Fixed Points of the Weyl Group action on a Maximal Torus and the Center of a Reductive Group

Question

Suppose $G$ is a connected reductive group over an algebraically closed field. Then given a maximal torus $T$, we can define a Weyl group $W$ and consider $T^W$, the Weyl-invariants of $T$. This clearly contains the center $Z(G)$ but can be larger. For instance, in $\mathrm{PGL}_2$, the element with $\mathrm{GL}_2$ representative $\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ is not central but is a Weyl-invariant.

Are the connected components of these groups the same? Is there a simple condition one can place on $G$ that forces these groups to be equal?

Answer 1

Notice that nothing in the problem is harmed by base change, so that it doesn't matter that the ground field is algebraically closed.

The identity component of $T^W$ is generated by the images of the $W$-invariant cocharacters of $T$, and the $W$-invariant part of the cocharacter lattice consists precisely of the central cocharacters. (Proof: try applying the reflection in a root to a cocharacter.) Thus we always have the desired equality of identity components.

In general $X^*(T^W) = X^*(T)_W$, whereas $X^*(\mathrm Z(G)) = X^*(T)/Q$, where $Q$ is the root lattice; so your question about equality $T^W = \mathrm Z(G)$ is asking whether $$ Q = X^*(T)(W) = \mathbb Z\{\langle X^*(T), \alpha^\vee\rangle\alpha \mid \alpha \in \Phi\}. $$ This certainly happens if $G$ is simply connected (since then the fundamental weights lie in $X^*(T)$), but not only then (for example, it also works if $G = T$ is a torus). I think that it is probably equivalent to the derived group of $G$ being simply connected, but I have not checked (and now @KasperAndersen points out that it is not true! Let's guess again: "the fundamental group of $\mathcal D(G)$ has odd order" might do, or perhaps "2 is not a torsion prime").

Answer 2

Let $X = Hom(T, G_m)$ and $Y = Hom(G_m, T)$ be the character and cocharacter lattices of $T$, respectively. Let $k$ be the (algebraically closed) ground field. Note that $T = Spec(k[X])$ (a canonical identification).

Let $Z$ be the center of $G$. Then $Z = Spec(k[X / R])$ where $R$ is the root lattice in $X$. On the other hand, $T^W = Spec(k[X_W])$, where $X_W$ denotes the $W$-coinvariants, i.e., $$X_W = X / \langle x - wx : x \in X, w \in W \rangle.$$ Let $D = \langle x - wx : x \in X, w \in W \rangle \subset X$. Note that $D$ is certainly contained in $R$, since $x - wx$ will always be a sum of roots. Tracking things through, the inclusion $D \hookrightarrow R$ corresponds to a surjection $X / D \twoheadrightarrow X/R$ and via $Spec(K[\bullet])$ to the inclusion $Z \hookrightarrow T^W$.

The short exact sequence $$1 \rightarrow R/D \rightarrow X/D \rightarrow X/R \rightarrow 1$$ corresponds to a short exact sequence of groups of multiplicative type, $$1 \rightarrow Z \rightarrow T^W \rightarrow Spec(k[R/D]) \rightarrow 1.$$ So you can see that the deviation between $Z$ and $T^W$ is precisely measured by $R/D$.

Note that $D$ contains $2 \alpha$ for every root $\alpha$ (note $x - wx = 2 \alpha$ if $x = \alpha$ and $w = s_\alpha$). Hence $D$ is a full-rank sublattice of $R$, and so $R/D$ is finite. It follows that the homomorphism $Z \rightarrow T^W$ restricts to an isomorphism on neutral components.

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Let's look at $D$ a bit more closely, to determine when $D = R$. Fix a system of simple roots $\Delta$. Then $$D = \langle x - wx : x \in X, w \in W = \langle x - s_\alpha(x) : x \in X, \alpha \in \Delta \rangle.$$ But $x - s_\alpha(x) = \langle x, \alpha^\vee \rangle \alpha$. So while the root lattice is given by $R = {\mathbb Z} \Delta$, the sublattice $D$ is generated by $\langle x, \alpha^\vee \rangle \alpha$ for all $\alpha \in \Delta$.

