# Affine equivalence of Coxeter permutahedra?

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?

• Note that the term reflection group already refers to this "standard realization". You probably meant Coxeter group. – M. Winter Jan 24 '20 at 14:45

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.