## Towards integration on matrix Lie groups #2

In the last post I discussed some preliminaries from differential geometry which we’ll need to define the integral $\int_G f$ of a function $f : G \to \mathbb{C}$, where $G$ is a matrix Lie group. Naturally, we want to do this in a left-invariant way, namely if $L_g : G \to G$ is the map $h \mapsto gh$ (left multiplication by $g$), then we want $\int_G f \circ L_g = \int_G f$. This is analogous to the case of (say) the group $(\mathbb{R}, +)$ where “translation invariance” of an integral would translate (no pun intended) into a condition like $\int_{\mathbb{R}} f(x) \; dx = \int_{\mathbb{R}} f(x+g) \; dx$. The group $(\mathbb{R}, +)$ is abelian of course, so left and right invariance are the same concept.

Let us now return to our friend $\mathrm{Alt}^d(G)$, the set of volume forms on $G$. In the language of differential geometry, the elements of $\mathrm{Alt}^d(G)$ are smooth sections of $\Lambda^d( T^* G )$, or in other words, top-degree differential forms on $G$. One might thus prefer to write $\Omega^d(G)$ rather than $\mathrm{Alt}^d(G)$. Recall that $T_g^* G = (T_g G)^* = (g \cdot \mathfrak{g})^*$. There is no reason, a priori, that $\mathrm{Alt}^d(G)$ should be nonempty. Yes, we know $\Lambda^d( T_g^* G )$ has dimension 1 for every $g \in G$, but why should there exist a smooth section of this bundle? Well, it turns out that if you fix some basis $\{ X_1, \ldots, X_d \}$ for the Lie algebra $\mathfrak{g}$ and simply define, for each $g \in G$, $\eta_g \in \Lambda^d((g \cdot \mathfrak{g})^*)$ to be the thing that assigns a “volume” of 1 to the parallelotope formed by the “$g$-translated basis” $\{ gX_1, \ldots, gX_d \} \subseteq g \cdot \mathfrak{g}$, and then piece all these together by declaring

$\displaystyle \eta(\xi_1,\ldots,\xi_d)(g) = \eta_g(\underbrace{\xi_1(g)}_{\in g \cdot \mathfrak{g}},\ldots,\underbrace{\xi_d(g)}_{\in g \cdot \mathfrak{g}})$

then you actually find that $\eta \in \mathrm{Alt}^d(G)$.

### Absolute values

Let me now correct some errata from last time. I said that an $\omega \in \mathrm{Alt}^d(G)$ is usually referred to as a density, and also claimed that $\omega_g$ is a way of assigning “unsigned volumes” to ordered parallelotopes in $T_g G$. This is slightly incorrect. The actual situation is that these guys $\omega$ in $\mathrm{Alt}^d(G)$ really are (as mentioned above) legit top-degree differential forms on $G$. The objects that we would call densities are the absolute values $|\omega|$ of some such $\omega$. Notice that $\omega$ eats $d$ vector fields on $G$ and produces a smooth function on $G$. After we compose this with the absolute value, I believe we merely get a map $|\omega| : \Xi(G)^d \to \mathcal{C}(G)$; note the modesty of the codomain. Indeed, $x \mapsto x$ is a smooth function, but $x \mapsto |x|$ is not even continuously differentiable, although it is continuous. However, we don’t care, since any continuous function is surely still good enough to integrate.

It should be noted that unless one chooses a basis $\{ X_1, \ldots, X_d \}$ for the Lie algebra ahead of time, the Haar integral (which we will soon define) is only going to make sense up to scaling by a positive factor. This is because in order to fully determine a left-invariant form $\eta$ as desired, we have to say “which” parallelotopes in the Lie algebra we want to take as fundamental, that is, which ones we want to assign a volume of 1.

