Prove that a simplex is a manifold with corners

Question

First of all we remember some elementary definitions and results about manifolds and simplexes.

Definition

A function $f$ defined in a subset $S$ of $\Bbb R^k$ is said of class $C^r$ if it can be extended to a function $\phi$ (said $C^r$-extension) that is of class $C^r$ in a open neighborhood of $S$.

Lemma

If $f$ is a function defined in a subset $S$ of $\Bbb R^n$ such that for any $x\in S$ there exists a function $f_x$ defined in a neighborhood of $x$ that is of class $C^r$ and compatible with $f$ on $U_x\cap S$ then $f$ is of class $C^r$.

Lemma

If $U$ is an open set of $H_n:=[0,+\infty)^n$ then the derivatives of two different extensions $\phi$ and $\varphi$ of a $C^r$-function $f$ agree in $U$.

Definition

A $k$-manifold with corners in $\Bbb R^n$ of class $C^r$ is a subspace $M$ of $\Bbb R^n$ whose points have a neighborhood $V$ in $M$ that is the immage of a homeomorphism $\phi$ of calss $C^r$ defined an open set $U$ of $\Bbb R^k$ or of $H_k$ and whose derivative has rank $k$.

Definition

If $x_0,...,x_k$ are $(k+1)$ affinely indipendent points of $\Bbb R^n$ (which means that the vectors $(x_1-x_0),...,(x_k-x_0)$ are linearly independent) then simplex determined by them is the set $$ \mathcal S:=\Biggl\{x\in\Bbb R^n: x=x_0+\alpha^i\vec v_i\,\,\,\text{and}\,\,\, \sum_{i=1}^k\alpha^i\le1\,\,\,\text{and}\,\,\,\alpha^i\ge0\,\,\,\text{for all}\,i=1,\dots,k\Biggl\} $$ where $\vec v_i:=(x_i-x_0)$ for each $i=1,\dots,k$.

With the previous definition we let try to prove that any simplex is a $k$-manifold with corners. First of all we define the function $f$ from $\Bbb R^k$ to $\Bbb R^n$ by putting $$ f(x):=x_0+x^1\vec v_1+\dots+x^k\vec v_k $$ for each $x\in\Bbb R^k$ and thus we observe that it is a smooth homeomorphism because it is the composition of a translation with a linear and injective map betweeen finite dimensional topological vector spaces -for details about this statement we refer to to the text Functional Analysis of Walter Rudin. So if we show that $f^{-1}[S]$ is a manifold with corners then the statement follows immediately. Then we observe that $$ f^{-1}[\mathcal S]=\Big\{x\in\Bbb R^k:x^1+\dots+x^k\le 1\,\,\,\text{and}\,\,\,x^i\ge0\,\,\,\text{for all}\,\,\,i=1,\dots,k\Big\} $$ and thus we conclude that the set $f^{-1}[\mathcal S]$ is the standard $k$-simplex $\mathcal E_k$ generated by the point $O,O+\hat e_1,\dots,O+\hat e_k$ which (if this can interest) is equal to the intersection of a $k$ parallelotopes $\mathcal P_1,\dots,\mathcal P_k$ with the not negative space $H^k_k$ due to analogous arguments here advanced. Now we find a coordinate patch about any point $x$ of $\mathcal E_k$ and we are doing this assuming that it is $$ x^1+\dots+x^k\lneq 1\,\,\,\,\,\,\text{or either}\,\,\,x^1+\dots+x^k=1 $$ Now the function scalar $g$ defined in $\Bbb R^k$ through the equation $$ g(x):=x^1+\dots+x^k $$ for any $x\in\Bbb R^k$ is trivially continuous (indeed it is a liner map between finite dimensional topological vector spaces or alternatively is finite sum of continuous functions) so that the set $g^{-1}\big[(0,1)\big]$ is open and thus observing that $$ g^{-1}\big[(0,1)\big]\cap\mathcal E_k=\big\{x\in\Bbb R^k:0\lneq x^1+\dots+x^k\lneq 1\big\}\cap\big\{x\in\Bbb R^k:x^1+\dots+x^k\le 1\,\,\,\text{and}\,\,\,x^i\ge0\,\,\,\text{for all}\,\,\,i=1,\dots,k\big\}=\\ \big\{x\in\Bbb R^k:0\lneq x^1+\dots+x^k\lneq 1\big\}\cap\big\{x\in\Bbb R^k:x^i\ge0\,\,\,\text{for all}\,\,\,i=1,\dots,k\big\}=g^{-1}\big[(0,1)\big]\cap H^k_k $$ we conclude that the restriction of the identity map to the set $g^{-1}\big[(0,1)\big]\cap H^k_k$ is a coordinate patch about $x$ in an open set of the not negative space $H^k_k$ when in the first case it is $$ x^1+\dots+x^k\neq 0 $$ too. Otherwise $\mathcal E_k$ is contained into the not negative space $H^k_k$ so that if $x\in\mathcal E_k$ is such that $$ x^1+\dots+x^k=0 $$ then $x$ is just the origin $O$ and thus if we find an open cube centered at $O$ whose intersection with $H^k_k$ is contained in $\mathcal E_k$ then the restriction of the identity map to the intersection between $H^k_k$ and this open cube is a coordinate patch about $O$ defined in an open set of $H^k_k$. So observing that if $x$ is an element lying in the intersection between $H^k_k$ and the open cube $C\Big(O,\frac1k\Big)$ centered at $O$ and of radius $\frac1k$ then $$ x^1+\dots+x^k\le k\cdot\max\big\{|x^i|:i=1,\dots,k\big\}=k\cdot\| x-O\|_\infty<k\cdot\frac1 k=1 $$ so that the statement follows immediately. Now we have to find a coordinate patch supposing that $x$ is an element of $\mathcal E_k$ such that $$ x^1+\dots+x^k=1 $$ but unfortuantely I have some difficulties and thus I ask to do it. Could someone help me, please?

