The Unapologetic Mathematician

Mathematics for the interested outsider

The Lie Algebra of a Lie Group

Since a Lie group G is a smooth manifold we know that the collection of vector fields \mathfrak{X}G form a Lie algebra. But this is a big, messy object because smoothness isn’t a very stringent requirement on a vector field. The value can’t vary too wildly from point to point, but it can still flop around a huge amount. What we really want is something more tightly controlled, and hopefully something related to the algebraic structure on G to boot.

To this end, we consider the “left-invariant” vector fields on G. A vector field X\in\mathfrak{X}G is left-invariant if the diffeomorphism L_h:G\to G of left-translation intertwines X with itself for all h\in G. That is, X must satisfy L_{h*}\circ X=X\circ L_h; or to put it another way: L_{h*}\left(X(g)\right)=X(hg). This is a very strong condition indeed, for any left-invariant vector field is determined by its value at the identity e\in G. Just set g=e and find that X(h)=L_{h*}\left(X(e)\right)

The really essential thing for our purposes is that left-invariant vector fields form a Lie subalgebra. That is, if X and Y are left-invariant vector fields, then so is their sum X+Y, scalar multiples cX — where c is a constant and not a function varying as we move around M — and their bracket [X,Y]. And indeed left-invariance of sums and scalar multiples are obvious, using the formula X(h)=L_{h*}\left(X(e)\right) and the fact that L_{h*} is linear on individual tangent spaces. As for brackets, this follows from the lemma we proved when we first discussed maps intertwining vector fields.

So given a Lie group G we get a Lie algebra we’ll write as \mathfrak{g}. In general Lie groups are written with capital Roman letters, while their corresponding Lie algebras are written with corresponding lowercase fraktur letters. When G has dimension n, \mathfrak{g} also has dimension n — this time as a vector space — since each vector field in \mathfrak{g} is uniquely determined by a single vector in \mathcal{T}_eG.

We should keep in mind that while \mathfrak{g} is canonically isomorphic to \mathcal{T}_eG as a vector space, the Lie algebra structure comes not from that tangent space itself, but from the way left-invariant vector fields interact with each other.

And of course there’s the glaring asymmetry that we’ve chosen left-invariant vector fields instead of right-invariant vector fields. Indeed, we could have set everything up in terms of right-invariant vector fields and the right-translation diffeomorphism R_g:G\to G. But it turns out that the inversion diffeomorphism i:G\to G interchanges left- and right-invariant vector fields, and so we end up in the same place anyway.

How does the inversion i act on vector fields? We recognize that i^{-1}=i, and find that it sends the vector field X to i_*\circ X\circ i. Now if X is left-invariant then L_{h*}\circ X=X\circ L_h for all h\in G. We can then calculate

\displaystyle\begin{aligned}R_{h*}\circ\left(i_*\circ X\circ i\right)&=\left(R_h\circ i\right)_*\circ X\circ i\\&=\left(i\circ L_{h^{-1}}\right)_*\circ X\circ i\\&=i_*\circ L_{h^{-1}*}\circ X\circ i\\&=i_*\circ X\circ L_{h^{-1}}\circ i\\&=\left(i_*\circ X\circ i\right)\circ R_h\end{aligned}

where the identities R_h\circ i=i\circ L_{h^{-1}} and L_h^{-1}\circ i=i\circ R_h reflect the simple group equations g^{-1}h=\left(h^{-1}g\right)^{-1} and h^{-1}g^{-1}=\left(gh\right)^{-1}, respectively. Thus we conclude that if X is left-invariant then i_*\circ X\circ i is right-invariant. The proof of the converse is similar.

The one thing that’s left is proving that if X and Y are left-invariant then their right-invariant images have the same bracket. This will follow from the fact that i_*(X(e))=-X(e), but rather than prove this now we’ll just push ahead and use left-invariant vector fields.

June 8, 2011 Posted by | Algebra, Differential Topology, Lie Groups, Topology | 8 Comments

   

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