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Let $\mathcal C$ denote the commuting variety of pairs of matrices in $M_n(\mathbb{C})$, defined as:

$$ \mathcal C = \{ (A, B) \in M_n(\mathbb{C})^2 \mid [A, B] = 0 \}. $$

It is well known that $\mathcal C$ is irreducible of dimension $n^2 + n$.

Now, let $U$ be a vector subspace of $M_n(\mathbb{C})$. Consider the set of commuting pairs where the first matrix is restricted to $U$:

$$ \mathcal C_U = \mathcal C \cap (U \times M_n(\mathbb{C})). $$

Observations

  1. Certainly $\mathcal C_U$ contains the subset $\{ 0 \} \times M_n(\mathbb{C})$, so $\dim \mathcal C_U \geq n^2$.
  2. From computational experiments for small $n$ and generic $U$, I observed that:
    • When $\mathrm{codim} \, U \leq n$, we have $\dim \mathcal C_U = \dim \mathcal C - \mathrm{codim} \, U$, so with each lower dimension $\dim \mathcal C_U$ decreases.
    • For $\mathrm{codim} \, U \geq n$, the dimension of $\mathcal C_U$ stabilizes at $n^2$.

Questions

  1. Generality: Is it true that for generic choices of $U$, the dimension of $\mathcal C_U$ behaves as observed?
  2. Exceptions: Are there examples of particular subspaces $U$ where $\dim \mathcal C_U$ behaves unexpectedly?
  3. Stabilization: Do we always have $\dim \mathcal C_U = n^2$ for $\mathrm{codim} \, U = n$?
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1 Answer 1

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The answer to "Generality" seems to be yes. The answer to "Stabilization" is no (and therefore the answer to "Exceptions" is yes). Let's begin with "Generality".

Fix $0 \leq c \leq n \geq 2$ and let $\mathcal{D}$ be the variety of triples $(U, A, B)$, where $U$ is a codimension $c$ subspace of $M_n(\mathbf{C})$ and $(A, B) \in \mathcal{C}$ satisfies $A \in U$; this is a subvariety of $\operatorname{Gr}(M_n(\mathbf{C}), n^2 - c) \times M_n(\mathbf{C}) \times M_n(\mathbf{C})$.

Proposition. If $c < n$ then $\mathcal{D}$ is irreducible of dimension $(n^2 - c - 1)c + n^2 + n$, and if $c = n$ then $\mathcal{D}$ has two irreducible components, each of the same dimension $n^3 (= (n^2 - n - 1)n + n^2 + n)$.

This is enough to establish "Generality": indeed, there is a natural surjective map $\pi\colon \mathcal{D} \to \operatorname{Gr}(M_n(\mathbf{C}), n^2 - c)$ given by $\pi(U, A, B) = U$, and for a fixed $U$ the fiber $\pi^{-1}(U)$ is isomorphic to $\mathcal{C}_U$. So by generic flatness, it would follow that for generic $U$ we have $\dim \mathcal{C}_U = \dim \mathcal{D} - \dim \operatorname{Gr}(M_n(\mathbf{C}), n^2 - c) = n^2 + n - c$, as desired.

Claim 1. Every irreducible component of $\mathcal{D}$ is of dimension $\geq (n^2 - c - 1)c + n^2 + n$.

To see this, note that there is a Cartesian diagram $$\require{AMScd} \begin{CD} \mathcal{D} @>{}>> \operatorname{Gr}(M_n(\mathbf{C}), n^2 - c) \times \mathcal{C}\\ @VVV @VVV \\ \mathcal{I} @>{i}>> \operatorname{Gr}(M_n(\mathbf{C}), n^2 - c) \times M_n(\mathbf{C}) \end{CD}$$ where $\mathcal{I} = \{(U, A)\colon A \in U\}$ and the right vertical map is given by $(U, A, B) \mapsto (U, A)$. The map $i$ is a closed embedding of codimension $c$ between smooth varieties, so it is locally cut out by $c$ equations. In particular, $\mathcal{D}$ is locally cut out of $\operatorname{Gr}(M_n(\mathbf{C}), n^2 - c) \times \mathcal{C}$ by $c$ equations, and the claim follows.

