widetilde{M}_{\mu\nu} M μν ϵ \epsilon ϵ (61) , and thus the reconstructed gauge field A μ A_\mu A μ ϵ → 0 \epsilon\to0 ϵ → 0
We then analyzed the far-field regime r min ≫ N L 4 r_{\min} \gg NL_{4} r m i n ≫ N L 4 (77) with P 4 = C P_{4}=C P 4 = C [33 , 34 ] . The Abelianization in the far-field limit can be shown due to the suppression of the off-diagonal components as e − 2 π r min / ( N L 4 ) e^{-2\pi r_{\min}/(NL_{4})} e − 2 π r m i n / ( N L 4 ) γ m L 3 \gamma_m L_{3} γ m L 3 ρ ⃗ m \vec{\rho}_m ρ m A 4 A b e l A_{4}^{\mathrm{Abel}} A 4 Abel (102) with the N N N y m − y m − 1 = L 3 ( Re α m , Im α m ) \boldsymbol{y}_m-\boldsymbol{y}_{m-1}=L_{3}(\operatorname{Re}\alpha_m,\operatorname{Im}\alpha_m) y m − y m − 1 = L 3 ( Re α m , Im α m ) y 3 , m − y 3 , m − 1 = L 3 κ m y_{3,m}-y_{3,m-1}=L_{3}\kappa_m y 3 , m − y 3 , m − 1 = L 3 κ m κ m ≡ q / N ( m o d 1 ) \kappa_m \equiv q/N \pmod{1} κ m ≡ q / N ( mod 1 ) α m = 0 \alpha_m=0 α m = 0 L 3 = N ℓ 3 L_{3}=N\ell_{3} L 3 = N ℓ 3 L 4 = ℓ 4 L_{4}=\ell_{4} L 4 = ℓ 4 R 2 × T p / N -twisted 2 \mathbb{R}^{2} \times T^{2}_{p/N\text{-twisted}} R 2 × T p / N -twisted 2 ℓ 3 ≫ N ℓ 4 \ell_{3} \gg N\ell_{4} ℓ 3 ≫ N ℓ 4
Moduli space of the fractional instanton. Let us discuss the moduli space M 1 / N \mathcal{M}_{1/N} M 1/ N R 2 × T p / N -twisted 2 \mathbb{R}^{2} \times T^{2}_{p/N\text{-twisted}} R 2 × T p / N -twisted 2 dim M 1 / N = 4 N Q t o p = 4 \dim\mathcal{M}_{1/N}=4NQ_{\mathrm{top}}=4 dim M 1/ N = 4 N Q top = 4 ( y , y 3 , 0 ) (\boldsymbol{y},y_{3,0}) ( y , y 3 , 0 ) x 4 x_{4} x 4 δ ν A μ = F ν μ \delta_\nu A_\mu=F_{\nu\mu} δ ν A μ = F ν μ tr F μ λ F ν λ = 1 4 δ μ ν tr F ρ σ F ρ σ \operatorname{tr}F_{\mu\lambda}F_{\nu\lambda}=\frac{1}{4}\delta_{\mu\nu}\operatorname{tr}F_{\rho\sigma}F_{\rho\sigma} tr F μ λ F ν λ = 4 1 δ μν tr F ρ σ F ρ σ ⟨ a , b ⟩ = 2 g 2 ∫ tr a μ b μ \langle a,b\rangle= \frac{2}{g^2}\int\operatorname{tr}a_\mu b_\mu ⟨ a , b ⟩ = g 2 2 ∫ tr a μ b μ
g μ ν = 2 g 2 ∫ d 4 x tr F μ λ F ν λ = ( S I / N ) δ μ ν , (107) g_{\mu\nu}=\frac{2}{g^2}\int\mathrm{d}^{4}x\operatorname{tr}F_{\mu\lambda}F_{\nu\lambda}=(S_I/N)\delta_{\mu\nu}, \tag*{(107)} g μν = g 2 2 ∫ d 4 x tr F μ λ F ν λ = ( S I / N ) δ μν , (107) where S I = 8 π 2 g 2 S_I=\frac{8\pi^2}{g^2} S I = g 2 8 π 2
Thus, the nontrivial part is to determine the global structure of M 1 / N \mathcal{M}_{1/N} M 1/ N x a ↦ x a + ℓ a x_a\mapsto x_a+\ell_a x a ↦ x a + ℓ a A ↦ P a A P a − 1 A\mapsto P_aAP_a^{-1} A ↦ P a A P a − 1 P 3 P 4 P 3 − 1 = e − 2 π i p / N P 4 P_3P_4P_3^{-1}=e^{-2\pi ip/N}P_4 P 3 P 4 P 3 − 1 = e − 2 π i p / N P 4 x 3 ↦ x 3 + ℓ 3 x_3\mapsto x_3+\ell_3 x 3 ↦ x 3 + ℓ 3 P 4 P 3 P 4 − 1 = e 2 π i p / N P 3 P_4P_3P_4^{-1}=e^{2\pi ip/N}P_3 P 4 P 3 P 4 − 1 = e 2 π i p / N P 3 x 4 ↦ x 4 + ℓ 4 x_4\mapsto x_4+\ell_4 x 4 ↦ x 4 + ℓ 4 x 3 x_3 x 3 § 4 , whose diagonal components are shifted by P 3 P_3 P 3 y 3 , 0 ∈ [ 0 , N ℓ 3 ) = [ 0 , L 3 ) y_{3,0}\in[0,N\ell_3)=[0,L_3) y 3 , 0 ∈ [ 0 , N ℓ 3 ) = [ 0 , L 3 ) x 4 x_4 x 4 N N N
M 1 / N ≃ R 2 × T N ℓ 3 , N ℓ 4 2 , vol M 1 / N = N 2 ℓ 3 ℓ 4 Area ( R 2 ) , (108) \mathcal{M}_{1/N}\simeq\mathbb{R}^2\times T^2_{N\ell_3,N\ell_4},\qquad\operatorname{vol}\mathcal{M}_{1/N}=N^2\ell_3\ell_4\operatorname{Area}(\mathbb{R}^2), \tag*{(108)} M 1/ N ≃ R 2 × T N ℓ 3 , N ℓ 4 2 , vol M 1/ N = N 2 ℓ 3 ℓ 4 Area ( R 2 ) , (108) which is the N 2 N^2 N 2 (5) . This is the counterpart of the observation by Davies, Hollowood, Khoze, and Mattis [75 ] for monopole-instantons on R 3 × S 1 \mathbb{R}^3\times S^1 R 3 × S 1 U ( 1 ) U(1) U ( 1 ) Ω ∈ [ 0 , 2 π ) \Omega\in[0,2\pi) Ω ∈ [ 0 , 2 π ) x 4 x_4 x 4 X 4 ∈ [ 0 , 2 π / v ) = [ 0 , N β ) X_4\in[0,2\pi/v)=[0,N\beta) X 4 ∈ [ 0 , 2 π / v ) = [ 0 , N β ) [ 0 , β ) [0,\beta) [ 0 , β ) Λ 3 / N \Lambda^3/N Λ 3 / N R 2 × T 2 \mathbb{R}^2\times T^2 R 2 × T 2 N N N R 3 × S 1 \mathbb{R}^3\times S^1 R 3 × S 1 ℓ 3 \ell_3 ℓ 3 [21 ] .
