Introduction
The commutator problem is an early instance of the problem of describing polynomial images on matrix algebras. For matrices A A A B B B [ A , B ] = A B − B A [A,B]=AB-BA [ A , B ] = A B − B A tr ( A B ) = tr ( B A ) \operatorname{tr}(AB)=\operatorname{tr}(BA) tr ( A B ) = tr ( B A ) [1 Satz 3] . Albert and Muckenhoupt extended the result to arbitrary fields [2 ] . Thus a linear condition on matrix entries completely describes the image of this particular noncommutative polynomial. Replacing the commutator by a general polynomial leads to the image problem considered here.
Throughout this paper K K K M n ( K ) M_n(K) M n ( K ) n × n n\times n n × n I n I_n I n tr T = ∑ i = 1 n T i i \operatorname{tr} T=\sum_{i=1}^{n}T_{ii} tr T = ∑ i = 1 n T ii
s l n ( K ) = { T ∈ M n ( K ) : tr T = 0 } . \mathfrak{sl}_n(K)=\{T\in M_n(K):\operatorname{tr} T=0\}. sl n ( K ) = { T ∈ M n ( K ) : tr T = 0 } . The free associative algebra K ⟨ X 1 , … , X m ⟩ K\langle X_1,\ldots,X_m\rangle K ⟨ X 1 , … , X m ⟩ K K K m m m
p ( X 1 , … , X m ) = ∑ σ ∈ S m a σ X σ ( 1 ) ⋯ X σ ( m ) , a σ ∈ K . p(X_1,\ldots,X_m)=\sum_{\sigma\in S_m}a_\sigma X_{\sigma(1)}\cdots X_{\sigma(m)},\qquad a_\sigma\in K. p ( X 1 , … , X m ) = σ ∈ S m ∑ a σ X σ ( 1 ) ⋯ X σ ( m ) , a σ ∈ K . where S m S_m S m { 1 , … , m } \{1,\ldots,m\} { 1 , … , m } m m m n n n
p ( M n ( K ) ) = { p ( A 1 , … , A m ) : A i ∈ M n ( K ) } . p(M_n(K))=\{p(A_1,\ldots,A_m):A_i\in M_n(K)\}. p ( M n ( K )) = { p ( A 1 , … , A m ) : A i ∈ M n ( K )} . A polynomial identity has zero image. A central polynomial has nonzero image consisting of scalar matrices. These two possibilities arise naturally in polynomial identity theory: the standard identity of Amitsur and Levitzki [3 ] and the central polynomials of Formanek [4 ] give examples.
The L’vov–Kaplansky question asks whether p ( M n ( K ) ) p(M_n(K)) p ( M n ( K )) [7 Introduction] ; Kanel-Belov, Malev, and Rowen discuss its attribution to Kaplansky and its formulation by
2020 Mathematics Subject Classification. Primary 16R10; Secondary 15A24, 16R30.
Keywords. Multilinear polynomial, matrix algebra, L’vov–Kaplansky conjecture. L’vov [10 Introduction] . Over the present field, an affirmative answer says that the image is one of
{ 0 } , K I n , s l n ( K ) , M n ( K ) . \{0\},\quad K I_n,\quad\mathfrak{sl}_n(K),\quad M_n(K). { 0 } , K I n , sl n ( K ) , M n ( K ) . Each of these possibilities occurs among multilinear polynomials. The zero polynomial and the identity input give the first and fourth, the commutator gives the third, and multilinear central polynomials give the second.
The linear span of the image has a simpler description than the image itself. Brešar and Klep proved that this span is a Lie ideal and classified it by the polynomial’s identities and trace [5 Theorems 2.3 and 4.5] . For a multilinear polynomial, the underlying identity is
[ p ( A 1 , … , A m ) , T ] = ∑ i = 1 m p ( A 1 , … , [ A i , T ] , … , A m ) . [p(A_1,\ldots,A_m),T]=\sum_{i=1}^{m}p(A_1,\ldots,[A_i,T],\ldots,A_m). [ p ( A 1 , … , A m ) , T ] = i = 1 ∑ m p ( A 1 , … , [ A i , T ] , … , A m ) . Their tracial Nullstellensatz identifies polynomials of zero trace with sums of commutators modulo matrix identities. The image problem asks which matrices occur as single values. In particular, passing from traceless containment to full image requires attaining each conjugacy class of nonzero trace.
There are two useful ways to restrict the question: fix the matrix order and allow arbitrary polynomial degree, or fix the degree and allow arbitrary matrix order. Kanel-Belov, Malev, and Rowen proved the full image classification in matrix order two [10 Theorem 2] . Their work combines conjugation-invariant cones, generic matrices, and invariant theory. In matrix order three, their subsequent paper obtains image and density restrictions, with possible power-central behavior [8 ] .
Mesyan generalized the commutator theorem by proving that every nonzero multilinear polynomial of degree at most three contains all traceless matrices in its image, in the range of field sizes stated in [6 Theorem 13] . He proposed the corresponding containment for a polynomial of degree m m m n ≥ m − 1 n \ge m-1 n ≥ m − 1 n ≥ 3 n \ge3 n ≥ 3 [7 ] . A lemma in that argument required correction. Fagundes, de Mello, and dos Santos supplied the correction and established
s l n ( K ) ⊆ p ( M n ( K ) ) ( n ≥ 3 ) (1) \mathfrak{sl}_n(K) \subseteq p(M_n(K)) \qquad(n \ge3) \tag*{(1)} sl n ( K ) ⊆ p ( M n ( K )) ( n ≥ 3 ) (1) for every nonzero multilinear quartic [13 Theorem 8] . Thus the trace-zero quartic case follows from existing work. The trace-nonzero case requires attainment of the remaining conjugacy classes.
For the full conjecture in degree three, Dykema and Klep proved the result for every even matrix order and every odd order below 17 [11 Theorem 1.2] . Their proof uses Sylvester’s clock-and-shift basis and determinant conditions on the six word coefficients. Vitas proved the degree-three conjecture in all orders [12 Theorem B] , by constructing slices covering prescribed Jordan forms and solving the resulting coefficient conditions. We use his theorem when replacing one input of a quartic by the identity leaves a cubic whose trace is nonzero.
We prove the full degree-four conjecture. The additional conclusion beyond (1) is that every trace-nonzero quartic has image M n ( K ) M_n(K) M n ( K ) n ≥ 3 n \ge3 n ≥ 3
Theorem 1.1. Let p p p K K K n ≥ 3 n \ge3 n ≥ 3
p ( M n ( K ) ) = { { 0 } , p = 0 , s l n ( K ) , p ≠ 0 and tr p is identically zero , M n ( K ) , tr p is not identically zero . p(M_n(K))=
\begin{cases}
\{0\}, & p=0,\\
\mathfrak{sl}_n(K), & p\ne0\ \text{and}\ \operatorname{tr}p\ \text{is identically zero},\\
M_n(K), & \operatorname{tr}p\ \text{is not identically zero}.
\end{cases} p ( M n ( K )) = ⎩ ⎨ ⎧ { 0 } , sl n ( K ) , M n ( K ) , p = 0 , p = 0 and tr p is identically zero , tr p is not identically zero . Consequently the image of p p p
Two words are cyclically equivalent when one is obtained by moving an initial segment of the other to its end. The formal trace of a polynomial is its class in the vector space with cyclic word classes as a basis. For a quartic the six class coefficients are the sums of the four word coefficients in each class. Matrix trace vanishes on the difference of cyclically equivalent words. Conversely, rotate each word to place X 4 X_4 X 4 W ( X 1 , X 2 , X 3 ) W(X_1, X_2, X_3) W ( X 1 , X 2 , X 3 )
tr p ( A 1 , A 2 , A 3 , A 4 ) = tr ( W ( A 1 , A 2 , A 3 ) A 4 ) . \operatorname{tr} p(A_1,A_2,A_3,A_4)=\operatorname{tr}\bigl(W(A_1,A_2,A_3)A_4\bigr). tr p ( A 1 , A 2 , A 3 , A 4 ) = tr ( W ( A 1 , A 2 , A 3 ) A 4 ) . The trace pairing is the bilinear form ( U , V ) ↦ tr ( U V ) = ∑ i , j U i j V j i (U,V) \mapsto\operatorname{tr}(UV)=\sum_{i,j}U_{ij}V_{ji} ( U , V ) ↦ tr ( U V ) = ∑ i , j U ij V j i U U U V V V W W W M n ( K ) M_n(K) M n ( K ) 2 n 2n 2 n [3 ] , so W = 0 W=0 W = 0 n ≥ 3 n\geq3 n ≥ 3 Theorem 1.1 .
We fix all but one input and study the image of the resulting linear map. Weighted permutation inputs produce slices containing subspaces defined by prescribed zero entries. Scalar shifts of identity reductions produce slices of rank at least n 2 − 1 n^2-1 n 2 − 1 3 × 3 3\times3 3 × 3
Linear slices and conjugacy classes
For a chosen input position, a slice is the linear map obtained by fixing the other inputs. Its image is a subspace of M n ( K ) M_n(K) M n ( K ) n n n E i j E_{ij} E ij ( i , j ) (i,j) ( i , j )
A conjugacy class of T T T { g T g − 1 : g ∈ GL n ( K ) } \{gTg^{-1}:g\in\operatorname{GL}_n(K)\} { g T g − 1 : g ∈ GL n ( K )} GL n ( K ) \operatorname{GL}_n(K) GL n ( K ) C ∼ T C\sim T C ∼ T C C C H H H H ⊥ = { A : tr ( A X ) = 0 for every X ∈ H } H^\perp=\{A:\operatorname{tr}(AX)=0\text{ for every }X\in H\} H ⊥ = { A : tr ( A X ) = 0 for every X ∈ H } H H H dim H \dim H dim H
Lemma 2.1. For n ≥ 2 n\geq2 n ≥ 2 nonzero conjugation-invariant subspace of s l n ( K ) \mathfrak{sl}_n(K) sl n ( K ) s l n ( K ) \mathfrak{sl}_n(K) sl n ( K ) s l n ( K ) \mathfrak{sl}_n(K) sl n ( K ) s l n ( K ) \mathfrak{sl}_n(K) sl n ( K ) s l n ( K ) \mathfrak{sl}_n(K) sl n ( K )
Proof. For the first assertion, diagonal conjugation has one-dimensional weight spaces K E i j K E_{ij} K E ij i ≠ j i\ne j i = j E i j E_{ij} E ij ( K × ) n (K^\times)^n ( K × ) n E i j E_{ij} E ij I n + t E j i I_n+tE_{ji} I n + t E j i
