Orbit spaces of n-tuples of square matrices under simultaneous conjugation

Question

Let $n, p, \geq 1$ be integers. Denote the set of ordered partitions of $p$ by $\Pi$: each $\pi \in \Pi$ is an ordered $k$-tuple $(p_1,p_2, \dotsc, p_k)$ where $p_1+\dotsb+p_k = p$. Write $\pi \leq \pi'$ if $\pi$ is a refinement of $\pi'$ (or equal to it). For $\pi \in \Pi$ a partition of $p$ into $k$ parts and $\sigma \in \Sigma_k$, we write $\sigma(\pi) = (p_{\sigma(1)}, \dotsc, p_{\sigma(k)})$.

For $\pi \in \Pi$ we denote by $[\pi]$ the corresponding unordered partition $\{p_1,p_2, \dotsc, p_k\}$. For unordered partitions $[\pi], [\pi']$ into $k$ and $k'$ parts we write $[\pi] \leq [\pi']$ if $\pi \leq \sigma(\pi')$ for some permutation $\sigma \in \Sigma_{k'}$.

Now let $V$ denote the set of $n$-tuples of $p \times p$ matrices over $\mathbb{C}$. $G = \operatorname{GL}_p(\mathbb{C})$ acts on $V$ by simultaneous conjugation. For $\pi \in \Pi$ let $V_{\pi}$ denote the subspace of $V$ consisting of $n$-tuples of matrices which are $\pi$-blockwise upper triangular; that is, when partitioned into $k^2$ blocks such that the $i,j$th block is a $p_i \times p_j$ matrix, the $i,j$th block is zero if $i>j$. The orbit spaces $G\cdot V_{\pi}$ are closed and irreducible, and clearly \begin{equation} G \cdot V_{\pi} \subseteq G \cdot V_{\pi'} \tag{$*$}\label{481947_star}\end{equation} whenever $\pi \leq \pi'$.

Now if $n=1$ then every element of $V$ can be written in Jordan canonical form; in particular $G \cdot V_{\pi} = G \cdot V_{\sigma(\pi)}$ for every $\sigma \in \Sigma_k$. So the converse of \eqref{481947_star} cannot be true for $n=1$. I'd like to know whether it is true for $n \geq 2$: is it the case that $G \cdot V_{\pi} \subseteq G \cdot V_{\pi'}$ if and only if $\pi \leq \pi'$?

Here's why I think it might be: consider the set $\bigcup_{\sigma \in \Sigma_k} G \cdot V_{\sigma(\pi)}$. This doesn't depend on the ordering of $\pi$, so we'll call it $X_{[\pi]}$. In some sense these are more natural objects to study, since by the Artin–Voigt theorem one has $$X_{[\pi]} = \zeta^{-1}(V_{\pi})$$ where $\zeta: V \rightarrow V/G$ is the canonical quotient map, and these varieties can be defined by the vanishing of certain $G$-invariants. If I've understood it correctly, it's shown in [Lieven LeBruyn. Orbits of matrix tuples. Seminaires Congres, Société Mathématique de France, 2:245–261, 1997. UIA report 95–20] that $$X_{[\pi]} \subseteq X_{[\pi']}$$ if and only if $[\pi] \leq [\pi']$, for all $n \geq 1$.

The smallest possible counterexample to \eqref{481947_star} is where $p=3$: is it obvious that $G \cdot V_{(1,2)} \neq G \cdot V_{(2,1)}$?

Edit: in response to Lieven's answer below - I originally accepted but now I'm not convinced. The problem is with the phrase

"since representation type is an isomorphism invariant..." -This means that if $A \in G \cdot V_{\rho_{\pi}}$ and $g \in G$, then $g \cdot A \in G \cdot V_{\rho(\pi)}$ also. Which is obvious. Further, if $A' \in V_{\rho_{\sigma(\pi)}}$ has the same simple diagonal blocks as $A$ in a different order, then $f(A) = f(A')$ for all invariants.

But the second part of the phrase "... we have $G \cdot V_{\rho_{\pi}} = G \cdot V_{\rho_{\sigma(\pi)}}$" seems to be much stronger. It says that for all $A \in V_{\rho_{\pi}}$ we can find a $g \in G$ with $g \cdot A \in V_{\rho_{\sigma(\pi)}}$. Is that really true?

For example, suppose that $n \geq 2$ and $B = \begin{pmatrix} a_{11}&a_{12}\\a_{21}&a_{22} \end{pmatrix} \in M^n_2(\mathbb{C})$ is simple. We may choose $a_{13}, a_{23}, a_{33} \in \mathbb{C}^n$ so that $$A = \begin{pmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23} \\ 0 & 0 & a_{33} \end{pmatrix}$$ is not semisimple. Then $A \in V_{\rho_{(2,1)}}$. I claim that $g \cdot A \not \in V_{\rho_{(1,2)}}$ for any $g \in G$. Let $P_{\pi}$ denote the parabolic subgroup stabilising $V_{\pi}$. $G$ has Bruhat decomposition $$G = P_{(1,2)} \cup P_{(1,2)} (123) P_{(1,1,1)} \cup P_{(1,2)} (132) P_{(1,1,1)}$$ where $(132)$ and $(123)$ denote permutation matrices in $G$. So if $g \cdot A \in V_{(\rho_{(1,2)})}$ we can write $g = phb$ where $p \in P_{(1,2)}$, $h \in \{I,(123),(132)\}$ and $b \in B$. As $p$ fixes $V_{(\rho_{(1,2)})}$ we get $hb \cdot A \in V_{(\rho_{(1,2)})}$.

Now the top left submatrix of $b \cdot A$ is simple, so its entry in the $(2,1)$ position is nonzero, and $Ib \cdot A \not \in V_{(\rho_{(1,2)})}$. Further, the entry in position $(1,2)$ in $b \cdot A$ is also nonzero, and this is the entry in position $(3,1)$ of $(132)b \cdot A$, so $(132)b \cdot A \not \in V_{(\rho_{(1,2)})}$. Finally, the entries in positions $(1,3)$ and $(2,3)$ in $b \cdot A$ cannot both be zero, and these are the entries in positions $(2,1)$ and $(3,1)$ in $(123)b \cdot A$, so $(132)b \cdot A \not \in V_{(\rho_{(1,2)})}$. We conclude that $g \cdot A \not \in V_{(\rho_{(1,2)})}$ for any $g \in G$.

Answer 1

For $n=1$ your definition of $V_{\pi}$ implies that $G.V_{\pi}=M_p(\mathbb{C})$ for any partition $\pi$ of $p$, by the Jordan normal form.

For $n \geq 2$ one still has $G.V_{\pi} = G.V_{\sigma(\pi)}$ for any partition $\pi$ of $p$ and any permutation $\sigma$ on its components.

For $\pi=(p_1,\dots,p_k)$ we have that $V_{\pi}$ is the Zariski closure of the the open subset $V_{\rho_{\pi}}$ of $n$-tuples of matrices of 'representation type' $\rho_{\pi}=(1,p_1;\dots;1,p_k)$, that is $n$-tuples of the required block upper-triangular form satisfying in addition that the $n$ matrices in the $p_i \times p_i$ position generate $M_{p_i}(\mathbb{C})$ for all $p_i$, and if some of the $p_i$'s are equal, that the corresponding $p_i$-dimensional irreducibles are non-isomorphic.

As 'representation type' is an isomorphism invariant it follows that $G.V_{\rho_{\pi}} = G.V_{\rho_{\sigma(\pi)}}$, and hence that their Zariski closures $G.V_{\pi}$ resp. $G.V_{\sigma(\pi)}$ are the same.