As an application of this structure, let us revisit the computation of the gluino condensate of N = 1 \mathcal{N}=1 N = 1 S U ( N ) SU(N) S U ( N ) R 2 × T 2 \mathbb{R}^2\times T^2 R 2 × T 2 R 3 × S 1 \mathbb{R}^3\times S^1 R 3 × S 1 [75 ] also evaluate ⟨ tr λ λ ⟩ \langle\operatorname{tr}\lambda\lambda\rangle ⟨ tr λλ ⟩ [76 ] likewise obtains the condensate from the superpotential. Here, the ’t Hooft twist fixes the vacuum classically, so either approach reduces ⟨ tr λ λ ⟩ \langle\operatorname{tr}\lambda\lambda\rangle ⟨ tr λλ ⟩
In the N = 1 \mathcal{N}=1 N = 1 [76 ] , the bosonic action is deformed by the nilpotent-SUSY-exact term ⊃ t ∫ ∣ F − ⋆ F ∣ 2 \supset t\int\lvert F-\star F\rvert^2 ⊃ t ∫ ∣ F − ⋆ F ∣ 2 t → ∞ t\to\infty t → ∞ [75 ] , the condensate is then given as follows:
⟨ tr λ λ ( x ) ⟩ = M P V 3 e − S I / N + i θ / N ∫ M 1 / N d 4 X ( 2 π ) 2 ( g 2 S I N ) 2 ∫ d 2 χ 2 g 2 S I / N tr λ λ ( x ; X , χ ) = M P V 3 e − S I / N + i θ / N ( g 2 S I / N ) 2 ( 2 π ) 2 × ∫ R 2 × T N ℓ 3 , N ℓ 4 2 d 4 X ∫ d 2 χ tr λ λ ( x ; X , χ ) ∫ R 2 × T ℓ 3 , ℓ 4 2 d 4 X ∫ d 2 χ tr λ λ ( x ; X , χ ) = 16 π 2 Λ 3 e i θ / N N 2 × N 2 , (109) \begin{aligned}
\langle\operatorname{tr}\lambda\lambda(x)\rangle&= M_{\mathrm{PV}}^{3}e^{-S_{I}/N+i\theta/N}\int_{\mathcal{M}_{1/N}}\frac{\mathrm{d}^{4}X}{(2\pi)^{2}}\left(\frac{g^{2}S_{I}}{N}\right)^{2}\int\frac{\mathrm{d}^{2}\chi}{2g^{2}S_{I}/N}\operatorname{tr}\lambda\lambda(x;X,\chi) \\
&= M_{\mathrm{PV}}^{3}e^{-S_{I}/N+i\theta/N}\frac{(g^{2}S_{I}/N)^{2}}{(2\pi)^{2}}\times\frac{\int_{\mathbb{R}^{2}\times T^{2}_{N\ell_{3},N\ell_{4}}\!}\mathrm{d}^{4}X\int\mathrm{d}^{2}\chi\,\operatorname{tr}\lambda\lambda(x;X,\chi)}{\int_{\mathbb{R}^{2}\times T^{2}_{\ell_{3},\ell_{4}}\!}\mathrm{d}^{4}X\int\mathrm{d}^{2}\chi\,\operatorname{tr}\lambda\lambda(x;X,\chi)} \\
&= \frac{16\pi^{2}\Lambda^{3}e^{i\theta/N}}{N^{2}}\times N^{2}, \tag*{(109)}
\end{aligned} ⟨ tr λλ ( x )⟩ = M PV 3 e − S I / N + i θ / N ∫ M 1/ N ( 2 π ) 2 d 4 X ( N g 2 S I ) 2 ∫ 2 g 2 S I / N d 2 χ tr λλ ( x ; X , χ ) = M PV 3 e − S I / N + i θ / N ( 2 π ) 2 ( g 2 S I / N ) 2 × ∫ R 2 × T ℓ 3 , ℓ 4 2 d 4 X ∫ d 2 χ tr λλ ( x ; X , χ ) ∫ R 2 × T N ℓ 3 , N ℓ 4 2 d 4 X ∫ d 2 χ tr λλ ( x ; X , χ ) = N 2 16 π 2 Λ 3 e i θ / N × N 2 , (109) where the strong scale is Λ 3 e i θ / N = M P V 3 e − S I / N + i θ / N \Lambda^{3}e^{i\theta/N}=M_{\mathrm{PV}}^{3}e^{-S_{I}/N+i\theta/N} Λ 3 e i θ / N = M PV 3 e − S I / N + i θ / N written with the four translational collective coordinates $X_{\mu}$. The two gluino zero-modes are generated by supersymmetry, $\lambda_{\alpha}=\frac{1}{2}\chi_{\beta}(\sigma_{\mu}\bar{\sigma}_{\nu})_{\alpha}^{\ \beta}F_{\mu\nu}(x-X)$. Their Grassmann integral gives the gauge-invariant density $\int\mathrm{d}^{2}\chi\operatorname{tr}\lambda\lambda(x;X,\chi)=\operatorname{tr}F_{\mu\nu}F_{\mu\nu}(x-X)$, where the normalization of the fermionic zero-mode is given by $\int_{\mathbb{R}^{2}\times T^{2}_{\ell_{3},\ell_{4}}}\mathrm{d}^{4}X\int\mathrm{d}^{2}\chi\operatorname{tr}\lambda\lambda(x;X,\chi)=2g^{2}S_{I}/N$ (see also {citep}[75, Eq. (69) ]). Since tr λ λ \operatorname{tr}\lambda\lambda tr λλ ℓ 3 , 4 \ell_{3,4} ℓ 3 , 4 X 3 , 4 X_{3,4} X 3 , 4 N 2 N^{2} N 2 M 1 / N ≃ R 2 × T N ℓ 3 , N ℓ 4 2 \mathcal{M}_{1/N}\simeq\mathbb{R}^{2}\times T^{2}_{N\ell_{3},N\ell_{4}} M 1/ N ≃ R 2 × T N ℓ 3 , N ℓ 4 2 16 π 2 Λ 3 / N 2 16\pi^{2}\Lambda^{3}/N^{2} 16 π 2 Λ 3 / N 2 N 2 N^{2} N 2 N N N [77 ] based on the large-N N N [78 , 79 ] , which finds agreement within the present uncertainties.
