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Let $G$ be a complex, connected, semi-simple Lie group (throw in simply connected if you like) with Lie algebra $\mathfrak g$. Let $T \subseteq B$ be a maximal torus and choice of Borel, respectively. Let $\mathfrak t$ be the Lie algebra of the torus (ie a Cartan subalgebra), $R$ be a root system, and $R_+$ be a positive root system corresponding to our choice of Borel. Take $\mathfrak t_+^* \cong \mathfrak t_+$ to be the positive Weyl chamber.

I would like to see how both the Borel and the positive Weyl chamber change when, instead of considering the group $G$, I consider the group $G \times \mathbb C^\times$.

Certainly the group is no longer semi-simple, but it is still reductive. Gut intuition would suggest that the Borel $\tilde B$ of $G \times \mathbb C^\times$ should just correspond to "adding another torus" (something like $\tilde B = (T\times \mathbb C^\times) \rtimes U$, where $U$ is the unipotent), and the positive Weyl chamber would be$^1$ $\mathfrak t_+ \oplus \mathbb R$.

My effort is as follows:

  1. The Lie algebra of $G \times \mathbb C^\times$ is $\mathfrak g \oplus \mathbb C$, with the bracket acting on each component separately.
  2. There are no "new" roots. Let $\tilde \alpha: \mathfrak t \oplus \mathbb C \to \mathbb C$ be a root. Then for all $(H,z) \in \mathfrak t \oplus \mathbb C$, if $(X,w) \in \mathfrak g \oplus \mathbb C$ then the root condition is \begin{align*} [(H,z), (X,t)] &= ([H,X],0) \\ &=\tilde \alpha(H,z) (X,t). \end{align*} This can only happen if $\tilde\alpha(H,z) = \alpha(H), X \in \mathfrak g_\alpha,t=0$. Hence the $\tilde \alpha$ root space $\mathfrak g_{\tilde \alpha}$ is naturally isomorphic to $\mathfrak g_{\alpha}$. The only thing that changes is the 'zero root space;' i.e. the Cartan picks up an extra dimension.
  3. It is easy to check that if $\langle,\rangle$ is an $\text{Ad}$-invariant inner product of $\mathfrak g$, then $$((X,z),(Y,w)) = \langle X,Y \rangle zw$$ is an $\text{Ad}$-invariant inner product on $\mathfrak g\oplus \mathbb C$.
  4. The positive Weyl chamber is $$ C = \{ (X,z) \in \mathfrak t_{\mathbb R} \oplus \mathbb R: ((X,z), (\alpha, 0)) \geq 0\}.$$ But $((X,z), (\alpha, 0))=0$ regardless of choice of $(X,z)$.

I must have made a mistake somewhere.


[1] I have seen the positive Weyl chamber as both a subset of the full Cartan $\mathfrak t$ and as a subset of the real part of the Cartan $\mathfrak t_{\mathbb R}$ to which the Killing form is positive-definite. If anyone has any insight as to the difference, this would also be great.

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First, the formula in 3. does not define a bilinear form: if you multiply $(X,z)$ by $\lambda$, the RHS is multiplied by $\lambda^2$. You want a sum, not a product.

In Bourbaki, one of the axioms of root system is that the roots should generate the whole space. This reflects only semisimple groups. If you want to describe reductive groups, the good notion is that of root datum $(X, \Phi, Y, \Phi^\vee)$, where $X$ and $Y$ are two lattices in duality (characters and cocharacters), $\Phi$ is the set of roots and $\Phi^\vee$ is the set of coroots. See SGA3.

The root datum of a direct product of reductive groups is the direct product of the root data. So your question is really: what is the root datum of $\mathbb C^*$. The answer is $(\mathbb Z, \emptyset, \mathbb Z, \emptyset)$. A torus has no roots or coroots. The associated real vector space is $\mathbb R$. There are no reflection hyperplanes, there is only one Weyl chamber, the whole line. The condition of having positive pairing with simple coroots is vacuous since there are no coroots.

So if you do the direct product $G \times \mathbb C^*$, the Weyl chambers are just direct products of a Weyl chamber for $G$ by $\mathbb R$. No wonder they are "degenerate": the Killing form is now degenerate, since the group is not semisimple.

Hope that helps. It is more interesting to look at non-semisimple groups which are not direct products. Exercise: determine the root data of $GL(2)$ and compare it with that of $PGL(2) \times \mathbb C^*$.

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  • $\begingroup$ Hi Daniel! Don't you have anything better to do---like, tables or something? $\endgroup$ – Stephen Jun 20 '15 at 17:28

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