E i j + t ( E j j − E i i ) − t 2 E j i , E_{ij}+t(E_{jj}-E_{ii})-t^2E_{ji}, E ij + t ( E j j − E ii ) − t 2 E j i , so the diagonal differences belong as well. If the initial nonzero matrix is diagonal, it is nonscalar because its trace is zero; conjugation by I n + t E i j I_n+tE_{ij} I n + t E ij
The span of the conjugates of a nonscalar matrix is invariant. The linear span of differences of conjugates is a nonzero invariant subspace of s l n ( K ) \mathfrak{sl}_n(K) sl n ( K ) e i e_i e i i i i e i ∗ e_i^* e i ∗ E i j E_{ij} E ij i ≠ j i \ne j i = j
E i i − E j j = ( e i + e j ) ( e i ∗ − e j ∗ ) + E i j − E j i . E_{ii}-E_{jj}=(e_i+e_j)(e_i^*-e_j^*)+E_{ij}-E_{ji}. E ii − E j j = ( e i + e j ) ( e i ∗ − e j ∗ ) + E ij − E j i . The outer product in this expression is rank one and square zero, because ( e i ∗ − e j ∗ ) ( e i + e j ) = 0 (e_i^*-e_j^*)(e_i+e_j)=0 ( e i ∗ − e j ∗ ) ( e i + e j ) = 0 s l n ( K ) \mathfrak{sl}_n(K) sl n ( K )
Lemma 2.2. If an ordinary multilinear polynomial has a value of nonzero trace on M n ( K ) M_n(K) M n ( K )
Proof. Choose a primitive n n n ζ ∈ K \zeta\in K ζ ∈ K D D D 1 , ζ , … , ζ n − 1 1,\zeta,\ldots,\zeta^{n-1} 1 , ζ , … , ζ n − 1 R R R R e i = e i + 1 Re_i=e_{i+1} R e i = e i + 1 D R = ζ R D DR=\zeta RD D R = ζ R D
( D a R b ) ( D c R d ) = ζ − b c D a + c R b + d . (D^aR^b)(D^cR^d)=\zeta^{-bc}D^{a+c}R^{b+d}. ( D a R b ) ( D c R d ) = ζ − b c D a + c R b + d . Exponents are taken modulo n n n b b b a a a ( ζ a i ) a , i = 0 n − 1 (\zeta^{ai})_{a,i=0}^{n-1} ( ζ ai ) a , i = 0 n − 1 b b b n 2 n^2 n 2 b ≠ 0 b \ne0 b = 0 b = 0 b=0 b = 0 ∑ i = 0 n − 1 ζ a i \sum_{i=0}^{n-1}\zeta^{ai} ∑ i = 0 n − 1 ζ ai a = 0 a=0 a = 0 n ≠ 0 n \ne0 n = 0
Lemma 2.3. Every hyperplane in M n ( K ) M_n(K) M n ( K ) s l n ( K ) \mathfrak{sl}_n(K) sl n ( K )
Proof. Write the hyperplane as { X : tr ( A X ) = 0 } \{X:\operatorname{tr}(AX)=0\} { X : tr ( A X ) = 0 } A A A T T T C C C A A A T T T Lemma 2.1 shows that the span of conjugates of A A A s l n ( K ) \mathfrak{sl}_n(K) sl n ( K ) Lemma 2.1 , rank-one square-zero matrices span s l n ( K ) \mathfrak{sl}_n(K) sl n ( K ) N N N tr ( [ C , T ] N ) ≠ 0 \operatorname{tr}([C,T]N)\ne0 tr ([ C , T ] N ) = 0
t ⟼ tr ( T ( I n + t N ) C ( I n − t N ) ) t\longmapsto\operatorname{tr}\bigl(T(I_n+tN)C(I_n-tN)\bigr) t ⟼ tr ( T ( I n + tN ) C ( I n − tN ) ) has nonzero derivative at zero. It has a root in K K K I n − t N I_n-tN I n − tN I n + t N I_n+tN I n + tN C C C T T T T T T
We shall also need a codimension-two version. We use the following consequence of de Bondt’s block theorem [14 Theorem 3.3 and Corollary 3.2] . If S ⊂ M n ( K ) S\subset M_n(K) S ⊂ M n ( K ) 0 < dim S < n 0<\dim S<n 0 < dim S < n I n ∉ S I_n\notin S I n ∈ / S
r + 1 = dim K ( x ) span K ( x ) { C x : C ∈ S + K I n } . r+1=\dim_{K(x)}\operatorname{span}_{K(x)}\{Cx:C\in S+KI_n\}. r + 1 = dim K ( x ) span K ( x ) { C x : C ∈ S + K I n } . Here x x x x 1 , … , x n x_1,\ldots,x_n x 1 , … , x n K ( x ) K(x) K ( x ) S S S r × ( n − r ) r\times(n-r) r × ( n − r ) 1 ≤ r < n 1\le r<n 1 ≤ r < n
Theorem 2.4. Suppose n ≥ 3 n\ge3 n ≥ 3 I n , A , B I_n,A,B I n , A , B O \mathcal{O} O
{ ( tr ( A C ) , tr ( B C ) ) : C ∈ O } = K 2 . \{(\operatorname{tr}(AC),\operatorname{tr}(BC)):C\in\mathcal{O}\}=K^2. {( tr ( A C ) , tr ( B C )) : C ∈ O } = K 2 . Proof. First suppose n = 3 n=3 n = 3 A , B ∈ M 3 ( K ) A,B\in M_3(K) A , B ∈ M 3 ( K ) I 3 , A , B I_3,A,B I 3 , A , B T ∈ M 3 ( K ) T\in M_3(K) T ∈ M 3 ( K ) a , b a,b a , b c c c τ = tr ( T + c I 3 ) ≠ 0 \tau=\operatorname{tr}(T+cI_3)\ne0 τ = tr ( T + c I 3 ) = 0
A ′ = A − a + c tr A τ I 3 , B ′ = B − b + c tr B τ I 3 . A'=A-\frac{a+c\operatorname{tr}A}{\tau}I_3,\qquad B'=B-\frac{b+c\operatorname{tr}B}{\tau}I_3. A ′ = A − τ a + c tr A I 3 , B ′ = B − τ b + c tr B I 3 . The block theorem applied to span { A ′ , B ′ } \operatorname{span}\{A',B'\} span { A ′ , B ′ } r = 1 r=1 r = 1 2 2 2 T T T r , 3 − r r,3-r r , 3 − r D 1 , D 2 D_1,D_2 D 1 , D 2 E E E Z = ( D 1 0 E D 2 ) Z=\begin{pmatrix}D_1&0\\E&D_2\end{pmatrix} Z = ( D 1 E 0 D 2 ) tr ( A ′ Z ) = tr ( B ′ Z ) = 0 \operatorname{tr}(A'Z)=\operatorname{tr}(B'Z)=0 tr ( A ′ Z ) = tr ( B ′ Z ) = 0 E E E S = span { A ′ , B ′ } S=\operatorname{span}\{A',B'\} S = span { A ′ , B ′ } E ↦ ( T ↦ tr ( T 12 E ) ) E\mapsto(T\mapsto\operatorname{tr}(T_{12}E)) E ↦ ( T ↦ tr ( T 12 E )) S ∗ S^* S ∗ T ↦ T 12 T\mapsto T_{12} T ↦ T 12
X D 1 − D 2 X = E . XD_1-D_2X=E. X D 1 − D 2 X = E . The diagonal D 1 , D 2 D_1,D_2 D 1 , D 2 ( I 0 X I ) \begin{pmatrix}I&0\\X&I\end{pmatrix} ( I X 0 I ) diag ( D 1 , D 2 ) \operatorname{diag}(D_1,D_2) diag ( D 1 , D 2 ) ( D 1 0 E D 2 ) \begin{pmatrix}D_1&0\\E&D_2\end{pmatrix} ( D 1 E 0 D 2 ) T + c I 3 T+cI_3 T + c I 3 Z ∼ T + c I 3 Z\sim T+cI_3 Z ∼ T + c I 3 tr Z = τ \operatorname{tr}Z=\tau tr Z = τ
tr ( A Z ) = a + c tr A τ tr Z = a + c tr A , \operatorname{tr}(AZ)=\frac{a+c\operatorname{tr}A}{\tau}\operatorname{tr}Z=a+c\operatorname{tr}A, tr ( A Z ) = τ a + c tr A tr Z = a + c tr A , and the same equation holds for B B B Z − c I 3 ∼ T Z-cI_3\sim T Z − c I 3 ∼ T a , b a,b a , b
For arbitrary n ≥ 3 n\ge3 n ≥ 3 I n , A , B I_n,A,B I n , A , B e = max x dim span { x , A x , B x } e=\max_x\dim\operatorname{span}\{x,Ax,Bx\} e = max x dim span { x , A x , B x } e = 3 e=3 e = 3 x x x e = 1 e=1 e = 1 A , B A,B A , B e = 2 e=2 e = 2 S = span { A , B } S=\operatorname{span}\{A,B\} S = span { A , B } r = 1 r=1 r = 1 W W W P P P e = 2 e=2 e = 2 e 1 , e j , e k e_1,e_j,e_k e 1 , e j , e k 2 × 2 2\times2 2 × 2 j , k j,k j , k A , B A,B A , B W W W A W = P A ∣ W A_W=P A|_W A W = P A ∣ W B B B
For a nonscalar semisimple T T T n n n 3 × 3 3\times3 3 × 3 W W W D D D ker P \ker P ker P W W W
a − tr ( A 22 D ) , b − tr ( B 22 D ) , a-\operatorname{tr}(A_{22}D),\qquad b-\operatorname{tr}(B_{22}D), a − tr ( A 22 D ) , b − tr ( B 22 D ) , where A 22 , B 22 A_{22},B_{22} A 22 , B 22 T T T
The minimal nonzero nilpotent orbit also attains every pair. Seek C = u v C=uv C = uv u ≠ 0 u\ne0 u = 0 v ≠ 0 v\ne0 v = 0
v ( u ) = 0 , v ( A u ) = a , v ( B u ) = b . v(u)=0,\qquad v(Au)=a,\qquad v(Bu)=b. v ( u ) = 0 , v ( A u ) = a , v ( B u ) = b . For ( a , b ) ≠ ( 0 , 0 ) (a,b)\ne(0,0) ( a , b ) = ( 0 , 0 ) e = 3 e=3 e = 3 u u u v v v e = 2 e=2 e = 2 v = e 1 ∗ v=e_1^* v = e 1 ∗ u ∈ span { e 2 , … , e n } u\in\operatorname{span}\{e_2,\ldots,e_n\} u ∈ span { e 2 , … , e n } ( a , b ) = ( 0 , 0 ) (a,b)=(0,0) ( a , b ) = ( 0 , 0 ) u u u A A A u , A u , B u u,Au,Bu u , A u , B u C C C
Finally, let O O O d = dim O d = \dim O d = dim O X = O ‾ X = \overline{O} X = O F F F O O O X X X F F F λ \lambda λ λ I n \lambda I_n λ I n λ I n + N \lambda I_n+N λ I n + N λ I n \lambda I_n λ I n X ∩ F X \cap F X ∩ F
There are finitely many boundary orbits in X ∖ O X \setminus O X ∖ O Z ( T ) = { C : C T = T C } Z(T)=\{C : CT=TC\} Z ( T ) = { C : C T = T C } λ \lambda λ r 1 , r 2 , … r_1,r_2,\ldots r 1 , r 2 , … r i , r j r_i,r_j r i , r j min ( r i , r j ) \min(r_i,r_j) min ( r i , r j ) K [ t ] K[t] K [ t ] K [ t ] / ( t r i ) K[t]/(t^{r_i}) K [ t ] / ( t r i ) K [ t ] / ( t r j ) K[t]/(t^{r_j}) K [ t ] / ( t r j ) c λ , j c_{\lambda,j} c λ , j λ \lambda λ j j j
dim Z ( T ) = ∑ λ ∑ i , k min ( r i , r k ) = ∑ λ , j c λ , j 2 , dim O T = n 2 − dim Z ( T ) . \dim Z(T)=\sum_{\lambda}\sum_{i,k}\min(r_i,r_k)=\sum_{\lambda,j}c_{\lambda,j}^{2},\qquad\dim O_T=n^2-\dim Z(T). dim Z ( T ) = λ ∑ i , k ∑ min ( r i , r k ) = λ , j ∑ c λ , j 2 , dim O T = n 2 − dim Z ( T ) . The positive c λ , j c_{\lambda,j} c λ , j n n n c λ , j 2 ≡ c λ , j ( m o d 2 ) c_{\lambda,j}^{2}\equiv c_{\lambda,j}\pmod2 c λ , j 2 ≡ c λ , j ( mod 2 ) X X X d d d GL n ( K ) \operatorname{GL}_n(K) GL n ( K ) d − 2 d-2 d − 2 C ↦ tr ( A C ) C\mapsto\operatorname{tr}(AC) C ↦ tr ( A C ) s l n ( K ) \mathfrak{sl}_n(K) sl n ( K ) s l n ( K ) \mathfrak{sl}_n(K) sl n ( K ) A A A
dim ( ( X ∖ O ) ∩ F ) ≤ max { 0 , d − 3 } . \dim((X\setminus O)\cap F)\leq\max\{0,d-3\}. dim (( X ∖ O ) ∩ F ) ≤ max { 0 , d − 3 } . Every component of X ∩ F X\cap F X ∩ F d − 2 d-2 d − 2 T T T c λ , j c_{\lambda,j} c λ , j n − 1 n-1 n − 1 n n n n n n ( n − 1 ) 2 + 1 (n-1)^2+1 ( n − 1 ) 2 + 1 d ≥ 2 n − 2 ≥ 4 d\geq2n-2\geq4 d ≥ 2 n − 2 ≥ 4 X ∩ F X\cap F X ∩ F O ∩ F ≠ ∅ O\cap F\ne\varnothing O ∩ F = ∅
Corollary 2.5. Suppose n ≥ 3 n\geq3 n ≥ 3 p p p p p p n 2 − 2 n^2-2 n 2 − 2 p ( M n ( K ) ) = M n ( K ) p(M_n(K))=M_n(K) p ( M n ( K )) = M n ( K )
Proof. In the same frozen-input space, intersect the rank open set with the open set on which the slice has nonzero trace. Its image H H H s l n ( K ) \mathfrak{sl}_n(K) sl n ( K ) I n I_n I n Theorem 2.4 , or Lemma 2.3 in codimension one, shows that H H H Lemma 2.2 supplies the scalar values.