The phase e i θ / N e^{i\theta/N} e i θ / N Z 2 N \mathbb{Z}_{2N} Z 2 N Z N \mathbb{Z}_{N} Z N [80 , 81 ] . Since the twisted compactification preserves the anomaly [1 , 82 , 83 , 84 ] , the N N N R 2 \mathbb{R}^{2} R 2 θ \theta θ R 2 × T 2 \mathbb{R}^{2}\times T^{2} R 2 × T 2 N N N R 2 \mathbb{R}^{2} R 2 ⟨ tr λ λ ⟩ k = 16 π 2 Λ 3 e i ( θ + 2 π k ) / N \langle\operatorname{tr}\lambda\lambda\rangle_{k}=16\pi^{2}\Lambda^{3}e^{i(\theta+2\pi k)/N} ⟨ tr λλ ⟩ k = 16 π 2 Λ 3 e i ( θ + 2 π k ) / N θ → θ + 2 π k \theta\to\theta+2\pi k θ → θ + 2 π k R 2 × T p / N -twist 2 \mathbb{R}^{2}\times T^{2}_{p/N\text{-twist}} R 2 × T p / N -twist 2 R 2 \mathbb{R}^{2} R 2 [85 ] for the analogous case of the multi-charge Schwinger model).
We would like to comment on the relation of our above computation with the recent discussion on the gluino condensate at the twisted T 4 T^{4} T 4 [86 , 87 , 88 , 89 ] . In their earlier works [86 , 87 ] , the moduli space was taken with N N N N × 16 π 2 Λ 3 N\times16\pi^{2}\Lambda^{3} N × 16 π 2 Λ 3 N N N R 4 \mathbb{R}^{4} R 4 [88 ] by noticing the mismatch of the natural normalizations between the path integral and the Hamiltonian interpretation, which suggests the N N N [89 ] , they emphasized the importance of summing over twisted sectors to select a single superselection sector in the infinite volume limit, which was correctly taken but a bit implicit in earlier works, and it has a natural counterpart here. In the R 2 × T 2 \mathbb{R}^{2} \times T^{2} R 2 × T 2 R 2 \mathbb{R}^{2} R 2 N Z + q N\mathbb{Z}+q N Z + q R 2 \mathbb{R}^{2} R 2 T 2 T^{2} T 2 q q q R 2 \mathbb{R}^{2} R 2 q ∈ Z N q \in\mathbb{Z}_{N} q ∈ Z N θ \theta θ
An important open question is the fluctuation determinant around the center-vortex fractional instanton, which controls the center-vortex fugacity in non-supersymmetric theories (see [90 ] for calorons). It determines the magnitude of the topological susceptibility and of the string tension in the dilute vortex gas of pure Yang–Mills theory [1 ] , and that of the fractionalized Kobayashi–Maskawa–’t Hooft vertex for the η ′ \eta' η ′ [1 , 5 ] . So far, its N N N [23 footnote 6] . Even if a closed expression cannot be derived, its parametric dependence on N N N p p p
D-brane interpretation. The fractional instanton turns out to have a simple D-brane picture, which extends the one used by Lee, Lu and Yi for the KvBLLY caloron [24 , 25 ] . Let us consider N N N x 1 , 2 , 3 , 4 x_{1,2,3,4} x 1 , 2 , 3 , 4 U ( N ) U(N) U ( N ) F U ( N ) F^{U(N)} F U ( N ) U ( N ) U(N) U ( N ) n 34 n_{34} n 34 x 1 , 2 x_{1,2} x 1 , 2 ∫ T 2 tr F 34 U ( N ) = 2 π n 34 \int_{T^{2}}\operatorname{tr}F_{34}^{U(N)}=2\pi n_{34} ∫ T 2 tr F 34 U ( N ) = 2 π n 34 n 12 n_{12} n 12 x 3 , 4 x_{3,4} x 3 , 4 ∫ R 2 tr F 12 U ( N ) = 2 π n 12 \int_{\mathbb{R}^{2}}\operatorname{tr}F_{12}^{U(N)}=2\pi n_{12} ∫ R 2 tr F 12 U ( N ) = 2 π n 12 U ( N ) U(N) U ( N ) U ( 1 ) U(1) U ( 1 )
1 8 π 2 ∫ tr ( F U ( N ) ∧ F U ( N ) ) = Q t o p + n 12 n 34 N ∈ Z , (110) \frac{1}{8\pi^{2}}\int\operatorname{tr}\left(F^{U(N)} \wedge F^{U(N)}\right)=Q_{\mathrm{top}}+\frac{n_{12}n_{34}}{N}\in\mathbb{Z}, \tag*{(110)} 8 π 2 1 ∫ tr ( F U ( N ) ∧ F U ( N ) ) = Q top + N n 12 n 34 ∈ Z , (110) so that the S U ( N ) SU(N) S U ( N ) Q t o p ≡ − n 12 n 34 / N m o d 1 Q_{\mathrm{top}}\equiv-n_{12}n_{34}/N\ \mathrm{mod}\ 1 Q top ≡ − n 12 n 34 / N mod 1 [91 , 92 , 93 ] . Let us focus on n 12 n 34 = − p q n_{12}n_{34}=-pq n 12 n 34 = − pq p q ≡ 1 m o d N pq\equiv1\ \mathrm{mod}\ N pq ≡ 1 mod N Q t o p = 1 / N Q_{\mathrm{top}}=1/N Q top = 1/ N ( 1 − p q ) / N (1-pq)/N ( 1 − pq ) / N p = q = 1 p=q=1 p = q = 1
We now T-dualize along x 4 x_{4} x 4 x ~ 4 \widetilde{x}_{4} x 4 N N N x 1 , 2 , 3 , 4 x_{1,2,3,4} x 1 , 2 , 3 , 4 p p p x 1 , 2 x_{1,2} x 1 , 2 N N N x 1 , 2 , 3 x_{1,2,3} x 1 , 2 , 3 p p p x 1 , 2 , 4 x_{1,2,4} x 1 , 2 , 4 ( N , − p ) (N,-p) ( N , − p ) ( x 3 , x ~ 4 ) (x_{3},\widetilde{x}_{4}) ( x 3 , x 4 ) [94 ] because gcd ( N , p ) = 1 \gcd(N,p)=1 g cd( N , p ) = 1 x 3 x_{3} x 3 N N N 2 π / ( N ℓ 4 ) 2\pi/(N\ell_{4}) 2 π / ( N ℓ 4 ) U ( 1 ) U(1) U ( 1 ) P 3 = S − p P_{3}=S^{-p} P 3 = S − p P 4 = C P_{4}=C P 4 = C
The D2-branes along x 3 , 4 x_{3,4} x 3 , 4 q q q x 3 x_{3} x 3 q q q N N N F 12 F_{12} F 12 2 π v ~ n ( q ) 2\pi\widetilde{v}_{n}^{(q)} 2 π v n ( q ) x 3 → x 3 + ℓ 3 x_{3}\to x_{3}+\ell_{3} x 3 → x 3 + ℓ 3 p p p x 1 , 2 , 4 x_{1,2,4} x 1 , 2 , 4 q q q ℓ 3 \ell_{3} ℓ 3 x ~ 4 \widetilde{x}_{4} x 4 [24 ] . Its length, 1 / N 1/N 1/ N Q t o p = 1 / N Q_{\mathrm{top}}=1/N Q top = 1/ N S = 8 π 2 / ( N g 2 ) S=8\pi^{2}/(Ng^{2}) S = 8 π 2 / ( N g 2 ) Figure 1 .
In this paper, we have used the ADHM construction for the doubly-periodic instanton with Q t o p = 1 Q_{\mathrm{top}}=1 Q top = 1 R 2 × T 2 \mathbb{R}^{2}\times T^{2} R 2 × T 2 1 / N 1/N 1/ N [95 , 96 ] ) and requires further study.
Acknowledgments
The authors thank Erich Poppitz and Antonio González-Arroyo for reading the early draft and having the useful discussions. Y.H. thanks Takeshi Morita for discussions on the D-brane picture. This work was completed during the long-term workshop at the Yukawa Institute for Theoretical Physics (YITP), “New Frontiers in QCD 2026” (YITP-T-26-01), and the discussion there was quite valuable. Discussions during RIKEN iTHEMS-YITP workshop: “Resurgence, Matrix model, String field theory and Instantons (or Really Marvelous Sparkling Ideas)” (YITP-W-26-11) were also useful in developing this work. This work was partially supported by Japan Society for the Promotion of Science (JSPS) Research Fellowship for Young Scientists Grant No. 23KJ1161 (Y.H.), by JSPS KAKENHI Grant No. 26K07106 (Y.T.), and by Center for Gravitational Physics and Quantum Information (CGPQI) at Yukawa Institute for Theoretical Physics.
