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Let $A = \mathcal M_n(k)$ be the full matrix algebra over a field $k$. If $\phi:A\to A$ is a nonzero endomorphism of $A$ as a Lie algebra, must it automatically be an endomorphism of $A$ as a unital $k$-algebra?

If not, what would be necessary conditions?

EDIT As TorstenSchoeneberg pointed out in the comments, a necessary condition is that $\phi$ is the identity on $k$. Could this also be a sufficient condition?

doetoe
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    If $\phi$ is an endormorphism of $M_n(k)$ then $x\mapsto \phi(x)+tr(x) I$ is an endomorphism of the Lie algebra but not of $M_n(k)$. – reuns Jan 28 '19 at 01:01
  • @reuns of course, thanks! – doetoe Jan 28 '19 at 06:39
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    I think automorphisms of $\mathfrak{sl}_n$ indeed do extend uniquely to automorphisms of the unital algebra $M_n$. The problem, as can be seen in reuns' construction, is the centre of the latter, i.e. the scalar multiples of $I$, on which a unital automorphism must be the identity, whereas a Lie algebra endomorphism can be any scalar multiplication. – Torsten Schoeneberg Jan 28 '19 at 07:20
  • @TorstenSchoeneberg Thanks! Do you know if that is a sufficient condition ($\phi$ preserving $k$, the center of $M_n(k)$)? – doetoe Jan 28 '19 at 08:40
  • @TorstenSchoeneberg I edited the question to incorporate your observation. – doetoe Jan 28 '19 at 08:45
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    @reuns: I guess to preserve $k$-linearity we need $\frac1n \cdot tr$ instead of just $tr$. -- doetoe: Which leads me to $\phi:= \frac1n tr$ as a counterexample to the sufficiency hypothesis. Also, I have to admit I think the first sentence in my earlier comment is wrong in general ($\phi(X) = -X^t$ for $n \ge 3$?); but I still think the question can be settled along those lines. Thinking about it. – Torsten Schoeneberg Jan 28 '19 at 18:02

1 Answers1

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Short answer: This is true (at least in characteristic zero) if and only if $\phi$ is given by conjugation with some matrix.

Let $k$ be a field of characteristic $0$, and let's call $\mathfrak{gl}_n(k)$ the Lie algebra, but $M_n(k)$ the associative unital algebra of $n \times n$-matrices over $k$ (i.e. it's the same set and even $k$-vector space, but the multiplicative structure is different.) Let $\phi$ be an endomorphism of $\mathfrak{gl}_n(k)$.

With $\mathfrak{sl}_n(k)$ denoting those matrices in $\mathfrak{gl}_n(k)$ of trace $0$, observe that we have a decomposition

$$\mathfrak{gl}_n(k) \simeq \mathfrak{sl}_n(k) \oplus k \cdot Id$$ given by $x \mapsto (x - \frac1n tr(x), \frac1n tr(x))$. This is a decomposition of $k$-vector spaces and even of Lie algebras, but (for $n \ge 2$) not of associative algebras.

Because $\mathfrak{sl}_n(k)$ is the derived Lie algebra, and $k \cdot Id$ is the centre of $\mathfrak{gl}_n(k)$, both spaces are actually invariant under $\phi$, and it suffices to investigate the restrictions $\phi_{\mathfrak{sl}_n(k)}$ and $\phi_{k\cdot Id}$. As discussed in the comments, $\phi_{k\cdot Id}$ can be any scalar multiplication, but for $\phi$ to be an endomorphism of $M_n(k)$, it necessarily has to be the identity, so let's assume that from now on.

Now since $\mathfrak{sl}_n(k)$ is simple, we either have $\phi_{\mathfrak{sl}_n(k)} = 0$, or $\phi_{\mathfrak{sl}_n(k)}$ is an automorphism.

  • If $\phi_{\mathfrak{sl}_n(k)} =0$, i.e. $\phi = \frac1n tr$ is just the projection on the second summand, $\phi$ is not an endomorphism of $M_n(k)$ for $n \ge 2$ (because e.g. there are matrices in $\mathfrak{sl}_n(k)$ whose associative product has non-zero trace).

  • If $\phi_{\mathfrak{sl}_n(k)}$ is an automorphism, it is either an inner or an outer one.

    • If it is an inner one, by definition it is given by conjugation with some $g \in GL_n(k)$, and hence obviously is also an automorphism of $M_n(k)$ given by $\phi(x) = gxg^{-1}$.

    • If $\phi_{\mathfrak{sl}_n(k)}$ is an outer automorphism, we have $n \ge 3$ and up to conjugation like above we have $\phi_{\mathfrak{sl}_n(k)}$ being the negative of the transpose: $\phi_{\mathfrak{sl}_n(k)}(x) = -x^t$. But e.g. $E_{12}\cdot E_{23} = E_{13}$ whereas $-E_{21}\cdot -E_{31} = 0$, so this has no chance to be multiplicative even on the traceless matrices.

Torsten Schoeneberg
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  • Thanks! Is it easy to see that the only outer automorphism up to conjugation is the negative transpose? – doetoe Jan 29 '19 at 09:19
  • It's kind of a standard result, cf. https://math.stackexchange.com/q/1865062/96384, https://mathoverflow.net/q/14735/27465. A proof from scratch might be non-trivial. – Torsten Schoeneberg Jan 29 '19 at 17:28