# The Unapologetic Mathematician

## 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.

## 7 Comments	»

1. “we get a Lie algebra we’ll write as \mathfrak{G}” – this should be lowercase, given the context.

Comment by Andrei | June 8, 2011 | Reply

2. Sorry, typo. Fixed.

Comment by | June 8, 2011 | Reply

3. [...] In particular, we can identify it with the tangent space at the identity matrix , and thus with the Lie algebra of [...]

Pingback by | June 9, 2011 | Reply

4. [...] a Vector Field, Vector fields on compact manifolds are complete, Maps Intertwining Vector Fields, The Lie algebra of a Lie group, General linear groups are Lie groups, The Lie algebra of a general linear [...]

Pingback by | June 12, 2011 | Reply

5. [...] space at back to itself: . But we know that this tangent space is canonically isomorphic to the Lie algebra . That is, . So now we can define by . We call this the “adjoint representation” of [...]

Pingback by | June 13, 2011 | Reply

6. [...] The Lie Algebra of a Lie Group (unapologetic.wordpress.com) [...]

Pingback by | June 15, 2011 | Reply

7. [...] and Flows When we discussed the Lie algebra of a Lie group we discussed “left-invariant” vector fields. More generally than this if is a [...]

Pingback by | June 17, 2011 | Reply

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