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Two-cycle slices
Fix all but two inputs, denoted B , C B,C B , C s s s z i ∈ K s z_i\in K^s z i ∈ K s F ( u , v , w ) F(u,v,w) F ( u , v , w ) B B B C C C u u u B B B v v v B , C B,C B , C w w w C C C G G G C C C B B B 1 ≤ i , j ≤ n 1 \leq i,j \leq n 1 ≤ i , j ≤ n
p i j = ∑ h = 1 n F ( z i , z h , z j ) B i h C h j + ∑ h = 1 n G ( z i , z h , z j ) C i h B h j . (2) p_{ij}=\sum_{h=1}^{n}F(z_i,z_h,z_j)B_{ih}C_{hj}+\sum_{h=1}^{n}G(z_i,z_h,z_j)C_{ih}B_{hj}. \tag*{(2)} p ij = h = 1 ∑ n F ( z i , z h , z j ) B ih C hj + h = 1 ∑ n G ( z i , z h , z j ) C ih B hj . (2) For example, with two diagonal inputs A , D A,D A , D A B D C ABDC A B D C a i d h a_i d_h a i d h F F F A C D B ACDB A C D B a i d h a_i d_h a i d h G G G u , v ∈ K s u,v \in K^s u , v ∈ K s
P ( u , v ) = F ( u , v , u ) , Q ( u , v ) = G ( v , u , v ) , Δ ( u , v ) = Q ( u , v ) Q ( v , u ) − P ( u , v ) P ( v , u ) . P(u,v)=F(u,v,u),\qquad Q(u,v)=G(v,u,v),\qquad\Delta(u,v)=Q(u,v)Q(v,u)-P(u,v)P(v,u). P ( u , v ) = F ( u , v , u ) , Q ( u , v ) = G ( v , u , v ) , Δ ( u , v ) = Q ( u , v ) Q ( v , u ) − P ( u , v ) P ( v , u ) . Lemma 3.1. Every nonscalar T ∈ M 3 ( K ) T \in M_3(K) T ∈ M 3 ( K )
Proof. Choose w , ϕ w,\phi w , ϕ ϕ ( w ) = 1 \phi(w)=1 ϕ ( w ) = 1 ϕ ( T w ) = 0 \phi(Tw)=0 ϕ ( T w ) = 0 K w ⊕ ker ϕ Kw \oplus\ker\phi K w ⊕ ker ϕ
T = ( 0 r c S ) T=\begin{pmatrix}0&r\\c&S\end{pmatrix} T = ( 0 c r S ) in this decomposition. Then ϕ ( T 2 w ) = r c \phi(T^2w)=rc ϕ ( T 2 w ) = r c tr ( T 2 ) = 2 r c + tr ( S 2 ) \operatorname{tr}(T^2)=2rc+\operatorname{tr}(S^2) tr ( T 2 ) = 2 r c + tr ( S 2 ) tr S = tr T \operatorname{tr}S=\operatorname{tr}T tr S = tr T e 2 ( T ) e_2(T) e 2 ( T )
det ( t I 3 − T ) = t 3 − ( tr T ) t 2 + e 2 ( T ) t − det T , \det(tI_3-T)=t^3-(\operatorname{tr}T)t^2+e_2(T)t-\det T, det ( t I 3 − T ) = t 3 − ( tr T ) t 2 + e 2 ( T ) t − det T , we have e 2 ( T ) = ( ( tr T ) 2 − tr ( T 2 ) ) / 2 e_2(T)=((\operatorname{tr}T)^2-\operatorname{tr}(T^2))/2 e 2 ( T ) = (( tr T ) 2 − tr ( T 2 )) /2 det S = ( ( tr S ) 2 − tr ( S 2 ) ) / 2 \det S=((\operatorname{tr}S)^2-\operatorname{tr}(S^2))/2 det S = (( tr S ) 2 − tr ( S 2 )) /2
tr S = tr T , det S = e 2 ( T ) + ϕ ( T 2 w ) . \operatorname{tr}S=\operatorname{tr}T,\qquad\det S=e_2(T)+\phi(T^2w). tr S = tr T , det S = e 2 ( T ) + ϕ ( T 2 w ) . When the minimal polynomial has degree three, choose w w w S S S ( t − λ ) ( t − μ ) (t-\lambda)(t-\mu) ( t − λ ) ( t − μ ) λ ≠ μ \lambda\ne\mu λ = μ T 2 = ( λ + μ ) T − λ μ I 3 T^2=(\lambda+\mu)T-\lambda\mu I_3 T 2 = ( λ + μ ) T − λ μ I 3 μ 2 \mu^2 μ 2 S S S μ ≠ 0 \mu\ne0 μ = 0 w , T w w,Tw w , T w ϕ \phi ϕ μ = 0 \mu=0 μ = 0 K w Kw K w λ \lambda λ T = λ I 3 + N T=\lambda I_3+N T = λ I 3 + N N 2 = 0 ≠ N N^2=0\ne N N 2 = 0 = N λ 2 \lambda^2 λ 2 λ = 0 \lambda=0 λ = 0 K w = im N Kw=\operatorname{im}N K w = im N w w w
Lemma 3.2. Fix a pairing of all coordinates in even dimension n ≥ 4 n\ge4 n ≥ 4 n ≥ 3 n\ge3 n ≥ 3
Proof. For n = 2 r ≥ 4 n=2r\ge4 n = 2 r ≥ 4 i , i + r i,i+r i , i + r n ≥ 5 n\ge5 n ≥ 5 w , T w w,Tw w , T w ϕ ( w ) = 1 \phi(w)=1 ϕ ( w ) = 1 ϕ ( T w ) = 0 \phi(Tw)=0 ϕ ( T w ) = 0 K w Kw K w n = 3 n=3 n = 3 Lemma 3.1 and pair the two eigenvectors of the complementary compression.
Proposition 3.3. If Δ \Delta Δ (2) , then p ( M n ( K ) ) = M n ( K ) p(M_n(K))=M_n(K) p ( M n ( K )) = M n ( K ) n ≥ 3 n\ge3 n ≥ 3
Proof. First suppose F , G F,G F , G B i , σ ( i ) = b i ≠ 0 B_{i,\sigma(i)}=b_i\ne0 B i , σ ( i ) = b i = 0 σ \sigma σ n n n Z = B C Z=BC Z = B C
p i j = F ( z i , z σ ( i ) , z j ) Z i j + b σ − 1 ( j ) b σ − 1 ( i ) G ( z i , z σ − 1 ( j ) , z j ) Z σ − 1 ( i ) , σ − 1 ( j ) . (3) p_{ij}=F(z_i,z_{\sigma(i)},z_j)Z_{ij}+\frac{b_{\sigma^{-1}(j)}}{b_{\sigma^{-1}(i)}}G(z_i,z_{\sigma^{-1}(j)},z_j)Z_{\sigma^{-1}(i),\sigma^{-1}(j)}. \tag*{(3)} p ij = F ( z i , z σ ( i ) , z j ) Z ij + b σ − 1 ( i ) b σ − 1 ( j ) G ( z i , z σ − 1 ( j ) , z j ) Z σ − 1 ( i ) , σ − 1 ( j ) . (3) Thus the operator preserves orbits of the simultaneous permutation of the two indices. The diagonal orbit of each two-cycle has determinant Δ \Delta Δ
Give all weights in each cycle a common value, with independent nonzero values for different cycles. On an entry orbit of length d d d w 1 , w 2 w_1,w_2 w 1 , w 2 ( w 2 / w 1 ) (w_2/w_1) ( w 2 / w 1 )
∏ ( i , j ) F ( z i , z σ ( i ) , z j ) + ( − 1 ) d − 1 ( w 2 / w 1 ) d ∏ ( i , j ) G ( z i , z σ − 1 ( j ) , z j ) , \prod_{(i,j)} F(z_i,z_{\sigma(i)},z_j) + (-1)^{d-1}(w_2/w_1)^d \prod_{(i,j)} G(z_i,z_{\sigma^{-1}(j)},z_j), ( i , j ) ∏ F ( z i , z σ ( i ) , z j ) + ( − 1 ) d − 1 ( w 2 / w 1 ) d ( i , j ) ∏ G ( z i , z σ − 1 ( j ) , z j ) , where each product runs over that entry orbit. Indeed, in the determinant expansion a row can select its diagonal or cyclic entry; selecting one cyclic entry forces that choice around the entire orbit. Between two two-cycles, every factor is a nonzero polynomial on three independent vertex tuples. For a two-cycle with tuples u , v u,v u , v w w w F ( u , v , w ) F ( v , u , w ) ≠ 0 F(u,v,w)F(v,u,w) \ne0 F ( u , v , w ) F ( v , u , w ) = 0 G ( w , u , v ) G ( w , v , u ) ≠ 0 G(w,u,v)G(w,v,u) \ne0 G ( w , u , v ) G ( w , v , u ) = 0 Lemma 3.2 , proving coverage of nonscalar targets. The invertible two-cycle diagonal map gives a value of nonzero trace already in size two. Corner embedding and Lemma 2.2 give every scalar target in all sizes.
If F = 0 F=0 F = 0 Δ \Delta Δ Q ≠ 0 Q \ne0 Q = 0 σ ( i ) = i − 1 \sigma(i)=i-1 σ ( i ) = i − 1 n n n G ( u , v , u ) G(u,v,u) G ( u , v , u ) n ≥ 3 n \ge3 n ≥ 3 G = 0 G=0 G = 0 F F F
Identity reductions and scalar shifts
Write ad A ( Y ) = [ A , Y ] \operatorname{ad}_A(Y)=[A,Y] ad A ( Y ) = [ A , Y ] p ( X i = I ) p(X_i=I) p ( X i = I ) I n I_n I n proper if every such substitution is zero. For a cubic, write the coefficient vector in the order A B C , A C B , B A C , B C A , C A B , C B A ABC, ACB, BAC, BCA, CAB, CBA A B C , A C B , B A C , B C A , C A B , C B A ( a , b , c , d , e , f ) (a,b,c,d,e,f) ( a , b , c , d , e , f )
a + c + d = b + e + f = a + b + c = d + e + f = a + b + e = c + d + f = 0. a+c+d=b+e+f=a+b+c=d+e+f=a+b+e=c+d+f=0. a + c + d = b + e + f = a + b + c = d + e + f = a + b + e = c + d + f = 0. Solving gives ( a , b , c , d , e , f ) = ( a , − a − c , c , − a − c , c , a ) (a,b,c,d,e,f)=(a,-a-c,c,-a-c,c,a) ( a , b , c , d , e , f ) = ( a , − a − c , c , − a − c , c , a ) a [ A , [ B , C ] ] + c [ B , [ A , C ] ] a[A,[B,C]]+c[B,[A,C]] a [ A , [ B , C ]] + c [ B , [ A , C ]]
span { [ A , [ B , Y ] ] , [ B , [ A , Y ] ] } . (4) \operatorname{span}\{[A,[B,Y]],[B,[A,Y]]\}. \tag*{(4)} span {[ A , [ B , Y ]] , [ B , [ A , Y ]]} . (4) We use the Lie algebra s l 2 ( K ) \mathfrak{sl}_2(K) sl 2 ( K ) h = diag ( 1 , − 1 ) h=\operatorname{diag}(1,-1) h = diag ( 1 , − 1 ) e = E 12 e=E_{12} e = E 12 f = E 21 f=E_{21} f = E 21 n n n ρ n \rho_n ρ n w 0 , … , w n − 1 w_0,\ldots,w_{n-1} w 0 , … , w n − 1
ρ n ( h ) w i = ( n − 1 − 2 i ) w i , ρ n ( e ) w i = i ( n − i ) w i − 1 , ρ n ( f ) w i = w i + 1 , \rho_n(h)w_i=(n-1-2i)w_i,\qquad\rho_n(e)w_i=i(n-i)w_{i-1},\qquad\rho_n(f)w_i=w_{i+1}, ρ n ( h ) w i = ( n − 1 − 2 i ) w i , ρ n ( e ) w i = i ( n − i ) w i − 1 , ρ n ( f ) w i = w i + 1 , with vectors outside the stated range equal to zero. These formulas satisfy the three commutator relations of $h,e,f.
