The Unapologetic Mathematician

Mathematics for the interested outsider

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 GL(V) 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 g\in G. As we might expect, this is the map \tau_g=L_g\circ R_{g^{-1}}:G\to G — that is, \tau_g(h)=ghg^{-1}. This is a diffeomorphism from G back to itself, and in particular it has the identity e\in G as a fixed point: \tau_g(e)=e. Thus the derivative sends the tangent space at e back to itself: \tau_{g*e}:\mathcal{T}_eG\to\mathcal{T}_eG. But we know that this tangent space is canonically isomorphic to the Lie algebra \mathfrak{g}. That is, \tau_g\in GL(\mathfrak{g}). So now we can define \mathrm{Ad}:G\to GL(\mathfrak{g}) by \mathrm{Ad}(g)=\tau_g. We call this the “adjoint representation” of G.

To get even more specific, we can consider the adjoint representation of GL_n(\mathbb{R}) on its Lie algebra \mathfrak{gl}_n(\mathbb{R})\cong M_n(\mathbb{R}). I say that \mathrm{Ad}_g is just \tau_g itself. That is, if we view GL_n(\mathbb{R}) as an open subset of M_n(\mathbb{R}) then we can identify \mathcal{I}_e:M_n(\mathbb{R})\cong\mathfrak{gl}_n(\mathbb{R}). The fact that \tau_g and \mathrm{Ad}(g) both commute means that \mathcal{I}_e\circ\tau_g=\mathrm{Ad}(g)\circ\mathcal{I}_e, meaning that \tau_g and \mathrm{Ad}(g) are “the same” transformation, under this identification of these two vector spaces.

Put more simply: to calculate the adjoint action of g\in GL_n(\mathbb{R}) on the element of \mathfrak{gl}_n(\mathbb{R}) corresponding to A\in M_n(\mathbb{R}), it suffices to calculate the conjugate gAg^{-1}; then

\displaystyle\left[\mathrm{Ad}(g)\right](\mathcal{I}_e(A))=\mathcal{I}_e(\tau_g(A))=\mathcal{I}_e(gAg^{-1})

June 13, 2011 Posted by | Algebra, Differential Topology, Lie Groups, Topology | Leave a comment

The Lie Algebra of a General Linear Group

Since GL_n(\mathbb{R}) is an open submanifold of M_n(\mathbb{R}), the tangent space of GL_n(\mathbb{R}) at any matrix A is the same as the tangent space to M_n(\mathbb{R}) at A. And since M_n(\mathbb{R}) is (isomorphic to) a Euclidean space, we can identify M_n(\mathbb{R}) with \mathcal{T}_AM_n(\mathbb{R}) using the canonical isomorphism \mathcal{I}_A:M_n(\mathbb{R})\to\mathcal{T}_AM_n(\mathbb{R}). In particular, we can identify it with the tangent space at the identity matrix I, and thus with the Lie algebra \mathfrak{gl}_n(\mathbb{R}) of GL_n(\mathbb{R}):

\displaystyle M_n(\mathbb{R})\cong\mathcal{T}_IGL_n(\mathbb{R})\cong\mathfrak{gl}_n(\mathbb{R})

But this only covers the vector space structures. Since M_n(\mathbb{R}) is an associative algebra it automatically has a bracket: the commutator. Is this the same as the bracket on \mathfrak{gl}_n(\mathbb{R}) under this vector space isomorphism? Indeed it is.

To see this, let A be a matrix in M_n(\mathbb{R}) and assign X(I)=\mathcal{I}_I(A)\in\mathcal{T}_IGL_n(\mathbb{R}). This specifies the value of the vector field X at the identity in GL_n(\mathbb{R}). We extend this to a left-invariant vector field by setting

\displaystyle X(g)=L_{g*}X(I)=L_{g*}\mathcal{I}_I(A)=\mathcal{I}_{g}(gA)

where we subtly slip from left-translation by g within GL_n(\mathbb{R}) to left-translation within the larger manifold M_n(\mathbb{R}). We do the same thing to go from another matrix B to another left-invariant vector field Y.

Now that we have our hands on two left-invariant vector fields X and Y coming from two matrices A and B. We will calculate the Lie bracket [X,Y] — we know that it must be left-invariant — and verify that its value at I indeed corresponds to the commutator AB-BA.

Let u^{ij}:GL_n(\mathbb{R})\to\mathbb{R} be the function sending an n\times n matrix to its (i,j) entry. We hit it with one of our vector fields:

\displaystyle Yu^{ij}(g)=Y_gu^{ij}=\mathcal{I}_g(gB)u^{ij}=u^{ij}(gB)

That is, Yu^{ij}=u^{ij}\circ R_B, where R_B is right-translation by B. To apply the vector X_I=A to this function, we must take its derivative at I in the direction of A. If we consider the curve through I defined by c(t)=I+tA we find that

\displaystyle X_IYu^{ij}=\dot{c}(0)(u^{ij}\circ R_B)=\frac{d}{dt}(u^{ij}(B+tAB))\Big\vert_{t=0}=(AB)_{i,j}

Similarly, we find that Y_IXu^{ij}=(BA)_{i,j}. And thus

\displaystyle [X,Y]_Iu^{ij}=(AB-BA)_{i,j}=\mathcal{I}_I(AB-BA)(u^{ij})

Of course, for any Q\in M_n(\mathbb{R}) we have the decomposition

\displaystyle\mathcal{I}_IQ=\sum\limits_{i,j=1}^n\mathcal{I}_IQ(u^{ij})\frac{\partial}{\partial u^{ij}}\Big\vert_I

Therefore, since we’ve calculated [\mathcal{I}_IA,\mathcal{I}_IB](u^{ij})=\mathcal{I}_I(AB-BA)(u^{ij}) 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 \mathfrak{gl}_n(\mathbb{R}) agrees with the commutator on M_n(\mathbb{R}), and thus that these two are isomorphic as Lie algebras.

June 9, 2011 Posted by | Algebra, Differential Topology, Lie Groups, Topology | 2 Comments

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

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 G is a smooth n-dimensional manifold equipped with a multiplication G\times G\to G and an inversion G\to G 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 G\times G we mean the product manifold.

We can use these to construct some other useful maps. For instance, if h\in G is any particular element we know that we have a smooth inclusion G\to G\times G defined by g\mapsto (h,g). Composing this with the multiplication map we get a smooth map L_h:G\to G defined by L_h(g)=hg, which we call “left-translation by h“. Similarly we get a smooth right-translation R_h(g)=gh.

June 6, 2011 Posted by | Algebra, Differential Topology, Lie Groups, Topology | 2 Comments

   

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