Fourier transform of the doubly-periodic ADHM data
In this appendix, we derive the dual description on T ^ 2 \widehat{T}^{2} T 2 § 3.2 , starting from the doubly-periodic ADHM data (32) with the ADHM index I = ( n 3 , n 4 ) ∈ Z 2 I=(n_{3},n_{4})\in\mathbb{Z}^{2} I = ( n 3 , n 4 ) ∈ Z 2
Dictionary and the ADHM data on the dual torus
The lattice index I = ( n 3 , n 4 ) I=(n_{3},n_{4}) I = ( n 3 , n 4 ) ξ = ( ξ 3 , ξ 4 ) \xi=(\xi_{3},\xi_{4}) ξ = ( ξ 3 , ξ 4 ) T ^ 2 \widehat{T}^{2} T 2 ξ a ∼ ξ a + 2 π / L a \xi_{a}\sim\xi_{a}+2\pi/L_{a} ξ a ∼ ξ a + 2 π / L a ( a = 3 , 4 ) (a=3,4) ( a = 3 , 4 ) V ^ = 4 π 2 / ( L 3 L 4 ) \widehat{V}=4\pi^{2}/(L_{3}L_{4}) V = 4 π 2 / ( L 3 L 4 ) I ⋅ ( L ξ ) : = n 3 L 3 ξ 3 + n 4 L 4 ξ 4 I\cdot(L\xi):=n_{3}L_{3}\xi_{3}+n_{4}L_{4}\xi_{4} I ⋅ ( L ξ ) := n 3 L 3 ξ 3 + n 4 L 4 ξ 4 X I X_{I} X I M I J M_{IJ} M I J
X ^ ( ξ ) = ∑ I ∈ Z 2 e i I ⋅ ( L ξ ) X I , M ^ ( ξ , ξ ′ ) = ∑ I , J ∈ Z 2 e i ( I ⋅ ( L ξ ) − J ⋅ ( L ξ ′ ) ) M I J . (111) \widehat{X}(\xi)=\sum_{I\in\mathbb{Z}^{2}}e^{iI\cdot(L\xi)}X_{I},\qquad\widehat{M}(\xi,\xi')=\sum_{I,J\in\mathbb{Z}^{2}}e^{i(I\cdot(L\xi)-J\cdot(L\xi'))}M_{IJ}. \tag*{(111)} X ( ξ ) = I ∈ Z 2 ∑ e i I ⋅ ( L ξ ) X I , M ( ξ , ξ ′ ) = I , J ∈ Z 2 ∑ e i ( I ⋅ ( L ξ ) − J ⋅ ( L ξ ′ )) M I J . (111) One can easily check the following rules:
the unit matrix δ I , J \delta_{I,J} δ I , J V ^ δ ( 2 ) ( ξ − ξ ′ ) \widehat{V}\,\delta^{(2)}(\xi-\xi') V δ ( 2 ) ( ξ − ξ ′ )
matrix products become M N ^ ( ξ , ξ ′ ) = V ^ − 1 ∫ T ^ 2 d 2 ξ ′ ′ M ^ ( ξ , ξ ′ ′ ) N ^ ( ξ ′ ′ , ξ ′ ) \widehat{MN}(\xi,\xi')=\widehat{V}^{-1}\int_{\widehat{T}^{2}}d^{2}\xi''\,\widehat{M}(\xi,\xi'')\widehat{N}(\xi'',\xi') M N ( ξ , ξ ′ ) = V − 1 ∫ T 2 d 2 ξ ′′ M ( ξ , ξ ′′ ) N ( ξ ′′ , ξ ′ ) M X ^ ( ξ ) = V ^ − 1 ∫ T ^ 2 d 2 ξ ′ M ^ ( ξ , ξ ′ ) X ^ ( ξ ′ ) \widehat{MX}(\xi)=\widehat{V}^{-1}\int_{\widehat{T}^{2}}d^{2}\xi'\,\widehat{M}(\xi,\xi')\widehat{X}(\xi') M X ( ξ ) = V − 1 ∫ T 2 d 2 ξ ′ M ( ξ , ξ ′ ) X ( ξ ′ )
