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

The Lie group of invertible matrices GLn(C)GL_n(\mathbb{C}) acts by conjugation on its Lie algebra gln(C)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 GLn(C)GL_n(\mathbb{C}) is already non-trivial for me.

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

Ring RR is a PID, and M≅⨁i,jR/(t−λi)nijM \cong \bigoplus_{i,j} R/(t-\lambda_i)^{n_{ij}}. To compute the endomorphism ring we use HomR(R/f,R/g)≅R/g.c.d.(f,g)\text{Hom}_R(R/f,R/g) \cong R/\text{g.c.d.}(f,g). So
Hom(⨁i,jR/(t−λi)nij,⨁k,lR/(t−λk)nkl)≅⨁i,j,lR/(t−λi)g.c.d.(nij,nil)
\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
EndR(M)∗≅⨁i,j,l(R/(t−λi)g.c.d.(nij,nil))∗
\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−λ)mR/(t-\lambda)^m are truncated polynomials with non-zero constant term.

=================

=================

1 Answer
1

=================

I’m not sure your question has an easy explicit answer but I can offer some comments. Let A∈GLn(C)A \in GL_n(\mathbb{C}) be a matrix with distinct eigenvalues λ1,…,λk\lambda_1, \dots, \lambda_k of algebraic multiplicity n1,…,nkn_1, \dots, n_k.

If PP commutes with AA then both the eigenspaces and the generalized eigenspaces of AA must be PP-invariant and so PP is conjugate to a block-diagonal matrix. This implies that the stabilizer of AA will be isomorphic a subgroup of GLn1(C)×⋯×GLnk(C)GL_{n_1}(\mathbb{C}) \times \dots \times GL_{n_k}(\mathbb{C}).
If A∈GLn(C)A \in GL_n(\mathbb{C}) is already diagonal, A=diag(λ1,…,λ1⏟n1 times,…,λk,…,λk⏟nk times)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 GLn1(C)×⋯×GLnk(C)GL_{n_1}(\mathbb{C}) \times \dots \times GL_{n_k}(\mathbb{C}) (considered as a subgroup of GLn(C)GL_n(\mathbb{C}) in the obvious way). This follows from the fact that AA acts as a multiplication by a scalar on each eigenspace which commutes with everything.
If A=λI+NA = \lambda I + N where NN is nilpotent then PP commutes with AA if and only if PP commutes with NN so it is enough to understand the stabilizers of nilpotent matrices.
If AA is nilpotent and PP commutes with AA, the subspaces ker(Ai)\ker(A^i) must be PP-invariant and so PP stabilizes the flag
0=ker(A0)≤ker(A1)≤⋯≤ker(Ar)=V 0 = \ker(A^0) \leq \ker(A^1) \leq \dots \leq \ker(A^r) = V
where rr is the nilpotency index of AA. This implies that the stabilizer of AA will be conjugate to a subgroup of the group of invertible block upper triangular matrices with block sizes dimker(Ai)−dimker(Ai−1)\dim \ker(A^i) – \dim \ker(A^{i-1}).
This is where things get hairy. Assume AA is in Jordan canonical form , of nilpotency index rr and with the blocks ordered by increasing size. Denote by mim_i the number of Jordan blocks of size ii in AA. A collection vk↦vk−1↦⋯↦v1↦v0=0v_k \mapsto v_{k-1} \mapsto \dots \mapsto v_1 \mapsto v_0 = 0 will be called a (non-trivial AA-)chain if Av_i = Av_{i-1}Av_i = Av_{i-1} and v_1 \neq 0v_1 \neq 0 (so that the chain cannot be shortened). The fact that AA is in Jordan canonical form means that the standard basis vectors (e_1, \dots, e_n)(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. 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 PP will commute with AA if and only if it will map each such chain of length kk to another chain of length kk (which might not consist of vectors that are part of the standard basis). A chain of length kk is determined uniquely by the vector v_kv_k as long as v_k \in \ker(A^k) \setminus \ker(A^{k-1})v_k \in \ker(A^k) \setminus \ker(A^{k-1}). A collection of chains (v_i^j)(v_i^j) will be linearly independent if and only if the vectors v_1^jv_1^j are linearly independent and thus we arrive at the following description of the stabilizer of AA: For each 1 \leq i \leq r1 \leq i \leq r choose m_im_i vectors v_{i}^{i,j} \in \ker(A^i)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} 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 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 PP we obtain a matrix that commutes with AA. 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 44 \times 4 nilpotent matrices:

Assume
A = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}. 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 AA are e_1 \mapsto 0e_1 \mapsto 0 and e_4 \mapsto e_3 \mapsto e_2 \mapsto 0e_4 \mapsto e_3 \mapsto e_2 \mapsto 0. Thus, to get a matrix that commutes with AA we need to choose a vector v_3^{3,1} = ae_1 + be_2 + ce_3 + de_4 \in \ker(A^3)v_3^{3,1} = ae_1 + be_2 + ce_3 + de_4 \in \ker(A^3) that will generate a 3 \times 33 \times 3 block, a vector v_1^{1,1} = xe_1 + ye_2 \in \ker(A)v_1^{1,1} = xe_1 + ye_2 \in \ker(A) that will generate a 1 \times 11 \times 1 block such that v_1^{1,1},A^2v_3^{3,1}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}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 AA 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 \} \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\dim \ker(A) + \dim \ker(A^3) = 6 dimensional subgroup of GL_4(\mathbb{C})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}^4.0 \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.
Assume
A = \begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}.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 AA are e_1 \mapsto 0e_1 \mapsto 0, e_2 \mapsto 0e_2 \mapsto 0 and e_4 \mapsto e_3 \mapsto 0e_4 \mapsto e_3 \mapsto 0. Thus, to get a matrix that commutes with AA we need to choose a vector v_2^{2,1} = ae_1 + be_2 + ce_3 + de_4 \in \ker(A^2)v_2^{2,1} = ae_1 + be_2 + ce_3 + de_4 \in \ker(A^2) that will generate a 2 \times 22 \times 2 block and two vectors v_1^{1,1} = xe_1 + ye_2 + ze_3 \in \ker(A)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)v_1^{1,2} = ue_1 + ve_2 + we_3 \in \ker(A) that will generate two 1 \times 11 \times 1 blocks such that v_1^{1,1}, v_1^{1,2}, v_2^{2,1}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}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 AA 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 \} \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) = 102\dim \ker(A) + \dim \ker(A^2) = 10 dimensional subgroup of GL_4(\mathbb{C})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}^40 \leq \ker(A) = \operatorname{span} \{ e_1, e_2, e_3 \} \leq \ker(A^2) = \mathbb{C}^4.
Assume
A = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}.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 AA are e_2 \mapsto e_1 \mapsto 0e_2 \mapsto e_1 \mapsto 0 and e_4 \mapsto e_3 \mapsto 0e_4 \mapsto e_3 \mapsto 0. Thus, to get a matrix that commutes with AA we need to choose two vectors v_2^{2,1} = ae_1 + be_2 + ce_3 + de_4 \in \ker(A^2)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_4v_2^{2,2} = we_1 + xe_2 + ye_3 + ze_4 that will generate 2 \times 22 \times 2 blocks such that v_2^{2,1}, v_2^{2,2}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}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 AA 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 \} \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)2 \dim \ker(A^2)-dimensional subgroup of GL_4(\mathbb{C})GL_4(\mathbb{C}). By using the ordered basis (e_1,e_3,e_2,e_4)(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}^40 \leq ker(A) = \operatorname{span} \{ e_1, e_3 \} \leq \ker(A^2) = \mathbb{C}^4, we see that \operatorname{stab}(A)\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 \}. \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 \}.
Finally, assume that
A = \begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 &0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix} 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 AA is a single n \times nn \times n Jordan block. Since we only have a single chain e_4 \mapsto \dots \mapsto e_1 \mapsto 0e_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)v_4^{4,1} = ae_1 + be_2 + ce_3 + de_4 \in \ker(A^4) that will generate a 4 \times 44 \times 4 block such that A^3v_4^{4,1} = de_1A^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 \} \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\dim \ker(A^4) = 4-dimensional subgroup of \operatorname{GL}_4(\mathbb{C})\operatorname{GL}_4(\mathbb{C}) (the minimum possible for a stabilizer of a 4 \times 44 \times 4 matrix).

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

  

 

Thanks for your answer! I also was thinking about this and added an update to my post. It looks like my update is a way to reduce your case 5 to your case 3 and 4.
– Alex
yesterday