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orangeskid
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It's enough to check in even dimension, that is for the unit ball $B_n$ in $\mathbb{C}^n$. The map $z\mapsto \pi |z|^2$ from $\mathbb{C}$ to $[0, \infty)$ is measure preserving, and so is the map $\mathbb{C}^n \to [0, \infty)^n$, $(z_1, \ldots, z_n) \mapsto (\pi |z_1|^2, \ldots, \pi|z_n|^2)$. The product of disks $D^n$ maps to $[0,\pi]^n$, and the unit ball maps to the simplex $\{\sum_{k=1}^n t_k \le \pi\}$, which is $\frac{1}{n!}$ of the cube $[0,\pi]^n$. So that's why

$$\operatorname{ Vol}(B_n) = \frac{\pi^n}{n!}$$

Obs: why the simplex $\{\sum_1^n t_k \le 1\}$ has volume $\frac{1}{n!}$ is discussed here.

$\bf{Added:}$ A similar argument shows that if $d$ is a natural number then the volume of the body

$$\{ \sum_{k=1}^{d n} |x_k|^d \le R^d \}$$ is of the form

$$\frac{\kappa^n R^{d n}}{n!}$$ ( $\kappa$ instead of $\pi$) and so decreases to $0$ as $n\to \infty$. The crucial obervation again is that a map of the form

$$\mathbb{R}^d \to [0, \infty), \ \ (x_1, \ldots, x_d) \mapsto \kappa \cdot F(x_1, \ldots, x_d)$$ where $F$ is a form of degree $d$, is measure preserving( for some $\kappa >0$).

$\bf{Added:}$ This "reduction to the simplex" also handles some "concentration of measure around equator".

It's enough to check in even dimension, that is for the unit ball $B_n$ in $\mathbb{C}^n$. The map $z\mapsto \pi |z|^2$ from $\mathbb{C}$ to $[0, \infty)$ is measure preserving, and so is the map $\mathbb{C}^n \to [0, \infty)^n$, $(z_1, \ldots, z_n) \mapsto (\pi |z_1|^2, \ldots, \pi|z_n|^2)$. The product of disks $D^n$ maps to $[0,\pi]^n$, and the unit ball maps to the simplex $\{\sum_{k=1}^n t_k \le \pi\}$, which is $\frac{1}{n!}$ of the cube $[0,\pi]^n$. So that's why

$$\operatorname{ Vol}(B_n) = \frac{\pi^n}{n!}$$

Obs: why the simplex $\{\sum_1^n t_k \le 1\}$ has volume $\frac{1}{n!}$ is discussed here.

$\bf{Added:}$ A similar argument shows that if $d$ is a natural number then the volume of the body

$$\{ \sum_{k=1}^{d n} |x_k|^d \le R^d \}$$ is of the form

$$\frac{\kappa^n R^{d n}}{n!}$$ ( $\kappa$ instead of $\pi$) and so decreases to $0$ as $n\to \infty$. The crucial obervation again is that a map of the form

$$\mathbb{R}^d \to [0, \infty), \ \ (x_1, \ldots, x_d) \mapsto \kappa \cdot F(x_1, \ldots, x_d)$$ where $F$ is a form of degree $d$, is measure preserving( for some $\kappa >0$).

It's enough to check in even dimension, that is for the unit ball $B_n$ in $\mathbb{C}^n$. The map $z\mapsto \pi |z|^2$ from $\mathbb{C}$ to $[0, \infty)$ is measure preserving, and so is the map $\mathbb{C}^n \to [0, \infty)^n$, $(z_1, \ldots, z_n) \mapsto (\pi |z_1|^2, \ldots, \pi|z_n|^2)$. The product of disks $D^n$ maps to $[0,\pi]^n$, and the unit ball maps to the simplex $\{\sum_{k=1}^n t_k \le \pi\}$, which is $\frac{1}{n!}$ of the cube $[0,\pi]^n$. So that's why

$$\operatorname{ Vol}(B_n) = \frac{\pi^n}{n!}$$

Obs: why the simplex $\{\sum_1^n t_k \le 1\}$ has volume $\frac{1}{n!}$ is discussed here.