$$ \underline{\text{**ACHTUNG**}} $$

Courteously I ask not to prove the statement by showing that the simplex $\mathcal E_k$ is a trasversal intersection between manifolds with corners because I studied trasversality ONLY for manifold WITHOUT boundary and unfortunately I did NOT REALLY understand the proof of trasversality theorems for manifold with corners: moreover I have to prove the result for pratical purposes (I am studying Continuum Mechanics) and thus I really need to find a coordinate patch about the points of the obliqual face of $\mathcal E_k$, that's all. Thanks for your attention if you have read what is written above.

Answer 1

Above we showed that the subset $\overset{\circ}{\mathcal E}_k$ of $\mathcal E_k$ whose elements have value in $[0,1)$ with respect the scalar function $g$ is open in $H^k_k$ since it is union of the sets $$ g^{-1}\big[\,(0,1)\,\big]\cap H^k_k\,\,\,\text{and}\,\,\,C\Big(O,\frac 1k\Big)\cap H^k_k $$ that are open in $H^k_k$. Now provided that $$ \xi^1+\dots+\xi^k=1 $$ for any $\xi\in\mathcal E_k$ then surely it must be $$ \xi^j\gneq0 $$ for any $j=1,\dots,k$ and in particular for now we assume that this happens for the last coordiante. So in this case we define the map $\varphi$ from $\Bbb R^k$ into $\Bbb R^k$ through the equation $$ \varphi(x):=\big(x^1,\dots,x^{k-1},1-(x^1+\dots+x^{k-1}+x^k)\big) $$ for any $x\in\Bbb R^k$ and thus we are going to prove that this map can be restricted to a coordinate patch about $\xi\in\mathcal E_k$ whose last coordinate is not zero, i.e. we let to prove that the restriction $\phi$ of $\varphi$ at the set $\overset{\circ}{\mathcal E}_k$ is the researched coordinate patch. First of all we observe that $\varphi$ is a diffeomorphism of $\Bbb R^k$ onto $\Bbb R^k$ being an affine map, i.e. it is compostion of a translation with a linear map between finite dimensional topological vector spaces and both these maps are diffeomorphism. So if we prove that $\varphi$ maps $\overset{\circ}{\mathcal E}_k$ into an open set of $\mathcal E_k$ containing $\xi$ then we will proved that $\varphi$ is a coordinate patch about $\xi$ so that we let to do this. Now if $x$ is an elemen of $\overset{\circ}{\mathcal F}$ then it must be $$ x^i\in[0,1) $$ for each $i=1,\dots,k$ since otherwise it would be $$ x^1+\dots+x^{i-1}+1+x^{i+1}+\dots+x^k\le x^1+\dots+x^{i-1}+x^i+x^{i+1}+\dots+x^k<1\Rightarrow\\ x^1+\dots+x^{i-1}+x^{i+1}+\dots+x^k<0 $$ that is impossible if $x$ lies in $\overset{\circ}{\mathcal E_k}$ and thus in $H^k_k$ and so we can conclude that $$ x\in\overset{\circ}{\mathcal E}_k\Rightarrow x\in[0,1)^k\wedge x^1+\dots+x^k\lneq1\Rightarrow\\\varphi(x)\in H^k_k\wedge\varphi^1(x)+\dots+\varphi^k(x)\le1\Rightarrow\varphi(x)\in\mathcal E_k $$ and this proves that $\varphi$ maps $\overset{\circ}{\mathcal E}_k$ into $\mathcal E_k$. So if the last coordinate of $\xi$ is not zero and the others are not negative then surely $$ g(\xi^1,\dots,\xi^{k-1},0)\ge0\,\,\,\text{and}\,\,\,g(\xi^1,\dots,\xi^{k-1},0)=\xi^1+\dots+\xi^{k-1}<\xi^1+\dots+\xi^{k-1}+\xi^k=1 $$ so that $(\xi^1,\dots,\xi^{k-1},0)$ is an element of $\overset{\circ}{\mathcal E}_k$ and in particular it is such that $$ \varphi(\xi^1,\dots,\xi^{k-1},0)=\big(\xi^1,\dots,\xi^{k-1},1-(\xi^1+\dots+\xi^{k-1})\big)=\big(\xi^1,\dots,\xi^{k-1},\xi^k\big)=\xi $$ and thus it is proved that $\xi$ is an element of $\varphi\big[\,\overset{\circ}{\mathcal E}_k\,\big]$. Now we previously define the set $$ \tilde{\mathcal E}_k:=\{x\in\mathcal E_k:x^k\neq 0\} $$ and thus we let prove that $\varphi$ carries $\overset{\circ}{\mathcal E}_k$ onto $\tilde{\mathcal E}_k$. So we first observe that $$ x\in\overset{\circ}{\mathcal E}_k\Rightarrow 0\le g(x)\lneq1\Rightarrow0\lneq1-g(x)\le1\Rightarrow\varphi^k(x)>0\Rightarrow\varphi(x)\in\tilde{\mathcal E}_k $$ and so we conclude that $$ \varphi\big[\,\overset{\circ}{\mathcal E}_k\,\big]\subseteq\tilde{\mathcal E}_k $$ so that we let to prove the opposite inclusion. Now for any element $\xi$ of $\tilde{\mathcal E}_k$ it must be $$ g(\xi)=1\,\,\,\text{or either}\,\,\,g(\xi)<1 $$ but effectively we just proved above that if $g(\xi)=1$ and $\xi^k>0$ then surely $$ \xi=\varphi(x) $$ for any $x\in\overset{\circ}{\mathcal E}$ so that to follow we are supposing that $g(\xi)$ is strictly less than $1$. So we observe that $$ \begin{cases}\xi\in\tilde{\mathcal E}_k\\ g(\xi)\lneq1\end{cases}\Rightarrow \begin{cases}\xi^i\ge0\,\,\,\forall\,i=1,\dots,(k-1)\\\xi^k\gneq0\\ \xi^1+\dots+\xi^{k-1}+\xi^k\lneq1\end{cases}\Rightarrow\\ \begin{cases}\xi^i\ge0\,\,\,\forall\,i=1,\dots,(k-1)\\0\lneq\xi^k\lneq1\\ 0\lneq1-(\xi^1+\dots+\xi^{k-1}+\xi^k)\lneq1\end{cases}\Rightarrow \begin{cases}\xi^i\ge0\,\,\,\forall\,i=1,\dots,(k-1)\\0\lneq1-\xi^k\lneq1\\ 0\lneq1-(\xi^1+\dots+\xi^{k-1}+\xi^k)\lneq1\end{cases}\Rightarrow \begin{cases}\xi^i\ge0\,\,\,\forall\,i=1,\dots,(k-1)\\ 1-(\xi^1+\dots+\xi^{k-1}+\xi^k)\gneq0\\ 