Stabilizers of the adjoint representation of $GL_n(\mathbb{C})$

Question

The Lie group of invertible matrices $GL_n(\mathbb{C})$ acts by conjugation on its Lie algebra $gl_n(\mathbb{C})$ of all matrices. Orbits of this action are in bijection with canonical forms of matrices: rational canonical forms or Jordan forms. But how one can compute stabilizers of this action? In concrete terms I want to find all invertible matrices commuting with a given matrix in its canonical form.

Of course, this question has a general version for any Lie group and adjoint representation on its Lie algebra. Perhaps, it is possible to solve it in this generality, but the case of $GL_n(\mathbb{C})$ is already non-trivial for me.

Update: Perhaps, it is easier to understand in terms of modules over $R=\mathbb{C}[t]$. Given an $n \times n$ matrix $A$ we can consider $M = \mathbb{C}^n$ as a $\mathbb{C}[t]$ module on length $n$. Matrices commuting with $A$ are $\text{End}_R(M)$, invertible matrices are invertible elements of this ring $\text{End}_R(M)^*$.

Ring $R$ is a PID, and $M \cong \bigoplus_{i,j} R/(t-\lambda_i)^{n_{ij}}$. To compute the endomorphism ring we use $\text{Hom}_R(R/f,R/g) \cong R/\text{g.c.d.}(f,g)$. So $$ \text{Hom}(\bigoplus_{i,j} R/(t-\lambda_i)^{n_{ij}}, \bigoplus_{k,l} R/(t-\lambda_k)^{n_{kl}}) \cong \bigoplus_{i,j,l} R/(t-\lambda_i)^{\text{g.c.d.}(n_{ij}, n_{il})} $$ Then $$ \text{End}_R(M)^* \cong \bigoplus_{i,j,l} (R/(t-\lambda_i)^{\text{g.c.d.}(n_{ij}, n_{il})})^* $$ Invertible elements of $R/(t-\lambda)^m$ are truncated polynomials with non-zero constant term.

Answer 1

I'm not sure your question has an easy explicit answer but I can offer some comments. Let $A \in GL_n(\mathbb{C})$ be a matrix with distinct eigenvalues $\lambda_1, \dots, \lambda_k$ of algebraic multiplicity $n_1, \dots, n_k$.

1. If $P$ commutes with $A$ then both the eigenspaces and the generalized eigenspaces of $A$ must be $P$-invariant and so $P$ is conjugate to a block-diagonal matrix. This implies that the stabilizer of $A$ will be isomorphic a subgroup of $GL_{n_1}(\mathbb{C}) \times \dots \times GL_{n_k}(\mathbb{C})$.
2. If $A \in GL_n(\mathbb{C})$ is already diagonal, $$A = \operatorname{diag}(\underbrace{\lambda_1, \dots, \lambda_1}_{n_1 \text{ times}}, \dots, \underbrace{\lambda_k, \dots, \lambda_k}_{n_k \text{ times}})$$ then the stabilizer is the whole of $GL_{n_1}(\mathbb{C}) \times \dots \times GL_{n_k}(\mathbb{C})$ (considered as a subgroup of $GL_n(\mathbb{C})$ in the obvious way). This follows from the fact that $A$ acts as a multiplication by a scalar on each eigenspace which commutes with everything.
3. If $A = \lambda I + N$ where $N$ is nilpotent then $P$ commutes with $A$ if and only if $P$ commutes with $N$ so it is enough to understand the stabilizers of nilpotent matrices.
4. If $A$ is nilpotent and $P$ commutes with $A$, the subspaces $\ker(A^i)$ must be $P$-invariant and so $P$ stabilizes the flag $$ 0 = \ker(A^0) \leq \ker(A^1) \leq \dots \leq \ker(A^r) = V$$ where $r$ is the nilpotency index of $A$. This implies that the stabilizer of $A$ will be conjugate to a subgroup of the group of invertible block upper triangular matrices with block sizes $\dim \ker(A^i) - \dim \ker(A^{i-1})$.
5. This is where things get hairy. Assume $A$ is in Jordan canonical form , of nilpotency index $r$ and with the blocks ordered by increasing size. Denote by $m_i$ the number of Jordan blocks of size $i$ in $A$. A collection $v_k \mapsto v_{k-1} \mapsto \dots \mapsto v_1 \mapsto v_0 = 0$ will be called a (non-trivial $A$-)chain if $Av_i = Av_{i-1}$ and $v_1 \neq 0$ (so that the chain cannot be shortened). The fact that $A$ is in Jordan canonical form means that the standard basis vectors $(e_1, \dots, e_n)$ split into a disjoint union of chains $$ e_1 \mapsto 0, \\ \vdots \\ e_{m_1} \mapsto 0, \\ e_{m_1 + 2} \mapsto e_{m_1 + 1} \mapsto 0, \\ \vdots \\ e_{n-r+1} \mapsto e_{n - r + 2} \mapsto \dots \mapsto e_n \mapsto 0. $$ and each chain contributes a single block of the appropriate size to the Jordan form. An invertible matrix $P$ will commute with $A$ if and only if it will map each such chain of length $k$ to another chain of length $k$ (which might not consist of vectors that are part of the standard basis). A chain of length $k$ is determined uniquely by the vector $v_k$ as long as $v_k \in \ker(A^k) \setminus \ker(A^{k-1})$. A collection of chains $(v_i^j)$ will be linearly independent if and only if the vectors $v_1^j$ are linearly independent and thus we arrive at the following description of the stabilizer of $A$: For each $1 \leq i \leq r$ choose $m_i$ vectors $v_{i}^{i,j} \in \ker(A^i)$ such that the vectors $$ v_1^{1,1}, \dots, v_1^{1,m_1}, Av_2^{2,1}, \dots, Av_2^{2,m_2}, \dots, A^{r-1}v_r^{r,1}, \dots, A^{r-1}v_r^{r,m_r} $$ are linearly independent. Each such vector defines a chain $$ v_{i}^{i,j} \mapsto Av_{i}^{i,j} = v_{i-1}^{i,j} \mapsto \dots \mapsto A^{i - 1} v_{i}^{i,j} = v_1^{i,j} \mapsto A^{i} v_{i}^{i,j} = 0$$ and by placing the vectors as columns into an invertible matrix $P$ we obtain a matrix that commutes with $A$. All invertible commuting matrices are obtained in this way.

