It is well known that for a matrix $A$ in $\mathfrak{sl}_n(\mathbb{C})$, we have the following equivalence: $$\dim Z(A) \text{ is minimal} \leftrightarrow A \text{ is cyclic}$$ where $Z(A)$ is the centralizer of $A$ (elements commuting with $A$). The minimal dimension is the rank of the Lie algebra, so $n-1$ in this case. Cyclic means that there is a vector generating the whole vector space under the action of $A$.

My question is whether there is a "doubled" version of this: Is it true that for a pair of commuting matrices $(A,B)$ in $\mathfrak{sl}_n(\mathbb{C})$ we have $$\dim Z(A,B) \text{ is minimal} \leftrightarrow (A,B) \text{ is cyclic}$$ where $Z(A,B)=Z(A)\cap Z(B)$ is the common centralizer (elements commuting with $A$ and $B$)? Here the centralizer is meant in $\mathfrak{sl}_n$, and minimal means minimal among commuting pairs. Remark that the minimal dimension of the centralizer of a commuting pair is also the rank ($n-1$ here) of the Lie algebra.

It is not so difficult to show that $(A,B)$ cyclic implies that the centralizer is of minimal dimension. For the direct implication, in the case of one matrix, one can use the Frobenius decomposition, itself based on the invariant factor decomposition which only works over euclidean rings. For one matrix, we have the ring $\mathbb{C}[x]$ of polynomials in one variable, which is euclidean. But for the doubled version, we need $\mathbb{C}[x,y]$ which is not euclidean (two commuting matrices gives the vector space the structure of a $\mathbb{C}[x,y]$-module). Is there another argument for the direct implication?

Edit: $(A,B)$ is cyclic means that $\mathbb{C}^n$, as module over the subalgebra generated by $\{A,B\}$, is generated by a single element. For commuting $A,B$, this means here that there exists $v\in\mathbb{C}^n$ such that $\mathbb{C}[A,B]v=\mathbb{C}^n$.


2 Answers 2


Edit: I've just realised that I didn't read your post properly. I took "$Z(A, B)$ is cyclic" to mean that the subalgebra generated by $A$, $B$, is a cyclic $k$-algebra, i.e. isomorphic to $k[x]/I$ for some ideal $I$. Your question is obviously quite different, but I'll leave this answer up for interest.

You are asking if all maximal abelian subalgebra (edit: of dimension = the rank of ${\mathfrak g}$) are cyclic. Probably the easiest counterexample is (edit: not) the following one in ${\mathfrak{gl}}_4$:

$$\left\{ \begin{pmatrix} a & b & c & d \\ 0 & a & 0 & c \\ 0 & 0 & a & b \\ 0 & 0 & 0 & a \end{pmatrix}\right\}.$$

The key to this is the pair of commuting nilpotent elements $$\begin{pmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}.$$ This is a principal nilpotent pair, as defined by Ginzburg (reference to follow). We obtain a maximal abelian subalgebra (always of dimension $n$) for each principal nilpotent pair; up to conjugacy, there is one principal nilpotent pair for each partition of $n$ (where $\lambda$ and $\lambda^t$ produce the same pair up to reordering); the corresponding subalgebra is only cyclic (in my sense) if $\lambda$ or $\lambda^t$ is the trivial partition (in which case one of the elements of the pair equals zero).

Edit: all the principal nilpotent pairs $(A,B)$ are cyclic in your sense.

  • $\begingroup$ Indeed you misunderstood the question. With $(A,B)$ cyclic, I mean that there exists a vector v such that the span of $P(A,B)v$ where $P$ runs over all polynomials gives $\mathbb{C}^n$. In the case with only one matrix, there is a theorem, not so difficult (see for instance book of Steinberg on conjugacy classes in Lie algebras) stating that for $g \in \mathfrak{sl}_n$ the following are equivalent: $$g \text{ is regular} \leftrightarrow g \text{ is cyclic} \leftrightarrow \text{minimal polynomial = characteristic polynomial} \leftrightarrow \dim Z(A) \text{ is minimal}.$$ $\endgroup$
    – AThomas
    Sep 26, 2019 at 7:58

I just found a counter-example for $\mathfrak{sl}_3$. Take $A= \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 0 & 1 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix}$.

You can check that the commun centralizer is of dimension 2, but the couple $(A,B)$ does not admit any cyclic vector.

  • $\begingroup$ Please accept your own answer so that the question is not bumped as unanswered. $\endgroup$
    – YCor
    Oct 27, 2019 at 11:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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