inner products become ∑ I X I † Y I = V ^ − 1 ∫ T ^ 2 d 2 ξ X ^ ( ξ ) † Y ^ ( ξ ) \sum_{I}X_{I}^{\dagger}Y_{I}=\widehat{V}^{-1}\int_{\widehat{T}^{2}}d^{2}\xi\,\widehat{X}(\xi)^{\dagger}\widehat{Y}(\xi) ∑ I X I † Y I = V − 1 ∫ T 2 d 2 ξ X ( ξ ) † Y ( ξ )
In particular, an operator of the form M ^ ( ξ , ξ ′ ) = V ^ O ξ δ ( 2 ) ( ξ − ξ ′ ) \hat{M}(\xi,\xi')=\hat{V}\mathcal{O}_\xi\delta^{(2)}(\xi-\xi') M ^ ( ξ , ξ ′ ) = V ^ O ξ δ ( 2 ) ( ξ − ξ ′ ) O ξ \mathcal{O}_\xi O ξ T ^ 2 \hat{T}^2 T ^ 2 X ^ \hat{X} X ^ O ξ \mathcal{O}_\xi O ξ V ^ O ξ O ξ ′ δ ( 2 ) ( ξ − ξ ′ ) \hat{V}\mathcal{O}_\xi\mathcal{O}'_\xi\delta^{(2)}(\xi-\xi') V ^ O ξ O ξ ′ δ ( 2 ) ( ξ − ξ ′ )
About B ( x ) B(x) B ( x ) Inserting (32) , B ( x ) I J = ( n 4 L 4 σ 4 + n 3 L 3 σ 3 − x ) δ I , J + A ^ I − J B(x)_{IJ}=(n_4L_4\sigma_4+n_3L_3\sigma_3-x)\delta_{I,J}+\hat{A}_{I-J} B ( x ) I J = ( n 4 L 4 σ 4 + n 3 L 3 σ 3 − x ) δ I , J + A ^ I − J
B ^ ( x ) ( ξ , ξ ′ ) = ∑ J e i J ⋅ L ( ξ − ξ ′ ) [ n 4 J L 4 σ 4 + n 3 J L 3 σ 3 − x ] + ∑ J e i J ⋅ L ( ξ − ξ ′ ) ∑ n ∈ Z 2 e i n ⋅ ( L ξ ) A ^ n = V ^ [ − i σ 4 ∂ ξ 4 − i σ 3 ∂ ξ 3 + A ^ ( ξ ) − x ] δ ( 2 ) ( ξ − ξ ′ ) = V ^ D x δ ( 2 ) ( ξ − ξ ′ ) , (112) \begin{aligned}
\hat{B}(x)(\xi,\xi')&=\sum_J e^{iJ\cdot L(\xi-\xi')}\left[n_4^JL_4\sigma_4+n_3^JL_3\sigma_3-x\right]+\sum_J e^{iJ\cdot L(\xi-\xi')}\sum_{n\in\mathbb{Z}^2}e^{in\cdot(L\xi)}\hat{A}_n\\
&=\hat{V}\left[-i\sigma_4\partial_{\xi_4}-i\sigma_3\partial_{\xi_3}+\hat{A}(\xi)-x\right]\delta^{(2)}(\xi-\xi')\\
&=\hat{V}D_x\delta^{(2)}(\xi-\xi'), \tag*{(112)}
\end{aligned} B ^ ( x ) ( ξ , ξ ′ ) = J ∑ e i J ⋅ L ( ξ − ξ ′ ) [ n 4 J L 4 σ 4 + n 3 J L 3 σ 3 − x ] + J ∑ e i J ⋅ L ( ξ − ξ ′ ) n ∈ Z 2 ∑ e in ⋅ ( L ξ ) A ^ n = V ^ [ − i σ 4 ∂ ξ 4 − i σ 3 ∂ ξ 3 + A ^ ( ξ ) − x ] δ ( 2 ) ( ξ − ξ ′ ) = V ^ D x δ ( 2 ) ( ξ − ξ ′ ) , (112) where ( n 3 J , n 4 J ) = J (n_3^J,n_4^J)=J ( n 3 J , n 4 J ) = J A ^ ( ξ ) = ∑ n e i n ⋅ ( L ξ ) A ^ n = σ μ A ^ μ ( ξ ) \hat{A}(\xi)=\sum_n e^{in\cdot(L\xi)}\hat{A}_n=\sigma_\mu\hat{A}_\mu(\xi) A ^ ( ξ ) = ∑ n e in ⋅ ( L ξ ) A ^ n = σ μ A ^ μ ( ξ ) D x = σ μ ( D μ − x μ ) D_x=\sigma_\mu(D_\mu-x_\mu) D x = σ μ ( D μ − x μ ) D D D (34) , i.e.
