The Lu–Wenzel Conjecture

Introduction

For a square complex matrix $X$, write $X^\ast$ for its conjugate transpose and $\lVert X\rVert_F^2 = \operatorname{Tr}(X^\ast X)$, where $\operatorname{Tr}$ is the sum of the diagonal entries. The commutator of matrices of the same size is $[X,Y] = XY-YX$. Submultiplicativity and the triangle inequality give

In 2005, Böttcher and Wenzel asked for the optimal constant and conjectured the sharper inequality

for real square matrices [3]. They proved the two-dimensional case. Vong and Jin established the inequality in every real dimension [14], and Lu obtained an independent proof as part of his work on normal scalar curvature [9]. Böttcher and Wenzel subsequently proved the complex inequality and studied related estimates for unitarily invariant norms [2]. Lu’s eigenvalue proof [10] and the account by Lu and Wenzel [12] explain several connections among these approaches.

A second source of commutator inequalities is submanifold geometry. De Smet, Dillen, Verstraelen and Vrancken proved a curvature inequality in codimension two and proposed its extension to arbitrary codimension in 1999 [5]. This became the normal scalar curvature conjecture. Through the Gauss and Ricci equations, it reduces to the matrix inequality

for every family of real symmetric matrices $B_1,\ldots,B_m$ of the same size, with $m \ge1$. Ge and Tang proved this inequality and classified its equality cases [7]. Lu independently proved the inequality and used it to obtain geometric pinching results [9]. Thus (1) controls one commutator for arbitrary matrices, whereas (2) controls the total commutator energy of a symmetric family.

Lu and Wenzel brought these questions together in their study of the normal Ricci curvature inequality [11]. Their starting point was an estimate arising in a Simons–Bochner formula, in which one symmetric matrix is distinguished from an orthogonal family. For real square matrices $X,Y_1,\ldots,Y_m$ of the same size, with $m \ge1$, they imposed the bilinear conditions

Two matrices $A,B$ are Frobenius-orthogonal when $\operatorname{Tr}(A^\ast B)=0$. For mutually orthogonal $X,Y_1,\ldots,Y_m$, their fundamental conjecture predicts

This is Conjecture 8 of [11]. Proposition 10 there shows that the fundamental conjecture implies (2) for arbitrary real matrix families satisfying $\operatorname{Tr}(B_a[B_b,B_c])=0$ for all indices. Symmetric matrices satisfy these constraints because the trace of a product of a symmetric matrix and a skew-symmetric matrix is zero. The constraints therefore preserve the geometric case while allowing nonsymmetric matrices.

Ge, Li, Lu and Zhou [6] reformulated the fundamental conjecture in terms of the spectrum of the commutator operator. For an integer $n \ge2$, equip $\mathcal{H}=M_n(\mathbb{C})$ with

linear in the second variable. The adjoint of a complex linear operator on $\mathcal{H}$ is also denoted by a star. For $X\in\mathcal{H}$, set

Cyclicity of trace gives the adjoint formula, so $T_X$ is Hermitian and positive semidefinite. Write $\lambda_1(T_X)\ge\cdots\ge\lambda_{n^2}(T_X)\ge0$ for its eigenvalues counted with multiplicity. The largest eigenvalue is the maximum of $\lVert[X,Y]\rVert_F^2$ over unit $Y$, so (1) gives $\lambda_1(T_X)\le2$ for unit $X$.

Ge, Li, Lu and Zhou proved that every positive eigenvalue occurs with even multiplicity. They showed that the fundamental conjecture is equivalent to bounding the first $2k$ eigenvalues by $2k+2$, and extended these equivalences to complex matrices [6]. They established the spectral inequalities when $X$ is normal, when $X$ has rank one, and when $n=2$ or $3$. In general they obtained

for $\lVert X\rVert_F=1$ and $1\le k\le\lfloor n^2/2\rfloor$. These results and their geometric context are reviewed in [13 Section 4.3]. The spectral conjecture asks for the bound $2k+2$ in every dimension. A preprint of Liu computes the spectrum when $X^2=0$ and obtains all the predicted partial sums for this class and its scalar translates [8]. Carptopus proves the inequalities on coordered alignment and isoclinic subfamilies of rank-two matrices, together with stability estimates under a positive lower bound on the smaller singular value [4].

The squared norm on the right is \lVert X\rVert_F^2-\lvert\operatorname{Tr}X\rvert^2/n. Theorem 1 proves the complex version of [6 Conjecture 4], numbered Conjecture 7 in the preprint. For real X$, the complex operator is the complexification of the real one, with the same eigenvalues and multiplicities. When $n=1$, all commutators vanish.

