Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This question is obviously related to that recent question of mine, but I feel it’s sufficiently different to be posted as a separate question. Let $V$ be a finite-dimensional space. Let ${\cal L}(V)$ denote the space of all endomorphisms of $V$. Say that an endomorphism $\phi$ of ${\cal L}(V)$ is invariant when it satisfies $$ \phi (gfg^{-1})=g\phi(f)g^{-1} $$ for any $f,g \in {\cal L}(V)$ with $g$ invertible.

Prove or find a counterexample or provide a reference : $\phi$ is invariant iff there are two constants $a,b$ such that $\phi(f)=af+b{\sf tr}(f){\bf id}_V$ for all $f$. With the help of a PARI-GP program, I have checked that this is true when ${\sf dim}(V) \leq 5$. Intuitively, the similitude invariants of a matrix are functions of the coefficients of the characteristic polynomial, and the second largest coefficient, the trace, is the only linear one.

share|cite|improve this question
up vote 6 down vote accepted

Let $G=\operatorname{GL}(V)$ be the group of automorphisms of $V$. Then as a $G$-module, $\mathcal L(V)$ is isomorphic to $V\otimes V^*$, and $\hom_k(\mathcal L(V),\mathcal L(V))$ is isomorphic to $V^{\otimes 2}\otimes V^{*\otimes 2}$. Your question is, in this language:

what is the dimension of the $G$-invariant subspace of $V^{\otimes 2}\otimes V^{*\otimes 2}$?

You will find this done in treatements of classical invariant theory.If I recall correctly, this particular case is done in detail in Kraft+Procesi's notes on Classical Invariant Theory, which you can find online

Notice that $V^{\otimes 2}\otimes V^{*\otimes 2}$ is isomorphic as a $G$-module to $hom_k(V^{\otimes 2},V^{\otimes 2})$, so its invariant subspace is actually the space of $G$-equivariant maps, $hom_G(V^{\otimes 2},V^{\otimes 2})$. The fact that there are exactly two linear invariants in your sense is the case $m=2$ of the claim stated on page 31 of those notes —this is part of what's called Schur-Weyl duality.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.