D a = − i ∂ ξ a + A ^ a ( ξ ) ( a = 3 , 4 ) , D 1 , 2 = A ^ 1 , 2 ( ξ ) . (113) D_a=-i\partial_{\xi_a}+\hat{A}_a(\xi)\quad(a=3,4),\qquad D_{1,2}=\hat{A}_{1,2}(\xi). \tag*{(113)} D a = − i ∂ ξ a + A ^ a ( ξ ) ( a = 3 , 4 ) , D 1 , 2 = A ^ 1 , 2 ( ξ ) . (113) Setting x = 0 x=0 x = 0 § 3.2 . The hermiticity of B μ B_\mu B μ ( A ^ μ ) n † = ( A ^ μ ) − n (\hat{A}_\mu)_n^\dagger=(\hat{A}_\mu)_{-n} ( A ^ μ ) n † = ( A ^ μ ) − n A ^ μ ( ξ ) \hat{A}_\mu(\xi) A ^ μ ( ξ ) − i ∂ ξ a -i\partial_{\xi_a} − i ∂ ξ a T ^ 2 \hat{T}^2 T ^ 2 D μ D_\mu D μ B ( x ) † B(x)^\dagger B ( x ) † D x † = σ ˉ μ ( D μ − x μ ) D_x^\dagger=\bar{\sigma}_\mu(D_\mu-x_\mu) D x † = σ ˉ μ ( D μ − x μ )
About λ \lambda λ By (32) and (26) , λ I † = ζ † ( P 3 n 3 P 4 n 4 ) † = ∑ m e − i I ⋅ ( L μ m ) ζ † Π m \lambda_I^\dagger=\zeta^\dagger(P_3^{n_3}P_4^{n_4})^\dagger=\sum_m e^{-iI\cdot(L\mu_m)}\zeta^\dagger\Pi_m λ I † = ζ † ( P 3 n 3 P 4 n 4 ) † = ∑ m e − i I ⋅ ( L μ m ) ζ † Π m μ m = ( μ 3 , m , μ 4 , m ) \mu_m=(\mu_{3,m},\mu_{4,m}) μ m = ( μ 3 , m , μ 4 , m )
λ ^ † ( ξ ) = ∑ m = 1 N ∑ I ∈ Z 2 e i I ⋅ L ( ξ − μ m ) ζ † Π m = V ^ ∑ m = 1 N δ ( 2 ) ( ξ − μ m ) ζ † Π m . (114) \hat{\lambda}^\dagger(\xi)=\sum_{m=1}^{N}\sum_{I\in\mathbb{Z}^2}e^{iI\cdot L(\xi-\mu_m)}\zeta^\dagger\Pi_m=\hat{V}\sum_{m=1}^{N}\delta^{(2)}(\xi-\mu_m)\zeta^\dagger\Pi_m. \tag*{(114)} λ ^ † ( ξ ) = m = 1 ∑ N I ∈ Z 2 ∑ e i I ⋅ L ( ξ − μ m ) ζ † Π m = V ^ m = 1 ∑ N δ ( 2 ) ( ξ − μ m ) ζ † Π m . (114) which is (35) : the holonomies P 3 , 4 P_{3,4} P 3 , 4 μ m ∈ T ^ 2 \mu_m\in\hat{T}^2 μ m ∈ T ^ 2 ∑ J e − i J ⋅ ( L ξ ′ ) λ J = λ ^ † ( ξ ′ ) † \sum_J e^{-iJ\cdot(L\xi')}\lambda_J=\hat{\lambda}^\dagger(\xi')^\dagger ∑ J e − i J ⋅ ( L ξ ′ ) λ J = λ ^ † ( ξ ′ ) † Π m Π m ′ = δ m m ′ Π m \Pi_m\Pi_{m'}=\delta_{mm'}\Pi_m Π m Π m ′ = δ m m ′ Π m
λ ^ † λ ( ξ , ξ ′ ) = λ ^ † ( ξ ) λ ^ † ( ξ ′ ) † = V ^ δ ( 2 ) ( ξ − ξ ′ ) × V ^ ∑ m = 1 N δ ( 2 ) ( ξ − μ m ) ζ † Π m ζ . (115) \hat{\lambda}^\dagger\lambda(\xi,\xi')=\hat{\lambda}^\dagger(\xi)\hat{\lambda}^\dagger(\xi')^\dagger=\hat{V}\delta^{(2)}(\xi-\xi')\times\hat{V}\sum_{m=1}^{N}\delta^{(2)}(\xi-\mu_m)\zeta^\dagger\Pi_m\zeta. \tag*{(115)} λ ^ † λ ( ξ , ξ ′ ) = λ ^ † ( ξ ) λ ^ † ( ξ ′ ) † = V ^ δ ( 2 ) ( ξ − ξ ′ ) × V ^ m = 1 ∑ N δ ( 2 ) ( ξ − μ m ) ζ † Π m ζ . (115) Thus λ † λ \lambda^\dagger\lambda λ † λ T ^ 2 \hat{T}^2 T ^ 2 V ^ ∑ m δ ( 2 ) ( ξ − μ m ) ζ † Π m ζ \hat{V}\sum_m\delta^{(2)}(\xi-\mu_m)\zeta^\dagger\Pi_m\zeta V ^ ∑ m δ ( 2 ) ( ξ − μ m ) ζ † Π m ζ χ m † \chi_m^\dagger χ m † m m m N × 2 N\times2 N × 2 ζ \zeta ζ
