5
$\begingroup$

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?

$\endgroup$
5
$\begingroup$

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").

$\endgroup$
  • $\begingroup$ Oh -- you beat me to it! Well, I'm glad that our answers are equivalent. $\endgroup$ – Marty Aug 9 '18 at 3:35
  • $\begingroup$ @Marty, well, you sling the language better than I do. :-) I like your description of why the identity components are equal. $\endgroup$ – LSpice Aug 9 '18 at 10:44
6
$\begingroup$

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.

------------------update-------------------

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.

$\endgroup$
  • $\begingroup$ Many thanks for your answer Marty. I feel like I should've figured this out, but sometimes it's nice to see how someone else thinks of these things. I'll see you around next time you're in Berkeley or at BAANTAG. $\endgroup$ – Alexander Aug 9 '18 at 5:42
  • $\begingroup$ The "$D = R$ if for all $\alpha$ …" part is clear, but is the converse really obvious? $\endgroup$ – LSpice Aug 9 '18 at 18:39
  • 1
    $\begingroup$ I think so, since the $\alpha^\vee$ are linearly independent (over $Q$)... I'm taking simple roots. Or am I misunderstanding or making a dumb mistake? $\endgroup$ – Marty Aug 9 '18 at 20:46
  • $\begingroup$ Of course, sorry. I missed that you were taking simple roots. $\endgroup$ – LSpice Aug 10 '18 at 1:42
  • $\begingroup$ You are right. Let me reformulate what JMO and Osse shows in more algebraic terms: First of all everything behaves well with respect to direct products so we can assume that the root datum is indecomposable (note I'm not saying simple here). Then we have two disjoint possibilities: 1) The root datum is simple and adjoint of type $B_n$ ($n=1$ is also allowed) OR 2) we have R=D in the notation above. $\endgroup$ – Kasper Andersen Aug 10 '18 at 7:58
2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ $T^W$ doesn't equal $\mathrm Z(G)$ for $G = \operatorname{PGL}_2$ even if $p = 2$. Maybe you are passing to reduced subschemes? (Anyway, this is a nice answer!) $\endgroup$ – LSpice Aug 9 '18 at 10:40
  • $\begingroup$ Also, in the language of algebraic groups, there is no such thing as $\operatorname{PSL}_3$, or at least it's misleading; assuming you mean $\operatorname{SL}_3/\mathrm Z(\operatorname{SL}_3)$, it's the same as $\operatorname{PGL}_3$, so one usually calls it that to avoid confusion with the rational-points quotient $\operatorname{SL}_3(k)/\mathrm Z(\operatorname{SL}_3(k))$. $\endgroup$ – LSpice Aug 9 '18 at 10:42
  • 1
    $\begingroup$ I'm unfortunately somewhat ignorant when it comes to algebraic groups. At least the statement is true for compact Lie groups and the proofs just use root system yoga. So is the following statement correct: $T^W = Z(G) \times (\mathbb{Z}/2)^r$ where $r$ is defined as before (ignoring the restriction on $\text{char}(k)$)? $\endgroup$ – Kasper Andersen Aug 9 '18 at 11:15
  • 1
    $\begingroup$ Quick answer: No the Matthey paper is unfortunately unpublished (but he refers to Osse). Both JMO and Osse answers the question: To what extent can one reconstruct the root datum from the Weyl group action? The short answer is that the only indeterminacy comes from direct factors of $G$ isomorphic to $\text{Sp}_n$ or $\text{SO}_{2n+1}$. From this one gets the center formula after some work. $\endgroup$ – Kasper Andersen Aug 9 '18 at 13:52
  • 1
    $\begingroup$ I've updated my answer, which might explain the connection to $SO_{2n+1}$-factors. $\endgroup$ – Marty Aug 9 '18 at 17:10
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Dear Prof. Humphreys, The question arose because a friend and I happened to be interested in the following situation. One has a reductive group $G$ defined over a number field $F$ with maximal torus $T$ and one wants to understand when the Galois invariants of the complex dual torus $\hat{T}^{\Gamma}$ are contained within $Z(\hat{G})$. In the case where $G$ is split, we concluded that the $\Gamma$-action on $\hat{T}$ factors through the Weyl action. Therefore in the particular case where $G=SL_2$, so that $\hat{G}= PGL_2$ and $T$ is an anisotropic maximal torus $\endgroup$ – Alexander Aug 12 '18 at 0:25
  • $\begingroup$ The $\Gamma$ invariants are not in the center for the reason discussed above. So even if it's not a natural question, it happened to be relevant! $\endgroup$ – Alexander Aug 12 '18 at 0:27
  • 1
    $\begingroup$ @Alexander: It's probably useful even at this point to revise your header and the inadequate subject tags. For example, "Weyl [group] invariants" basically have nothing to do with the actual question. The invariants were actually worked out by Chevalley. Note too that you are concerned with algebraic groups (in practice semisimple) over number fields of characteristic 0. (So a tag 'nt.nunber-theory' would perhaps be appropriate.) $\endgroup$ – Jim Humphreys Aug 15 '18 at 14:54
  • $\begingroup$ @Jim_Humphreys Ok thanks for the feedback. I hope the changes I have made are an improvement. $\endgroup$ – Alexander Aug 15 '18 at 18:42
  • $\begingroup$ @Alexander: The minor changes to your header are too weak, so I'll stick with my own answer. $\endgroup$ – Jim Humphreys Aug 18 '18 at 23:36

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.