As an aside, one of the nice things about (oriented) Riemannian manifolds is that their tangent spaces carry inner products, at which point Hodge duality phenomena, which yield natural isomorphisms $\Omega^k(M) \cong \Omega^{n-k}(M)$, would give us a distinguished element of $\Omega^d(M)$ ($\cong \Omega^0(M) = \mathcal{C}^\infty(M)$); merely pick the one corresponding to the constant function 1. In less high-brow parlance, we could use the orientation and inner product on each tangent space to establish some isometric isomorphism to $\mathbb{R}^n$ and then invoke the determinant over there. However we’re not so lucky to have such Riemannian structure here.

### Integrating a function supported on one chart

Having now discussed the geometric intuition behind the machinery, we are ready to define the integral. In general, to do this, we will choose some coordinate system $\{ (V_\alpha, \varphi_\alpha) \}_{\alpha \in A}$ on our Lie group $G$. The actual definition of the integral turns out to be independent of which coordinate system we use; it depends only on the density $|\omega|$. The simplest case, of course, is when the function $f : G \to \mathbb{C}$ we are integrating actually vanishes outside of a single $V = V_\alpha$. In this case we write $\varphi = \varphi_\alpha$ and put

$\displaystyle \int_G f \; |\omega| = \int_{\varphi(V)} f(\varphi^{-1}(x)) \left| \omega_{\varphi^{-1}(x)} \left( \frac{\partial}{\partial x_1} \varphi^{-1}(x), \ldots, \frac{\partial}{\partial x_d} \varphi^{-1}(x) \right) \right| \; dx.$

As disgusting as the above may look on first sight, I claim this is the natural thing to do. Parsing the right-hand side, we see that we are now integrating over a region $\varphi(V) \subseteq \mathbb{R}^d$. The integrand, of course, is merely the function that takes points in $\varphi(V)$ to their corresponding points in $G$, and thereafter feeds them to $f$. Well… almost. The huge expression involving $\omega$ that we’re multiplying by actually turns out to be very important. Intuitively, this should not be surprising: what if we chose charts $\varphi$ such that $\varphi(V)$ were larger and larger subsets of $\mathbb{R}^d$? Our integral could be made arbitrarily large. However, this is where the second factor comes in, and keeps things under control. Let’s look at it a little more closely:

$\displaystyle \left| \omega_{\varphi^{-1}(x)} \left( \underbrace{\frac{\partial}{\partial x_1} \varphi^{-1}(x)}_{\in T_{\varphi^{-1}(x)}G}, \ldots, \underbrace{\frac{\partial}{\partial x_d} \varphi^{-1}(x)}_{\in T_{\varphi^{-1}(x)}G} \right) \right|.$

I guess you can probably interpret this scaling factor as some kind of Radon-Nikodym derivative. If we think again about the signature of $\omega_{\varphi^{-1}(x)}$, it wants to take in $d$ elements of the tangent space at $\varphi^{-1}(x) \in G$. Here is perhaps where the PMATH 365 point of view comes in handy again: $T_g G = g \cdot \mathfrak{g}$ can also be interpreted as consisting of derivations at $g$ of the algebra $\mathcal{C}^\infty(G)$ of smooth functions. The tangent space at $x$ in $\mathbb{R}^d$ admits the natural coordinate basis

$\displaystyle \left\{ \left.\frac{\partial}{\partial x_1} \right|_x, \ldots, \left.\frac{\partial}{\partial x_d} \right|_x \right\}$.

Now $\varphi$ is a map from $V \subseteq G$ to $\varphi(V) \subseteq \mathbb{R}^d$, so the things

$\displaystyle \frac{\partial}{\partial x_i} \varphi^{-1}(x)$ which we might write $\displaystyle \left. \frac{\partial}{\partial x_i} \right|_{\varphi^{-1}(x)}$

which appear above are nothing more than the images of the coordinate basis elements $\left. \partial / \partial x_i \right|_x$ under the pushforward map $((\varphi^{-1})_*)_x : T_x \mathbb{R}^n \to T_{\varphi^{-1}(x)} G$. Notice that if $\varphi$ was mapping $V$ onto some HUGE set $\varphi(V)$, then it seems reasonable to expect that $\varphi^{-1}$ would have very small partial derivatives, which would cause the scaling factor given by $|\omega|$ to be a very small positive number.