The proof maximizes the total excess of the eigenvalues above one on the traceless unit sphere. At a maximizer, a strict gap separates the eigenvalues above one from the others. Their sum is smooth in a neighborhood, and its second derivative contains the variation of its spectral projection. Contracting this derivative against commutator directions gives a nonnegative sum of squares. A local maximum principle then permits minimization of the commutator energy within the maximum set. The critical equation, together with membership in the selected spectral subspace, reduces the maximizer to a normal or rank-one matrix.

Normalization and spectral pairs

Subtracting $(\operatorname{Tr} X)I_n/n$ does not change $D_X$. If the resulting matrix is nonzero, dividing it by its Frobenius norm divides $T_X$ by the square of that norm. For scalar matrices the theorem is immediate. We therefore work on the sphere

with the real metric $\operatorname{Re}\langle\cdot,\cdot\rangle$. Use $\operatorname{Id}$ for the identity on $\mathcal{H}$ and $\operatorname{Tr}_{\mathcal{H}}$ for the trace of a complex linear operator on $\mathcal{H}$. From now on, norms of matrices are Frobenius norms. For a Hermitian operator $A$, $A_+$ is obtained by replacing each eigenvalue of $A$ by its maximum with zero. Set

We shall prove $G(X)\le2$ on $S$. This implies the theorem since each sum of $2k$ eigenvalues minus $2k$ is bounded above by $G(X)$.

Unitary conjugation preserves the spectrum, since

for a unitary $n\times n$ matrix $U$, and $Y\mapsto UYU^*$ is an isometry. Let $E_{ij}$, $1\le i,j\le n$, denote the matrix unit with a one in position $(i,j)$ and zeros elsewhere. The trace identity

follows from

Here $\sum_{i,j}E_{ji}X^*E_{ij}=(\operatorname{Tr}X^*)I_n$.

Let $\mathcal{J}(Y)=Y^*$, an antiunitary involution of $\mathcal{H}$, so it is conjugate linear and $\langle\mathcal{J}U,\mathcal{J}V\rangle=\langle V,U\rangle$. When the base matrix $X$ is fixed or understood as the variable in a local spectral neighborhood, write $D=D_X$ and $T=T_X$. Since $([X,Y^*])^*=-[X^*,Y]$, one has

Consequently $K$ commutes with $T$. On the eigenspace $E_\alpha$ for a positive eigenvalue $\alpha$ of $T$, one has $\langle Ku,Kv\rangle=\alpha\langle v,u\rangle$. Thus $K/\sqrt{\alpha}$ is antiunitary there and squares to $-\operatorname{Id}$. An orthonormal basis can therefore be chosen in pairs $e,f$ with

For an antiunitary map $Q$ with $Q^2=-\operatorname{Id}$, antiunitarity gives $\langle Qe,Q^2e\rangle=\langle Qe,e\rangle$, and $Q^2e=-e$ makes both sides zero. Thus $e\perp Qe$. The orthogonal complement of their span is invariant under $Q$, so induction produces the pairs in (6) and proves even multiplicity.

Conversely, the spectral inequalities imply $G(X)\le2$ on $S$. Indeed, the number of eigenvalues above one is even; when it is positive, apply the inequality at this cutoff.

Fix an integer $1\le r<n^2$. Suppose there is a strict gap between the $r$-th and $(r+1)$-st eigenvalues of $T$. Let $P$ be the orthogonal projection onto the eigenspaces for the $r$ largest eigenvalues, which are positive. With an orthonormal eigenbasis write

The sets $I$ and $O$ partition the $n^2$ basis indices into selected and omitted indices. A sum over all $j$ means $j\in I\cup O$. Entire repeated eigenspaces are selected. For a complex subspace $U \subset\mathcal{H}$, write $U^* = \{u^* : u \in U\}$. Put $W = \operatorname{ran} P$. Then $D(W) = W^*$, and (6) proves the following identity in the complex tensor product $\mathcal{H} \otimes_{\mathbb{C}} \mathcal{H}$

is skew under interchange of its factors.

Indeed each pair contributes $\sqrt{\alpha}(f^* \otimes e^* - e^* \otimes f^*)$. The tensor is independent of the orthonormal selected basis, since the second factor transforms with the conjugate change of coordinates.

The Hessian of a separated spectral sum

Work on a neighborhood in $\mathbb{S}$ where the strict cut (7) persists. At each point choose a fixed positively oriented contour $\Gamma$ enclosing the selected eigenvalues and excluding the others, also at nearby points. Diagonalizing $T_X$ and integrating the scalar resolvents gives

The formula gives smoothness even at repeated eigenvalues. Define

Fix an orthonormal basis $e_1,\ldots,e_{n^2}$ of $\mathcal{H}$. Expanding the selected basis vectors in this basis gives

Thus $L$ is independent of the selected frame and smooth as a function of $P$. For a tangent vector $H$, differentiating $P^2=P$ shows that $P'$ has zero diagonal blocks in $\operatorname{ran}P\oplus\ker P$. Since $T$ is block diagonal, $\operatorname{Tr}_{\mathcal{H}}(P'T)=0$. The term with $P$ fixed is the derivative of $\sum_{i\in I}\lVert[X,q_i]\rVert^2$, so $dF(H)=2\operatorname{Re}\langle LX,H\rangle$. The identity $\langle X,LX\rangle=\sum_{i\in I}\lVert[q_i,X]\rVert^2=F$ gives the radial component of the gradient. Each value of $D_{q_i}^*$ is a commutator and has trace zero, so $LX$ is traceless. Therefore the real sphere gradient is

Define the complex orthogonal projection $\Pi_X$, written $\Pi$ at a fixed base matrix $X$, by

Its image lies in the real tangent space of $\mathbb{S}$ and is closed under multiplication by $i$. For $Y\in\operatorname{ran}\Pi$, differentiation gives $T'_Y=D^*D_Y+D_Y^*D$. For $i\in I,j\in O$, put

Then $\langle q_j,T'_{q_i}\rangle=A_{ji}(Y)+C_{ji}(Y)$. The adjoint formula and $D_Yq=-D_qY$ give these two expressions. The first term is complex linear in $Y$, the second conjugate linear. The sphere Hessian $\operatorname{Hess}_{\mathbb{S}}F(Y,Y)$ is the second derivative of $F$ along the sphere geodesic starting with velocity $Y$. It is

Indeed, differentiating $T_XP=PT_X$ gives the off-diagonal entry

The two off-diagonal blocks give

Along the sphere geodesic with initial velocity $Y$, the acceleration is $-\lVert Y\rVert^2X$, and $T''=2D_Y^*D_Y-2\lVert Y\rVert^2T$. Thus $\operatorname{Tr}_{\mathcal H}(PT'')=2\langle Y,LY\rangle-2F\lVert Y\rVert^2$. Differentiating $F'=\operatorname{Tr}_{\mathcal H}(PT')$ proves (10).

The identities $A_{ji}(iY)=iA_{ji}(Y)$ and $C_{ji}(iY)=-iC_{ji}(Y)$ give

Fix a selected eigenvalue $\alpha>0$ and an omitted eigenvalue $\beta\geq0$, and sum over $i,j$ in their respective eigenspaces $E_\alpha,E_\beta$. The vectors $Dq_i/\sqrt{\alpha}$ form an orthonormal basis of $E_\alpha^*$ by (6). For $\beta>0$, the vectors $Dq_j/\sqrt{\beta}$ form one of $E_\beta^*$. Changing orthonormal basis in these two sums gives

For $\beta=0$, $Dq_j=0$, so the first identity still holds, and the second follows by changing the selected basis alone. The denominator $\alpha-\beta$ is constant in each such sum. For $U\in\mathcal H$, let $\lvert U\rangle\langle U\rvert$ denote the operator $H\mapsto U\langle U,H\rangle$. Define the Hermitian operator

Then

Contraction with the commutator covariance

Let $A_i,B_i$ be two matrix families indexed by a common finite set, with all sums over $a,i$ in that set. Suppose that $\sum_i B_i\otimes A_i$ is skew. Put

Then

Set $R=\sum_iB_iA_i$. Skewness implies $\sum_iA_iB_i=-R$ and $\sum_i[B_i,A_i]=2R$. Expand the two commutators in $I_0$ into four terms. The terms with a minus sign are

For the two remaining terms, skewness in the $a$ index and cyclicity give

They therefore cancel. Thus $I_0=2\lVert R\rVert^2=\frac{1}{2}\left\lVert\sum_i[B_i,A_i]\right\rVert^2$.

Apply (13) with $A_i=q_i^*$, $B_i=Dq_i$. Identity (8) supplies skewness, and

For any three basis indices $a,i,j$, define

We restrict $a,i$ to $I$ in the following sums, but the definition also gives $u_{jai}$ when $j$ is omitted. Using $u_{aij}=\operatorname{Tr}([Dq_a,q_i]q_j)$, Parseval over the orthonormal basis $q_j^*$ gives

Therefore (13), (14), and (9) imply the exact identity

Since $D$ is a derivation and every commutator has trace zero,

The last two terms equal $-u_{iaj}$ and $u_{jai}$, respectively, by cyclicity. Thus

On $I^3$, permutation of the indices makes the sums of $\lvert u_{aij}\rvert^2$, $\lvert u_{iaj}\rvert^2$, and $\lvert u_{jai}\rvert^2$ equal. Summing $\lvert u_{jai}\rvert^2=\lvert u_{iaj}-u_{aij}\rvert^2$ gives