ζ = ( χ 1 † ⋮ χ N † ) , ζ † Π m ζ = χ m χ m † , ρ m j = − π χ m † τ j χ m , ∣ ρ ⃗ m ∣ = π χ m † χ m . (116) \zeta=\begin{pmatrix}\chi_1^\dagger\\ \vdots\\ \chi_N^\dagger\end{pmatrix},\qquad\zeta^\dagger\Pi_m\zeta=\chi_m\chi_m^\dagger,\qquad\rho_m^j=-\pi\chi_m^\dagger\tau_j\chi_m,\qquad|\vec{\rho}_m|=\pi\chi_m^\dagger\chi_m. \tag*{(116)} ζ = χ 1 † ⋮ χ N † , ζ † Π m ζ = χ m χ m † , ρ m j = − π χ m † τ j χ m , ∣ ρ m ∣ = π χ m † χ m . (116) where ρ ⃗ m \vec{\rho}_m ρ m (39) ; the last relation follows from tr 2 ( χ m χ m † ) = ∣ χ m ∣ 2 \operatorname{tr}_2(\chi_m\chi_m^\dagger)=|\chi_m|^2 tr 2 ( χ m χ m † ) = ∣ χ m ∣ 2
The ADHM constraint as the Hitchin system with impurities
By the product rule, Δ † Δ = λ † λ + B ( x ) † B ( x ) \Delta^\dagger\Delta=\lambda^\dagger\lambda+B(x)^\dagger B(x) Δ † Δ = λ † λ + B ( x ) † B ( x ) V ^ K x δ ( 2 ) ( ξ − ξ ′ ) \widehat{V}K_x\delta^{(2)}(\xi-\xi') V K x δ ( 2 ) ( ξ − ξ ′ )
K x = D x † D x + V ^ ∑ m = 1 N δ ( 2 ) ( ξ − μ m ) ζ † Π m ζ . (117) K_x=D_x^\dagger D_x+\widehat{V}\sum_{m=1}^{N}\delta^{(2)}(\xi-\mu_m)\zeta^\dagger\Pi_m\zeta. \tag*{(117)} K x = D x † D x + V m = 1 ∑ N δ ( 2 ) ( ξ − μ m ) ζ † Π m ζ . (117) With σ ˉ μ σ ν = δ μ ν σ 0 + η ˉ μ ν l σ l \bar{\sigma}_\mu\sigma_\nu=\delta_{\mu\nu}\sigma_0+\bar{\eta}^{l}_{\mu\nu}\sigma_l σ ˉ μ σ ν = δ μν σ 0 + η ˉ μν l σ l η ˉ 4 j l = − η ˉ j 4 l = δ l j \bar{\eta}^{l}_{4j}=-\bar{\eta}^{l}_{j4}=\delta_{lj} η ˉ 4 j l = − η ˉ j 4 l = δ l j η ˉ j k l = ϵ l j k \bar{\eta}^{l}_{jk}=\epsilon_{ljk} η ˉ j k l = ϵ l j k D μ D_\mu D μ (37) , in which the constants x μ x_\mu x μ (41) and the antisymmetry of η ˉ μ ν l \bar{\eta}^{l}_{\mu\nu} η ˉ μν l
1 2 η ˉ μ ν l [ D μ , D ν ] = − i 2 η ˉ μ ν l F ^ μ ν = − i ( F ^ 4 l + 1 2 ϵ l j k F ^ j k ) . (118) \frac{1}{2}\bar{\eta}^{l}_{\mu\nu}[D_\mu,D_\nu]=-\frac{i}{2}\bar{\eta}^{l}_{\mu\nu}\widehat{F}_{\mu\nu}=-i\left(\widehat{F}_{4l}+\frac{1}{2}\epsilon_{ljk}\widehat{F}_{jk}\right). \tag*{(118)} 2 1 η ˉ μν l [ D μ , D ν ] = − 2 i η ˉ μν l F μν = − i ( F 4 l + 2 1 ϵ l j k F j k ) . (118) On the other hand, by (39) and τ l = − i σ l \tau_l=-i\sigma_l τ l = − i σ l ζ † Π m ζ = 1 2 π ( σ 0 ∣ ρ ⃗ m ∣ + i σ l ρ m l ) \zeta^\dagger\Pi_m\zeta=\frac{1}{2\pi}(\sigma_0|\vec{\rho}_m|+i\sigma_l\rho_m^l) ζ † Π m ζ = 2 π 1 ( σ 0 ∣ ρ m ∣ + i σ l ρ m l ) K x = σ 0 K x 0 + σ l K x l K_x=\sigma_0K_x^0+\sigma_lK_x^l K x = σ 0 K x 0 + σ l K x l
