3
$\begingroup$

Let $L$ be a finite-dimensional Lie algebra over a field of characteristic zero. It is not difficult to see (and also follows from Theorem 4.4 of [G. Hochschild: Semi-simple algebras and generalized derivations, Amer. J. Math. 64, (1942), 677–694]) that the derivation algebra $D(L)$ of $L$ is simple if and only if so is $L$.

Now suppose that $L$ is defined over an algebraically closed field $\mathbb{F}$ of positive characteristic. If $D(L)$ is simple then all derivations of $L$ are inner (in particular, $L$ is restricted), and $L$ is either simple or a central extension of a simple Lie algebra. It is clear that such an extension cannot be split. Thus my question is the following:

Suppose that $L$ is a non-split finite-dimensional central extension of a restricted simple Lie algebra $\mathfrak{g}$ over a field $\mathbb{F}$ of characteristic $p>0$. Is it possible that every derivation of $L$ is inner?

Improve this question
$\endgroup$
5
  • $\begingroup$ This is a somewhat tricky question, since it inevitably involves the classification of finite dimensional simple Lie algebras (restricted or not) in prime characteristic. Presumably some case-by-case work is needed, using the results obtained over the years by a number of people for $p>3$ (Block-Wilson, Premet-Strade especially). For $p=2,3$ much less is known about the classification. Even though the book by G.B. Seligman Modular Lie Algebras (Springer, 1967) is now half a century old, you might look at his survey V.5 on derivations, along with the references. $\endgroup$
    – Jim Humphreys
    Commented Mar 8, 2017 at 20:28
  • $\begingroup$ P.S. It seems the only interesting case with $\mathfrak{g}$ coming from a "classical" root system of type A-G for $p>3$ is the quotient of a special linear algebra $\mathfrak{sl}_n$ by the scalars when $p|n$. Here the full general linear Lie algebra of $n \times n$ matrices acts on the special linear algebra by outer derivations. But there are also restricted simple Lie algebras of "Cartan type" to consider, not closely connected with algebraic groups. $\endgroup$
    – Jim Humphreys
    Commented Mar 8, 2017 at 20:37
  • $\begingroup$ @Jim Humphreys: Thank you very much for your comments. Indeed, if $\mathfrak{g}$ is classical then $H^2(\mathfrak{g}, \mathbb{F})=0$ and so any central extension is split. This is also the case when $\mathfrak{g}\cong W(n)$ $(n>1)$, $\mathfrak{g}\cong K(n)$ $(n\not\equiv 3 \,\rm{mod} \,p)$, or $\mathfrak{g}$ is the Melikian algebra in characteristic 5. $\endgroup$
    – Salvatore Siciliano
    Commented Mar 9, 2017 at 10:14
  • $\begingroup$ @Jim Humphreys: Suppose $\mathfrak{g}$ is a perfect and centreless Lie (super)algebra over a ring. Then the derivation algebra of $\mathfrak{g}$ is canonically isomorphic to the derivation algebra of its universal central extension, under an isomorphism that preserves inner derivations. For more general coverings there are conditions for lifting, but the inner derivations always lift. For details see Theorem 2.6 of my 2001 paper in universal central extensions of Lie superalgebras $\endgroup$
    – Erhard Neher
    Commented Mar 10, 2017 at 1:34
  • $\begingroup$ @Erhard Neher: One can easily see that $L$ must be perfect when $D(L)$ is simple. If I could prove that $L$ is the universal central extension of $\mathfrak{g}$ then I would be done. $\endgroup$
    – Salvatore Siciliano
    Commented Mar 12, 2017 at 17:16

1 Answer 1

Reset to default
4
$\begingroup$

I realized that it is indeed possible. Consider e.g. the restricted simple Lie algebra $W(1)$ over a field $\mathbb{F}$ of characteristic $p>3$ and let $\hat{W}(1)$ denote its universal central extension. As $H^2(W(1), \mathbb{F})\neq 0$, $\hat{W}(1)$ is a non-trivial central extension of $W(1)$. Moreover, as all derivations of $W(1)$ are inner, denoted by $Z(\hat{W}(1))$ the center of $\hat{W}(1)$, one has

$$D(\hat{W}(1))\cong D(W(1))\cong W(1)\cong \frac{\hat{W}(1)}{Z(\hat{W}(1))},$$

thus all derivations of $\hat{W}(1)$ are inner.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .