The Adjoint Representation

Since Lie groups are groups, they have representations — homomorphisms to the general linear group of some vector space or another. But since is a Lie group, we can use this additional structure as well. And so we say that a representation of a Lie group should not only be a group homomorphism, but a smooth map of manifolds as well.

As a first example, we define a representation that every Lie group has: the adjoint representation. To define it, we start by defining conjugation by . As we might expect, this is the map — that is, . This is a diffeomorphism from back to itself, and in particular it has the identity as a fixed point: . Thus the derivative sends the tangent 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 .

To get even more specific, we can consider the adjoint representation of on its Lie algebra . I say that is just itself. That is, if we view as an open subset of then we can identify . The fact that and both commute means that , meaning that and are “the same” transformation, under this identification of these two vector spaces.

Put more simply: to calculate the adjoint action of on the element of corresponding to , it suffices to calculate the conjugate ; then

The Lie Algebra of a General Linear Group

Since is an open submanifold of , the tangent space of at any matrix is the same as the tangent space to at . And since is (isomorphic to) a Euclidean space, we can identify with using the canonical isomorphism . In particular, we can identify it with the tangent space at the identity matrix , and thus with the Lie algebra of :

But this only covers the vector space structures. Since is an associative algebra it automatically has a bracket: the commutator. Is this the same as the bracket on under this vector space isomorphism? Indeed it is.

To see this, let be a matrix in and assign . This specifies the value of the vector field at the identity in . We extend this to a left-invariant vector field by setting

where we subtly slip from left-translation by within to left-translation within the larger manifold . We do the same thing to go from another matrix to another left-invariant vector field .

Now that we have our hands on two left-invariant vector fields and coming from two matrices and . We will calculate the Lie bracket — we know that it must be left-invariant — and verify that its value at indeed corresponds to the commutator .

Let be the function sending an matrix to its entry. We hit it with one of our vector fields:

That is, , where is right-translation by . To apply the vector to this function, we must take its derivative at in the direction of . If we consider the curve through defined by we find that

Similarly, we find that . And thus

Of course, for any we have the decomposition

Therefore, since we’ve calculated we know these two vectors have all the same components, and thus are the same vector. And so we conclude that the Lie bracket on agrees with the commutator on , and thus that these two are isomorphic as Lie algebras.

The Lie Algebra of a Lie Group

Since a Lie group is a smooth manifold we know that the collection of vector fields 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 to boot.

To this end, we consider the “left-invariant” vector fields on . A vector field is left-invariant if the diffeomorphism of left-translation intertwines with itself for all . That is, must satisfy ; or to put it another way: . This is a very strong condition indeed, for any left-invariant vector field is determined by its value at the identity . Just set and find that

The really essential thing for our purposes is that left-invariant vector fields form a Lie subalgebra. That is, if and are left-invariant vector fields, then so is their sum , scalar multiples — where is a constant and not a function varying as we move around — and their bracket . And indeed left-invariance of sums and scalar multiples are obvious, using the formula and the fact that 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 we get a Lie algebra we’ll write as . In general Lie groups are written with capital Roman letters, while their corresponding Lie algebras are written with corresponding lowercase fraktur letters. When has dimension , also has dimension — this time as a vector space — since each vector field in is uniquely determined by a single vector in .

We should keep in mind that while is canonically isomorphic to 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 . But it turns out that the inversion diffeomorphism interchanges left- and right-invariant vector fields, and so we end up in the same place anyway.

How does the inversion act on vector fields? We recognize that , and find that it sends the vector field to . Now if is left-invariant then for all . We can then calculate

where the identities and reflect the simple group equations and , respectively. Thus we conclude that if is left-invariant then is right-invariant. The proof of the converse is similar.

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

Lie Groups

Now we come to one of the most broadly useful and fascinating structures on all of mathematics: Lie groups. These are objects which are both smooth manifolds and groups in a compatible way. The fancy way to say it is, of course, that a Lie group is a group object in the category of smooth manifolds.

To be a little more explicit, a Lie group is a smooth -dimensional manifold equipped with a multiplication and an inversion which satisfy all the usual group axioms (wow, it’s been a while since I wrote that stuff down) and are also smooth maps between manifolds. Of course, when we write we mean the product manifold.

We can use these to construct some other useful maps. For instance, if is any particular element we know that we have a smooth inclusion defined by . Composing this with the multiplication map we get a smooth map defined by , which we call “left-translation by “. Similarly we get a smooth right-translation .