0\lneq\xi^1+\dots+\xi^{k-1}+\big(1-(\xi^1+\dots+\xi^{k-1}+\xi^k)\big)\lneq1\end{cases} $$ and thus we conclude that $\big(\xi^1,\dots,\xi^{k-1},1-(\xi^1+\dots+\xi^{k-1}+\xi^k)\big)$ is an element of $\overset{\circ}{\mathcal E}_k$ and so observing that $$ \varphi\big(\xi^1,\dots,\xi^{k-1},1-(\xi^1+\dots+\xi^{k-1}+\xi^k)\biggl)=\\ \biggl(\xi^1,\dots,\xi^{k-1},1-\Big(\xi^1+\dots+\xi^{k-1}+\big(1-(\xi^1+\dots+\xi^{k-1}+\xi^k)\big)\Big)\biggl)=\\ \Biggl(\xi^1,\dots,\xi^{k-1},1-\Big((\xi^1+\dots+\xi^{k-1})+1-(\xi^1+\dots+\xi^{k-1})-\xi^k\Big)\Biggl)=\\ \big(\xi^1,\dots,\xi^{k-1},1-(1-\xi^k)\big)=(\xi^1,\dots,\xi^{k-1},\xi^k)=\xi$$ we conclude that $\varphi$ maps $\overset{\circ}{\mathcal E}_k$ onto $\tilde{\mathcal E}_k$. So we let to prove that $\tilde{\mathcal E}_k$ is open in $\mathcal E_k$ and we are doing this proving that the set $\mathcal E_k\setminus\tilde{\mathcal E}_k$ is closed in $\mathcal E_k$, that is it is the intersection of a closed set of $\Bbb R^k$ with $\mathcal E_k$. Now we remember that any symplex is closed (see here for details) so that the $(k-1)$ standard simplex $\mathcal E_{k-1}$ is closed: so observing that $$ \mathcal E_k\setminus\tilde{\mathcal E}_k=\{x\in\mathcal E_k:x^k=0\}=\\ \{x\in\Bbb R^k:x^1+\dots x^{k-1}\le1\wedge x^i\ge0,\,\forall\,i=1,\dots,(k-1)\wedge x^k=0\}=\\ \{x\in\Bbb R^{k-1}:x^1+\dots+x^{k-1}\le1\wedge x^i\,\forall\,i=1,\dots,(k-1)\ge0\}\times\{0\}=\mathcal E_{k-1}\times\{0\} $$ we conclude that $\mathcal E_k\setminus\tilde{\mathcal E}_k$ is closed in $\Bbb R^k$ (ideed it is product of closed set) and so $\mathcal E_k$ too. So we proved the existence of a coordinate patch when the last coordinate of $\xi$ is not zero. Otherwise if the last coordinate of $\xi$ is zero then surely (remember what observed above) there must exist $j=1,\dots,k-1$ such that $$ \xi^j\neq0 $$ so that let be $\psi$ the diffeormosphism of $\Bbb R^k$ onto $\Bbb R^k$ that interchanges the $j$-th coordinate with the last, i.e. $$ [\psi(x)](i):=\begin{cases}x^k,\,\,\,\text{if}\,\,\,i=j\\ x^j,\,\,\,\text{if}\,\,\,i=k\\ x^i\,\,\,\text{otherwise}\end{cases} $$ for any $x\in\Bbb R^k$. Now $\psi$ is an involution that maps $\mathcal E_k$ onto $\mathcal E_k$ and in particular if the last coordinate of $\xi$ is zero then this does not happen for $\psi^{-1}(\xi)$ so that if $\varphi$ is a coordinate patch about $\psi^{-1}(\xi)$ then the composition $\psi\circ\varphi$ of $\psi$ with $\varphi$ is just a coordinate patch about $\xi$ and so the statement finally holds.