Can this binary operation of vectors on a simplex be mapped to addition in a vector space in general? I'm working with a problem where I have $N$-dimensional vectors in the first orthant confined to the unit simplex, i.e. their components satisfy
\begin{align}
     v_i & > 0 \ \forall\, i\text{ and} \\
     \sum_{i=1}^N v_i & = 1.
\end{align}
Call the space these vectors are in  $\Delta^{N-1}$ for the simplex it is (excluding the boundary).
Define the binary operation for $v$, $w\in \Delta^{N-1}$
\begin{align}
    v\odot w & = \frac{v_i w_i}{\sum_j v_j w_j} \\
     &\equiv u.
\end{align}
This operation defines an abelian group. Clearly $u\in \Delta^{N-1}\ \forall \ v,\ w$, so it is closed. It is obviously commutative. It is also associative
\begin{align}
   u\odot(v\odot w) & = \frac{u_i \frac{v_i w_i}{\sum_j v_j w_j}}{\sum_k u_k \frac{v_k w_k}{\sum_j v_j w_j}} \\
   & = \frac{u_i v_i w_i}{\sum_k u_k v_k w_k} \\
   & = (u\odot v)\odot w.
\end{align}
The identity element is obvious $e_i = \frac{1}{N}\ \forall\ i$. The inverse element is likewise obvious $[v^{-1}]_i = \frac{v_i^{-1}}{\sum_{j=1}^N v_j^{-1}}$. (The inverse element requirement is the reason for excluding the boundary vectors).
Is there a mapping from $\Delta^{N-1}$ to $\mathbb{R}^{N-1}$ that maps the $\odot$ operation to vector addition? 
The case for $N=2$ is actually pretty straightforward. If we map vectors to
\begin{align}
    \phi_v &= \ln\left(\frac{v_1}{v_2}\right)
\end{align}
then $\phi_v + \phi_w$ will have the same value as $\ln(u_1/u_2)$. How can this be generalized?
 A: Here is how I see this example: we will define an operation $S: \mathbb R^N \to \Delta^{N-1}$, given by
$$
S(x)_i = \frac{\exp(x_i)}{\sum_{j=1}^N \exp(x_j)}.
$$
You can check for yourself that this is a surjective mapping (onto the interior of the simplex) and that $S(x+y) = S(x) \odot S(y)$. Thus, it becomes a surjection of vector spaces if you port over the scalar multiplication via $S$. The kernel of this surjection is precisely the constant vectors: the subspace $\{\langle c, \ldots, c \rangle \in \mathbb R^N \mid c \in \mathbb R\}$. Now you can take any complementary subspace -- in this case that just means any subspace that does not contain a non-zero constant vector -- of this kernel, and then $S$ restricts to an isomorphism on this subspace. For example, you could take $\{x \in \mathbb R^N \mid x_N = 0\}$, or $\{x \in \mathbb R^N \mid \sum_{j=1}^N x_j = 0\}$. The latter is elegant, but the former looks more like $\mathbb R^{N-1}$, which is what you're looking for.
Now all that remains is to compute an inverse. If $v \in \Delta^{N-1}$, then the pre-images of $v$ under $S$ are precisely
$$
(\log(v_1) + \alpha, \ldots, \log(v_N) + \alpha).
$$
We need the one where $\log(v_N) + \alpha = 0$, so the result is exactly
$$
(\log(v_1) - \log(v_N), \ldots, \log(v_{N-1}) - \log(v_N), 0) = \left(\log\left(\frac{v_1}{v_N}\right), \ldots, \log\left(\frac{v_{N-1}}{v_N}\right), 0\right).
$$
N.B.: If you are interested in machine learning, you will recognize $S$ as the so-called $\mathrm{SoftMax}$ function.