Let $G'$ be the derived subgroup, and $T' = T \cap G'$, with cocharacter lattice $Y'$ and character lattice $X'$. Then $Y' = Y \cap (R^\vee \otimes {\mathbb Q})$, where $R^\vee$ is the coroot lattice, and $$X' = X / \{ x \in X : \forall \alpha \in \Delta, \langle x, \alpha^\vee \rangle = 0 \}.$$

Thus $D$ only depends on $G'$.
$$D = \langle \langle x, \alpha^\vee \rangle \alpha : x \in X', \alpha \in \Delta \rangle.$$ Now $D = R$ if and only if for all $\alpha^\vee$, the exists $x \in X'$ such that $\langle x, \alpha^\vee \rangle = 1$. Now what to do...

Decompose the root system into irreducible subroot systems, yielding $\Delta = \Delta_1 \sqcup \cdots \Delta_m$. Suppose that $\alpha \in \Delta_j$ and $\Delta_j$ does not have type $B$. Avoid $A_1 = B_1$ too! Then, there exists a root $\beta$ such that $\langle \beta, \alpha^\vee \rangle$ is odd. It follows that there exists $x \in X'$ such that $\langle x, \alpha^\vee \rangle = 1$.

On the other hand, if $\Delta_j$ does have type $B$ (or type $A_1 = B_1$), then $R = D$ implies the existence of $x \in X'$ such that $\langle x, \alpha^\vee \rangle = 1$ for the unique long simple coroot $\alpha^\vee$.

So, what does this all mean? Consider the simply-connected cover of $G'$, which fits into a short exact sequence, $$1 \rightarrow \pi_1(G') \rightarrow G_{sc} \rightarrow G' \rightarrow 1.$$
The simply-connected group $G_{sc}$ is a direct product of simple factors; I think that $R = D$ is equivalent to the statement that for every type $B$ factor $Spin_{2n+1}$ in $G_{sc}$, the induced map $Spin_{2n+1} \rightarrow G'$ is injective (i.e., it doesn't factor through $SO_{2n+1}$). I haven't checked the details... but it seems consistent with Kasper Andersen's discussion for compact Lie groups.

Answer 3

In general one has $T^W = Z(G) \times (\mathbb{Z}/2)^r$ where $r$ is the number of direct factors in $G$ isomorphic to $\text{SO}_{2n+1}$ ($n\geq 1$) except for the case $\text{char}(k)=2$ where $r=0$. The case $\text{PGL}_2$ considered above is isomorphic to $\text{SO}_{3}$ (note $A_1=B_1$) so $r=1$ (unless $\text{char}(k)=2$ where $r=0$). Note also that $r=0$ if the derived group of $G$ is simply connected. However the converse doesnt hold, ie consider $G=PSL_3$ ($\text{char}(k)\neq 2$). References (for the compact Lie group case) are Proposition 3.2(iv) (Update: this should be 3.2(vi)) in Jackowski, McClure and Oliver - Self-homotopy equivalences of classifying spaces of compact, connected Lie groups (MSN), Section 4 in Osse - $\lambda$-structures and representation rings of compact, connected Lie groups (MSN) or Theorem 1.6 in the paper "Normalizers of maximal tori and cohomology of Weyl groups" by M. Matthey.

Answer 4

To me the question itself (and the answers) are out of focus, starting with the claim that the ring of Weyl group invariants is somehow central. Chevalley's 1955 argument does show that this ring is a polynomial ring, generalizing the classical ring generated by elementary symmetric polynomials (equal in number to the rank of the derived group). But the extra arguments by Harish-Chandra to realize the center of the universal enveloping algebra of the Lie algebra involve a more subtle $\rho$-shift. (There are other arguments needed in prime characteristic.)

It's helpful here to avoid the magic word reductive, since the main issue is the semisimple case. This group always has a finite center, trivial if the group is of adjoint type. In general a reductive group is the almost-direct product of a central torus and a connected semisimple group.