Lemma 4.1. Let Q = ( q a b ) Q=(q_{ab}) Q = ( q ab ) 3 × 3 3 \times3 3 × 3 K K K ( Q + Q T ) / 2 (Q+Q^{\mathsf{T}})/2 ( Q + Q T ) /2 n ≥ 2 n \ge2 n ≥ 2 x 1 , x 2 , x 3 ∈ s l 2 ( K ) x_1,x_2,x_3 \in\mathfrak{sl}_2(K) x 1 , x 2 , x 3 ∈ sl 2 ( K ) n n n ρ n \rho_n ρ n s l 2 ( K ) \mathfrak{sl}_2(K) sl 2 ( K )
M = ∑ a , b = 1 3 q a b ad ρ n ( x a ) ad ρ n ( x b ) M=\sum_{a,b=1}^{3}q_{ab}\operatorname{ad}_{\rho_n}(x_a)\operatorname{ad}_{\rho_n}(x_b) M = a , b = 1 ∑ 3 q ab ad ρ n ( x a ) ad ρ n ( x b ) is invertible on s l n ( K ) \mathfrak{sl}_n(K) sl n ( K )
Proof. View Q Q Q Q ( u , v ) = u T Q v Q(u,v)=u^{\mathsf{T}}Qv Q ( u , v ) = u T Q v K 3 K^3 K 3 u u u Q ( u , u ) = 1 Q(u,u)=1 Q ( u , u ) = 1 V = { v : Q ( u , v ) = 0 } V=\{v:Q(u,v)=0\} V = { v : Q ( u , v ) = 0 } span { u , v } \operatorname{span}\{u,v\} span { u , v } Q ( v , v ) = 0 Q(v,v)=0 Q ( v , v ) = 0 v ∈ V v\in V v ∈ V Q ∣ V Q|_V Q ∣ V Q ( v , w ) = b ≠ 0 Q(v,w)=b\ne0 Q ( v , w ) = b = 0 span { u + t v , w } \operatorname{span}\{u+tv,w\} span { u + t v , w } − t b ( Q ( w , u ) − t b ) -tb(Q(w,u)-tb) − t b ( Q ( w , u ) − t b ) Q ∣ V = 0 Q|_V=0 Q ∣ V = 0 Q = 1 Q=1 Q = 1
For x , y ∈ s l 2 ( K ) x,y\in\mathfrak{sl}_2(K) x , y ∈ sl 2 ( K ) u , v u,v u , v x a = u a x + v a y x_a=u_ax+v_ay x a = u a x + v a y Q Q Q
( 0 α β 0 ) , α β ( α + β ) ≠ 0 ; ( s k − k 0 ) , s k ≠ 0. \begin{pmatrix}0&\alpha\\ \beta&0\end{pmatrix},\quad\alpha\beta(\alpha+\beta)\ne0;\qquad
\begin{pmatrix}s&k\\-k&0\end{pmatrix},\quad sk\ne0. ( 0 β α 0 ) , α β ( α + β ) = 0 ; ( s − k k 0 ) , s k = 0. These follow by choosing an isotropic basis when the symmetric part has rank two, and by diagonalizing it when it has rank one. It suffices to treat the corresponding operators α A B + β B A \alpha AB+\beta BA α A B + β B A s A 2 + k [ A , B ] sA^2+k[A,B] s A 2 + k [ A , B ]
Let V j V_j V j 2 j + 1 2j+1 2 j + 1 j ∈ Z ≥ 1 j\in\mathbb{Z}_{\ge1} j ∈ Z ≥ 1 h , e , f h,e,f h , e , f H , E , F H,E,F H , E , F A , B A,B A , B s l 2 ( K ) \mathfrak{sl}_2(K) sl 2 ( K ) v m v_m v m − j ≤ m ≤ j -j\le m\le j − j ≤ m ≤ j
H v m = 2 m v m , E v m = ( j − m ) v m + 1 , F v m = ( j + m ) v m − 1 , Hv_m=2mv_m,\qquad Ev_m=(j-m)v_{m+1},\qquad Fv_m=(j+m)v_{m-1}, H v m = 2 m v m , E v m = ( j − m ) v m + 1 , F v m = ( j + m ) v m − 1 , and with vectors beyond the endpoints zero. In the first case take A = H A=H A = H B = H + t ( E + F ) B=H+t(E+F) B = H + t ( E + F ) t = 0 t=0 t = 0 ( α + β ) H 2 (\alpha+\beta)H^2 ( α + β ) H 2
− 2 α β j ( j + 1 ) α + β t 2 -\frac{2\alpha\beta j(j+1)}{\alpha+\beta}t^2 − α + β 2 α β j ( j + 1 ) t 2 at order two. The first-order map sends v 0 v_0 v 0 2 α j ( v 1 − v − 1 ) 2\alpha j(v_1-v_{-1}) 2 α j ( v 1 − v − 1 ) v ± 1 v_{\pm1} v ± 1 v 0 v_0 v 0 ± 2 β ( j + 1 ) \pm2\beta(j+1) ± 2 β ( j + 1 ) 1 / ( 4 ( α + β ) ) 1/(4(\alpha+\beta)) 1/ ( 4 ( α + β )) v 1 , v − 1 v_1,v_{-1} v 1 , v − 1
In the second case take A = H A=H A = H B = t ( E + F ) B=t(E+F) B = t ( E + F ) s H 2 + 2 k t ( E − F ) sH^2+2kt(E-F) s H 2 + 2 k t ( E − F ) 2 k 2 j ( j + 1 ) t 2 / s 2k^2j(j+1)t^2/s 2 k 2 j ( j + 1 ) t 2 / s 2 k j , − 2 k j 2kj,-2kj 2 k j , − 2 k j − 2 k ( j + 1 ) , 2 k ( j + 1 ) -2k(j+1),2k(j+1) − 2 k ( j + 1 ) , 2 k ( j + 1 ) 1 / ( 4 s ) 1/(4s) 1/ ( 4 s ) x a x_a x a
Under the action x ⋅ Y = [ ρ n ( x ) , Y ] x\cdot Y=[\rho_n(x),Y] x ⋅ Y = [ ρ n ( x ) , Y ] ρ n ( e ) j \rho_n(e)^j ρ n ( e ) j 0 ≤ j < n 0\le j<n 0 ≤ j < n 2 j 2j 2 j
E F r v = r ( 2 j − r + 1 ) F r − 1 v EF^rv=r(2j-r+1)F^{r-1}v E F r v = r ( 2 j − r + 1 ) F r − 1 v for E v = 0 Ev=0 E v = 0 H v = 2 j v Hv=2jv H v = 2 j v 2 j + 1 2j+1 2 j + 1 F 2 j + 1 v = 0 F^{2j+1}v=0 F 2 j + 1 v = 0 − 2 j − 2 -2j-2 − 2 j − 2 0 ≤ r ≤ 2 j 0 \le r \le2j 0 ≤ r ≤ 2 j F r v ≠ 0 F^{r}v \ne0 F r v = 0 E E E v v v ∑ j = 0 n − 1 ( 2 j + 1 ) = n 2 \sum_{j=0}^{n-1}(2j+1)=n^2 ∑ j = 0 n − 1 ( 2 j + 1 ) = n 2
M n ( K ) = V 0 ⊕ V 1 ⊕ ⋯ ⊕ V n − 1 . M_n(K)=V_0\oplus V_1\oplus\cdots\oplus V_{n-1}. M n ( K ) = V 0 ⊕ V 1 ⊕ ⋯ ⊕ V n − 1 . Here V 0 = K I n V_0=K I_n V 0 = K I n s l n ( K ) \mathfrak{sl}_n(K) sl n ( K )
Lemma 4.2. Suppose a quartic p p p Y Y Y x 1 , x 2 , x 3 x_1,x_2,x_3 x 1 , x 2 , x 3 λ 1 , λ 2 , λ 3 ∈ K \lambda_1,\lambda_2,\lambda_3\in K λ 1 , λ 2 , λ 3 ∈ K t t t x a + t λ a I x_a+t\lambda_a I x a + t λ a I
∑ a , b = 1 3 q a b ad x a ad x b ( Y ) , \sum_{a,b=1}^{3}q_{ab}\operatorname{ad}_{x_a}\operatorname{ad}_{x_b}(Y), a , b = 1 ∑ 3 q ab ad x a ad x b ( Y ) , where the scalar matrix Q = ( q a b ) Q=(q_{ab}) Q = ( q ab ) p p p p ( M n ( K ) ) = M n ( K ) p(M_n(K))=M_n(K) p ( M n ( K )) = M n ( K ) n ≥ 3 n\ge3 n ≥ 3
α [ A , [ B , Y ] ] + β [ B , [ A , Y ] ] , α β ( α + β ) ≠ 0. \alpha[A,[B,Y]]+\beta[B,[A,Y]],\qquad\alpha\beta(\alpha+\beta)\ne0. α [ A , [ B , Y ]] + β [ B , [ A , Y ]] , α β ( α + β ) = 0. Proof. After the shifts the slice is exactly L 0 + t M L_0+tM L 0 + tM x a ′ ∈ s l 2 ( K ) x'_a\in\mathfrak{sl}_2(K) x a ′ ∈ sl 2 ( K ) Lemma 4.1 and freeze x a = ρ n ( x a ′ ) x_a=\rho_n(x'_a) x a = ρ n ( x a ′ ) M M M s l n ( K ) \mathfrak{sl}_n(K) sl n ( K ) s l n ( K ) \mathfrak{sl}_n(K) sl n ( K ) T ↦ T − ( tr T / n ) I n T\mapsto T-(\operatorname{tr}T/n)I_n T ↦ T − ( tr T / n ) I n ( n 2 − 1 ) × ( n 2 − 1 ) (n^2-1)\times(n^2-1) ( n 2 − 1 ) × ( n 2 − 1 ) det ( M ∣ s l n ( K ) ) ≠ 0 \det(M|_{\mathfrak{sl}_n(K)})\ne0 det ( M ∣ sl n ( K ) ) = 0 t t t n 2 − 1 n^2-1 n 2 − 1 s l n ( K ) \mathfrak{sl}_n(K) sl n ( K ) Lemmas 2.3 and 2.2 . For the final assertion, the cubic itself gives a rank-n 2 − 1 n^2-1 n 2 − 1 Lemma 4.1 , using the matrix with the two off-diagonal entries α , β \alpha,\beta α , β
Lemma 4.3. Suppose a quartic p p p
p ( A , B , C , I ) = λ [ A , [ B , C ] ] , p ( I , B , C , D ) = μ [ D , [ B , C ] ] , λ ≠ 0 , p(A,B,C,I)=\lambda[A,[B,C]],\qquad p(I,B,C,D)=\mu[D,[B,C]],\qquad\lambda\ne0, p ( A , B , C , I ) = λ [ A , [ B , C ]] , p ( I , B , C , D ) = μ [ D , [ B , C ]] , λ = 0 , and its restriction with A , D A,D A , D M n ( K ) M_n(K) M n ( K ) n ≥ 3 n\ge3 n ≥ 3
Proof. For diagonal A , D A,D A , D B i j C j i B_{ij}C_{ji} B ij C j i i i i α ( a i − a j ) ( d i − d j ) \alpha(a_i-a_j)(d_i-d_j) α ( a i − a j ) ( d i − d j ) C i j B j i C_{ij}B_{ji} C ij B j i β ( a i − a j ) ( d i − d j ) \beta(a_i-a_j)(d_i-d_j) β ( a i − a j ) ( d i − d j ) d i = d j d_i=d_j d i = d j A A A a i = a j a_i=a_j a i = a j a a a d d d ( a i − a j ) ( d i − d j ) (a_i-a_j)(d_i-d_j) ( a i − a j ) ( d i − d j ) α + β ≠ 0 \alpha+\beta\ne0 α + β = 0 α 2 ≠ β 2 \alpha^2\ne\beta^2 α 2 = β 2 α = β ≠ 0 \alpha=\beta\ne0 α = β = 0
Take B = ∑ i = 1 n − 1 E i , i + 1 B=\sum_{i=1}^{n-1}E_{i,i+1} B = ∑ i = 1 n − 1 E i , i + 1 C = ∑ i = 1 n − 1 3 i E i + 1 , i C=\sum_{i=1}^{n-1}3^iE_{i+1,i} C = ∑ i = 1 n − 1 3 i E i + 1 , i c i = 3 i c_i=3^i c i = 3 i 1 ≤ i ≤ n 1\le i\le n 1 ≤ i ≤ n c 0 = c n = 0 c_0=c_n=0 c 0 = c n = 0 [ B , C ] [B,C] [ B , C ] c i − c i − 1 c_i-c_{i-1} c i − c i − 1 D D D d i ≠ d i + 1 d_i\ne d_{i+1} d i = d i + 1 A A A i i i α c i ( d i − d i + 1 ) ( a i − a i + 1 ) \alpha c_i(d_i-d_{i+1})(a_i-a_{i+1}) α c i ( d i − d i + 1 ) ( a i − a i + 1 ) n − 1 n-1 n − 1 ( 1 , − 1 , 1 , − 1 , … ) (1,-1,1,-1,\ldots) ( 1 , − 1 , 1 , − 1 , … )
Replacing D D D D + t I n D+tI_n D + t I n t λ [ A , [ B , C ] ] t\lambda[A,[B,C]] t λ [ A , [ B , C ]] r r r − ( n − 1 ) ≤ r ≤ n − 1 -(n-1) \le r \le n-1 − ( n − 1 ) ≤ r ≤ n − 1 span { E i , i + r : 1 ≤ i , i + r ≤ n } \operatorname{span}\{E_{i,i+r}: 1 \le i, i+r \le n\} span { E i , i + r : 1 ≤ i , i + r ≤ n } B , C , D B,C,D B , C , D 1 , − 1 , 0 1,-1,0 1 , − 1 , 0 [ B , C ] [B,C] [ B , C ] t t t n 2 − 1 n^2-1 n 2 − 1 Lemmas 2.3 and 2.2 finish the proof.
Proper quartics
Proposition 5.1. Every proper multilinear quartic with nonzero formal trace is surjective on M n ( K ) M_n(K) M n ( K )
Proof. Let sgn ( σ ) \operatorname{sgn}(\sigma) sgn ( σ ) − 1 -1 − 1 ( X 1 , X 2 , X 3 , X 4 ) = ( A , B , C , D ) (X_1,X_2,X_3,X_4)=(A,B,C,D) ( X 1 , X 2 , X 3 , X 4 ) = ( A , B , C , D )
s 4 ( A , B , C , D ) = ∑ σ ∈ S 4 sgn ( σ ) X σ ( 1 ) X σ ( 2 ) X σ ( 3 ) X σ ( 4 ) . s_4(A,B,C,D)=\sum_{\sigma\in S_4}\operatorname{sgn}(\sigma)X_{\sigma(1)}X_{\sigma(2)}X_{\sigma(3)}X_{\sigma(4)}. s 4 ( A , B , C , D ) = σ ∈ S 4 ∑ sgn ( σ ) X σ ( 1 ) X σ ( 2 ) X σ ( 3 ) X σ ( 4 ) . Cyclic rotation of four distinct letters changes the sign and preserves matrix trace, so tr s 4 = 0 \operatorname{tr}s_4=0 tr s 4 = 0 p p p p p p 4 ! 4! 4 ! s 4 s_4 s 4 p p p
A proper polynomial of multidegree ( 2 , 1 , 1 ) (2,1,1) ( 2 , 1 , 1 ) A , B , C A,B,C A , B , C
g = α [ A , B ] [ A , C ] + β [ A , C ] [ A , B ] + γ [ A , [ A , [ B , C ] ] ] + δ [ [ A , [ A , B ] ] , C ] . (5) g=\alpha[A,B][A,C]+\beta[A,C][A,B]+\gamma[A,[A,[B,C]]]+\delta[[A,[A,B]],C]. \tag*{(5)} g = α [ A , B ] [ A , C ] + β [ A , C ] [ A , B ] + γ [ A , [ A , [ B , C ]]] + δ [[ A , [ A , B ]] , C ] . (5) For a word, its central-translation derivative in a letter is the sum of the words obtained by deleting one occurrence of that letter. This is the coefficient of t t t t I n tI_n t I n (5) .