$\bf{Added:}$ A similar argument shows that if $d$ is a natural number then the volume of the body

$$\{ \sum_{k=1}^{d n} |x_k|^d \le R^d \}$$ is of the form

$$\frac{\kappa^n R^{d n}}{n!}$$ ( $\kappa$ instead of $\pi$) and so decreases to $0$ as $n\to \infty$. The crucial obervation again is that a map of the form

$$\mathbb{R}^d \to [0, \infty), \ \ (x_1, \ldots, x_d) \mapsto \kappa \cdot F(x_1, \ldots, x_d)$$ where $F$ is a form of degree $d$, is measure preserving( for some $\kappa >0$).

$\bf{Added:}$ This "reduction to the simplex" also handles some "concentration of measure around equator".

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orangeskid
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It's enough to check in even dimension, that is for the unit ball $B_n$ in $\mathbb{C}^n$. The map $z\mapsto \pi |z|^2$ from $\mathbb{C}$ to $[0, \infty)$ is measure preserving, and so is the map $\mathbb{C}^n \to [0, \infty)^n$, $(z_1, \ldots, z_n) \mapsto (\pi |z_1|^2, \ldots, \pi|z_n|^2)$. The product of disks $D^n$ maps to $[0,\pi]^n$, and the unit ball maps to the simplex $\{\sum_{k=1}^n t_k \le \pi\}$, which is $\frac{1}{n!}$ of the cube $[0,\pi]^n$. So that's why

$$\operatorname{ Vol}(B_n) = \frac{\pi^n}{n!}$$

Obs: why the simplex $\{\sum_1^n t_k \le 1\}$ has volume $\frac{1}{n!}$ is discussed here.

$\bf{Added:}$ A similar argument shows that if $d$ is a natural number then the volume of the body

$$\{ \sum_{k=1}^{d n} |x_k|^d \le R^d \}$$ is of the form

$$\frac{\kappa^n R^{d n}}{n!}$$ ( $\kappa$ instead of $\pi$) and so decreases to $0$ as $n\to \infty$. The crucial obervation again is that a map of the form

$$\mathbb{R}^d \to [0, \infty), \ \ (x_1, \ldots, x_d) \mapsto \kappa \cdot F(x_1, \ldots, x_d)$$ where $F$ is a form of degree $d$, is measure preserving( for some $\kappa >0$).

It's enough to check in even dimension, that is for the unit ball $B_n$ in $\mathbb{C}^n$. The map $z\mapsto \pi |z|^2$ from $\mathbb{C}$ to $[0, \infty)$ is measure preserving, and so is the map $\mathbb{C}^n \to [0, \infty)^n$, $(z_1, \ldots, z_n) \mapsto (\pi |z_1|^2, \ldots, \pi|z_n|^2)$. The product of disks $D^n$ maps to $[0,\pi]^n$, and the unit ball maps to the simplex $\{\sum_{k=1}^n t_k \le \pi\}$, which is $\frac{1}{n!}$ of the cube $[0,\pi]^n$. So that's why

$$\operatorname{ Vol}(B_n) = \frac{\pi^n}{n!}$$

Obs: why the simplex $\{\sum_1^n t_k \le 1\}$ has volume $\frac{1}{n!}$ is discussed here.

It's enough to check in even dimension, that is for the unit ball $B_n$ in $\mathbb{C}^n$. The map $z\mapsto \pi |z|^2$ from $\mathbb{C}$ to $[0, \infty)$ is measure preserving, and so is the map $\mathbb{C}^n \to [0, \infty)^n$, $(z_1, \ldots, z_n) \mapsto (\pi |z_1|^2, \ldots, \pi|z_n|^2)$. The product of disks $D^n$ maps to $[0,\pi]^n$, and the unit ball maps to the simplex $\{\sum_{k=1}^n t_k \le \pi\}$, which is $\frac{1}{n!}$ of the cube $[0,\pi]^n$. So that's why

$$\operatorname{ Vol}(B_n) = \frac{\pi^n}{n!}$$

Obs: why the simplex $\{\sum_1^n t_k \le 1\}$ has volume $\frac{1}{n!}$ is discussed here.