The description above is somewhat technical but allows you in practice to calculate easily the stabilizer explicitly. You can also deduce the dimension of the stabilizer. I'll demonstrate this for $4 \times 4$ nilpotent matrices:

1. Assume $$ A = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}. $$ The corresponding chains of $A$ are $e_1 \mapsto 0$ and $e_4 \mapsto e_3 \mapsto e_2 \mapsto 0$. Thus, to get a matrix that commutes with $A$ we need to choose a vector $v_3^{3,1} = ae_1 + be_2 + ce_3 + de_4 \in \ker(A^3)$ that will generate a $3 \times 3$ block, a vector $v_1^{1,1} = xe_1 + ye_2 \in \ker(A)$ that will generate a $1 \times 1$ block such that $v_1^{1,1},A^2v_3^{3,1}$ are linearly independent. Then the matrix defined by $Pe_1 = v_1^{1,1}, Pe_4 = v_3^{3,1}, Pe_3 = Av_3^{3,1}, Pe_2 = A^2v_3^{3,1}$ will commute with $A$ and any such matrix will be of this form. Thus, $$ \operatorname{stab}(A) = \left \{ \begin{pmatrix} x & 0 & 0 & a \\ y & d & c & b \\ 0 & 0 & d & c \\ 0 & 0 & 0 & d \end{pmatrix} : xd \neq 0 \right \} $$ which is a $\dim \ker(A) + \dim \ker(A^3) = 6$ dimensional subgroup of $GL_4(\mathbb{C})$. You can also see the block structure forced by the stabilizing of the flag $$0 \leq \ker(A) = \operatorname{span} \{ e_1, e_2 \} \leq \ker(A^2) = \operatorname{span} \{ e_1, e_2, e_3 \} \leq \ker(A^3) = \mathbb{C}^4.$$
2. Assume $$A = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}.$$ The corresponding chains of $A$ are $e_1 \mapsto 0$, $e_2 \mapsto 0$ and $e_4 \mapsto e_3 \mapsto 0$. Thus, to get a matrix that commutes with $A$ we need to choose a vector $v_2^{2,1} = ae_1 + be_2 + ce_3 + de_4 \in \ker(A^2)$ that will generate a $2 \times 2$ block and two vectors $v_1^{1,1} = xe_1 + ye_2 + ze_3 \in \ker(A)$ and $v_1^{1,2} = ue_1 + ve_2 + we_3 \in \ker(A)$ that will generate two $1 \times 1$ blocks such that $v_1^{1,1}, v_1^{1,2}, v_2^{2,1}$ are linearly independent. Then the matrix defined by $Pe_1 = v_1^{1,1}, Pe_2 = v_1^{1,2}, Pe_4 = v_2^{2,1}, Pe_3 = Av_2^{2,1}$ will commute with $A$ and any such matrix will be of this form. Thus, $$ \operatorname{stab}(A) = \left \{ \begin{pmatrix} x & u & 0 & a \\ y & v & 0 & b \\ z & w & d & c \\ 0 & 0 & 0 & d \end{pmatrix} : d(yw - vz) \neq 0 \right \} $$ which is a $2\dim \ker(A) + \dim \ker(A^2) = 10$ dimensional subgroup of $GL_4(\mathbb{C})$. You can also see the block structure forced by the stabilizing of the flag $0 \leq \ker(A) = \operatorname{span} \{ e_1, e_2, e_3 \} \leq \ker(A^2) = \mathbb{C}^4$.
