# The Unapologetic Mathematician

## 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

## 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

## 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

## 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