If $\sigma:M_3\rightarrow S_3$ is a linear map such that $\sigma(PMP^{-1})=P\sigma(M)P^{-1}$, then $\sigma(M)=\ldots$

Question

In page 32 of the book A Mathematical Introduction to Fluid Mechanics, by Alexandre Chorin and Jerrold E. Marsden, it is used and commented the following property: let $M_3$ be the space of $3\times 3$ matrices and $S_3$ be the subspace of symmetric matrices. Let $\sigma:M_3\rightarrow S_3$ be a linear map, with the property that $\sigma(PMP^{-1})=P\sigma(M)P^{-1}$ for each orthogonal matrix $P$ and each $M\in M_3$. Then there exist constants $\lambda$ and $\mu$ such that $\sigma(M)=\lambda\text{Trace}(M)I_3+\mu(M+M^T)$. I do not understand the brief proof from the book. Could you provide a detailed proof?

Answer 1

Observe that any $M \in M_3$ can be uniquely decomposed as $$ M = \frac{1}{3}\operatorname{Tr}(M)I_3 + \frac{1}{2}(M-M)^T + \left(\frac{1}{2}(M+M^T)-\frac{1}{3}\operatorname{Tr}(M)I_3 \right), $$ where

• $\frac{1}{3}\operatorname{Tr}(M)I_3 \in \mathbb{R}I_3$, the subspace of all scalar multiples of the $3 \times 3$ identity $I_3$,
• $\frac{1}{2}(M-M)^T \in A_3$, the subspace of all $3 \times 3$ skew-symmetric matrices, and
• $\frac{1}{2}(M+M^T)-\frac{1}{3}\operatorname{Tr}(M)I_3 \in S_3^0$, the subspace of all $3 \times 3$ symmetric matrices of trace $0$;

in particular, it follows that $\sigma = \sigma_1 + \sigma_2$, where $\sigma_1 : M_3 \to \mathbb{R}I_3$ is defined by $$ \sigma_1(M) := \frac{1}{3}\operatorname{Tr}(\sigma(M))I_3, $$ and where $\sigma_2 := \sigma - \sigma_1 : M_3 \to S_3^0$. It therefore suffices to find constants $C_1$ and $C_2$, such that $$ \forall M \in M_3, \quad \sigma_1(M) = C_1 \left(\frac{1}{3}\operatorname{Tr}(M)I_3\right), \quad \sigma_2(M) = C_2 \left(\frac{1}{2}(M+M^T) - \frac{1}{3}\operatorname{Tr}(M)I_3\right). $$

Now, let $O_3$ be the group of $3 \times 3$ orthogonal matrices. A subspace $V$ of $M_3$ is called $O_3$-invariant if $$ \forall M \in V, \; \forall P \in O_3, \quad PMP^{-1} \in V; $$ an $O_3$-invariant subspace is called irreducible if the only $O_3$-invariant subspaces of $V$ are $\{0\}$ and $V$ itself. It's easy to check that $\mathbb{R}I_3$, $A_3$, and $S_3^0$ are all $O_3$-invariant, and it's trivial that $1$-dimensional $O_3$-invariant subspace $\mathbb{R}I_3$ is irreducible; it's a fairly non-trivial fact that the $3$-dimensional $O_3$-invariant subspace $A_3$ and the $5$-dimensional $O_3$-invariant subspace $S_3^0$ are also irreducible. Because $\sigma$ is $O_3$-equivariant, in the sense that $$ \forall M \in M_3, \; \forall P \in O_3, \quad \sigma(PMP^{-1}) = P\sigma(M)P^{-1}, $$ one can show that $\sigma_1$ and $\sigma_2$ are $O_3$-equivariant as well, so we can flesh out a variant of Chorin and Marsden's argument as follows.

Let's first find a constant $C_1$, such that $$ \forall M \in M_3, \quad \sigma_1(M) = C_1 \left(\frac{1}{3}\operatorname{Tr}(M)I_3\right). $$ First, let's show that $\sigma_1$ vanishes on $A_3 \oplus S_3^0$, so that $\sigma_1$ is completely determined by its restriction to $\mathbb{R}I_3$. On the one hand, by $O_3$-equivariance of $\sigma_1$ and the rank-nullity theorem, the set $\{M \in A_3 \mid \sigma_1(M) = 0\}$ is an $O_3$-invariant subspace of $A_3$ with dimension at least $2$; by irreducibility of $A_3$, it follows that $\{M \in A_3 \mid \sigma_1(M) = 0\} = A_3$, so that $\sigma_1$ vanishes on $A_3$. On the other hand, by exactly the same argument, the $O_3$-equivariance of $\sigma_1$, the rank-nullity theorem, and the irreducibility of $S_3^0$ together imply that $\{M \in S_3^0 \mid \sigma_1(M) = 0\} = S_3^0$, so that $\sigma_1$ also vanishes on $S_3^0$. But now, for any $\alpha \in \mathbb{R}$, $$ \sigma_1(\alpha I_3) = \frac{1}{3}\operatorname{Tr}(\sigma(\alpha I_3)) I_3 = \frac{1}{3}\operatorname{Tr}(I_3) \alpha I_3 = \left(\frac{1}{3}\operatorname{Tr}(\sigma(I_3))\right) \frac{1}{3}\operatorname{Tr}(\alpha I_3) I_3, $$ so that $C_1 := \tfrac{1}{3}\operatorname{Tr}(\sigma(I_3))$ does the job.