3. Assume $$A = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}.$$ The corresponding chains of $A$ are $e_2 \mapsto e_1 \mapsto 0$ and $e_4 \mapsto e_3 \mapsto 0$. Thus, to get a matrix that commutes with $A$ we need to choose two vectors $v_2^{2,1} = ae_1 + be_2 + ce_3 + de_4 \in \ker(A^2)$ and $v_2^{2,2} = we_1 + xe_2 + ye_3 + ze_4$ that will generate $2 \times 2$ blocks such that $v_2^{2,1}, v_2^{2,2}$ are linearly independent. Then the matrix defined by $Pe_2 = v_2^{2,1}, Pe_1 = Av_2^{2,1}, Pe_4 = v_2^{2,2}, Pe_3 = Av_2^{2,2}$ will commute with $A$ and any such matrix will be of this form. Thus, $$ \operatorname{stab}(A) = \left \{ \begin{pmatrix} b & a & x & w \\ 0 & b & 0 & x \\ 0 & c & 0 & y \\ d & d & z & z \end{pmatrix} : bz - xd \neq 0 \right \} $$ which is a $2 \dim \ker(A^2)$-dimensional subgroup of $GL_4(\mathbb{C})$. By using the ordered basis $(e_1,e_3,e_2,e_4)$ which is more adapted to the flag $0 \leq ker(A) = \operatorname{span} \{ e_1, e_3 \} \leq \ker(A^2) = \mathbb{C}^4$, we see that $\operatorname{stab}(A)$ is conjugate to the subgroup $$ \left \{ \begin{pmatrix} b & x & a & w \\ d & z & c & y \\ 0 & 0 & b & x \\ 0 & 0 & d & z \end{pmatrix} : bz - xd \neq 0 \right \} = \left \{ \begin{pmatrix} A_{2 \times 2} & B_{2 \times 2} \\ 0_{2 \times 2} & A_{2 \times 2} \end{pmatrix} : A \in GL_2(\mathbb{C}), B \in M_2(\mathbb{C})) \right \}. $$
4. Finally, assume that $$ A = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 &0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix} $$ consists of a single Jordan block. What follows generalizes easily to the case where $A$ is a single $n \times n$ Jordan block. Since we only have a single chain $e_4 \mapsto \dots \mapsto e_1 \mapsto 0$, we need to find a single vector $v_4^{4,1} = ae_1 + be_2 + ce_3 + de_4 \in \ker(A^4)$ that will generate a $4 \times 4$ block such that $A^3v_4^{4,1} = de_1$ will be linearly independent. Thus, we get $$ \operatorname{stab}(A) = \left \{ \begin{pmatrix} d & c & b & a \\ 0 & d & c & b \\ 0 & 0 & d & c \\ 0 & 0 & 0 & d \end{pmatrix} : d \neq 0 \right \} = \left \{ g(A) \, | \, g \in \mathbb{C}[X], g(A) \text{ is invertible} \right \}$$ which is a $\dim \ker(A^4) = 4$-dimensional subgroup of $\operatorname{GL}_4(\mathbb{C})$ (the minimum possible for a stabilizer of a $4 \times 4$ matrix).

I can't see how much more can be said in general.