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I'm just studying Lie algebras. If $A$ is a $k$-algebra (not necessarily Lie or associative, just a bilinear law), it is straightforward to check that any derivation algebra of $A$ is a Lie algebra. I suppose that the converse is not true, but I can't find a counterexample.

  1. Is there any example of a Lie algebra which is not isomorphic to the derivation algebra of any $k$-algebra (again, not necessarily associative or unital)?

  2. Is there any example of a Lie algebra which is not isomorphic to the derivation algebra of any Lie algebra?

For the the second question the user YCor gives a positive answer. However, I am more interested in the first (and actually my original) question. Thank you!

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    $\begingroup$ en.wikipedia.org/wiki/Ado%27s_theorem $\endgroup$
    – Sam Hopkins
    Commented Oct 14, 2021 at 17:52
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    $\begingroup$ Is "$k$-algebra" meant to be "associative unital $k$-algebra"? $\endgroup$
    – YCor
    Commented Oct 14, 2021 at 18:37
  • $\begingroup$ No, for me a $k$-algebra $A$ is only a $k$-vector space endowed with a bilinear map $A \times A \rightarrow A$. I should have mentioned it in the question. $\endgroup$
    – cos_dm_math21
    Commented Oct 14, 2021 at 18:41
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    $\begingroup$ With so much flexibility I would be tempted to believe this might be false, i.e., it might be true that every Lie algebra is isomorphic to some derivation algebra, although it might be quite tricky to prove. $\endgroup$
    – YCor
    Commented Oct 14, 2021 at 18:51

2 Answers 2

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$\newcommand{\r}{\mathfrak{r}}\newcommand{\h}{\mathfrak{h}}\newcommand{\g}{\mathfrak{g}}\newcommand{\a}{\mathfrak{a}}$The 2-dimensional abelian Lie algebra $\a_2$ is not isomorphic to the derivation Lie algebra of any Lie algebra.

Suppose otherwise. Let $\g$ be such a Lie algebra. We discuss according to whether the inner derivation algebra (that is $\g$ modulo its center) has dimension $d$ equal to 2, 1, or 0.

  • if $d=2$, $\g$ modulo its center has dimension 2. So the derived subalgebra has dimension 1. If $[\g,\g]$ is contained in the center we deduce that $\g$ is 2-step-nilpotent, and isomorphic to the direct product of the Heisenberg Lie algebra $\h_3$ with some abelian Lie algebra. In this case, the derivation algebra is much larger (for $\h_3$ it is 6-dimensional). Otherwise $\g$ is the direct product of the two-dimensional non-abelian Lie algebra $\r$ with an abelian Lie algebra. But then its derivation algebra contains $\r$, hence is not abelian.

  • $d=1$ is impossible, because the center in a Lie algebra can't be of codimension 1 (for the same reason the quotient of a group by its center can't be cyclic)

  • $d=0$ means that $\g$ is abelian. But then its derivation algebra is either infinite, or has dimension some square.

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    $\begingroup$ Isn't the OP asking about derivations of an associative algebra? $\endgroup$
    – Sam Hopkins
    Commented Oct 14, 2021 at 18:13
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    $\begingroup$ Ah indeed. The word "associative" would have been welcome earlier :) especially as the OP has only used the lie-algebra tag. $\endgroup$
    – YCor
    Commented Oct 14, 2021 at 18:15
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    $\begingroup$ So Ado's theorem answers the question in the affirmative, no? (Sorry, "affirmative" here is ambiguous: I mean it shows every finite-dimensional Lie algebra is realized as derivations of an associative algebra.) $\endgroup$
    – Sam Hopkins
    Commented Oct 14, 2021 at 18:20
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    $\begingroup$ I guess there is actually an issue of whether the derivations of the matrix algebra in question are all inner (see math.stackexchange.com/questions/4205733/… for a related question) $\endgroup$
    – Sam Hopkins
    Commented Oct 14, 2021 at 18:37
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    $\begingroup$ @SamHopkins but this would work equally for Lie algebras. I understand now the question as whether every $\mathfrak{g}$ is isomorphic to $\mathrm{Der}(A)$ for some unital associative algebra $A$ [or maybe just associative, or even arbitrary, OP should clarify]. This sounds quite clear when OP says "quite often defined as derivation algebras". A given Lie algebra such as $\mathfrak{gl}_n$ does not "define" all its subalgebras. $\endgroup$
    – YCor
    Commented Oct 14, 2021 at 18:41
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If $A$ is an arbitrary algebra over a field of characteristic $p>0$, then its derivation algebra is a restricted Lie algebra under the usual Lie bracket $[D_1,D_2]=D_1D_2-D_2D_1$ and the ordinary $p$-exponentation. This shows that, in positive characteristic, a non-restrictable Lie algebra cannot be the derivation algebra of an algebra of any kind.

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  • $\begingroup$ Nice. What's an example of non-restrictable Lie algebra (say, over an algebraically closed field of char $p>0$)? $\endgroup$
    – YCor
    Commented Nov 26, 2021 at 15:09
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    $\begingroup$ Consider the $(p+2)$-dimensional Lie algebra $L=Fx+Fy_1+\cdots+Fy_{p+1}$ with structure constants given by $[x,y_i]=y_{i+1}$ for $i=1,\ldots,p$, $[x,y_{p+1}]=0$, and $[y_i,y_j]=0$ for all $i,j=1,\ldots,p+1$. As $(\mathrm{ad}x)^p$ is not an inner derivation, it follows that $L$ cannot be restrictable. $\endgroup$
    – Salvatore Siciliano
    Commented Nov 26, 2021 at 18:41
  • $\begingroup$ By the way, I should mention, in char $p>0$, the following 3-dimensional non-restrictable Lie algebra: for a nonzero scalar $c$, the Lie algebra $\mathfrak{g}(c)$ with basis $(X,Y,Z)$ and bracket $[X,Y]=Y$, $[X,Z]=cZ$, $[Y,Z]=0$. Then $\mathfrak{c}$ is restrictable if and only if $c^p=c$ (i.e., if and only if $c$ belongs to the prime subfield $\mathbf{F}_p$ (indeed otherwise $\mathrm{ad}(X)^p$ is not an inner derivation). $\endgroup$
    – YCor
    Commented Dec 22, 2022 at 10:18
  • $\begingroup$ @Yves. Sure, this is a nice example as well! $\endgroup$
    – Salvatore Siciliano
    Commented Dec 24, 2022 at 17:01

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