Introduce

For fixed $i$ and every $H\in\mathcal{H}$, cyclicity and Parseval give

Summing over $i\in I$ proves the operator identity $\sum_{i\in I,j\in I\cup O}\lvert[q_i^*,q_j^*]\rangle\langle[q_i^*,q_j^*]\rvert=L$, by equality of Hermitian quadratic forms. Using $(\alpha_i+\beta_j)/(\alpha_i-\beta_j)=1+2\beta_j/(\alpha_i-\beta_j)$, we can rewrite (11) as

Use the direct covariance expression $B=\sum_{a\in I}\lvert Dq_a\rangle\langle Dq_a\rvert$. Taking its trace against the terms in (18) gives

For the first identity, expand $\sum_{a,i}\lVert[q_i,Dq_a]\rVert^2$ in the complete orthonormal starred basis $q_j^*$. For the other two use $\operatorname{Tr}_{\mathcal{H}}(\lvert U\rangle\langle U\rvert\lvert V\rangle\langle V\rvert)=\lvert\langle U,V\rangle\rvert^2$ and $\langle[q_i^*,q_j^*],Dq_a\rangle=u_{aij}$. Also $\operatorname{Tr}_{\mathcal{H}}B=F$. Thus

Subtracting $\lVert J\rVert^2$ in (20), substituting (15), and canceling the terms with $j \in I$ by (17) gives

In the last line we interchanged $a,i$ in one half of the first sum. Define

Expanding each factor $\alpha/(\alpha-\beta)$ as $1+\beta/(\alpha-\beta)$ in the preceding expression gives

The strict cut and $\beta_j\geq0$ suffice for nonnegativity.

For $i\in I,j\in O$,

since $Dq_i\in W^*$. Consequently $R_{IO}X=0$, $K_{II}X=\mathcal L X$, and $\mathcal L X=J$. The projection onto the scalar matrices makes no contribution because $B(I_n)=0$ and $\mathcal L(I_n)=-F I_n$.

Expanding the projection off $X$, with $\lVert X\rVert=1$, gives

since $\mathcal L X=J$ and $\langle X,J\rangle=0$. Hence

Using the fixed basis $e_1,\ldots,e_{n^2}$, define the real tangent fields $Z_a(X)=\Pi_XD_XP(X)e_a$, $1\leq a\leq n^2$. Their covariance satisfies $\sum_{a=1}^{n^2}|Z_a\rangle\langle Z_a|=\Pi B\Pi$. Therefore

Together with (12) and (23), these give

Set $C(X)=[X^*,X]$ on $\mathcal S$, writing $C$ at the current base matrix. Suppose that $X$ is a local maximum of $F$ on $\mathcal S$, with $F>0$. Then $J=0$ and the real Hessian is nonpositive. Equation (24) forces $\mathcal S=0$. For every omitted $\beta_j>0$, (21) gives $u_{aij}=u_{iaj}=0$, and (16) gives $u_{jai}=0$. Let $O_+$ be the orthogonal sum of the omitted eigenspaces for positive eigenvalues. For subspaces $U,V$, write $[U,V]$ for the linear span of the commutators $[u,v]$, $u\in U,v\in V$. The vectors $(Dq_j)^*$ for omitted $\beta_j>0$ span $O_+$, by (5)–(6), and $\langle(Dq_j)^*,[q_a,q_i]\rangle=u_{jai}=0$. Hence

Taking matrix adjoints gives $[W^*,W^*]\perp O_+^*$. By (14) and stationarity, $F X=\sum_{i\in I}[Dq_i,q_i^*]$ belongs to $[W^*,W^*]$, so $X\perp O_+^*$. The output singular-space decomposition is

The map $D^*$ takes $W^*$ into $W$ and kills $\ker D^*$. It follows that the Hermitian matrix satisfies

Extremal matrices

Proof of Theorem 1. On a real coordinate neighborhood of $S$ with a strict spectral gap, list the $2n^2$ fields $Z_a,iZ_a$, $1 \leq a \leq n^2$, from (24) as $Z_s$, $1 \leq s \leq2n^2$. For smooth tangent fields $U,V$, the sphere’s Levi–Civita derivative $\nabla_U V$ is the real orthogonal projection of the ambient derivative of $V$ in direction $U$ onto the tangent space of $S$. Define

For a real $C^2$ function $h$ on this neighborhood, set

Vector fields act on functions by directional differentiation. Since $\operatorname{Hess}_S h(Z_s,Z_s)=Z_s(Z_sh)-(\nabla_{Z_s}Z_s)h$, (24) gives

Lemma 5 below applies to $\mathcal{A}$ in the chosen coordinates.

