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
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.
One of the most important examples of a Lie group we’ve already seen: the general linear group of a finite dimensional vector space . Of course for the vector space this is the same as — or at least isomorphic to — the group of all invertible real matrices, so that’s a Lie group we can really get our hands on. And if has dimension , then , and thus .
So, how do we know that it’s a Lie group? Well, obviously it’s a group, but what about the topology? The matrix group sits inside the algebra of all matrices, which is an -dimensional vector space. Even better, it’s an open subset, which we can see by considering the (continuous) map . Since is the preimage of — which is an open subset of — is an open subset of .
So we can conclude that is an open submanifold of , which comes equipped with the standard differentiable structure on . Matrix multiplication is clearly smooth, since we can write each component of a product matrix as a (quadratic) polynomial in the entries of and . As for inversion, Cramer’s rule expresses the entries of the inverse matrix as the quotient of a (degree ) polynomial in the entries of and the determinant of . So long as is invertible these are two nonzero smooth functions, and thus their quotient is smooth at .
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.
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 .
To put it another way, is the vector specifies for the point . On the other hand, is the image of the vector specifies for the point . If these two vectors are the same for every , then we say that “intertwines” the two vector fields, or that and are “-related”. The latter term is a bit awkward, which is why I prefer the former, especially since it does have that same commutative-diagram feel as intertwinors between representations.
Anyway, in the case that is a diffeomorphism we can actually use this to transfer vector fields from one manifold to the other. Given a point , which point should it be compared to? the inverse image , of course. This point gets the vector , which then gets sent to . That is, if we define , then is guaranteed to intertwine and .
Since is a linear map on each stalk it’s clear that if intertwines and , as well as and , then intertwines and . But we’ve just seen that vector fields form a Lie algebra, and it would be nice if we could say the same for and . The catch is that we don’t just compute these point-by-point.
Let’s pick a text function and a point . We first check that
Now we can calculate
So intertwines and , as we asserted. In the case where is a diffeomorphism, this means that the construction above gives us a homomorphism of Lie algebras from to .
We know that any vector field can act as an endomorphism on the space of smooth functions on . What happens if we act by one vector field followed by another? To really make things explicit, let’s say that is a coordinate patch, so we can write
where is a coefficient function and measures how fast is changing as we increase the th coordinate function through . Now if we hit this with another vector field we find
At each point the field gives a vector , which acts as a derivation on the ring of smooth functions at . That is
Now, this is obviously an endomorphism on since it’s the composite of two endomorphisms. But it is not a vector field, since at a given point we don’t get a derivation of the ring of smooth functions at . Indeed, what happens if we give it the product of two functions?
We’ve got a bunch of terms left over at the end! But one thing is nice about it: the leftover terms are symmetric between and :
So what would happen if instead of using the regular composition product of these endomorphisms, we used the associated Lie bracket? We’d find
That is, the Lie bracket of and is another vector field! Indeed, let’s see what it looks like in coordinates:
where we can cancel off the two second partial derivatives because we’re assuming that is “smooth”, which in this case entails “has mixed second partial derivatives which commute” in any local coordinate system.
And so we might appropriately write
Of course, even where we don’t have local coordinates we can still write or and get a vector field. We may also find it useful to write down the value of this field at a point: . Indeed we can check that this behaves like a vector at :
And so the space of smooth vector fields on forms a Lie subalgebra of the Lie algebra of endomorphisms of the vector space .
It turns out that any vector field on a compact manifold is complete. That is, starting at any point we can follow the vector field on construct its integral curve as far forward or as far backward as we want. I’ll show this two ways.
First of all, let’s say that there is some open interval containing a closed interval such that is contained in the from the maximal flow of . That is, assume that no matter where we start we can flow forward or backward by . But if we start at this means we can flow forward by to reach a point
But then by assumption we can flow forward again by to reach
And we can keep flowing forward to reach for arbitrarily large . Similarly we can flow backward as far as we want.
So, is there such an interval? In general there doesn’t have to be, but if is compact there is. Indeed, at each point we can define the interval as the maximal open interval on which the integral curve based at is defined. Here, each and is a positive real number or . Since is compact, each of these functions has a minimum — and . And then the interval is contained in each , as asserted.
A little more analytically, let be an integral curve of a vector field , and suppose that there is a sequence increasing to for which . Then it can only make sense to define by defining for and , for continuity. But now if is the maximal integral curve of with then uniqueness tells us that for .
So what happens when is compact? In this case, any sequence of converging to gives rise to a sequence of converging to some common . Thus we can always extend any integral curve on a right-open interval to the closed interval , and then past to a somewhat longer right-open interval, and so on as far as we want to go. And thus, again, every integral curve can be extended forever in either direction, which makes complete.