Affine equivalence of Coxeter permutahedra?

Question

Suppose that $W=\langle s_1,\ldots, s_d\mid (s_is_j)^{m_{ij}}=e\rangle$ is a finite reflection group and consider its standard $d$-dimensional geometric realization (i.e., the Tits representation) $\rho: W\rightarrow V$.

For any $a\in V$, one can consider the convex hull $P(W;a)$ of the orbit $W\cdot a$, often called a (Coxeter) permutahedron in the literature. It is a standard fact that if $a$ and $b$ are generic points in $V$ (i.e., neither is fixed by any reflection), then the resulting permutahedra are necessarily combinatorially equivalent (essentially, this is because, without loss of generality, $a$ and $b$ can be taken to lie in the same fundamental region/Coxeter chamber).

My question is this: in the literature I have read, authors seem to say that $P(W;a)$ and $P(W;b)$ are only combinatorially equivalent. Why aren't they actually affinely equivalent (and in fact $W$-equivariantly so)? It would seem that sending $a$ to $b$ and each $s_i\cdot a$ to $s_i\cdot b$ would fit the bill for this, since $\{s_i\cdot x\mid 1\leq i\leq d\}$ forms a basis for $V$ for all generic $x$. What am I missing?

Answer 1

Consider the image below. These polygons are permutahedra of the reflection group $I_2(3)$ (symmetry group of the triangle), but they are not affinely equivalent.

Let $T$ be the affine transformation that maps $a\mapsto b$ and $s_i\cdot a\mapsto s_i\cdot b$. Usually there are further $s\in W$ other than the generators $s_1,...,s_d$. The vertex $s\cdot a$ is mapped to $s\cdot b$ if $s$ and $T$ commute, but this is most often not the case.