The function $G$ is continuous, by continuity of the ordered eigenvalues, and attains its maximum $M$ on $S$. Suppose $M>2$. Among the matrices at which $G=M$, choose $X_0$ minimizing $E(X)=\lVert[X^*,X]\rVert^2$. This is possible because the maximum set is compact. Let $r$ count the eigenvalues strictly above one at $X_0$. Then $0<r<n^2$, since $G(X_0)>0$ and $D_{X_0}(I_n)=0$. Choose a connected real coordinate neighborhood $\Omega$ of $X_0$ on which the gap at $r$ remains strict and $\lambda_r(T_X)>1$. Take $P$ and $F$ for this cutoff. On $\Omega$,

and equality holds throughout this chain at $X_0$. Thus the set where $F=M+r$ in the neighborhood lies in the maximum set of $G$. At each of its points (26) gives $PC=C$, with $C=[X^*,X]$.

Define the smooth real tangent field $V_0=-\Pi_DPC$ on the whole neighborhood. It is a smooth real linear combination of the fields $Z_a,iZ_a$. Indeed, if $C=\sum_a c_ae_a$, its coefficients are $-\operatorname{Re}c_a$ and $-\operatorname{Im}c_a$, respectively. By Lemma 5, the short integral curve of $V_0$ through $X_0$ stays in the maximum set of $F$. On that set $PC=C$ and $\langle X,DC\rangle=\langle D^*X,C\rangle=E$, so

For a tangent vector $H$, the derivative of $C$ is $[H^*,X]+[X^*,H]$. Cyclicity gives

Since $C=C^*$, these two scalars are conjugates, and hence

Moreover,

If $V_0(X_0)\ne0$, this derivative is strictly negative at time zero along the propagated curve. For small positive time the curve remains in $\{G=M\}$ and decreases $E$, contradicting the choice of $X_0$. Thus $V_0(X_0)=0$. In particular, at $X_0$,

All matrix-dependent quantities in the two cases below are evaluated at $X_0$. If $C=0$, then $X_0$ is normal. In a unitary basis in which it is diagonal, write $x_1,\ldots,x_n$ for its diagonal entries and $x$ for their column vector. The eigenvalues of $T_{X_0}$ are $|x_i-x_j|^2$ over ordered pairs $(i,j)$. Form the simple graph with an edge $ij$ when $|x_i-x_j|^2>1$, and let $m$ be its edge count. For $m=0$ we have $G(X_0)=0$. For $m>0$, orient the edges arbitrarily and let $A$ be the vertex-edge incidence matrix, with entries $+1$, $-1$ at the endpoints of each column and zero elsewhere. Then

Write $\deg i$ for the number of edges incident with vertex $i$. For the row indexed by an edge $e=ij$, the diagonal entry of $A^*A$ is two. The absolute off-diagonal entry is one exactly for edges sharing an endpoint with $e$, and, summing over all graph edges $f$,

The last bound follows because the two incident edge sets overlap only in $ij$. For an eigenvector of $A^*A$, evaluating its eigenvalue equation at a coordinate of largest absolute value shows that the absolute eigenvalue is at most the largest absolute row sum. Multiplication by $A$ and $A^*$ transfers the positive eigenspaces of $A^*A$ and $AA^*$ to each other, so their largest eigenvalues coincide. Hence $\lVert A^*x\rVert^2=\langle x,AA^*x\rangle\le(m+1)\lVert x\rVert^2=m+1$, and $G(X_0)\le2$.

Suppose instead that $C\ne0$. By (27), $TC=ED^*X_0=EC$. The eigenvalue $E$ exceeds one because $\operatorname{ran}P$ at $X_0$ consists of the eigenspaces above one. Diagonalize $C$ unitarily, with real diagonal entries $c_1,\ldots,c_n$. Each nonzero entry $(X_0)_{ij}$ satisfies $c_j-c_i=E$, by the commutator equation in (27). Write $s=\operatorname{rank}X_0$, and choose a nonzero minor of size $s$ with row set $R$ and column set $S$. A nonzero term of its determinant matches the rows to the columns along nonzero entries. Summing the corresponding relations gives

The coefficient of each $c_i$ in this difference is $0$, $1$, or $-1$, and at most $2s$ coefficients are nonzero. Since $\sum_i c_i^2=E$, Cauchy–Schwarz gives $s^2E^2\le2sE$. Thus $sE\le2$, and $E>1$ forces $s=1$.

