湖南有什么好玩的

Question

Let $A,B$ two $n\times n$ matrices over a field $F$ and consider the usual Lie commutator $[A,B]=AB-BA$. As $[A,B]$ has trace zero, if the characteristic $p\geq 0$ of $F$ does not divide $n$ then $[A,B]$ cannot be the identity matrix. This is false when $p\vert n$, as the case $p=n=2$ easily shows.

In general, is it always possible to obtain the identity as a Lie commutator of two matrices, provided $p\vert n$?

Answer 1

This is just a community wiki answer to mark this question as answered.

As user te4 noted in a comment, in an answer to a MSE question, a simple construction of two $p\times p$ matrices $A$ and $B$ with $[A,B]=I$ over any field of characteristic $p$ was provided: $$A=\pmatrix{0&1\\&0&2\\&&\ddots&\ddots\\&&&0&p-1\\&&&&0}, B=\pmatrix{0\\1&0\\&\ddots&\ddots\\&&1&0\\&&&1&0}$$

As Najib Idrissi notes in another comment, a solution for $n=p$ provides one for $n = kp$ for any integer $k \geq 1$ via a block matrix construction.

Answer 2

Here is a conceptual approach. As Mariano did on math.SE, we start with the observation that this is equivalent to finding finite-dimensional modules over the Weyl algebra

$$A = K\langle x, \partial \rangle/([\partial, x] - 1)$$

in positive characteristic. In characteristic $0$, the Weyl algebra is simple and has an infinite-dimensional simple module given by polynomials $K[x]$, on which it acts by differential operators. Simplicity implies it has no nonzero finite-dimensional quotients and hence no nonzero finite-dimensional modules.

In positive characteristic, $K[x]$ is no longer a simple module. The proof that it's simple in characteristic $0$ goes like this: starting from a nonzero polynomial $f \in K[x]$, we can repeatedly apply the derivative $\partial$ until we get a nonzero scalar, then act on this scalar by $x$ until we get every polynomial. In characteristic $p$, repeatedly applying the derivative is no longer guaranteed to produce a nonzero scalar since we now have

$$\partial(x^p) = px^{p-1} = 0$$

and in fact the ideal $I = (x^p)$ of polynomials divisible by $x^p$ is now a maximal proper submodule of $K[x]$. So we can quotient by it, obtaining $K[x]/x^p$, which is now a $p$-dimensional simple module on which $\partial$ and $x$ act by the matrices $A$ and $B$ in Sam Hopkins' answer, as Fedor Petrov points out.

The proof that the Weyl algebra is simple in characteristic $0$ is very similar and breaks down in the same way; when we inspect how that proof breaks down we learn that in characteristic $p$ the center of the Weyl algebra is $K[x^p, \partial^p]$, and we can now produce finite-dimensional quotients of $A$ by tensoring with morphisms $K[x^p, \partial^p] \to K$. The above module comes from tensoring with evaluation at the origin, which sets $x^p = \partial^p = 0$. In general we can set $x^p = a, \partial^p = b$ for any scalars $a, b \in K$ and, as Mariano explains, we get a $p^2$-dimensional simple algebra with center $K$, which therefore must be isomorphic to $M_p(K)$, so has a unique $p$-dimensional simple module. Conversely, if $K$ is algebraically closed, the center acts by scalars on any finite-dimensional simple module by Schur's lemma, so in this case this construction actually produces all finite-dimensional simple modules.