I was reading this paper on word vectorization recently and the author says this:
F be a homomorphism between the groups ($\mathbb{R}$;+) and ($\mathbb{R}_{>0}$; $\times$), i.e.,
$F((w_i-w_j)^T\tilde{w}_k) = \frac{F(w_i^T\tilde{w}_k)}{F(w_j^T\tilde{w}_k)}$
Here $w_i$, $w_k$ and $\tilde{w}_k$ are all matrix with real values. And F is an undefined function
I can understand this if he said F is a homomorphism between ($\mathbb{R}$;-) and ($\mathbb{R}_{>0}$; /). But he did not say that. And I can't figure out a way to prove that if F is a homomorphism from plus to multiply it can transform subtraction to division.
Can someone help me prove the author is right?
Here's a link to the paper: GloVe: Global Vectors for Word Representation.
First note that $w_i, w_j, \tilde{w}_k$ are elements of $\mathbb R^d$, i.e. vectors. Obviously they are regard as column vectors (a column vector is a $d \times 1$-matrix). In this interpretation, for $a, b \in \mathbb R^d$ the expression $a^Tb$ is in fact a matrix product - but it is nothing else than the usual dot product $a \cdot b$ in $\mathbb R^d$.
The authors begin by considerung very general functions $F$, but then
A little later they get even more restrictive:
This means that we get $$F((w_i - w_j)^T\tilde{w}_k ) = F(w^T_i \tilde{w}_k - w^T_j \tilde{w}_k ) = \frac{F(w^T_i \tilde{w}_k)}{F(w^T_j \tilde{w}_k)} .$$ Here we used $F(a-b) = F(a+(-b)) = F(a)(F(-b)$ and $F(-b) = 1/F(b)$. The latter holds because $1 = F(0) = F(b + (-b)) = F(b)F(-b)$.