A traceless rank-one unit matrix is unitarily similar to $E_{12}$. In fact, writing it as $uv^*$ with $\lVert u\rVert=\lVert v\rVert=1$, the trace condition gives $v^*u=0$. Completing $u,v$ to an orthonormal basis gives the stated unitary similarity. In the resulting unitary basis $X_0=E_{12}$, and multiplication gives

The displayed inputs are orthonormal and their outputs are orthogonal. All remaining matrix units except $E_{11},E_{22}$ are killed by $D$, and $D(E_{11}+E_{22})=0$. The resulting spectrum, in decreasing order, is

In particular $G(X_0)=2$ in the rank-one case. Both cases contradict $M>2$. The normalization in § 2 gives the inequality, and (28) gives equality at $E_{12}$ for the stated cutoffs. □

Inequalities for matrix families

Let $X$ be any unit matrix. The number $r$ of eigenvalues of $T_X$ strictly above one is even by (6). If $r=0$, the positive-part trace is zero. Otherwise Theorem 1, applied with $2k=r$, gives

The first eigenvalue is at most two by the pairing and the theorem at $k=1$. At every cutoff $\ell\ge2$, the partial sum is at most $\ell+2$ by the positive-part bound and at most $2n$ by (4). The bounds $2$ at cutoff one and $\min\{\ell+2,2n\}$ thereafter are the partial sums of (28). Append zero eigenvalues at indices beyond $n^2$; the same bounds then apply.

Proof. If $X=0$ or all $Y_a=0$, the inequalities hold directly. Otherwise take $\lVert X\rVert=1$ by homogeneity, remove the zero members, and relabel the remaining family as $Y_1,\ldots,Y_m$. Set $w_a=\lVert Y_a\rVert^2$ and $e_a=Y_a/\sqrt{w_a}$, with $w_1\ge\cdots\ge w_m>0$ after reordering. The operator $A=\sum_a w_a|e_a\rangle\langle e_a|$ is positive semidefinite and satisfies $A\le w_1\operatorname{Id}$, in the order of Hermitian operators. For positive semidefinite operators $R,S$, $\operatorname{Tr}_{\mathcal{H}}(RS)=\operatorname{Tr}_{\mathcal{H}}(S^{1/2}RS^{1/2})\ge0$. Apply this to $A$, $w_1\operatorname{Id}-A$, and the spectral positive part below. Since $T_X\le\operatorname{Id}+(T_X-\operatorname{Id})_+$, we obtain

Under the trace conditions, the span of the $Y_a$ is orthogonal to its image under $K=JD_X$, since $\langle Ku,v\rangle=-\operatorname{Tr}(u[X,v])$. Let $W$ be a subspace of this span of dimension $d\ge1$. Write $P_W$ for its orthogonal projection. Choose an orthonormal basis $u_1,\ldots,u_d$ diagonalizing the compression $P_WT_X|_W$, the operator that applies $T_X$ and projects back to $W$. Let $\mu_i\ge0$ be the corresponding eigenvalues. For each $\mu_i>0$, set $v_i=Ku_i/\sqrt{\mu_i}$. The vectors $v_i$ are orthonormal because $\langle Ku_i,Ku_j\rangle=\langle u_j,T_Xu_i\rangle$. They lie in $W^\perp$ and satisfy

using $KT_X=T_XK$ and Cauchy–Schwarz. For any orthonormal set of size $b$, the sum of its diagonal entries for a positive semidefinite operator is at most the sum of the first $b$ eigenvalues. Indeed, expanding the set in an eigenbasis gives weights in $[0,1]$ that sum to $b$, and their weighted eigenvalue sum is maximized by filling the largest entries first. Apply this to the $u_i$ and the available $v_i$, at most $2d$ vectors. The partial sum bound then yields $2\sum_i\mu_i\le2d+2$.

Use this estimate on the spans of $e_1,\ldots,e_j$ for each $1\le j\le m$. Setting $w_{m+1}=0$, summation by parts gives

Rescaling proves (30).

For the converse to (29), use $2k$ orthonormal eigenvectors for the largest eigenvalues. For (30), use one vector from each of the first $k$ positive spectral pairs in (6), or from every positive pair if there are fewer. Their span is orthogonal to its image under $K$, and the sum of the energies is half of the first $2k$ eigenvalue sum. With $h$ chosen vectors, the bound is $h+1\leq k+1$, and when $h=0$ all energies vanish. These chosen vectors belong to positive eigenspaces and are orthogonal to $X$, since $T_X X=0$. □

Proof. The all-zero family satisfies the inequality. Otherwise normalize $\sum_a\lVert B_a\rVert^2=1$ and maximize the left side on the compact set of normalized families satisfying the trace conditions. For a unitary $m\times m$ matrix $U=(U_{ab})$, replace the family by $B'_a=\sum_b U_{ab}B_b$. Orthogonality of the columns of $U$ gives $\sum_a\lVert B'_a\rVert^2=\sum_a\lVert B_a\rVert^2$. The commutators transform by

so the same orthogonality on both indices preserves the total energy. Every triple trace transforms as a linear combination of the original triple traces, so it remains zero. If $\Gamma_{ab}=\langle B_a,B_b\rangle$ is the Gram matrix, then the new one is $U\Gamma U^{T}$, where $T$ is transpose. The spectral theorem chooses a unitary $U$ making this diagonal. Thus a maximizing family may be chosen orthogonal. Reorder it so that $w_1\geq w_2\geq\cdots$, where $w_a=\lVert B_a\rVert^2$. If only one matrix is nonzero the energy is zero.

