1

Let $G$ be a connected compact Lie group and let $H$ be a connected Lie subgroup of $G$ such that $G$ and $H$ have the same rank.

I've come across a formula (given under the above assumptions ) Wich contain the expression

$(-2\pi)^{-\operatorname {dim}(G/H)/2}$,

So I thought that maybe the dimension of $G/H$ is even, is this true ?

Aaron Hendrickson
  • 6,042
Mira
  • 940
  • 1
    @Dietrich: Then $\dim G/H=0$, which is even. To the OP: I don't have time to write up a full answer, but yes it is even. One approach is to note that in the equal rank case, there is a formula for the Euler characteristic in terms of the orders of the respective Weyl groups. In particular, it easily follows that the Euler characteristic is non-zero, which implies the dimension is even via Poincare duality. – Jason DeVito - on hiatus Jun 02 '21 at 11:43
  • @Jason De Vito, thanks a lot for your interesting comment! – Mira Jun 06 '21 at 22:10

1 Answers1

3

If the two compact Lie groups have the same rank they have a common maximal torus, because of maximal torus in $H$ must also be maximal in $G$.

Now if $G$ is compact $G/T$ is always even dimensional, so the result follows.

In fact $G/T$ admits a complex structure as the Lie algebra $\cal G$ splits $\cal G= \cal T + \sum E_{\alpha}$ , and $E_{\alpha}$ is a one dimensionla complex representation of $T$, the sum being taken over the roots of $\cal G$.

To prove this, note that the action of $T$ on the orthogonal of $\cal T$ has no fixed vector..

Thomas
  • 7,470