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$W$ is the Weyl group of $G$ an algebraic group, $B$ is a Borel subgroup of $G$.

How do elements $w\in W$ correspond to elements of $G/B$?

I have seen it written 'we shall write $w$ for the element of the Weyl group and the corresponding element of $G/B$. I haven't been able to find anywhere that explains such a claim.

I know that $G/B\leftrightarrow \mathfrak{B}$, where the latter is the variety of all Borels - and I suppose then we can think of $w\in W$ as acting on full flags with bases adapted to the symmetric group $W\leftrightarrow S_m$.

alggeo
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  • https://en.wikipedia.org/wiki/Bruhat_decomposition – Qiaochu Yuan Jun 02 '17 at 04:56
  • @QiaochuYuan It seemed they meant an element $w\in W$ corresponded to an element of $G/B$. So you are saying take the Bruhat decomposition of $G$ and then quotient by $B$, and this will yield a one-to-one point correspondence $W\leftrightarrow G/B$?

    Doesn't that yield the Bruhat-cells, which are not simply points?

     – alggeo Jun 02 '17 at 05:23
  • It is simply not the case that elements of $W$ correspond exactly to elements of $G/B$, at least not without more data. The Bruhat decomposition gets you close, at least. Can you give more context for where you read this? – Qiaochu Yuan Jun 02 '17 at 05:56
  • @QiaochuYuan It was written in a seminar. I imagine what was meant was what is on page 4 end : https://arxiv.org/pdf/1501.00034.pdf – alggeo Jun 02 '17 at 06:31
  • If you assume connected, there is an injection of $W$ in $G/B$. See my answer. However, this is never a bijection. By rigidity, $W$ is finite, but $G/B$ is infinite, being irreducible and of dimension $\geq 1$. – D_S Jun 11 '17 at 16:26
  • You can identify $G/B$ with the set of Borel subgroups of $G$. Under this identification, the injection $W \rightarrow G/B$ gives a bijection between $W$ and the Borel subgroups of $G$ containing $T$. Another way to say this is that $W$ acts simply transitively on such groups. – D_S Jun 11 '17 at 16:30

2 Answers2

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Right, might as well assume $G$ is smooth, connected, affine, reductive, and has a Borel subgroup $B$ containing a maximal torus $T$.

There are two definitions for the Weyl group of $G$: one is as the automorphisms of the root system associated to $G$, one is as the quotient $N_G(T)/T$. They're the same (more precisely there is an isomorphism between the two).

From your previous question we know that given a set of rational points $S\subset G(k)$ we can associate some elements of $G/B$ (with the chosen Borel $B$ we have an isomorphism $G/B\cong \mathfrak{B}$, the variety of Borel subgroups of G; the bijection is then actually an equality between the fixed points of the conjugation action $S\times \mathfrak{B}\rightarrow \mathfrak{B}$ and those points corresponding to Borels containing $S$ in their rational points. To get the elements of $G/B$ that correspond to this, we start with a Borel $B'\supset S$ containing $S$, we need to find an element $g\in G(k)$ so that $gBg^{-1}=B'$, and then we associate $gB\rightsquigarrow B'$).

From here the association of $W=N_G(T)/C_G(T)=N_G(T)/T$ with points in $G/B$ isn't so difficult, almost never useful, but not super difficult. (It's usually more useful to find some rational point in $G(k)$ which represents a point in $G/B$).

Anyways, we have an action of $T$ on $\mathfrak{B}$ by conjugation. We take points $\mathfrak{B}^T$, and I claim these are in bijection with $W$. To see this, we first observe we can extend the action of $T$ on the fixed set $\mathfrak{B}^T$ to an action of $N_G(T)$ on $\mathfrak{B}^T$ (also conjugation).

The action of $N_G(T)$ on $\mathfrak{B}^T$ is transitive: if $T\subset B$ and $T\subset B'$ then find an element $g\in G(k)$ with $gBg^{-1}=B'$. We also have $gTg^{-1}=T'$ with $T'$ another maximal torus. But maximal tori are conjugate, even in $B'$, so we can find a $h\in B'(k)\subset G(k)$ with $hT'h^{-1}=T$. Then $hg$ takes $B$ to $B'$ and normalizes $T$, proving transitivity.

The action factors through $N_G(T)/C_G(T)$ and the induced action is simply transitive: the claim that it factors through the centralizer is an application of rigidity. We could write $B=T U$ for some unipotent group $U$; the action of $g\in C_G(T)(k)$ is trivial on $T$ by assumption, and trivial on $U$ by rigidity, hence trivial on $B$. Conversely, if $g\in N_G(T)$ fixes $B$, i.e. $gBg^{-1}=B$, then $g\in N_G(B)(k)=B(k)$. Hence $g\in N_B(T)$. The subgroup of $G$ generated by the maps $(x,t)\mapsto xtx^{-1}t^{-1}$ from $N_B(T)\times T$ has image in $T\cap [B,B]=T\cap U=\{e\}$, which is contained in the centralizer for obvious reasons.

Eoin
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  • Just a small point, but the Weyl group of an algebraic group is $N_G(T)/Z_G(T)$, which only coincides with $N_G(T)/T$ when $G$ is reductive. You implicitly note this later on but it hasn't been explicitly assumed that $G$ is reductive – Exit path Jun 03 '17 at 03:12
  • @leibnewtz Thanks, I'll just add that to the beginning in my list of assumptions. – Eoin Jun 03 '17 at 03:22
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Assume $G$ is connected. Let $B$ be a Borel subgroup of $G$, and $T$ a maximal torus of $B$. Then $B$ contains $Z_G(T)$ (Springer, 6.4.8). The Weyl group is defined to be $W = N_G(T)/Z_G(T)$.

We define an injection of $W$ into $G/B$ by $nZ_G(T) \mapsto nB$. This is well defined, because $Z_G(T) \subseteq B$.

To show that this is injective, suppose $n_1B = n_2B$. Then

$$n_2^{-1}n_1 \in N_G(T) \cap B = N_B(T) = Z_B(T) = Z_G(T) \cap B = Z_G(T)$$

where the second equality is 6.3.6(ii).

D_S
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