$\bf{Added:}$ A similar argument shows that if $d$ is a natural number then the volume of the body

$$\{ \sum_{k=1}^{d n} |x_k|^d \le R^d \}$$ is of the form

$$\frac{\kappa^n R^{d n}}{n!}$$ ( $\kappa$ instead of $\pi$) and so decreases to $0$ as $n\to \infty$. The crucial obervation again is that a map of the form

$$\mathbb{R}^d \to [0, \infty), \ \ (x_1, \ldots, x_d) \mapsto \kappa \cdot F(x_1, \ldots, x_d)$$ where $F$ is a form of degree $d$, is measure preserving( for some $\kappa >0$).

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orangeskid
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It's enough to check in even dimension, that is for the unit ball $B_n$ in $\mathbb{C}^n$. The map $z\mapsto \pi |z|^2$ from $\mathbb{C}$ to $[0, \infty)$ is measure preserving, and so is the map $\mathbb{C}^n \to [0, \infty)^n$, $(z_1, \ldots, z_n) \mapsto (\pi |z_1|^2, \ldots, \pi|z_n|^2)$. The product of disks $D^n$ maps to $[0,\pi]^n$, and the unit ball maps to the simplex $\{\sum_{k=1}^n x_k \le \pi\}$$\{\sum_{k=1}^n t_k \le \pi\}$, which is $\frac{1}{n!}$ of the cube $[0,\pi]^n$. So that's why

$$\operatorname{ Vol}(B_n) = \frac{\pi^n}{n!}$$

Obs: why the simplex $\{\sum_1^n x_k \le 1\}$$\{\sum_1^n t_k \le 1\}$ has volume $\frac{1}{n!}$ is discussed here.

It's enough to check in even dimension, that is for the unit ball $B_n$ in $\mathbb{C}^n$. The map $z\mapsto \pi |z|^2$ from $\mathbb{C}$ to $[0, \infty)$ is measure preserving, and so is the map $\mathbb{C}^n \to [0, \infty)^n$, $(z_1, \ldots, z_n) \mapsto (\pi |z_1|^2, \ldots, \pi|z_n|^2)$. The product of disks $D^n$ maps to $[0,\pi]^n$, and the unit ball maps to the simplex $\{\sum_{k=1}^n x_k \le \pi\}$, which is $\frac{1}{n!}$ of the cube $[0,\pi]^n$. So that's why

$$\operatorname{ Vol}(B_n) = \frac{\pi^n}{n!}$$

Obs: why the simplex $\{\sum_1^n x_k \le 1\}$ has volume $\frac{1}{n!}$ is discussed here.

It's enough to check in even dimension, that is for the unit ball $B_n$ in $\mathbb{C}^n$. The map $z\mapsto \pi |z|^2$ from $\mathbb{C}$ to $[0, \infty)$ is measure preserving, and so is the map $\mathbb{C}^n \to [0, \infty)^n$, $(z_1, \ldots, z_n) \mapsto (\pi |z_1|^2, \ldots, \pi|z_n|^2)$. The product of disks $D^n$ maps to $[0,\pi]^n$, and the unit ball maps to the simplex $\{\sum_{k=1}^n t_k \le \pi\}$, which is $\frac{1}{n!}$ of the cube $[0,\pi]^n$. So that's why

$$\operatorname{ Vol}(B_n) = \frac{\pi^n}{n!}$$

Obs: why the simplex $\{\sum_1^n t_k \le 1\}$ has volume $\frac{1}{n!}$ is discussed here.

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orangeskid
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