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Let $A$ be an algebra over a field $F$. A derivation of $A$ is an $F$-linear map $D : A\to A$ such that $D(ab) = aD(b) + D(a)b$ for all $a, b \in A$. The map $adx : L \to L$ is inner derivation. I'm looking for some examples non-inner derivations of Lie algebras.

Stahl
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negar
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    Given any algebra over $F$, define $[\cdot,\cdot]$ to always be $0$. Then any nonzero $D$ you cook up is an example. (I'm assuming this is too trivial to be worth actually writing as an answer.) – Jason DeVito - on hiatus Sep 10 '13 at 17:50
  • I'm very inexperienced with derivations. Is the usual formal derivation on polynomials $\Bbb R[x]$ inner? (By which I mean you use the power rule $D(x^n)=nx^{n-1}$ along with linearity to define $D$.) – rschwieb Sep 10 '13 at 17:55
  • @rschwieb: What lie bracket are you putting on $\mathbb{R}[x]$? Or am I misunderstanding? – Jason DeVito - on hiatus Sep 10 '13 at 18:01
  • @JasonDeVito I just mean the polynomials over the real numbers :) I guess the Lie bracket there is trivial since they are commutative? I guess that falls under your comment then. Sorry it was the first thing that came to mind. – rschwieb Sep 10 '13 at 18:02
  • @JasonDeVito Yes, for any associative ring that is the canonical Lie bracket on it, and for commutative rings it would be trivial. That's part of my tiny body of knowledge about Lie algebra :) – rschwieb Sep 10 '13 at 18:14

4 Answers4

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The easiest example is perhaps to consider all derivations of the Heisenberg Lie Algebra $\mathfrak{h}_3(K)$, i.e., all linear maps $D\colon \mathfrak{h}_3(K) \rightarrow \mathfrak{h}_3(K)$ satisfying $D([x,y])=[D(x),y]+[x,D(y)]$ for all $x,y$. Here the brackets are given by $[e_1,e_2]=-[e_2,e_1]=e_3$, where $(e_1,e_2,e_3)$ denotes a basis. The inner derivations are of the form $ad (x)$, and are linear combinations of $$ ad (e_1)=\begin{pmatrix} 0 & 0 & 0\cr 0 & 0 & 0\cr 0 & 1 & 0\cr \end{pmatrix},\; ad (e_2)=\begin{pmatrix} 0 & 0 & 0\cr 0 & 0 & 0\cr -1 & 0 & 0\cr \end{pmatrix},\, ad(e_3)=0. $$ However, the Heisenberg Lie Algebra has many other derivations (outer derivations). In fact, all linear maps of the form $$ D=\begin{pmatrix} d_1 & d_4 & 0\cr d_2 & d_5 & 0\cr d_3 & d_6 & d_1+d_5\cr \end{pmatrix} $$ are derivations of the Heisenberg Lie algebra.

Dietrich Burde
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I'll venture an example, at the risk of being too trivial.

The zero Lie bracket makes the real polynomials $\Bbb R[x]$ into a Lie algebra, and any inner derivation with respect to this bracket would have to be uniformly zero.

But ordinary differentiation is a nonzero derivation of real polynomials, so this would furnish an example.

rschwieb
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An instructive example is to pick a field $k$ and the commutative algebra $A=k[X]/(X^n)$. Since $A$ is commutative, its only inner derivation is the zero map, so any example of derivation you can come up with will be outer.

Write down explicitly the conditions for a linear map $A\to A$ to be a derivation, and solve them :-)

Mariano Suárez-Álvarez
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I just the same calculations and i get

\begin{pmatrix} d_1 & d_4 & 0\cr d_2 & d_2 & 0\cr d_3 & d_5 & d_1 + d_2\cr \end{pmatrix}

for the derivations of $\mathfrak{h}_3$.

I'm not sure who is right :P

Ioannis Tsartsaflis
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  • I am sure that there is a typo. The first two columns are arbitrary. So the second column must be $(d_4,d_5,d_6)$ instead of $(d_4,d_2,d_5)$. – Dietrich Burde Jun 29 '17 at 14:03