The Unapologetic Mathematician

Mathematics for the interested outsider

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.

August 10, 2012 - Posted by | 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 […]

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  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 Automorphisms of Lie Algebras « The Unapologetic Mathematician | 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 […]

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  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 All Derivations of Semisimple Lie Algebras are Inner « The Unapologetic Mathematician | September 11, 2012 | Reply


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