Define $sn: W\rightarrow \{1,-1\}$ by $sn(\sigma)=(-1)^{l(\sigma)}.$ Prove that $sn$ is a homomorphism (where $W$ is the Weyl group and $l$ is the length function of reduced expressions in $W$).
Given $\sigma_1,\sigma_2\in W$, write $\sigma_1=\sigma_{\alpha_1}\cdots\sigma_{\alpha_t}$ and $\sigma_2=\sigma_{\beta_1}\cdots\sigma_{\beta_t}$ reduced, for $\alpha_i,\beta_i\in \Delta$ (not necessarly distinct). Then the homomorphism condition ($sn(\sigma_1\sigma_2)=sn(\sigma_1)sn(\sigma_2) )$ would follow if we could show that $l(\sigma_1\sigma_2)=l(\sigma_1)+l(\sigma_2)$. But i can't see how this is true. For me i can only see that $\leq$, not equality. Any insights on how to prove this statement? Trying to write $\sigma_1\sigma_2$ using the above identites didn't help; and also the fact that $l(\sigma)=n(\sigma)$ didn't help too.
Since you have $W$ defined as a subgroup of $GL(V)$, you can simply consider the determinant map $\det:W\to\mathbb R$. This is a homomorphism, of course, and you can trivially show that on each reflection it takes the value $-1$. It follows, of course, that for all $\sigma\in W$ we have $\det\sigma=(-1)^{l(\sigma)}$. As an immediate consequence of this, your map $\operatorname{sn}$ is a homomorphism, since it coincides with $\det$.