Table 3.
For the derivative matrix, order the deleted letter as A , B , C A, B, C A , B , C 0 , 1 , 2 , 3 , 4 , 5 , 6 , 9 0,1,2,3,4,5,6,9 0 , 1 , 2 , 3 , 4 , 5 , 6 , 9 0 , 1 , 2 , 3 , 4 , 5 , 6 , 8 0,1,2,3,4,5,6,8 0 , 1 , 2 , 3 , 4 , 5 , 6 , 8 − 4 -4 − 4 1 1 1 0 , 1 , 2 , 4 0,1,2,4 0 , 1 , 2 , 4 (5) is ( α + β ) tr ( [ A , B ] [ A , C ] ) (\alpha+\beta)\operatorname{tr}([A,B][A,C]) ( α + β ) tr ([ A , B ] [ A , C ]) α + β ≠ 0 \alpha+\beta\ne0 α + β = 0
For a permutation σ \sigma σ { 1 , … , n } \{1,\ldots,n\} { 1 , … , n } A = diag ( a i ) A=\operatorname{diag}(a_i) A = diag ( a i ) B B B B i , σ ( i ) = b i ≠ 0 B_{i,\sigma(i)}=b_i\ne0 B i , σ ( i ) = b i = 0 Z = B C Z=BC Z = B C C C C
g i j = F i j Z i j − H i j b σ − 1 ( j ) b σ − 1 ( i ) Z σ − 1 ( i ) , σ − 1 ( j ) , g_{ij}=F_{ij}Z_{ij}-H_{ij}\frac{b_{\sigma^{-1}(j)}}{b_{\sigma^{-1}(i)}}Z_{\sigma^{-1}(i),\sigma^{-1}(j)}, g ij = F ij Z ij − H ij b σ − 1 ( i ) b σ − 1 ( j ) Z σ − 1 ( i ) , σ − 1 ( j ) , where
F i j = α ( a i − a σ ( i ) ) ( a σ ( i ) − a j ) + γ ( a i − a j ) 2 + δ ( a i − a σ ( i ) ) 2 , F_{ij}=\alpha(a_i-a_{\sigma(i)})(a_{\sigma(i)}-a_j)+\gamma(a_i-a_j)^2+\delta(a_i-a_{\sigma(i)})^2, F ij = α ( a i − a σ ( i ) ) ( a σ ( i ) − a j ) + γ ( a i − a j ) 2 + δ ( a i − a σ ( i ) ) 2 , H i j = γ ( a i − a j ) 2 + δ ( a σ − 1 ( j ) − a j ) 2 − β ( a i − a σ − 1 ( j ) ) ( a σ − 1 ( j ) − a j ) . H_{ij}=\gamma(a_i-a_j)^2+\delta(a_{\sigma^{-1}(j)}-a_j)^2-\beta(a_i-a_{\sigma^{-1}(j)})(a_{\sigma^{-1}(j)}-a_j). H ij = γ ( a i − a j ) 2 + δ ( a σ − 1 ( j ) − a j ) 2 − β ( a i − a σ − 1 ( j ) ) ( a σ − 1 ( j ) − a j ) . The two-cycle diagonal determinant is nonzero unless ( δ − α ) 2 = ( β + δ ) 2 (\delta-\alpha)^2=(\beta+\delta)^2 ( δ − α ) 2 = ( β + δ ) 2 α + β ≠ 0 \alpha+\beta\ne0 α + β = 0 2 δ = α − β 2\delta=\alpha-\beta 2 δ = α − β A A A
In the exceptional case retain the original coefficients, with 2 δ = α − β 2\delta=\alpha-\beta 2 δ = α − β α + β ≠ 0 \alpha+\beta\ne0 α + β = 0 ℓ ≥ 3 \ell\ge3 ℓ ≥ 3 x i = a i − a i + 1 x_i=a_i-a_{i+1} x i = a i − a i + 1 ∑ i x i = 0 \sum_i x_i=0 ∑ i x i = 0
( ( δ − α ) ℓ − ( β + δ ) ℓ ) ∏ i x i 2 = − 2 ( α + β 2 ) ℓ ∏ i x i 2 , \left((\delta-\alpha)^\ell-(\beta+\delta)^\ell\right)\prod_i x_i^2=-2\left(\frac{\alpha+\beta}{2}\right)^\ell\prod_i x_i^2, ( ( δ − α ) ℓ − ( β + δ ) ℓ ) i ∏ x i 2 = − 2 ( 2 α + β ) ℓ i ∏ x i 2 , a nonzero polynomial. Define binary quadratic polynomials by
f ( u , v ) = γ u 2 − α u v + ( δ − α ) v 2 , h ( u , v ) = ( β + δ ) u 2 + β u v + γ v 2 . f(u,v)=\gamma u^2-\alpha uv+(\delta-\alpha)v^2,\qquad h(u,v)=(\beta+\delta)u^2+\beta uv+\gamma v^2. f ( u , v ) = γ u 2 − α uv + ( δ − α ) v 2 , h ( u , v ) = ( β + δ ) u 2 + β uv + γ v 2 . Both are nonzero, since δ − α = − ( α + β ) / 2 \delta-\alpha=-(\alpha+\beta)/2 δ − α = − ( α + β ) /2 β + δ = ( α + β ) / 2 \beta+\delta=(\alpha+\beta)/2 β + δ = ( α + β ) /2 σ ( i ) = i − 1 \sigma(i)=i-1 σ ( i ) = i − 1
F i , i + 1 = f ( x i , x i − 1 ) , H i , i + 1 = h ( x i + 1 , x i ) . F_{i,i+1}=f(x_i,x_{i-1}),\qquad H_{i,i+1}=h(x_{i+1},x_i). F i , i + 1 = f ( x i , x i − 1 ) , H i , i + 1 = h ( x i + 1 , x i ) . For example, a i − a i − 1 = − x i − 1 a_i-a_{i-1}=-x_{i-1} a i − a i − 1 = − x i − 1 a i − 1 − a i + 1 = x i − 1 + x i a_{i-1}-a_{i+1}=x_{i-1}+x_i a i − 1 − a i + 1 = x i − 1 + x i F i , i + 1 F_{i,i+1} F i , i + 1 F F F
∏ i f ( x i + 1 , x i ) − ∏ i h ( x i + 1 , x i ) , ∑ i x i = 0. (6) \prod_i f(x_{i+1},x_i)-\prod_i h(x_{i+1},x_i),\qquad\sum_i x_i=0. \tag*{(6)} i ∏ f ( x i + 1 , x i ) − i ∏ h ( x i + 1 , x i ) , i ∑ x i = 0. (6) For ℓ ≥ 5 \ell\ge5 ℓ ≥ 5 K [ x 1 , … , x ℓ ] / ( ∑ i x i ) K[x_1,\ldots,x_\ell]/(\sum_i x_i) K [ x 1 , … , x ℓ ] / ( ∑ i x i ) ℓ − 1 \ell-1 ℓ − 1 K K K (6) would thus imply, for c ∈ K ∗ c\in K^* c ∈ K ∗ k k k
f ( u , v ) h ( u , v ) = c ( u / v ) k , − 2 ≤ k ≤ 2 , c ℓ = 1. (7) \frac{f(u,v)}{h(u,v)}=c(u/v)^k,\qquad-2\le k\le2,\qquad c^\ell=1. \tag*{(7)} h ( u , v ) f ( u , v ) = c ( u / v ) k , − 2 ≤ k ≤ 2 , c ℓ = 1. (7) The coefficient comparisons can be listed explicitly. For k = 1 k=1 k = 1 v f = c u h vf=cuh v f = c u h β + δ = 0 \beta+\delta=0 β + δ = 0 δ − α = 0 \delta-\alpha=0 δ − α = 0 α + β ≠ 0 \alpha+\beta\ne0 α + β = 0 k = 2 k=2 k = 2 v 4 v^{4} v 4 v 2 f = c u 2 h v^{2}f=cu^{2}h v 2 f = c u 2 h δ − α = 0 \delta-\alpha=0 δ − α = 0 k = − 2 k=-2 k = − 2 u 4 , u 3 v , u v 3 u^{4},u^{3}v,uv^{3} u 4 , u 3 v , u v 3 u 2 f = c v 2 h u^{2}f=cv^{2}h u 2 f = c v 2 h γ = α = β = 0 \gamma=\alpha=\beta=0 γ = α = β = 0 k = 0 k=0 k = 0 f = c h f=ch f = c h − α = c β -\alpha=c\beta − α = c β γ = c ( β + δ ) \gamma=c(\beta+\delta) γ = c ( β + δ ) δ − α = c γ \delta-\alpha=c\gamma δ − α = c γ
δ − α β + δ = c 2 = − 1. \frac{\delta-\alpha}{\beta+\delta}=c^{2}=-1. β + δ δ − α = c 2 = − 1. and c ℓ ≠ 1 c^{\ell}\ne1 c ℓ = 1 ℓ \ell ℓ k = − 1 k=-1 k = − 1 u f = c v h uf=cvh u f = c v h
γ = 0 , − α = c ( β + δ ) , δ − α = c β , δ ( 1 + c ) = 0. \gamma=0,\qquad-\alpha=c(\beta+\delta),\qquad\delta-\alpha=c\beta,\qquad\delta(1+c)=0. γ = 0 , − α = c ( β + δ ) , δ − α = c β , δ ( 1 + c ) = 0. If δ = 0 \delta=0 δ = 0 ( δ − α ) / ( β + δ ) (\delta-\alpha)/(\beta+\delta) ( δ − α ) / ( β + δ ) c = − 1 c=-1 c = − 1 c = − 1 c=-1 c = − 1 c ℓ ≠ 1 c^{\ell}\ne1 c ℓ = 1 ℓ ≥ 5 \ell\ge5 ℓ ≥ 5
For ℓ = 3 \ell=3 ℓ = 3 Φ ( u , v ) \Phi(u,v) Φ ( u , v ) (6) at ( x 1 , x 2 , x 3 ) = ( u , v , − u − v ) (x_1,x_2,x_3)=(u,v,-u-v) ( x 1 , x 2 , x 3 ) = ( u , v , − u − v ) u 6 u^{6} u 6
− ( α + β ) γ ( δ + γ ) . -(\alpha+\beta)\gamma(\delta+\gamma). − ( α + β ) γ ( δ + γ ) . Here and below a coefficient is taken in the ordinary commutative polynomial ring K [ u , v ] K[u,v] K [ u , v ] Φ = 0 \Phi=0 Φ = 0 γ = 0 \gamma=0 γ = 0 γ = − δ \gamma=-\delta γ = − δ 2 δ = α − β 2\delta=\alpha-\beta 2 δ = α − β u 5 v u^{5}v u 5 v
− ( α − β ) ( α + β ) 2 4 . -\frac{(\alpha-\beta)(\alpha+\beta)^{2}}{4}. − 4 ( α − β ) ( α + β ) 2 . It forces α = β \alpha=\beta α = β δ = γ = 0 \delta=\gamma=0 δ = γ = 0 u 4 v 2 u^{4}v^{2} u 4 v 2 2 α 3 ≠ 0 2\alpha^{3}\ne0 2 α 3 = 0 Φ \Phi Φ
Every odd n ≥ 3 n\ge3 n ≥ 3 n ≥ 6 n\ge6 n ≥ 6 3 , n − 3 3,n-3 3 , n − 3 n = 4 n=4 n = 4 d d d w 1 , w 2 w_1,w_2 w 1 , w 2
∏ ( i , j ) F i j − ( w 2 w 1 ) d ∏ ( i , j ) H i j , \prod_{(i,j)}F_{ij}-\left(\frac{w_2}{w_1}\right)^{d}\prod_{(i,j)}H_{ij}, ( i , j ) ∏ F ij − ( w 1 w 2 ) d ( i , j ) ∏ H ij , by the same cyclic determinant expansion as in Proposition 3.3. Between two nontrivial cycles the three eigenvalue arguments in each factor are independent. For a three-cycle and singleton, use the F F F H H H F , H F,H F , H δ − α = − ( α + β ) / 2 \delta-\alpha=-(\alpha+\beta)/2 δ − α = − ( α + β ) /2 β + δ = ( α + β ) / 2 \beta+\delta=(\alpha+\beta)/2 β + δ = ( α + β ) /2
For n ≥ 6 n\ge6 n ≥ 6 n = 4 n=4 n = 4 (5) in size three embed in every larger size, and Lemma 2.2 gives scalar targets for the original multilinear polynomial. This proves the proposition.