Write $Q$ for the maximum and $S_1=\sum_{b\neq1}\lVert[B_1,B_b]\rVert^2$ for the contribution with first index one. For $t>0$, multiply $B_1$ by $t$ and divide the whole family by $\sqrt{1+(t^2-1)w_1}$. The total squared norm becomes

Each triple trace is multiplied by the product of the scalars applied to its three matrices, so the trace constraints are preserved. The energy along this admissible path is

Its derivative at $t=1$ equals $4S_1-4Qw_1$. Maximality makes it zero, giving $S_1=Qw_1$. Applying (30) with $X=B_1$ gives $S_1\leq w_1(1-w_1+w_2)\leq w_1$. As $w_1>0$, this gives $Q\leq1$. □

A rank bound for the bracket on bivectors

Let $\mathfrak{g}=\mathfrak{sl}_n(\mathbb{C})$ be the subspace of traceless matrices with the Frobenius metric. The exterior square $\Lambda^2\mathfrak{g}$ is spanned by the bilinear alternating products $U\wedge V$, with $U\wedge V=-V\wedge U$. Choose a real orthonormal basis $F_1,\ldots,F_{n^2-1}$ of the traceless skew-Hermitian matrices, which satisfy $F_i^*=-F_i$. It is a complex orthonormal basis of $\mathfrak{g}$, since every $X\in\mathfrak{g}$ has the decomposition

into complex multiples of traceless skew-Hermitian matrices. The exterior metric is defined, for $U_1,U_2,V_1,V_2\in\mathfrak{g}$, by

Thus $F_i \wedge F_j$, $1 \le i < j \le n^2-1$, form an orthonormal basis, and the metric is preserved by a unitary change of basis. Write $b:\Lambda^2\mathfrak{g}\to\mathfrak{g}$ for the linear map $b(U\wedge V)=[U,V]$, and $b^*:\mathfrak{g}\to\Lambda^2\mathfrak{g}$ for its adjoint with respect to these metrics. For $\eta=\sum_{1\le i<j\le n^2-1}H_{ij}F_i\wedge F_j$, extend its coefficients to a skew-symmetric matrix $H$ by $H_{ji}=-H_{ij}$ and $H_{ii}=0$. The rank of $H$ is independent of the basis, since a change of basis acts by invertible congruence.

Proof. For $1\le i,j,\ell\le n^2-1$, put

These numbers are real because the Frobenius inner product of skew-Hermitian matrices is real. The first bracket makes them antisymmetric in $i,j$, and the last bracket makes them antisymmetric in $j,\ell$; thus they are fully alternating. For $X=\sum_{\ell=1}^{n^2-1}x_\ell F_\ell$, the full skew coefficient matrix $C$ of $b^*X$, with $C_{ji}=-C_{ij}$ and $C_{ii}=0$, has entries, for $i<j$,

The matrix of $D_X$ restricted to $\mathfrak{g}$ has $i,j$ entry $\sum_{\ell=1}^{n^2-1}c_{\ell ji}x_\ell=-C_{ij}$. Consequently the squared singular values of $C$, with one extra zero for the scalar direction, give the spectrum of $T_X$.

The exterior metric gives $\|H\|_F^2=2\|\eta\|^2$ and $\langle b^*X,\eta\rangle=\operatorname{Tr}(C^*H)/2$. If $P$ is the orthogonal projection onto the image of $H$, then $PH=H$, so

The last step bounds the trace of $CC^*$ on a subspace of dimension at most $2k$ by the corresponding eigenvalue sum, as in the proof of Corollary 2.

For fixed $X$ the estimate is attained on some unit $\eta$ of rank at most $2k$. To see this when $C\ne0$, the conjugate-linear map $v\mapsto\overline{C}v$ squares to $-C^*C$ since $C^{\mathsf T}=-C$, where $\mathsf T$ is the ordinary transpose. It commutes with $C^*C$, because its square is $-C^*C$, and its squared norm at $v$ is $\langle v,C^*Cv\rangle$. Thus, on an eigenspace of $C^*C$ for a positive eigenvalue $\alpha$, division by $\sqrt{\alpha}$ gives an antiunitary map of square $-\operatorname{Id}$. The construction leading to (6) supplies orthonormal pairs there. If $U$ is the unitary matrix with columns given by these pairs and an orthonormal kernel basis, $U^{\mathsf T}CU$ consists of two-by-two skew blocks and zeros. This follows because the $i,j$ entry of $U^{\mathsf T}CU$ is the conjugate of the $i$-th coordinate of $C U_j$ in the columns of $U$, with $U_j$ the $j$-th column. Retain the $k$ blocks of largest singular values, or all if fewer, and undo the congruence. The resulting skew matrix has squared Frobenius norm $\sum_{j=1}^{2k}\lambda_j(T_X)$, and its Frobenius inner product with $C$ equals that same sum. Multiplying this matrix by $\sqrt{2/\sum_{j=1}^{2k}\lambda_j(T_X)}$ makes it the coefficient matrix of a unit bivector $\eta$ for which

If $C=0$, the estimate is identically zero.