Next, define $\alpha\colon \mathcal{D} \to M_n(\mathbf{C})$ by $\alpha(U, A, B) = A$. For any $A \in M_n(\mathbf{C})$, let $Z(A)$ be the space of $B \in M_n(\mathbf{C})$ such that $[A, B] = 0$, and let $z_A = \dim Z(A)$. Note $n \leq z_A \leq n^2$. For each $n \leq z \leq n^2$, let $\Omega_z$ be the locally closed subvariety of $M_n(\mathbf{C})$ consisting of $A$ such that $z_A = z$, and note that $\Omega_n$ is open and dense in $M_n(\mathbf{C})$.

Claim 2. If $z > n$, then $\dim \Omega_z + z < n^2 + n$.

Indeed, let $\mathrm{pr}_1\colon \mathcal{C} \to M_n(\mathbf{C})$ be the first projection, and note that the closure of $\mathrm{pr}_1^{-1}(\Omega_z)$ is a proper subvariety of $\mathcal{C}$, and since $\mathcal{C}$ is irreducible of dimension $n^2 + n$ it follows that $\dim \mathrm{pr}_1^{-1}(\Omega_z) < n^2 + n$. The map $\mathrm{pr}_1^{-1}(\Omega_z) \to \Omega_z$ has fibers of dimension $z$, so the claim follows.

Claim 3. i) $\alpha^{-1}(\Omega_n)$ is irreducible of dimension $(n^2 - c - 1)c + n^2 + n$.

ii) If $n < z < n^2$, then $\dim(\alpha^{-1}(\Omega_z)) < (n^2 - c - 1)c + n^2 + n$.

iii) $\dim(\alpha^{-1}(\Omega_{n^2} - \{0\})) = (n^2 - c - 1)c + n^2 + 1$.

iv) $\dim(\alpha^{-1}(0)) = (n^2 - c - 1)c + n^2 + c$.

Note that for fixed $A \neq 0$, we have $\alpha^{-1}(A) \cong \operatorname{Gr}(M_n(\mathbf{C})/\mathbf{C}A, n^2 - c - 1) \times Z(A)$, of dimension $(n^2 - c - 1)c + z_A$. Moreover, $\alpha^{-1}(0) \cong \operatorname{Gr}(M_n(\mathbf{C}), n^2 - c) \times M_n(\mathbf{C})$, of dimension $(n^2 - c)c + n^2 = (n^2 - c - 1)c + n^2 + c$. If $n < z < n^2$, then $0 \not\in \Omega_z$, so it follows that $\dim \alpha^{-1}(\Omega_z) = \dim \Omega_z + (n^2 - c - 1)c + z$, and ii) follows by Claim 2. Points iii) and iv) follow similarly, as does the dimension calculation of i).

So it remains to prove that $\alpha^{-1}(\Omega_n)$ is irreducible. For this, since $\Omega_n$ is irreducible and $\alpha^{-1}(\Omega_n) \to \Omega_n$ has irreducible fibers, it suffices to show that this map admits sections locally on $\Omega_n$. So fix $A \in \Omega_n$ and let $V$ be a codimension $c + 1$ subspace of $M_n(\mathbf{C})$ not containing $A$. Let $W$ be a neighborhood of $A$ in $\Omega_n$ such that $A' \not\in V$ for all $A' \in W$, and note that there is a section $\sigma\colon W \to \alpha^{-1}(W)$ defined by $\sigma(A') = (V + \mathbf{C}A', A', 0)$.

Claim 4. If $X$ is an irreducible component of $\mathcal{D}$, then either $X$ is the closure of $\alpha^{-1}(\Omega_n)$ or $X = \alpha^{-1}(0)$ and $c = n$.

We have $X = \bigcup_{z = n}^{n^2} (X \cap \alpha^{-1}(\Omega_z))$, so there is a unique $z$ such that $X \cap \alpha^{-1}(\Omega_z)$ is open in $X$. In particular, we have $\dim X \leq \dim \alpha^{-1}(\Omega_z)$. If $n < z < n^2$, then Claim 3 ii) shows that $\dim X < (n^2 - c - 1)c + n^2 + n$, contradicting Claim 1. If $z = n^2$, then one of $X \cap \alpha^{-1}(\Omega_{n^2} - \{0\})$ and $X \cap \alpha^{-1}(0)$ is open in $X$. The former case cannot occur for the same reason as above (using Claim 3 iii)), while by Claim 1 the latter case can only occur if $\dim \alpha^{-1}(0) \geq (n^2 - c - 1)c + n^2 + n$. Since $c \leq n$, Claim 3 iv) shows that in this case $c = n$, and we see $X = \alpha^{-1}(0)$. Finally, if $z = n$ then $X$ is contained in the closure of $\alpha^{-1}(\Omega_n)$, and by Claim 3 i) this containment is an equality.