Now, let's find a constant $C_2$, such that $$ \forall M \in M_3, \quad \sigma_2(M) = C_2 \left(\frac{1}{2}(M+M^T) - \frac{1}{3}\operatorname{Tr}(M)I_3\right). $$ First, let's show that $\sigma_2$ vanishes on $\mathbb{R}I_3 \oplus A_3$, so that $\sigma_2$ is completely determined by its restriction to $S_3^0$. By $O_3$-equivariance of $\sigma_2$, the set $\{\sigma_2(M) \mid M \in \mathbb{R}I_3 \oplus A_3\}$ is an $O_3$-invariant subspace of $S_3^0$ with dimension at most $4$; by irreducibility of $S_3^0$, it follows that $\{\sigma_2(M) \mid M \in \mathbb{R}I_3 \oplus A_3\} = \{0\}$, so that $\sigma_2$ does indeed vanish on $\mathbb{R}I_3 \oplus A_3$. If $\sigma_2$ also vanishes on $S_3^0$, then we can take $C_2 = 0$ and be done. So let's take a closer look at the other case.

Suppose now that $\sigma_2$ does not vanish identically on $S_3^0$. Thus, by $O_3$-equivariance of $\sigma_2$, the set $\{M \in S_3^0 \mid \sigma_2(M) = 0\}$ defines an $O_3$-invariant subspace of $S_3^0$ with dimension strictly less than $5$; by irreducibility of $S_3^0$, it follows that $\{M \in S_3^0 \mid \sigma_2(M) = 0\} = \{0\}$, so that $\sigma_2$ is actually invertible as a map $S_3^0 \to S_3^0$. Now, let $C_2$ be any eigenvalue of $\sigma_2$ as a map $S_3^0 \to S_3^0$; by invertibility, it follows that $C_2 \neq 0$. By $O_3$-equivariance of $\sigma_2$, the set $\{M \in S_3^0 \mid \sigma_2(M) = C_2 M\}$ defines an $O_3$-invariant subspace of $S_3^0$ of dimension at least $1$; by irreducibility of $S_3^0$, it therefore follows that $\{M \in S_3^0 \mid \sigma_2(M) = C_2 M\} = S_3^0$, so that for any $M \in S_3^0$, $$ \sigma_2(M) = C_2 M = C_2\left(\frac{1}{2}(M+M^T)-\frac{1}{3}\operatorname{Tr}(M)I_3 \right), $$ as required.

If you're familiar with representation theory, you'll have noticed that I've used [the proof of] Schur's lemma left, right, and centre throughout. With a little more care, I suspect this can all be made to look a little more elementary looking, but there's no escaping representation theory; indeed, I honestly don't know how one could prove the vanishing of $\sigma$ on $A_3$ without it.

Answer 2

Presumably the matrices are real. Let $R_t$ denotes the $2\times2$ rotation matrix for angle $t$ and let $Q_t=(R_t\oplus1)\in SO(3)$.

We first show that $\sigma(K)=0$ when $K$ is skew-symmetric. We may consider only the case $K=Q_{\pi/2}$, because every $3\times3$ skew-symmetric matrix is orthogonally similar to a scalar multiple of this $K$. Let $S=\sigma(K)$. Then $Q_tKQ_t^\top=K$ and $Q_tSQ_t^\top=S$ for every $t$. Therefore $S=aI_2\oplus b$ for some scalars $a$ and $b$. However, for $D=\operatorname{diag}(1,-1,1)$, we also have $DSD=-S$ because $DKD=-K$. Hence $a=b=0$ and $S=\sigma(K)=0$.

Next, we show that $\sigma(Z)$ is a scalar multiple of $Z$ when $Z$ is a traceless symmetric matrix. Let us consider the case $Z=\operatorname{diag}(1,-1,0)$ first. Let $S=\sigma(Z)$. Then $\Lambda Z\Lambda=\Lambda$ and $\Lambda S\Lambda=S$ for every diagonal orthogonal matrix $\Lambda=\operatorname{diag}(\pm1,\pm1,\pm1)$. Hence $S$ is a diagonal matrix. Yet, we also have $Q_{\pi/2}ZQ_{\pi/2}^\top=-Z$ and $Q_{\pi/2}SQ_{\pi/2}^\top=-S$. Thus $S$ is a scalar multiple of $Z$. Let $S=\sigma(Z)=2\mu Z$.

It follows that $\sigma(Z)=2\mu Z$ for every traceless symmetric matrix $Z$, because $\sigma$ is a linear map that preserves orthogonal similarity and every traceless symmetric matrix $Z$ can be written as a finite weighted sum of the form $\sum_i c_i U_i\operatorname{diag}(1,-1,0)U_i^\top$ where each $U_i$ is an orthogonal matrix.

Finally, the given properties of $\sigma$ clearly imply that $\sigma(I_3)=\gamma I_3$ for some scalar $\gamma$. Since every matrix $M\in M_3(\mathbb R)$ can be written as $\frac13\operatorname{tr}(M)I_3+K+Z$, where $K=\frac12(M-M^\top)$ is skew-symmetric and $Z=\frac12(M+M^\top)-\frac13\operatorname{tr}(M)I_3$ is traceless and symmetric, we get \begin{align} \sigma(M) &=\frac13\operatorname{tr}(M)\sigma(I_3)+\sigma(K)+\sigma(Z)\\ &=\frac\gamma3\operatorname{tr}(M)I_3+0+2\mu Z\\ &=\frac{\gamma-2\mu}3\operatorname{tr}(M)I_3+\mu(M+M^\top). \end{align}