# The Unapologetic Mathematician

## Derivations

When first defining (or, rather, recalling the definition of) Lie algebras I mentioned that the bracket makes each element of a Lie algebra $L$ act by derivations on $L$ itself. We can actually say a bit more about this.

First off, we need an algebra $A$ over a field $\mathbb{F}$. This doesn’t have to be associative, as our algebras commonly are; all we need is a bilinear map $A\otimes A\to A$. In particular, Lie algebras count.

Now, a derivation $\delta$ of $A$ is firstly a linear map from $A$ back to itself. That is, $\delta\in\mathrm{End}_\mathbb{F}(A)$, where this is the algebra of endomorphisms of $A$ as a vector space over $\mathbb{F}$, not the endomorphisms as an algebra. Instead of preserving the multiplication, we impose the condition that $\delta$ behave like the product rule:

$\displaystyle\delta(ab)=\delta(a)b+a\delta(b)$

It’s easy to see that the collection $\mathrm{Der}(A)\subseteq\mathrm{End}_\mathbb{F}(A)$ is a vector subspace, but I say that it’s actually a Lie subalgebra, when we equip the space of endomorphisms with the usual commutator bracket. That is, if $\delta$ and $\partial$ are two derivations, I say that their commutator is again a derivation.

This, we can check:

$\displaystyle\begin{aligned} [\delta,\partial](ab)=&\delta(\partial(ab))-\partial(\delta(ab))\\=&\delta(\partial(a)b+a\partial(b)))-\partial(\delta(a)b+a\delta(b)))\\=&\delta(\partial(a)b)+\delta(a\partial(b)))-\partial(\delta(a)b)-\partial(a\delta(b)))\\=&\delta(\partial(a))b+\partial(a)\delta(b)+\delta(a)\partial(b)+a\delta(\partial(b))\\&-\partial(\delta(a))b-\delta(a)\partial(b)-\partial(a)\delta(b)-a\partial(\delta(b))\\=&[\delta,\partial](a)b+a[\delta,\partial](b)\end{aligned}$

We’ve actually seen this before. We identified the vectors at a point $p$ on a manifold with the derivations of the (real) algebra of functions defined in a neighborhood of $p$, so we need to take the commutator of two derivations to be sure of getting a new derivation back.

So now we can say that the mapping that sends $x\in L$ to the endomorphism $y\mapsto[x,y]$ lands in $\mathrm{Der}(L)$ because of the Jacobi identity. We call this mapping $\mathrm{ad}:L\to\mathrm{Der}(L)$ the “adjoint representation” of $L$, and indeed it’s actually a homomorphism of Lie algebras. That is, $\mathrm{ad}([x,y])=[\mathrm{ad}(x),\mathrm{ad}(y)]$. The endomorphism on the left-hand side sends $z\in L$ to $[[x,y],z]$, while on the right-hand side we get $[x,[y,z]]-[y,[x,z]]$. That these two are equal is yet another application of the Jacobi identity.

One last piece of nomenclature: derivations in the image of $\mathrm{ad}:L\to\mathrm{Der}(L)$ are called “inner”; all others are called “outer” derivations.

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Posted by John Armstrong | Algebra, Lie Algebras

## 4 Comments	»

1. [...] a group — which is the collection of all such that for all . That is, those for which the adjoint action is the zero derivation — the kernel of — which is clearly an [...]

Pingback by | August 13, 2012 | Reply

2. [...] that . In fact, while this case is very useful, all we need from is that it’s a nilpotent derivation of . The product rule for derivations generalizes [...]

Pingback by | August 18, 2012 | Reply

3. [...] -algebra — associative, Lie, whatever — and remember that contains the Lie algebra of derivations . I say that if then so are its semisimple part and its nilpotent part ; it’s enough to [...]

Pingback by | August 30, 2012 | Reply

4. [...] turns out that all the derivations on a semisimple Lie algebra are inner derivations. That is, they’re all of the form for [...]

Pingback by | September 11, 2012 | Reply

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