Proof of Proposition. If $c < n$, then Claim 3 i) and Claim 4 show that $\mathcal{D}$ is irreducible of the claimed dimension. If $c = n$, then Claim 3 i), iv) and Claim 4 show that $\mathcal{D} = \alpha^{-1}(0) \cup \overline{\alpha^{-1}(\Omega_n)}$, with both irreducible components of the claimed dimension. QED

Now for "Exceptions" and "Stabilization", retain the notation above. Let $U$ be a codimension $c$ subspace of $M_n(\mathbf{C})$. In general, $\pi^{-1}(U)$ is of dimension $n^2 + n - c$ if and only if $\pi^{-1}(U) \cap \alpha^{-1}(\Omega_z)$ is of dimension $\leq n^2 + n - c$ for all $z$. But $\dim \pi^{-1}(U) \cap \alpha^{-1}(\Omega_z) = \dim(U \cap \Omega_z) + z$, so the question is

(*) is $\dim(U \cap \Omega_z) + z \leq n^2 + n - c$ for all $z$?

I don't know the answer to (*) for general $c$, but the answer is no if $c = n$: suppose $U$ contains the identity matrix $I$, and note that $\Omega_{n^2} - \{0\} = \mathbf{C}^\times I$, so

$\dim(U \cap \Omega_{n^2}) + n^2 = 1 + n^2 > n^2 + n - c$

and thus $\dim \pi^{-1}(U) \geq n^2 + 1$ in this case.

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  • $\begingroup$ Wow, thank you very much for such a thorough answer! While considering this problem, we were mostly interested in the case when $U$ is a subspace of $\mathfrak{sl}_n(\mathbf{C})$, so fortunately, your counterexample for Stabilization does not apply in this case. Or do you think a counterexample might still exist under these assumptions? $\endgroup$
    – darko
    Commented 13 hours ago
  • $\begingroup$ An interesting fact we haven’t yet figured out how to use is that $\mathfrak{sl}_n(\mathbf{C})$ is an irreducible representation of $GL_n(\mathbf{C})$. $\endgroup$
    – darko
    Commented 13 hours ago
  • $\begingroup$ I'm not sure. One idea is to show that if $U$ is a subspace of $\mathfrak{sl}_n(\mathbf{C})$ of codimension $c \leq n - 1$, then $U$ intersects the $\mathrm{GL}_n(\mathbf{C})$-orbit $C$ of any non-central element of $\mathfrak{sl}_n(\mathbf{C})$ transversally. I believe this would imply that $\dim(U \cap \Omega_z) = \dim(\Omega_z) - c$ for any $z < n^2$, which would be enough for "Exceptions" in view of Claim 2 and the reduction to (*). When $c = 1$, this is true simply because otherwise $C \subset U$, which fails because $C$ generates $\mathfrak{sl}_n(\mathbf{C})$ (by irreducibility). $\endgroup$
    – SeanC
    Commented 12 hours ago
  • $\begingroup$ We were also considering transversality and agree that it would be sufficient in the case $c = n-1$. In fact, we believe we found a reference that addresses the case when $z$ is roughly at most $n^2/2$ (see Proposition I.2.16). Transversality seems to hold as long as the codimension of the variety which you take the linear sections of is small enough. However, we have no idea how to handle the case $z > n^2/2$. Do you perhaps know of any general methods for proving transversality? $\endgroup$
    – darko
    Commented 11 hours ago
  • $\begingroup$ I don't follow the reference -- if you mean to take $X$ to be the projectivization of $U$, then $X$ is degenerate (it is itself a linear subspace). Feel free to email me if you want to discuss further. $\endgroup$
    – SeanC
    Commented 11 hours ago

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