Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 111 / 3

Use the finite-dimensional vector-space topology on $V^{\ast }$: relative to the dual basis, it is the ordinary Euclidean topology on ${\mathbb{R}}^{|I|}$. Each coordinate map $f\mapsto f(e_i)$ is continuous, sois closed, while $A_i$, $A_i^{-}$, and the finite intersection $C={\bigcap }_i{A}_i$ are open. Their closures are

Every ${\sigma }^{\ast }(w)$ is an invertible linear map with inverse ${\sigma }^{\ast }(w^{-1})$. Linear maps between finite-dimensional topological vector spaces are continuous, so both it and its inverse are continuous. Hence every ${\sigma }^{\ast }(w)$ is a homeomorphism.

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For $f\in C$, duality givesThe positive-root criterion for Coxeter length saysis a positive root. Its basis coefficients are nonnegative and not all zero, so $f(\sigma (w^{-1})e_i)>0$ and ${\sigma }^{\ast }(w)(C)\subseteq A_i$. If the length decreases, that root is negative and the same calculation gives ${\sigma }^{\ast }(w)(C)\subseteq A_i^{-}$.

If ${\sigma }^{\ast }(w)$ is the identity and $w\ne 1$, choose a left descent $x_i$ from the first letter of a reduced expression for $w$. The preceding result givesalthough $C\subseteq A_i$. The two open half-spaces are disjoint, a contradiction. Thus the Dual geometric representation of a Coxeter group is faithful.

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Suppose ${\sigma }^{\ast }(w)(C)$ meets ${\sigma }^{\ast }(v)(C)$. Applying the homeomorphism ${\sigma }^{\ast }(v^{-1})$ shows that $C$ meets ${\sigma }^{\ast }(v^{-1}w)(C)$. If $v^{-1}w\ne 1$, choose one of its left descents. Part b places its image of $C$ in the corresponding negative half-space, while $C$ lies in the positive half-space. This is impossible, so $v=w$. The unionis disjoint.

When $W$ is finite, the closures ${\sigma }^{\ast }(w)(D)$ are the simplicial chambers cut out by the reflecting hyperplanes. Intersecting them with a sphere centred at the origin gives simplices. A face of type $J\subseteq I$ has stabilizer the standard parabolic subgroup $W_J$, so its translates are indexed by cosets $w{W}_J$, with reverse inclusion of cosets encoding incidence. This is precisely the Coxeter complex of $(W,I)$.

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If $W$ is finite, its finitely many reflecting hyperplanes divide $V^{\ast }$ into chambers, and the closures of these chambers are exactly the translates ${\sigma }^{\ast }(w)(D)$. Hence the Tits coneequals $V^{\ast }$.

Conversely, suppose $W$ is infinite and choose $f\in -C$, so $f(e_i)<0$ for every $i$. If $f\in U$, then $f\in {\sigma }^{\ast }(w)(D)$ for some $w$, and thereforefor every $i$. Since $f$ is strictly negative on every positive root and strictly positive on every negative root, each $\sigma (w)e_i$ must be negative. The length criterion givesfor every $i$. This contradicts the stated fact that in an infinite Coxeter group the length of every element can be increased by multiplication on the right by some simple generator. Thus $-C$ is not contained in $U$, and $U\ne V^{\ast }$.

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