# Volume of fundamental domain and Haar measure

In my research, I do need to know the Haar measure. I have spent some time on this subject, understanding theoretical part of the Haar measure, i.e existence and uniqueness, Haar measure on quotient. But I should confess I never felt confident with the Haar measure, essentially because theoretical part did not give me how to construct the Haar measure. Even when I was trying to understand Haar measure on $SL(n,\mathbb{R})$ via Iwasawa decomposition, I could not realize exactly how it works.

I think, or I should say I fell, I am not the only one how has problem with this delicate subject. So I thought I would want to share this with you and see how people think about the Haar measure and how you compute, for instance, the volume $${\rm Vol}(SL(n,\mathbb{Z})\backslash SL(n,\mathbb{R}))$$

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Do you mean SL(n,R)/SL(n,Z)? – David Roberts Jul 12 '11 at 3:21
I do not see difference. Would you please explain me the difference – M.B Jul 12 '11 at 3:37
@David: I think it is a matter of convention whether one writes quotients from the left or from the right. – Qiaochu Yuan Jul 12 '11 at 3:44
@Qiaochu Yuan: but surely if he's quotienting on the left, then the slash should go the other way . . . – John Pardon Jul 12 '11 at 3:50

In order to talk meaningfully about the volume of $SL(n,\mathbb{R})/SL(n,\mathbb{Z})$ you need to define a normalization for Haar measure.

One way to think about it is as follows: the space $M_n(\mathbb{R})$ of $n$ by $n$ matrices is just $\mathbb{R}^{n^2}$, so it has a natural notion of Lebesgue measure $\lambda$. We can normalize $\lambda$ so that $\lambda(M_n(\mathbb{R})/M_n(\mathbb{Z})) = 1$.

Now $SL(n,\mathbb{R})$ is a hypersurface in $M_n(\mathbb{R})$ given by $\det = 1$. So we need to restrict $\lambda$ from $M_n(\mathbb{R})$ to the hypersurface. One natural way to do that is to define for $E \subset SL(n,\mathbb{R})$,

$$\DeclareMathOperator{\Cone}{Cone} \mu( E ) = \lambda(\Cone(E)) ,$$

where $\Cone(E)$ is the Euclidean cone which is the union of all line segments starting at the origin and ending at $E$. Now $\mu$ is $SL(n,\mathbb{R})$ invariant, and therefore it is the Haar measure. This also defines a natural normalization for the measure. With this normalization, it makes sense to ask for $\mu(SL(n,\mathbb{R})/SL(n,\mathbb{Z}))$.

There is a beautiful formula due to Siegel:

$$\mu(SL(n,\mathbb{R})/SL(n,\mathbb{Z})) = \frac{1}{n} \zeta(2) \dots \zeta(n)$$

(I think I probably did not get all the factors right with my normalization). I will outline two completely elementary approaches to proving this formula. (Later on you see that the two approaches are really the same, and that this all has to do with Tamagawa numbers).

Approach 1: For a compactly supported function $f: \mathbb{R}^n \to \mathbb{R}$ we can define a function $\hat{f}: SL(n,\mathbb{R})/SL(n,\mathbb{Z}) \to \mathbb{R}$ by the formula

$$\hat{f}(\Delta) = \sum_{v \in \Delta'} f(v).$$ Here you think of $\Delta \in SL(n,\mathbb{R})/SL(n,\mathbb{Z})$ as a lattice in $\mathbb{R}^n$, and $\Delta'$ is the set of primitive vectors in $\Delta$. Then there is a formula due to Siegel (which you can prove by unfolding): $$\frac{1}{\mu(SL(n,\mathbb{R})/SL(n,\mathbb{Z}))}\int_{SL(n,\mathbb{R})/SL(n,\mathbb{Z})} \hat{f}(\Delta) \, d\mu(\Delta) = \frac{1}{\zeta(n)} \int_{\mathbb{R}^n} f.$$

You can now take $f$ to be the characteristic function of the ball of radius $\epsilon$ and then take $\epsilon \to 0$. The asymptotics of the integral on the left hand side is expressible in terms of $\mu(SL(n-1,\mathbb{R})/SL(n-1,\mathbb{Z}))$, so you get the formula for the volume by induction.

Approach 2: Let $E \subset SL(n,\mathbb{R})$ be the fundamental domain for the action of $SL(n,\mathbb{Z})$. Pick a large parameter $R > 0$. Then, $$\mu(SL(n,\mathbb{R})/SL(n,\mathbb{Z})) = \mu(E) = \lambda(\Cone(E)) = \frac{1}{R^{n^2}}\lambda(\Cone(R E)).$$

But the volume of the cone $\lambda(\Cone(RE))$ is asymptotic as $R \to \infty$ to the number of integer points in the cone, i.e. the cardinality of $\Cone(RE) \cap M_n(\mathbb{Z})$. Now points in $M_n(\mathbb{Z}) \cap \Cone(RE)$ parametrize integer lattices of covolume at most $R^n$. So if you count the number of sublattices of the standard lattice $\mathbb{Z}^n$ of index at most $R^n$ and take the leading term as $R \to \infty$ you also compute $\mu(SL(n,\mathbb{R})/SL(n,\mathbb{Z}))$. This will give the same answer as Approach 1.

It turns out that both approaches make sense in other situations, e.g. volumes of moduli spaces of holomorphic differentials. In that setting they both sort of work, but give different information.

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Thank you. I think I will learn a lot from your comment. Especially, since I have been working on your paper, where is giving a new proof of Siegel-Minkowski Mass formula. This might give me some ideas to be able to understand that work. – M.B Jul 12 '11 at 18:08

Supplementing the other answer (too long for a comment...) First, for examples of innocent-context computations, there is an on-line computation of volumes of $SL(n,\mathbb Z)\backslash SL(n,\mathbb R)$ and of $Sp(n,\mathbb Z)\backslash Sp(n,\mathbb R)$ here , written in essentially Siegel's style. The same style of computation can be done adelically, over arbitrary number fields, but still does effectively beg the question of normalization. Nevertheless, such computations show that the normalization can be determined inductively, and, in any case, that the global computation can be done nicely once we have a locally-everywhere normalization of measures.

Yet-another approach is (after Langlands) to look at suitable residues of Eisenstein series (e.g., as in the Boulder conference, AMS Proc Symp IX). In effect, such a computation back-handedly normalizes the Haar measure...

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