K x l = − i ( F ^ 4 l + 1 2 ϵ l j k F ^ j k ) + i V ^ 2 π ∑ m = 1 N ρ m l δ ( 2 ) ( ξ − μ m ) , (119) K_x^l=-i\left(\widehat{F}_{4l}+\frac{1}{2}\epsilon_{ljk}\widehat{F}_{jk}\right)+\frac{i\widehat{V}}{2\pi}\sum_{m=1}^{N}\rho_m^l\delta^{(2)}(\xi-\mu_m), \tag*{(119)} K x l = − i ( F 4 l + 2 1 ϵ l j k F j k ) + 2 π i V m = 1 ∑ N ρ m l δ ( 2 ) ( ξ − μ m ) , (119) K x 0 = ∑ μ = 1 4 ( D μ − x μ ) 2 + V ^ 2 π ∑ m = 1 N ∣ ρ ⃗ m ∣ δ ( 2 ) ( ξ − μ m ) . (120) K_x^0=\sum_{\mu=1}^{4}(D_\mu-x_\mu)^2+\frac{\widehat{V}}{2\pi}\sum_{m=1}^{N}|\vec{\rho}_m|\delta^{(2)}(\xi-\mu_m). \tag*{(120)} K x 0 = μ = 1 ∑ 4 ( D μ − x μ ) 2 + 2 π V m = 1 ∑ N ∣ ρ m ∣ δ ( 2 ) ( ξ − μ m ) . (120) The three self-duality conditions in (20) , tr 2 ( σ l Δ † Δ ) = 0 \operatorname{tr}_2(\sigma_l\Delta^\dagger\Delta)=0 tr 2 ( σ l Δ † Δ ) = 0 K x l = 0 K_x^l=0 K x l = 0 (40) . Note that K x l K_x^l K x l x x x Q t o p = 1 Q_{\mathrm{top}}=1 Q top = 1 A ^ μ ( ξ ) \widehat{A}_\mu(\xi) A μ ( ξ ) (40) reduces to (42) : the l = 3 l=3 l = 3 F ^ 43 = − F ^ 34 = V ^ 2 π ∑ m ρ m 3 δ ( 2 ) ( ξ − μ m ) \widehat{F}_{43}=-\widehat{F}_{34}=\frac{\widehat{V}}{2\pi}\sum_m\rho_m^3\delta^{(2)}(\xi-\mu_m) F 43 = − F 34 = 2 π V ∑ m ρ m 3 δ ( 2 ) ( ξ − μ m ) F ^ 12 = 0 \widehat{F}_{12}=0 F 12 = 0 l = 1 , 2 l=1,2 l = 1 , 2 ∂ ξ 4 A ^ 1 − ∂ ξ 3 A ^ 2 \partial_{\xi_4}\widehat{A}_1-\partial_{\xi_3}\widehat{A}_2 ∂ ξ 4 A 1 − ∂ ξ 3 A 2 ∂ ξ 3 A ^ 1 + ∂ ξ 4 A ^ 2 \partial_{\xi_3}\widehat{A}_1+\partial_{\xi_4}\widehat{A}_2 ∂ ξ 3 A 1 + ∂ ξ 4 A 2 1 2 ( ∂ ξ 3 − i ∂ ξ 4 ) H ^ \frac{1}{2}(\partial_{\xi_3}-i\partial_{\xi_4})\widehat{H} 2 1 ( ∂ ξ 3 − i ∂ ξ 4 ) H H ^ = A ^ 1 + i A ^ 2 \widehat{H}=\widehat{A}_1+i\widehat{A}_2 H = A 1 + i A 2
Once (40) holds, K x = σ 0 K x 0 K_x=\sigma_0K_x^0 K x = σ 0 K x 0 (20) identifies K x 0 K_x^0 K x 0 f x − 1 f_x^{-1} f x − 1 f x − 1 f x = 1 k f_x^{-1}f_x=\mathbbm{1}_k f x − 1 f x = 1 k K x 0 f ^ x ( ξ , ξ ′ ) = V ^ δ ( 2 ) ( ξ − ξ ′ ) K_x^0\widehat{f}_x(\xi,\xi')=\widehat{V}\delta^{(2)}(\xi-\xi') K x 0 f x ( ξ , ξ ′ ) = V δ ( 2 ) ( ξ − ξ ′ ) (43) . Finally, inverting (A.1), f I J = V ^ − 2 ∫ d 2 ξ d 2 ξ ′ e − i ( I ⋅ ( L ξ ) − J ⋅ ( L ξ ′ ) ) f ^ x ( ξ , ξ ′ ) f_{IJ}=\widehat{V}^{-2}\int d^2\xi\,d^2\xi'\,e^{-i(I\cdot(L\xi)-J\cdot(L\xi'))}\widehat{f}_x(\xi,\xi') f I J = V − 2 ∫ d 2 ξ d 2 ξ ′ e − i ( I ⋅ ( L ξ ) − J ⋅ ( L ξ ′ )) f x ( ξ , ξ ′ ) λ I = ∑ m e i I ⋅ ( L μ m ) Π m ζ \lambda_I=\sum_m e^{iI\cdot(L\mu_m)}\Pi_m\zeta λ I = ∑ m e i I ⋅ ( L μ m ) Π m ζ
λ [ σ μ ⊗ f x ] λ † = ∑ I , J λ I σ μ ( f x ) I J λ J † = ∑ m , m ′ = 1 N f ^ x ( μ m , μ m ′ ) Π m ζ σ μ ζ † Π m ′ . (121) \lambda[\sigma_\mu\otimes f_x]\lambda^\dagger=\sum_{I,J}\lambda_I\sigma_\mu(f_x)_{IJ}\lambda_J^\dagger=\sum_{m,m'=1}^{N}\widehat{f}_x(\mu_m,\mu_{m'})\Pi_m\zeta\,\sigma_\mu\,\zeta^\dagger\Pi_{m'}. \tag*{(121)} λ [ σ μ ⊗ f x ] λ † = I , J ∑ λ I σ μ ( f x ) I J λ J † = m , m ′ = 1 ∑ N f x ( μ m , μ m ′ ) Π m ζ σ μ ζ † Π m ′ . (121) whose μ = 4 \mu=4 μ = 4 μ = j \mu=j μ = j (24) , yield (44) and (45) , respectively.
