Update Math401_P1_3.md

content/Math401/Math401_P1_3.md

@@ -116,9 +116,69 @@ $$
 
 Hardcore computing may generates the bound but M. Gromov did not make the detailed explanation here.
 
-#### Detail proof by Takashi Shioya
+> Detail proof by Takashi Shioya.
+>
+> The central idea is to draw the connection between the given three topological spaces, $S^{2n+1}$, $CP^n$ and $\mathbb{R}$.
 
+First, we need to introduce the following distribution and lemmas/theorems:
 
+**OBSERVATION**
+
+consider the orthogonal projection from $\mathbb{R}^{n+1}$, the space where $S^n$ is embedded, to $\mathbb{R}^k$, we denote the restriction of the projection as $\pi_{n,k}:S^n(\sqrt{n})\to \mathbb{R}^k$. Note that $\pi_{n,k}$ is a 1-Lipschitz function (projection will never increase the distance between two points).
+
+We denote the normalized Riemannian volume measure on $S^n(\sqrt{n})$ as $\sigma^n(\cdot)$, and $\sigma^n(S^n(\sqrt{n}))=1$.
+
+#### Definition of Gaussian measure on $\mathbb{R}^k$
+
+We denote the Gaussian measure on $\mathbb{R}^k$ as $\gamma^k$.
+
+$$
+d\gamma^k(x)\coloneqq\frac{1}{\sqrt{2\pi}^k}\exp(-\frac{1}{2}\|x\|^2)dx
+$$
+
+$x\in \mathbb{R}^k$, $\|x\|^2=\sum_{i=1}^k x_i^2$ is the Euclidean norm, and $dx$ is the Lebesgue measure on $\mathbb{R}^k$.
+
+Basically, you can consider the Gaussian measure as the normalized Lebesgue measure on $\mathbb{R}^k$ with standard deviation $1$.
+
+#### Maxwell-Boltzmann distribution law
+
+> It is such a wonderful fact for me, that the projection of $n+1$ dimensional sphere with radius $\sqrt{n}$ to $\mathbb{R}^k$ is a Gaussian distribution as $n\to \infty$.
+
+For any natural number $k$,
+
+$$
+\frac{d(\pi_{n,k})_*\sigma^n(x)}{dx}\to \frac{d\gamma^k(x)}{dx}
+$$
+
+where $(\pi_{n,k})_*\sigma^n$ is the push-forward measure of $\sigma^n$ by $\pi_{n,k}$.
+
+In other words,
+
+$$
+(\pi_{n,k})_*\sigma^n\to \gamma^k\text{ weakly as }n\to \infty
+$$
+
+<details>
+<summary>Proof</summary>
+
+We denote the $n$ dimensional volume measure on $\mathbb{R}^k$ as $\operatorname{vol}_k$.
+
+Observe that $\pi_{n,k}^{-1}(x),x\in \mathbb{R}^k$ is isometric to $S^{n-k}(\sqrt{n-\|x\|^2})$, that is, for any $x\in \mathbb{R}^k$, $\pi_{n,k}^{-1}(x)$ is a sphere with radius $\sqrt{n-\|x\|^2}$ (by the definition of $\pi_{n,k}$).
+
+So,
+
+$$
+\begin{aligned}
+\frac{d(\pi_{n,k})_*\sigma^n(x)}{dx}&=\frac{\operatorname{vol}_{n-k}(\pi_{n,k}^{-1}(x))}{\operatorname{vol}_k(S^n(\sqrt{n}))}\\
+&=\frac{(n-\|x\|^2)^{\frac{n-k}{2}}}{\int_{\|x\|\leq \sqrt{n}}(n-\|x\|^2)^{\frac{n-k}{2}}dx}\\
+\end{aligned}
+$$
+
+as $n\to \infty$.
+
+$(n-\|x\|^2)^{\frac{n-k}{2}}=\left(n(1-\frac{\|x\|^2}{n})\right)^{\frac{n-k}{2}}\to n^{\frac{n-k}{2}}\exp(-\frac{\|x\|^2}{2})$
+
+</details>
 
 ## References