Taking the supremum over unit $X\in\mathfrak{g}$ recovers $\|b\eta\|^2$ on the left by duality. Conversely the inequality for $b\eta$ bounds the maximum over rank-constrained unit $\eta$ by $(k+1)\|X\|^2$. The attainment just proved gives the spectral inequality for traceless $X$, and subtracting the scalar part gives Theorem 1.

Appendix A. Local maximum propagation

The following local form of maximum propagation is part of the maximum principle of Bony [1].

Proof. Let $K=\{u=m\}$. Suppose an open ball of center $c$ and radius $R$ in $\Omega\setminus K$ touches $K$ at $p \in\Omega$. Replacing the center and radius by $p+t(c-p)$ and $tR$ with $0<t<1$ puts the new closed ball inside the original open ball except at $p$, by the triangle inequality and its equality case. The normal direction at $p$ is unchanged. Thus we may assume the closed ball lies in $\Omega$ and meets $K$ only at $p$.

In coordinates write $Z_{s}=\sum_{i=1}^{d}z_{s}^{i}\partial_{i}$ and $Y=\sum_{i=1}^{d}y^{i}\partial_{i}$, with $\partial_{i}=\partial/\partial x_{i}$. Set

Expanding $Z_{s}^{2}$ gives

The matrix $(a^{ij})$ is positive semidefinite. We show $Z_{s}(p)\cdot(p-c)=0$ for every $s$. If their squares sum to a positive number, take $v(x)=e^{-\kappa|x-c|^{2}}-e^{-\kappa R^{2}}$ for $\kappa>0$. Differentiation gives

Choose $\kappa$ large enough that $Av(p)>0$, and take a small ball $U$ centered at $p$ with $\overline{U}\subset\Omega$ and $Av>0$ on $\overline{U}$. The compact part of $\partial U$ in the closed exterior ball is disjoint from $K$. On it $u$ is uniformly below $m$ if this part is nonempty. For sufficiently small $\varepsilon>0$, $u+\varepsilon v<m$ there. Elsewhere on $\partial U$ we have $v<0$, giving the same strict inequality. Since $u(p)+\varepsilon v(p)=m$, the function $u+\varepsilon v$ attains an interior maximum in $U$. At that point $A(u+\varepsilon v)\le0$ by the vanishing gradient and the nonpositive Hessian, whereas $Au+\varepsilon Av>0$ on $U$. The contradiction proves the asserted tangency. In particular $W(p)\cdot(p-c)=0$ for every such exterior ball.

Fix $x_{0}\in K$, choose $\rho>0$ with $\overline{B(x_{0},2\rho)}\subset\Omega$, and set $K_{0}=K\cap B(x_{0},2\rho)$. Let $\gamma$ be the integral curve of $W$ from $x_{0}$ on a short interval $[0,T]$ with image in $B(x_{0},\rho/2)$. For each time choose a nearest point $p\in K_{0}$ to $\gamma(t)$, using compactness. Then $|\gamma(t)-p|\le|\gamma(t)-x_{0}|<\rho/2$, so

If $\gamma(t)\ne p$, the ball centered at $\gamma(t)$ of radius $|\gamma(t)-p|$ lies in $B(x_{0},\rho)$. Every point of $K$ in that ball would belong to $K_{0}$ and contradict minimality of the distance. Hence the ball avoids all of $K$. Thus $W(p)\cdot(\gamma(t)-p)=0$, also valid when $\gamma(t)=p$.

The squared distance $f(t)=\operatorname{dist}(\gamma(t),K_{0})^{2}$ is Lipschitz on this interval, since distance to a set is $1$-Lipschitz. Let $L$ be a Lipschitz constant for $W$ on $\overline{B(x_{0},2\rho)}$. At any point of differentiability of $f$, holding the nearest point $p$ fixed for positive increments gives

Since $f(0)=0$, the absolutely continuous function $e^{-2Lt}f(t)$ is nonincreasing and $f=0$. The same argument with $-W$ gives propagation in negative time. □

References

Zhili College, Tsinghua University, Beijing 100084, China

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