That’s all I have time to write for now. Next time we will discuss partitions of unity, which will permit the integration of functions even when they are not supported on a single chart, and finish by mentioning the modular function and what unimodularity has to do with right-invariance criteria.

just another guy trying to make the diagrams commute.
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### 7 Responses to Towards integration on matrix Lie groups #2

1. Erik Crevier says:

Nice. By the way, sometimes you actually *do* have a canonical choice of Riemannian metric on $G$: Use the Killing form (translated around appropriately). If $G$ is compact and semisimple, then this is negative-definite, so it’s basically a Riemannian metric up to choice of sign. So in that case you actually do get a sort of canonical choice of scaling for the haar integral. I wonder if that’s the same as choosing so that $\int_G 1dg=1$… I can’t see any obvious reason why that should be true, but it would be really nice. And if it’s not true, then I’m really curious about the ratio between the two choices.

Actually, even if $G$ is just semisimple but not necessarily compact, the Killing form is non-degenerate and so gives a pseudo-riemannian metric on $G$. I seem to recall that you can define hodge duals in that setting as well. Hmmmmm…..

• mlbaker says:

That is interesting. I mentioned this to Eeshan and he pointed out that the Killing form doesn’t seem like a “natural” choice of inner product on the Lie algebra; if this is true then I doubt the associated Haar measure will be a probability measure as you said.

2. Erik Crevier says:

Hmm. I don’t really understand the issue all to well, but I believe the killing form is more natural than it appears at first glance. The most compelling reason I can think of for this is the following: If $G$ is compact, then by Mashke’s theorem averaging there is an invariant inner product on $\mathfrak{g}$.

Now suppose that $G$ is a compact Lie group such that $\mathfrak{g}$ is simple. Then the adjoint representation of $G$ on $\mathfrak{g}$ is irreducible (because an invariant subspace would be an ideal by differentiating). In this case, Schur’s lemma tells you that the space of invariant inner products on $\mathfrak{g}$ is 1 dimensional. So, up to a choice of scaling, there is only one invariant inner product on $\mathfrak{g}$. Of course this is the killing form.

This doesn’t tell you anything directly about the semisimple case, but it does tell you that if you can come up with any universal way of assigning an invariant inner product which has nice functorial properties (ie it plays nice with direct sums), then it has to be the killing form up to scalars.

Now, I must confess that I don’t really find this compelling enough to motivate the killing form for me. I’m still trying to figure out exactly why it matters so much. But I suspect that there are stronger versions of this argument that pin it down pretty nicely. Also, the fact that it captures so much of the structure theory of $\mathfrak{g}$ is at least a very strong waggling pointer finger.

Also, this sort of argument doesn’t actually say anything about the issue at hand, because it only pins down the killing form up to a choice of scalar, and the choice of scalar is exactly what we are concerned with here. In fact, if it turns out that the killing form doesn’t give the right normalization to the haar integral, then I would *almost* be willing to take that as evidence that we picked the wrong choice of scalar for the killing form rather than the other way around.

• Erik Crevier says:

With the machinery we have set up (e.g. the $\mathrm{ad}$ maps and so on) the Killing form does seem “clean”, but I don’t know if that will translate into naturality. I mean, for a lot of the classical groups it reduces to a scalar multiple of the trace form — how does the trace form seem any less natural? That’s why I think it’s probably not going to give you the normalised integral.
So you are correct, you don’t get the normalized integral from the Killing form. Take $\mathrm{SO}(3)$. The last question on this assignment shows that using some easy choice of basis for $\mathfrak{so}(3)$ to define a volume form leads to a normalization factor of $\pi^2$ on the haar integral. I’m too lazy to work out what the Killing form looks like in this basis, but I am quite certain that you get an orthonormal basis for it by taking some rational combinations of those basis elements. In particular, there’s no way you’re going to get a factor of $\pi^2$ going from that volume element to the one defined by the Killing form.