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In order to talk meaningfully about the volume of $SL(n,R)/SL(n,Z)$ 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})$, $$\mu( E ) = \lambda(Cone(E)),$$ 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. 4 deleted 5 characters in body In order to talk meaningfully about the volume of$SL(n,R)/SL(n,Z)$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$E \subset SL(n,\mathbb{R})$, $$\mu( E ) = \lambda(Cone(E)),$$ 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$\mu(SL(n-1,\mathbb{R}))/\mu(SL(n-1,\mathbb{Z}))$, \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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In order to talk meaningfully about the volume of $SL(n,R)/SL(n,Z)$ 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 $E \subset SL(n,\mathbb{R})$,

$$\mu( E ) = \lambda(Cone(E)),$$

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 $\mu(SL(n,\mathbb{R})/\mathbb{Z})$. \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$\mu(SL(n-1,\mathbb{R}))/\mu(SL(n-1,\mathbb{Z}))$, 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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