Block transfer from sizes two and three
Lemma 6.1. Fix a free input position of an ordinary multilinear polynomial p p p p p p M min ( k , l ) ( K ) M_{\min(k,l)}(K) M m i n ( k , l ) ( K ) k , l k,l k , l k × l k\times l k × l
Proof. Let m m m p p p r r r diag ( A i , B i ) \operatorname{diag}(A_i,B_i) diag ( A i , B i ) A i ∈ M k ( K ) A_i \in M_k(K) A i ∈ M k ( K ) B i ∈ M l ( K ) B_i \in M_l(K) B i ∈ M l ( K ) Y ∈ M k × l ( K ) Y \in M_{k \times l}(K) Y ∈ M k × l ( K )
L k l ( Y ) = ∑ w a w A i 1 ⋯ A i s Y B j 1 ⋯ B j m − 1 − s , L_{kl}(Y)=\sum_w a_w A_{i_1}\cdots A_{i_s}YB_{j_1}\cdots B_{j_{m-1-s}}, L k l ( Y ) = w ∑ a w A i 1 ⋯ A i s Y B j 1 ⋯ B j m − 1 − s , where the word w w w i 1 , … , i s i_1,\ldots,i_s i 1 , … , i s r r r j 1 , … , j m − 1 − s j_1,\ldots,j_{m-1-s} j 1 , … , j m − 1 − s a w a_w a w p p p I k I_k I k I l I_l I l S S S r r r C S C_S C S C S C_S C S min ( k , l ) \min(k,l) min ( k , l ) S S S P P P Q Q Q C S ( X ) = c P X Q C_S(X)=cPXQ C S ( X ) = c P X Q c ≠ 0 c\ne0 c = 0 S S S t t t t t t T T T m − 1 − ∣ S △ T ∣ m-1-|S\mathbin{\triangle}T| m − 1 − ∣ S △ T ∣ △ \triangle △ t m − 1 C S t^{m-1}C_S t m − 1 C S det L k l \det L_{kl} det L k l det ( C S ) = c k l ( det P ) l ( det Q ) k ≠ 0 \det(C_S)=c^{kl}(\det P)^l(\det Q)^k\ne0 det ( C S ) = c k l ( det P ) l ( det Q ) k = 0
Lemma 6.2. Fix a free input position. Suppose p p p M 2 ( K ) M_2(K) M 2 ( K ) M 2 ( K ) / K I 2 M_2(K)/KI_2 M 2 ( K ) / K I 2 2 × 2 2 \times2 2 × 2
Proof. Let A A A F F F W ⊂ M 2 ( F ) W \subset M_2(F) W ⊂ M 2 ( F ) tr ∣ W ≠ 0 \operatorname{tr}|_W\ne0 tr ∣ W = 0 dim ( W + F I 2 ) / F I 2 ≥ 2 \dim(W+FI_2)/FI_2\ge2 dim ( W + F I 2 ) / F I 2 ≥ 2 det w = c ( tr w ) 2 \det w=c(\operatorname{tr}w)^2 det w = c ( tr w ) 2 W W W
For two independent generic free inputs Y , Z Y,Z Y , Z
T = tr ( p ( A , Z ) ) p ( A , Y ) − tr ( p ( A , Y ) ) p ( A , Z ) , T=\operatorname{tr}(p(A,Z))p(A,Y)-\operatorname{tr}(p(A,Y))p(A,Z), T = tr ( p ( A , Z )) p ( A , Y ) − tr ( p ( A , Y )) p ( A , Z ) , where A A A tr T = det T = 0 \operatorname{tr}T=\det T=0 tr T = det T = 0 T 2 = 0 T^2=0 T 2 = 0 [9 Proposition 1.31] . Both values and both traces defining T T T T = 0 T=0 T = 0
Choose U , V ∈ W U,V\in W U , V ∈ W tr U ≠ 0 \operatorname{tr}U\ne0 tr U = 0 U , V , I 2 U,V,I_2 U , V , I 2 U / tr U + z B U/\operatorname{tr}U+zB U / tr U + z B B = tr ( U ) V − tr ( V ) U B=\operatorname{tr}(U)V-\operatorname{tr}(V)U B = tr ( U ) V − tr ( V ) U U / tr U U/\operatorname{tr}U U / tr U W W W B B B U , V U,V U , V U , V , I 2 U,V,I_2 U , V , I 2 W W W tr U ≠ 0 \operatorname{tr}U\ne0 tr U = 0 det B \det B det B tr ( U B ) \operatorname{tr}(UB) tr ( U B )
det ( U tr U + z B ) = det U ( tr U ) 2 − z tr ( U B ) tr U + z 2 det B . \det\left(\frac{U}{\operatorname{tr}U}+zB\right)=\frac{\det U}{(\operatorname{tr}U)^2}-\frac{z\operatorname{tr}(UB)}{\operatorname{tr}U}+z^2\det B. det ( tr U U + z B ) = ( tr U ) 2 det U − tr U z tr ( U B ) + z 2 det B . The plane span { U , V } \operatorname{span}\{U,V\} span { U , V } I 2 I_2 I 2 K K K 2 × 2 2 \times2 2 × 2 U = L A ( Y 1 ) U=L_A(Y_1) U = L A ( Y 1 ) V = L A ( Y 2 ) V=L_A(Y_2) V = L A ( Y 2 ) L A L_A L A Y 1 , Y 2 Y_1,Y_2 Y 1 , Y 2 3 × 3 3 \times3 3 × 3 U , V , I 2 U,V,I_2 U , V , I 2 tr U ≠ 0 \operatorname{tr} U \ne0 tr U = 0 A , Y 1 , Y 2 A,Y_1,Y_2 A , Y 1 , Y 2 tr U \operatorname{tr} U tr U K K K Y 1 , Y 2 Y_1,Y_2 Y 1 , Y 2 K K K A A A
Lemma 6.3. If T ∈ M n ( K ) T \in M_n(K) T ∈ M n ( K ) ∑ i γ i = tr T \sum_i \gamma_i=\operatorname{tr} T ∑ i γ i = tr T T T T γ 1 , … , γ n \gamma_1,\ldots,\gamma_n γ 1 , … , γ n
Proof. This is Fillmore’s theorem [15 ] ; the following construction specifies the similarities. If T T T r r r s s s I n + t E r s I_n+tE_{rs} I n + t E r s t = ( T s s − T r r ) − 1 t=(T_{ss}-T_{rr})^{-1} t = ( T ss − T r r ) − 1 ( r , s ) (r,s) ( r , s ) T r s ≠ 0 T_{rs}\ne0 T r s = 0 r ≠ s r\ne s r = s B B B s s s
B s r = 0 , B s s = T r s , B s k = T r k − 1 ( k ≠ r , s ) . B_{sr}=0,\qquad B_{ss}=T_{rs},\qquad B_{sk}=T_{rk}-1\quad(k\ne r,s). B sr = 0 , B ss = T r s , B s k = T r k − 1 ( k = r , s ) . Its determinant is T r s T_{rs} T r s r r r B T B − 1 BTB^{-1} B T B − 1 s s s T r s / B s s T_{rs}/B_{ss} T r s / B ss k k k k ≠ r , s k\ne r,s k = r , s T r k − T r s B s k / B s s T_{rk}-T_{rs}B_{sk}/B_{ss} T r k − T r s B s k / B ss T T T u r = 0 u_r=0 u r = 0 u k = γ k − T k k u_k=\gamma_k-T_{kk} u k = γ k − T k k k ≠ r k\ne r k = r C = I n + u e r ∗ C=I_n+u e_r^* C = I n + u e r ∗ C − 1 = I n − u e r ∗ C^{-1}=I_n-u e_r^* C − 1 = I n − u e r ∗ k ≠ r k\ne r k = r
( C T C − 1 ) k k = T k k + u k T r k = γ k . (CTC^{-1})_{kk}=T_{kk}+u_kT_{rk}=\gamma_k. ( C T C − 1 ) k k = T k k + u k T r k = γ k . The remaining diagonal entry is T r r − ∑ k ≠ r u k = γ r T_{rr}-\sum_{k\ne r}u_k=\gamma_r T r r − ∑ k = r u k = γ r
Proposition 6.4. Suppose a quartic p p p M 2 ( K ) M_2(K) M 2 ( K ) M 2 ( K ) M_2(K) M 2 ( K ) M 3 ( K ) M_3(K) M 3 ( K ) p ( M n ( K ) ) = M n ( K ) p(M_n(K))=M_n(K) p ( M n ( K )) = M n ( K ) n ≥ 3 n\ge3 n ≥ 3
Proof. In size three, intersect the rank-seven and trace-nonzero opens. Theorem 2.4 makes each resulting slice meet every nonscalar orbit, and Corollary 2.5 gives full image. In size two use Lemma 6.2 .
For n ≥ 4 n\ge4 n ≥ 4 n n n Lemma 6.1 . This is possible since p p p M 2 ( K ) M_2(K) M 2 ( K )
For a nonscalar target T T T tr T \operatorname{tr} T tr T Lemma 2.2 supplies the scalar targets.
Reduction of the quartic coefficients
We now prove the trace-nonzero part of Theorem 1.1 . In the coefficient arguments, a trace identity is formal, that is, an identity modulo cyclic rotation of words.
If some single-identity reduction has nonzero trace, Vitas’s degree-three theorem [12 Theorem B] gives full image for that cubic, hence for p p p 2 n 2n 2 n n ≥ 3 n \ge3 n ≥ 3 r < s r < s r < s r , s r, s r , s u = ( a , d ) u = (a,d) u = ( a , d ) v = ( b , e ) v = (b,e) v = ( b , e ) B , C B, C B , C F , G F, G F , G (2) for this specialization. Set H ( u , v ) = − G ( v , u , v ) H(u,v) = -G(v,u,v) H ( u , v ) = − G ( v , u , v ) κ r s \kappa_{rs} κ r s a d ad a d P ( u , v ) − H ( u , v ) P(u,v) - H(u,v) P ( u , v ) − H ( u , v ) § 9 gives
κ 12 = κ 34 , κ 13 = κ 24 , κ 14 = κ 23 = − κ 12 − κ 13 . (8) \kappa_{12} = \kappa_{34}, \qquad\kappa_{13} = \kappa_{24}, \qquad\kappa_{14} = \kappa_{23} = -\kappa_{12} - \kappa_{13}. \tag*{(8)} κ 12 = κ 34 , κ 13 = κ 24 , κ 14 = κ 23 = − κ 12 − κ 13 . (8) For the same specialization,
P ( u , v ) − H ( u , v ) = κ r s ( a − b ) ( d − e ) , u = ( a , d ) , v = ( b , e ) , H ( u , v ) = − G ( v , u , v ) , (9) P(u,v) - H(u,v) = \kappa_{rs}(a-b)(d-e), \qquad u = (a,d), \ v = (b,e), \qquad H(u,v) = -G(v,u,v), \tag*{(9)} P ( u , v ) − H ( u , v ) = κ r s ( a − b ) ( d − e ) , u = ( a , d ) , v = ( b , e ) , H ( u , v ) = − G ( v , u , v ) , (9) Trace nonvanishing is ( κ 12 , κ 13 ) ≠ ( 0 , 0 ) (\kappa_{12}, \kappa_{13}) \ne(0,0) ( κ 12 , κ 13 ) = ( 0 , 0 ) § 9 shows. For a nonzero κ r s \kappa_{rs} κ r s B = E 12 B = E_{12} B = E 12 C = E 21 C = E_{21} C = E 21 κ r s ( a − b ) ( d − e ) \kappa_{rs}(a-b)(d-e) κ r s ( a − b ) ( d − e ) (9) . Hence trace nonvanishing is detected already in size two.
From (9) , the two-cycle determinant is
Δ = κ r s ( a − b ) ( d − e ) ( κ r s ( a − b ) ( d − e ) − P ( u , v ) − P ( v , u ) ) . \Delta= \kappa_{rs}(a-b)(d-e)\left(\kappa_{rs}(a-b)(d-e)-P(u,v)-P(v,u)\right). Δ = κ r s ( a − b ) ( d − e ) ( κ r s ( a − b ) ( d − e ) − P ( u , v ) − P ( v , u ) ) . Proposition 3.3 leaves only coefficients satisfying, for every pair,
κ r s = 0 or P ( u , v ) + P ( v , u ) = κ r s ( a − b ) ( d − e ) . (10) \kappa_{rs} = 0 \quad\text{or} \quad P(u,v) + P(v,u) = \kappa_{rs}(a-b)(d-e). \tag*{(10)} κ r s = 0 or P ( u , v ) + P ( v , u ) = κ r s ( a − b ) ( d − e ) . (10) At each label r r r κ \kappa κ r , s r,s r , s r r r (10) . The resulting linear polynomial in the two entries of the s s s (9) gives the same conclusion for the other ordering of B , C B,C B , C B , C B,C B , C B C , C B BC,CB B C , C B B = E 12 B = E_{12} B = E 12 C = E 21 C = E_{21} C = E 21 r , s r,s r , s μ 1 [ B , C ] \mu_1[B,C] μ 1 [ B , C ] μ 2 [ A , C ] \mu_2[A,C] μ 2 [ A , C ] μ 3 [ A , B ] \mu_3[A,B] μ 3 [ A , B ] μ 1 − μ 2 + μ 3 = 0 \mu_1 - \mu_2 + \mu_3 = 0 μ 1 − μ 2 + μ 3 = 0 a + d + e = b + c + f = 0 a+d+e = b+c+f = 0 a + d + e = b + c + f = 0
μ 1 = a + c + d , μ 2 = a + b + c , μ 3 = a + b + e . \mu_1 = a+c+d, \qquad\mu_2 = a+b+c, \qquad\mu_3 = a+b+e. μ 1 = a + c + d , μ 2 = a + b + c , μ 3 = a + b + e . Their indicated alternating sum is a + d + e = 0 a+d+e = 0 a + d + e = 0 r r r κ \kappa κ p ( X r = I ) p(X_r = I) p ( X r = I ) μ 1 , μ 2 , μ 3 \mu_1,\mu_2,\mu_3 μ 1 , μ 2 , μ 3 p p p (4) .
