Jtdylktuy

Definition of standard $n$-simplex

The standard $n$-simplex contains all points $vec{x} in mathbb{R}^{n + 1}$ such that $0 le x_i le 1$ and $vec{x} cdot vec{1} = 1$.

My definition of the area of $n$-simplex

$$A_{n} = frac{sqrt{n+1}}{n!}.$$

The standard $2$-simplex is an equilateral triangle with side length $sqrt{2}$ and vertices at $(1, 0, 0)$, $(0, 1, 0)$, and $(0, 0, 1)$. The area is $A_{2} = dfrac{sqrt{3}}2$.

The standard $1$-simplex is a line with vertices at $(1, 0)$ and $(0, 1)$. The length is $A_{1} = sqrt{2}$

I found the area of the standard $n$-simplex by first moving the simplex from $mathbb{R}^{n+1}$ to $mathbb{R}^{n}$.

Area of standard simplex

Question

How to directly integrate over the standard $n$-simplex with the coordinates of $mathbb{R}^{n+1}$?

I can't use the change of coordinate formula because the Jacobian of the transformation from $mathbb{R}^{n}$ to $mathbb{R}^{n+1}$ is rectangular.

Thanks.

Let $Delta^nsubseteq mathbb{R}^{n+1}$ be the standard $n$-simplex: $$Delta^n:=big{(x_0,x_1,x_2,ldots,x_n)inmathbb{R}_{geq 0}^{n+1},big|,x_0+x_1+x_2+ldots+x_n=1big}.$$ An arbitrary function $f:Delta^nto mathbb{R}$ can be considered as a function in $n$ free variables. That is, there exists a unique function $F:Sigma^nto mathbb{R}$ such that
$$f(x_0,x_1,x_2,ldots,x_n)=F(x_1,x_2,ldots,x_n)$$ for all $(x_0,x_1,x_2,ldots,x_n)inDelta^n$, where $$Sigma^n:=big{(x_1,x_2,ldots,x_n)subseteqmathbb{R}_{geq 0}^n,big|,x_1+x_2+ldots+x_nleq1big}subseteq mathbb{R}^n,.$$ In other words, $F$ is given by $$F(x_1,x_2,ldots,x_n):=fleft(1-x_1-x_2-ldots-x_n,x_1,x_2,ldots,x_nright),,$$ for all $(x_0,x_1,x_2,ldots,x_n)inDelta^n$. That is, if $nu_n$ is the volume measure on $Delta^n$ and $lambda_n$ is the Lebesgue measure on $mathbb{R}^n$, then
$$int_{Delta^n},f,text{d}nu_n=sqrt{n+1},int_{Sigma^n},F,text{d}lambda_n,.$$
This is because $$text{d}nu_n(x_0,x_1,x_2,ldots,x_n)=sqrt{n+1},text{d}lambda_n(x_1,x_2,ldots,x_n)$$
for all $(x_0,x_1,x_2,ldots,x_n)inDelta^n$ (this can be proven using the Jacobian determinant). In particular, if $fequiv 1$ so that $Fequiv 1$, then we get
$$text{vol}_nleft(Delta^nright)=sqrt{n+1},text{vol}_nleft(Sigma^nright)=frac{sqrt{n+1}}{n!},,$$
where $text{vol}_n$ is the $n$-dimensional volume.

To clarify some points, first note that there exists an isometry from the affine hyperplane $$H^n:=big{(x_0,x_1,x_2,ldots,x_n)inmathbb{R}^{n+1},big|,x_0+x_1+x_2+ldots+x_n=1big}$$
to $mathbb{R}^n$. We can take the isometry to be the unique affine map $varphi:H^ntomathbb{R}^{n}$ which sends $e_0,e_1,e_2,ldots,e_ninmathbb{R}^{n+1}$ to $$0,E_1+alpha_n, E,E_2+alpha_n, E,ldots,E_n+alpha_n, Einmathbb{R}^n,,$$ respectively, where $e_0,e_1,e_2,ldots,e_n$ are standard basis vectors of $mathbb{R}^{n+1}$, $E_1,E_2,ldots,E_n$ are standard basis vectors of $mathbb{R}^n$, $E:=E_1+E_2+ldots+E_n$, and
$$alpha_n:=frac{sqrt{n+1}-1}{n},.$$
Write $E_0:=-alpha_n,E$. Let $T:mathbb{R}^ntomathbb{R}^n$ be the unique linear transformation that sends $$E_1,E_2,ldots,E_ntext{ to }E_1-E_0,E_2-E_0,ldots,E_n-E_0,,$$ respectively. Prove that $$det(T)=1+n,alpha_n=sqrt{n+1},.$$
The volume measure $nu_n$ on $Delta^n$ is inherited from the volume measure on $H^n$, which is the pullback $varphi^*lambda_n$ of $lambda_n$. However, since $T$ maps the extreme points $0,E_1,E_2,ldots,E_n$ of $Sigma^n$ to the extreme points $0,E_1-E_0,E_2-E_0,ldots,E_n-E_0$ of $varphi(H^n)$, it is simpler to take the integral on $Sigma^n$, using the Change-of-Variables Theorem. That is,
$$begin{align}text{d}nu_n(x_0,x_1,ldots,x_n)&=text{d}(varphi^*lambda_n)left(x_0,x_1,ldots,x_nright)\&=text{d}lambda_nbig(varphileft(x_0,x_1,ldots,x_nright)big)\
&=text{d}lambda_nleft(x_1+alpha_n,sum_{i=1}^n,x_i,x_2+alpha_n,sum_{i=1}^n,x_i,ldots,x_n+alpha_n,sum_{i=1}^n,x_iright)
\&=text{d}lambda_nbig(T(x_1,x_2,ldots,x_n)big)\&=text{d}(T^*lambda_n)(x_1,x_2,ldots,x_n)\&=det(T),text{d}lambda_n(x_1,x_2,ldots,x_n)\&=sqrt{n+1},text{d}lambda_n(x_1,x_2,ldots,x_n)end{align}$$
for all $(x_0,x_1,ldots,x_n)inDelta^n$.