Chart to Interior of $n$-simplex

Question

Taken from here consider the $n$-simpliex $\Delta^n\subseteq\mathbb{R}^{n+1}$ defined by $$ \textstyle \Delta^n = \{x=(x_0,\dots,x_n)\in\Bbb R^{n+1}\mid \sum_0^n x_i=1,\,x_i\ge0\,\forall i \}, $$ whose interior, as discussed in this post, is given by $$ \textstyle \text{int}(\Delta^n) = \{x=(x_0,\dots,x_n)\in\Bbb R^{n+1}\mid \sum_0^n x_i=1,\,x_i>0\,\forall i \}. $$

It seems like this is a 1-chart smooth manifold (and $\Delta^n$ is a 1-chart manifold with boundary), but what exactly would be a smooth map $f:\mathbb{R}^n \rightarrow \text{int}(\Delta^n)$?

Answer 1

One mapping that works is as follows: interpret $\mathbb R^n$ as the subset of $\mathbb R^{n+1}$ of vectors whose final coordinate is 0. Then we can define $f: \mathbb R^n \to \mathrm{int}(\Delta^n)$ as $$ f(v)_i = \frac{\exp(v_i)}{\sum_{i = 0}^n \exp(v_i)}. $$

Answer 2

Hint

$\Delta^n$ is convex. And for any $a \in \mathbb R^n$ and $r>0$, there exists a smooth map $g : \mathbb R^n \mapsto B(a,r)$ where $B(a,r) \subseteq \mathbb R^n$ is the open ball centered on $a$ with radius equal to $r$.

Take $a \in \text{int}(\Delta^n)$ and $r> 0$ small enough such that the open ball $B(a,r)$ is included in $\text{int}(\Delta^n)$.

Because $\text{int}(\Delta^n)$ is convex, for any $x \in B(a,r) \setminus \{a\}$, the half line $l(a,x)$ starting on $a$ and passing through $x$ intersects the boundary of $\Delta^n$ at a single point $p_x$. Denote $$h(x) = a + \frac{\Vert x-a \Vert}{r}(p_x-a)$$ and $h(a) = a$. $h$ is a smooth invertible map with smooth inverse.

The map $f = h^{-1} \circ g$ satisfies the properties you're looking for.