In (4) , the excluded coefficients satisfy α β ( α + β ) = 0 \alpha\beta(\alpha+\beta) = 0 α β ( α + β ) = 0 α = 0 \alpha= 0 α = 0 β = 0 \beta= 0 β = 0 α + β = 0 \alpha+\beta= 0 α + β = 0 [ [ A , B ] , Y ] [[A,B],Y] [[ A , B ] , Y ]
[ X j , [ X k , X l ] ] , { j , k , l } = { 1 , 2 , 3 , 4 } ∖ { r } . [X_j,[X_k,X_l]], \qquad\{j,k,l\} = \{1,2,3,4\} \setminus\{r\}. [ X j , [ X k , X l ]] , { j , k , l } = { 1 , 2 , 3 , 4 } ∖ { r } . For a remaining nonzero reduction at input r r r r → j r \to j r → j f f f f f f f f f Q Q Q Lemma 4.1 ; a contributing arrow whose endpoint differs from f f f 2 × 2 2 \times2 2 × 2 Lemma 4.1 for some shifts. Lemma 4.2 resolves both patterns.
A path r → s → t r \to s \to t r → s → t t t t r ↔ s r \leftrightarrow s r ↔ s r r r s s s r → s r \to s r → s t → u t \to u t → u s s s κ r s ≠ 0 \kappa_{rs} \ne0 κ r s = 0 κ \kappa κ κ 12 , κ 13 , − κ 12 − κ 13 \kappa_{12}, \kappa_{13}, -\kappa_{12} - \kappa_{13} κ 12 , κ 13 , − κ 12 − κ 13 Proposition 5.1 .
After permuting variables to A , B , C , D A, B, C, D A , B , C , D
p ( C = I ) = p ( D = I ) = 0 , κ A B = 0 , p(C = I) = p(D = I) = 0,\qquad\kappa_{AB} = 0, p ( C = I ) = p ( D = I ) = 0 , κ A B = 0 , p ( A = I ) = μ [ B , [ C , D ] ] , p ( B = I ) = ν [ A , [ C , D ] ] , (11) p(A = I) = \mu[B,[C,D]],\qquad p(B = I) = \nu[A,[C,D]], \tag*{(11)} p ( A = I ) = μ [ B , [ C , D ]] , p ( B = I ) = ν [ A , [ C , D ]] , (11) with ( μ , ν ) ≠ ( 0 , 0 ) (\mu,\nu) \ne(0,0) ( μ , ν ) = ( 0 , 0 ) (10) hold.
For linear operators S , T S, T S , T { S , T } = S ∘ T + T ∘ S \{S,T\} = S \circ T + T \circ S { S , T } = S ∘ T + T ∘ S s 4 s_4 s 4 s 4 ( A , B , C , D ) s_4(A,B,C,D) s 4 ( A , B , C , D )
P 0 = A [ C , D ] B − B [ C , D ] A , P 1 = A B [ C , D ] − [ C , D ] B A , P 2 = [ A , B ] [ C , D ] + [ C , D ] [ A , B ] + P 0 , v 1 = 1 4 ( { a d B , a d D } [ A , C ] − { a d A , a d C } [ B , D ] + { a d B , a d C } [ A , D ] − { a d A , a d D } [ B , C ] ) + 1 2 s 4 , v 2 = 1 4 ( { a d B , a d D } [ A , C ] − { a d A , a d C } [ B , D ] − { a d B , a d C } [ A , D ] + { a d A , a d D } [ B , C ] ) − 1 2 s 4 . \begin{aligned}
P_0 &= A[C,D]B - B[C,D]A,\\
P_1 &= AB[C,D] - [C,D]BA,\\
P_2 &= [A,B][C,D] + [C,D][A,B] + P_0,\\
v_1 &= \frac{1}{4}\bigl(\{ad_B,ad_D\}[A,C] - \{ad_A,ad_C\}[B,D]\\
&\qquad\qquad+ \{ad_B,ad_C\}[A,D] - \{ad_A,ad_D\}[B,C]\bigr) + \frac{1}{2}s_4,\\
v_2 &= \frac{1}{4}\bigl(\{ad_B,ad_D\}[A,C] - \{ad_A,ad_C\}[B,D]\\
&\qquad\qquad- \{ad_B,ad_C\}[A,D] + \{ad_A,ad_D\}[B,C]\bigr) - \frac{1}{2}s_4.
\end{aligned} P 0 P 1 P 2 v 1 v 2 = A [ C , D ] B − B [ C , D ] A , = A B [ C , D ] − [ C , D ] B A , = [ A , B ] [ C , D ] + [ C , D ] [ A , B ] + P 0 , = 4 1 ( { a d B , a d D } [ A , C ] − { a d A , a d C } [ B , D ] + { a d B , a d C } [ A , D ] − { a d A , a d D } [ B , C ] ) + 2 1 s 4 , = 4 1 ( { a d B , a d D } [ A , C ] − { a d A , a d C } [ B , D ] − { a d B , a d C } [ A , D ] + { a d A , a d D } [ B , C ] ) − 2 1 s 4 . The conditions just obtained define the six-dimensional space
p = z 0 P 0 + z 1 P 1 + z 2 P 2 + a v 1 + b v 2 + d s 4 . (12) p = z_0P_0 + z_1P_1 + z_2P_2 + av_1 + bv_2 + ds_4. \tag*{(12)} p = z 0 P 0 + z 1 P 1 + z 2 P 2 + a v 1 + b v 2 + d s 4 . (12) The equations (11) , together with the four cross-pair equations, have rank 18 in the 24-word space, and the six displayed vectors are independent solutions. Appendix A prints their coefficients and a nonzero constraint minor. In these coordinates,
tr p = ( − z 0 + z 1 + z 2 ) tr ( [ A , B ] [ C , D ] ) . (13) \operatorname{tr} p = (-z_0 + z_1 + z_2)\operatorname{tr}([A,B][C,D]). \tag*{(13)} tr p = ( − z 0 + z 1 + z 2 ) tr ([ A , B ] [ C , D ]) . (13) Lemma 7.1. Suppose identifying two inputs with A A A A 2 = I n A^2 = I_n A 2 = I n V / 2 V/2 V /2 A V = − V A AV = -VA A V = − V A
L ( B ) = α ( A B A ) V + β V ( A B A ) + γ V B + δ B V . L(B) = \alpha(ABA)V + \beta V(ABA) + \gamma VB + \delta BV. L ( B ) = α ( A B A ) V + β V ( A B A ) + γ V B + δ B V . If
γ + δ − α − β ≠ 0 , ( α + β + γ + δ ) ( γ 2 − β 2 ) ( δ 2 − α 2 ) ≠ 0 , \gamma+ \delta- \alpha- \beta\ne0,\qquad(\alpha+ \beta+ \gamma+ \delta)(\gamma^2 - \beta^2)(\delta^2 - \alpha^2) \ne0, γ + δ − α − β = 0 , ( α + β + γ + δ ) ( γ 2 − β 2 ) ( δ 2 − α 2 ) = 0 , then the original polynomial is surjective in every order n ≥ 3 n \ge3 n ≥ 3
Proof. Choose V = diag ( v 1 , … , v n ) V = \operatorname{diag}(v_1,\ldots,v_n) V = diag ( v 1 , … , v n ) i ∗ i^* i ∗ i i i A A A ( i , j ) (i,j) ( i , j ) ( i ∗ , j ∗ ) (i^*,j^*) ( i ∗ , j ∗ )
( s t − t − s ) , s = γ v i + δ v j , t = β v i + α v j . \begin{pmatrix} s & t \\ -t & -s \end{pmatrix},\qquad s = \gamma v_i + \delta v_j,\qquad t = \beta v_i + \alpha v_j. ( s − t t − s ) , s = γ v i + δ v j , t = β v i + α v j . For a diagonal pair its determinant is ( ( α + β ) 2 − ( γ + δ ) 2 ) v i 2 ((\alpha+\beta)^2-(\gamma+\delta)^2)v_i^2 (( α + β ) 2 − ( γ + δ ) 2 ) v i 2 ( β 2 − γ 2 ) v i 2 (\beta^2-\gamma^2)v_i^2 ( β 2 − γ 2 ) v i 2 ( α 2 − δ 2 ) v j 2 (\alpha^2-\delta^2)v_j^2 ( α 2 − δ 2 ) v j 2 Lemma 3.2 . The diagonal two-cycle map has nonzero trace values, so Lemma 2.2 finishes the proof.
In (12) , identify C = A C=A C = A D = V / 2 D=V/2 D = V /2 Lemma 7.1 are
( α , β , γ , δ ) = ( z 1 + z 2 − b , z 1 + z 2 + b , z 0 + b , z 0 − b ) . (\alpha,\beta,\gamma,\delta)=(z_1+z_2-b,\ z_1+z_2+b,\ z_0+b,\ z_0-b). ( α , β , γ , δ ) = ( z 1 + z 2 − b , z 1 + z 2 + b , z 0 + b , z 0 − b ) . By (13) , z 0 − z 1 − z 2 ≠ 0 z_0-z_1-z_2\ne0 z 0 − z 1 − z 2 = 0
z 0 + z 1 + z 2 , z 0 + z 1 + z 2 + 2 b , z 0 + z 1 + z 2 − 2 b z_0+z_1+z_2,\qquad z_0+z_1+z_2+2b,\qquad z_0+z_1+z_2-2b z 0 + z 1 + z 2 , z 0 + z 1 + z 2 + 2 b , z 0 + z 1 + z 2 − 2 b to vanish. Interchanging the two outer inputs therefore leaves only
z 0 + z 1 + z 2 ∈ { 0 , 2 b , − 2 b } , z 1 − z 0 − z 2 ∈ { 0 , 2 b , − 2 b } . z_0+z_1+z_2\in\{0,2b,-2b\},\qquad z_1-z_0-z_2\in\{0,2b,-2b\}. z 0 + z 1 + z 2 ∈ { 0 , 2 b , − 2 b } , z 1 − z 0 − z 2 ∈ { 0 , 2 b , − 2 b } . The choice z 0 + z 2 = z 1 = 0 z_0+z_2=z_1=0 z 0 + z 2 = z 1 = 0
z 0 + z 2 = k b , z 1 = l b , z_0+z_2=kb,\qquad z_1=lb, z 0 + z 2 = k b , z 1 = l b , ( k , l ) ∈ { ( 1 , − 1 ) , ( − 1 , 1 ) , ( − 1 , − 1 ) , ( 0 , − 2 ) , ( − 2 , 0 ) , ( 1 , 1 ) , ( 2 , 0 ) , ( 0 , 2 ) } . (14) (k,l)\in\{(1,-1),(-1,1),(-1,-1),(0,-2),(-2,0),(1,1),(2,0),(0,2)\}. \tag*{(14)} ( k , l ) ∈ {( 1 , − 1 ) , ( − 1 , 1 ) , ( − 1 , − 1 ) , ( 0 , − 2 ) , ( − 2 , 0 ) , ( 1 , 1 ) , ( 2 , 0 ) , ( 0 , 2 )} . (14) For b = 0 b=0 b = 0 ( k , l ) (k,l) ( k , l ) u , a , d ∈ K u,a,d\in K u , a , d ∈ K
q k , l ( u , a , d ) = l P 1 + k P 2 + v 2 + u ( P 0 − P 2 ) + a v 1 + d s 4 . (15) q_{k,l}(u,a,d)=lP_1+kP_2+v_2+u(P_0-P_2)+av_1+ds_4. \tag*{(15)} q k , l ( u , a , d ) = l P 1 + k P 2 + v 2 + u ( P 0 − P 2 ) + a v 1 + d s 4 . (15) For the original coefficients in (12) with b ≠ 0 b\ne0 b = 0
p = b q k , l ( z 0 / b , a / b , d / b ) . p=bq_{k,l}(z_0/b,a/b,d/b). p = b q k , l ( z 0 / b , a / b , d / b ) . Trace nonvanishing for a member of this family is k + l − 2 u ≠ 0 k+l-2u\ne0 k + l − 2 u = 0
Eight quartic coefficient families
For the family ( 14 ) (14) ( 14 )
q k , l ( u , a , d ) ( B = I ) = ( k + l ) [ A , [ C , D ] ] , q_{k,l}(u,a,d)(B=I)=(k+l)[A,[C,D]], q k , l ( u , a , d ) ( B = I ) = ( k + l ) [ A , [ C , D ]] , q k , l ( u , a , d ) ( A = I ) = ( k − l ) [ [ C , D ] , B ] . q_{k,l}(u,a,d)(A=I)=(k-l)[[C,D],B]. q k , l ( u , a , d ) ( A = I ) = ( k − l ) [[ C , D ] , B ] . If k + l ≠ 0 k+l\ne0 k + l = 0 A A A B B B k − l ≠ 0 k-l\ne0 k − l = 0 C = E 12 C=E_{12} C = E 12 D = E 21 D=E_{21} D = E 21 [ C , D ] = diag ( 1 , − 1 ) [C,D]=\operatorname{diag}(1,-1) [ C , D ] = diag ( 1 , − 1 ) E 12 , E 21 E_{12},E_{21} E 12 , E 21 k + l − 2 u ≠ 0 k+l-2u\ne0 k + l − 2 u = 0 (13) also gives nonzero trace in size two: at A = C = E 12 A=C=E_{12} A = C = E 12 B = D = E 21 B=D=E_{21} B = D = E 21 2 ( k + l − 2 u ) 2(k+l-2u) 2 ( k + l − 2 u )
Lemma 8.1. For every pair in ( 13 ) (13) ( 13 ) u , a , d ∈ K u,a,d\in K u , a , d ∈ K ( 14 ) (14) ( 14 ) M 3 ( K ) M_3(K) M 3 ( K )
Proof. Use the matrices
H = diag ( 2 , 0 , − 2 ) , E = ( 0 2 0 0 0 1 0 0 0 ) , F = ( 0 0 0 1 0 0 0 2 0 ) . H=\operatorname{diag}(2,0,-2),\qquad
E=\begin{pmatrix}0&2&0\\0&0&1\\0&0&0\end{pmatrix},\qquad
F=\begin{pmatrix}0&0&0\\1&0&0\\0&2&0\end{pmatrix}. H = diag ( 2 , 0 , − 2 ) , E = 0 0 0 2 0 0 0 1 0 , F = 0 1 0 0 0 2 0 0 0 . Freeze the other three positions, in their original order, at
( H , E , F ) , ( H + I 3 , E , F ) , ( H + 2 I 3 , E , F ) , (H,E,F),\qquad(H+I_3,E,F),\qquad(H+2I_3,E,F), ( H , E , F ) , ( H + I 3 , E , F ) , ( H + 2 I 3 , E , F ) , ( H , F , E ) , ( H + I 3 , F , E ) . (H,F,E),\qquad(H+I_3,F,E). ( H , F , E ) , ( H + I 3 , F , E ) . The commutator relations are [ H , E ] = 2 E [H,E]=2E [ H , E ] = 2 E [ H , F ] = − 2 F [H,F]=-2F [ H , F ] = − 2 F [ E , F ] = H [E,F]=H [ E , F ] = H H 2 + 2 ( E F + F E ) = 8 I 3 H^2+2(EF+FE)=8I_3 H 2 + 2 ( E F + F E ) = 8 I 3 Y Y Y
s 4 ( Y , H , E , F ) = [ Y , H ] H + H [ Y , H ] + 2 [ Y , E ] F + 2 F [ Y , E ] + 2 [ Y , F ] E + 2 E [ Y , F ] = [ Y , H 2 + 2 ( E F + F E ) ] = 0. \begin{aligned}
s_4(Y,H,E,F)={}&[Y,H]H+H[Y,H]+2[Y,E]F+2F[Y,E]\\
&+2[Y,F]E+2E[Y,F]\\
={}&[Y,H^2+2(EF+FE)]=0.
