Timeline for connected compact semisimple lie group finite fundamental group

6 events

when toggle format	what		by	license	comment
May 2, 2012 at 1:48	comment	added	Igor Rivin		Thanks, I will check it out! Part of my question is which proof is the quickest from nothing...
May 2, 2012 at 1:04	comment	added	Faisal		The first 'equality' follows from the observation that the complex of left invariant forms computes $H^\ast_{\rm dR}(G)$ (not difficult to prove) together with the fact that said complex can be identified (via evaluation at the identity) with the complex $\wedge^q\mathfrak g^\ast$. You can find all the details in, e.g., Chevalley--Eilenberg.
May 2, 2012 at 1:03	comment	added	Faisal		Depends on how you set things up. E.g. let's view $H^\ast(\mathfrak g;\mathbb R)$ as being computed by the complex $(\wedge^q\mathfrak g^\ast,d)$ (I'll omit the formula for $d$...). Then the second equality is trivial: $f\in\mathfrak g^\ast$ is in $H^1$ iff $df=0$ (there are no 1-coboundaries). The formula for $d$ here is $df(X,Y)=f([X,Y])$, whence $f\in H^1 \iff f\in (\mathfrak g/[\mathfrak g,\mathfrak g])^\ast$.
May 1, 2012 at 23:34	comment	added	Igor Rivin		How easy are the first two equalities?
May 1, 2012 at 15:18	comment	added	Faisal		By the way, this argument is most likely due to Chevalley--Eilenberg. I would give a more precise reference but my erratic internet connection is making this difficult.
May 1, 2012 at 14:51	history	answered	Faisal	CC BY-SA 3.0