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Let $V$ be a finite dimensional vector space over $\mathbb{R}$. Let \begin{equation} \left\langle\:,\:\right\rangle:\mbox{End}(V)\otimes\mbox{End}(V)\rightarrow \mathbb{R}\end{equation} denote the pairing given by $\left\langle A,B\right\rangle = \mbox{Tr}(AB)$. We know that this pairing is $\mbox{GL}(V)$-invariant, symmetric, non-degenerate, but not definite (i,e. neither positive definite nor negative definite).

Supposed that $\beta:V\otimes V\rightarrow \mathbb{R}$ is a non-degenerate symmetric pairing. Recall $\mathfrak{so}(V,\beta)\subset \mathfrak{gl}(V)$ is defined by $$\mathfrak{so}(V,\beta)=\left\{A\in\mathfrak{gl}(V)\:|\: \forall v,w\in V \: \beta(Av,w)+\beta(v,Aw)=0\right\}.$$ We know that $\mathfrak{so}(V,\beta)$ is a Lie subalgebra of $\mathfrak{gl}(V)$.

The question: Establish conditions on $\beta$ under which the restriction of the trace pairing (i,e. $\left\langle\:,\:\right\rangle$ restricted to $\mathfrak{so}(V,\beta)$) is positive (also conditions to be negative) definite.

My attempt: We know that given $\beta:V\times V\rightarrow \mathbb{R}$ a non-degenerate symmetric form, for all $A\in \mbox{End}(V)$ there exist a unique $B\in \mbox{End}(V)$ which satisfies $\beta(Av,w)=\beta(v,Bw)$ for all $v,w\in V$.

The unique $B$ as above will be denoted by $A^{\dagger}$ and called the transpose of $A$ (with respect to $\beta$).

The assignment $A\rightarrow A^{\dagger}$ defines a linear map $(\cdot)^{\dagger}\mbox{End}(V)\rightarrow \mbox{End}(V)$ called transposition (with respect to $\beta$).

Therefore, we can consider $$\mathfrak{so}(V,\beta)=\left\{A\in\mathfrak{gl}(V)\:|\: A=-A^{\dagger} \: (\mbox{transpose with respect to }\beta)\right\}.$$ In this sense, we have

  1. Restriction the trace pairing is positive definite if $\mbox{Tr}(A^{\dagger}A)<0$ for all $A\in \mathfrak{so}(V,\beta)$ with $A\neq 0$.
  2. Restriction the trace pairing is negative definite if $\mbox{Tr}(A^{\dagger}A)>0$ for all $A\in \mathfrak{so}(V,\beta)$ with $A\neq 0$.

My question: Conditions 1 and 2 depend on $\beta$. Does the condition 1 and 2 answer the question? Can I characterize $A^{\dagger}$ in terms of $\beta$ of a better way? Are there more specific conditions on $\beta$ to answer the question?

Diego Fonseca
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1 Answers1

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$\DeclareMathOperator{\tr}{tr}$ Symmetric non-degenerate bilinear forms are determined by their signature $(q,p)$, where $q$ and $p$ are nonnegative integers with $q+p = \dim V$. More precisely, there exists a basis of $V$ such that the matrix representing $\beta$ in this basis has entries $+1$ or $-1$ along the diagonal and $0$ elsewhere: $$ I_{p,q} = \begin{pmatrix} I_p & 0 \\ 0 & - I_q \end{pmatrix} $$ where $I_n$ denotes the $n \times n$ identity matrix. Sylvester's Law of Inertia states that the integers $(p,q)$ that give the number of $+1$ and $-1$ entries are uniquely determined by the bilinear form.

From now on, let's identify $V$ with $\mathbb{R}^n$ (where $n = \dim V$) using this basis. Under this identification, the form $\beta$ is given by $(v,w) \mapsto v^T I_{p,q} \, w$. Given a linear map $A$, the condition $\beta(Av, w)+\beta(v,Aw) =0$ defining $\mathfrak{so}(V, \beta)$ translates to the following condition on the matrix representing $A$ (which I also denote $A$): $$ (Av)^T I_{p,q} \, w + v^T I_{p,q} \, (A w) \qquad \text{for all } v,w. $$ This is equivalent to $A= - (I_{p,q})^{-1} A^T \, I_{p,q}$. For a given signature $(p,q)$ with $p+q=n$, the set of $n\times n$ matrices satisfying this equality is denoted $\mathfrak{so}(p,q)$. (The above construction actually gives a lie algebra isomorphism between $\mathfrak{so}(V,\beta)$ and $\mathfrak{so}(p,q)$, where $(p,q)$ is the signature of $\beta$.)

There is an explicit characterization of the elements of $\mathfrak{so}(p,q)$, which will be useful later. For $p,q \geq 1$, write an arbitrary $n\times n$ matrix $A$ in block form as: $$ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} $$ where $a$ is a $p \times p$ matrix and $d$ is a $q\times q$ matrix. A quick calculation shows that the condition $A= - (I_{p,q})^{-1} A^T \, I_{p,q}$ is satisfied if and only if $a$ and $d$ are antisymmetric and $b= c^T$. When $p$ or $q$ is zero, the elements of $\mathfrak{so}(p,q)$ are the antisymmetric matrices.

Now that everything is on the table, we can prove the following:

Claim: the trace pairing on $\mathfrak{so}(p,q)$ is always indefinite when $n=p+q \geq 3$ and $p,q \geq 1$.

The proof is brute force: we exhibit matrices $A,B \in \mathfrak{so}(p,q)$ having $\tr(A^2) <0$ and $\tr(B^2) > 0$, respectively.

Since $p+q \geq 3$, at least one of $p$ or $q$ is greater than $1$. Suppose it is $p$ (the other case is done similarly). Consider the matrix $A$ which has a $2 \times 2$ block $\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$ in the upper-left corner and $0$ elsewhere. The preceding characterization shows that this is in $\mathfrak{so}(p,q)$ and $\tr(A^2)=-2$.

Now, consider the matrix $B$ which has entry $+1$ at the positions $(1,n)$ and $(n,1)$, and $0$ elsewhere. This is in $\mathfrak{so}(p,q)$ and we have $\tr(B^2) = 2$.

In dimension $2$ or when either of $p$ or $q$ is zero, the answer is positive:

Claim : For $(p,q)= (1,1)$, the trace pairing is positive-definite. For $(p,q) = (n,0)$ the trace pairing is negative-definite.

The $(1,1)$ case follows from the above characterization: matrices in $\mathfrak{so}(1,1)$ are of the form $\left( \begin{smallmatrix} 0 & a \\ a & 0 \end{smallmatrix} \right)$. The $(n,0)$ case follows from the result of this post: for any antisymmetric matrix $A$ we have $\tr(A^2) \leq 0$ with equality if and only if $A=0$.

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