\end{aligned} s 4 ( Y , H , E , F ) = = [ Y , H ] H + H [ Y , H ] + 2 [ Y , E ] F + 2 F [ Y , E ] + 2 [ Y , F ] E + 2 E [ Y , F ] [ Y , H 2 + 2 ( E F + F E )] = 0. Moving the free input or interchanging E , F E,F E , F s 4 ( Y , I 3 , E , F ) = 0 s_4(Y,I_3,E,F)=0 s 4 ( Y , I 3 , E , F ) = 0 H H H 3 ( i − 1 ) + j − 1 3(i-1)+j-1 3 ( i − 1 ) + j − 1 E i j E_{ij} E ij 1 ≤ i , j ≤ 3 1\le i,j\le3 1 ≤ i , j ≤ 3 § 10 , use rows and columns 0 , … , 6 0,\ldots,6 0 , … , 6 § 10 proves directly that these five minors have no common zero: on every factor of the first minor, two specified univariate specializations or resultants are coprime. Hence one of the five slices has rank at least seven for every ( u , a ) (u,a) ( u , a ) d d d
Proof of Theorem 1.1 . For a trace-nonzero quartic, the reductions of § 7 either give surjectivity directly, give a proper quartic covered by Proposition 5.1 , or give a nonzero scalar multiple of a member of ( 14 ) (14) ( 14 ) Lemma 8.1 and the size-two argument above satisfy Proposition 6.4. Thus every trace-nonzero quartic is surjective for n ≥ 3 n\ge3 n ≥ 3 s l n ( K ) \mathfrak{sl}_n(K) sl n ( K ) [13 Theorem 8] . The zero polynomial has zero image. For n = 2 n=2 n = 2 [10 ] , allowing { 0 } \{0\} { 0 } K 2 K_2 K 2 s l 2 ( K ) \mathfrak{sl}_2(K) sl 2 ( K ) M 2 ( K ) M_2(K) M 2 ( K ) n = 1 n=1 n = 1 { 0 } \{0\} { 0 } K K K
Quartic coefficient matrices
We specify the coefficient matrices used in § 7 . Order the words lexicographically as permutations of A , B , C , D A, B, C, D A , B , C , D I I I (4) .
Let c 1 , … , c 6 c_1,\ldots,c_6 c 1 , … , c 6
A B C D , A B D C , A C B D , A C D B , A D B C , A D C B , ABCD,\quad ABDC,\quad ACBD,\quad ACDB,\quad ADBC,\quad ADCB, A B C D , A B D C , A C B D , A C D B , A D B C , A D C B , in that order. The eight trace equations of the four single-identity reductions have coefficient matrix
( 1 0 0 1 1 0 0 1 1 0 0 1 1 0 1 1 0 0 0 1 0 0 1 1 1 1 1 0 0 0 0 0 0 1 1 1 1 1 0 0 1 0 0 0 1 1 0 1 ) . \begin{pmatrix}
1 & 0 & 0 & 1 & 1 & 0 \\
0 & 1 & 1 & 0 & 0 & 1 \\
1 & 0 & 1 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 & 1 & 1 \\
1 & 1 & 1 & 0 & 0 & 0 \\
0 & 0 & 0 & 1 & 1 & 1 \\
1 & 1 & 0 & 0 & 1 & 0 \\
0 & 0 & 1 & 1 & 0 & 1
\end{pmatrix}. 1 0 1 0 1 0 1 0 0 1 0 1 1 0 1 0 0 1 1 0 1 0 0 1 1 0 1 0 0 1 0 1 1 0 0 1 0 1 1 0 0 1 0 1 0 1 0 1 . Its reduced equations are
c 1 = c 6 , c 2 = − c 5 − c 6 , c 3 = c 5 , c 4 = − c 5 − c 6 . c_1=c_6,\quad c_2=-c_5-c_6,\quad c_3=c_5,\quad c_4=-c_5-c_6. c 1 = c 6 , c 2 = − c 5 − c 6 , c 3 = c 5 , c 4 = − c 5 − c 6 . Thus the trace quotient has dimension two. The definitions below give κ 12 = − c 5 \kappa_{12}=-c_5 κ 12 = − c 5 κ 13 = − c 6 \kappa_{13}=-c_6 κ 13 = − c 6 c i c_i c i
For a diagonal pair r , s r,s r , s B , C B,C B , C F i j , H i j F_{ij},H_{ij} F ij , H ij i , j ∈ { 0 , 1 } i,j\in\{0,1\} i , j ∈ { 0 , 1 } B B B C C C F i j F_{ij} F ij i i i r r r B , C B,C B , C j j j s s s C C C B B B H i j H_{ij} H ij i , j i,j i , j
P = F 00 a d + F 01 a e + F 10 b d + F 11 b e , H = H 00 a d + H 01 a e + H 10 b d + H 11 b e . P=F_{00}ad+F_{01}ae+F_{10}bd+F_{11}be,\qquad H=H_{00}ad+H_{01}ae+H_{10}bd+H_{11}be. P = F 00 a d + F 01 a e + F 10 b d + F 11 b e , H = H 00 a d + H 01 a e + H 10 b d + H 11 b e . The common notation in these formulas means evaluation of each coefficient functional on p p p κ r s = F 00 − H 00 \kappa_{rs}=F_{00}-H_{00} κ r s = F 00 − H 00
( F i j − H i j ) i j = κ r s ( 1 , − 1 , − 1 , 1 ) . (F_{ij}-H_{ij})_{ij}=\kappa_{rs}(1,-1,-1,1). ( F ij − H ij ) ij = κ r s ( 1 , − 1 , − 1 , 1 ) . which proves (9) . The coefficient equations for the second alternative in (10) are
F 00 + F 11 = κ r s , F 01 + F 10 = − κ r s ; (16) F_{00}+F_{11}=\kappa_{rs},\qquad F_{01}+F_{10}=-\kappa_{rs}; \tag*{(16)} F 00 + F 11 = κ r s , F 01 + F 10 = − κ r s ; (16) the other two are duplicates under exchanging the two vertices. This specifies the needed coefficient rows with integer entries.
To verify (12) , impose the deletion conditions (11) ; allow independent scalars μ , ν \mu,\nu μ , ν κ A B = 0 \kappa_{AB}=0 κ A B = 0 (16) for A C , A D , B C , B D AC,AD,BC,BD A C , A D , B C , B D (13) .
The six columns, in the lexicographic word order, are as follows.
caption:
:::
Their minor in rows 0, 1, 2, 3, 6, 8 has determinant 1, so they are independent.
Here are explicit indices for the rank-18 calculation. Number the 24 words from zero. Let D r D_r D r r r r r = A , B r = A, B r = A , B q r q_r q r [ B , [ C , D ] ] [B, [C, D]] [ B , [ C , D ]] [ A , [ C , D ] ] [A, [C, D]] [ A , [ C , D ]]
( D r ) j , ∗ − ( q r ) j ( D r ) 0 , ∗ , 0 ≤ j < 6. (D_r)_{j,*} - (q_r)_j (D_r)_{0,*}, \qquad0 \le j < 6. ( D r ) j , ∗ − ( q r ) j ( D r ) 0 , ∗ , 0 ≤ j < 6. Follow these twelve rows by the six rows of D C D_C D C D D D_D D D κ A B \kappa_{AB} κ A B (16) for each of A C AC A C A D AD A D B C BC B C B D BD B D 33 × 24 33 \times24 33 × 24
1 , 2 , 3 , 4 , 5 , 7 , 8 , 10 , 12 , 13 , 16 , 18 , 19 , 24 , 25 , 27 , 29 , 31 1, 2, 3, 4, 5, 7, 8, 10, 12, 13, 16, 18, 19, 24, 25, 27, 29, 31 1 , 2 , 3 , 4 , 5 , 7 , 8 , 10 , 12 , 13 , 16 , 18 , 19 , 24 , 25 , 27 , 29 , 31 and columns
0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 12 , 13 , 14 , 15 , 16 , 17 , 18 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 14, 15, 16, 17, 18 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 , 12 , 13 , 14 , 15 , 16 , 17 , 18 has determinant 4. Multiplication by the six printed columns gives zero. Thus its rank is both at least 18 and at most 24 − 6 = 18 24 - 6 = 18 24 − 6 = 18 (12) .
Rank-seven minors
For each pair ( k , l ) (k, l) ( k , l ) f 0 , … , f 4 f_0, \ldots, f_4 f 0 , … , f 4 Lemma 8.1 , after division by their nonzero rational factors. Table 1 gives these factorizations in the original parameters u , a u, a u , a (15) . The determinant zero sets are unaffected by the removed factors. For ( k , l ) = (k, l) = ( k , l ) = ( − 1 , − 1 ) (-1,-1) ( − 1 , − 1 ) f 0 f_0 f 0 ( k , l ) = ( 0 , − 2 ) (k,l)=(0,-2) ( k , l ) = ( 0 , − 2 ) f 1 f_1 f 1
Table 1 (PDF p. 22). The five minors for each coefficient family.
Table 1 (PDF p. 24).
For each fixed ( k , l ) (k,l) ( k , l ) Q \mathbb{Q} Q Table 2 lists such square-free parts, with rational constants removed.
Table 2. Elimination on each factor of the first minor.
Fix a pair ( k , l ) (k,l) ( k , l ) g g g f 0 f_0 f 0 g g g a a a a 0 a_0 a 0 Table 2 gives two coprime square-free parts of f i ( u , a 0 ) f_i(u,a_0) f i ( u , a 0 ) f j ( u , a 0 ) f_j(u,a_0) f j ( u , a 0 ) g = f i = f j = 0 g=f_i=f_j=0 g = f i = f j = 0
If g g g u u u f i f_i f i u u u Q [ a ] \mathbb{Q}[a] Q [ a ] Res u ( g , f i ) \operatorname{Res}_u(g,f_i) Res u ( g , f i ) r , s r,s r , s u u u s s s g g g r r r f i f_i f i 1 , u , … , u s − 1 1,u,\ldots,u^{s-1} 1 , u , … , u s − 1 1 , u , … , u r − 1 1,u,\ldots,u^{r-1} 1 , u , … , u r − 1 u u u g g g u u u a a a Table 2 have coprime square-free parts. Therefore a common zero of g , f i , f j g,f_i,f_j g , f i , f j a a a
Every factor of f 0 f_0 f 0 Q \mathbb{Q} Q K K K 1 1 1 Lemma 8.1 . The accompanying supplement additionally verifies the determinant calculations from matrix units and supplies polynomial combinations of the five original minors equal to $1.
Table 26.
Table 4.
Zhili College, Tsinghua University, Beijing 100084, China
Email address: tangqy24@mails.tsinghua.edu.cn
