 6:09 PM
Does anyone know of an example of a compact Lie group G, where the rational cohomology of BG not a polynomial ring? The rational cohomology of BG is always polynomial when G is connected. 6:41 PM
@OmarAntolín-Camarena It's fine, it's useful reference. But odd they don't even give a reference for that fact (because it is certainly not a triviality, as it involves showing that a certain alternating sum of elements of homotopy groups (at different basepoints!) somehow adds up to 0).
@NiallTaggart How about \$G=U(1)^3\rtimes C_3\$, where \$C_3\$ cyclically permutes the coordinates of \$U(1)^3\$? 7:14 PM
@CharlesRezk How do I take the action of \$C_3\$ into account here? My naive guess is that the cohomology of \$BG\$, in this case, is three copies of the polynomial ring on one generator, which is still polynomial? @NiallTaggart the cohomology of a semi-direct product of \$G\$ with a finite group \$H\$ is the invariants of \$H\$ acting on the cohomology of \$G\$. In particular, using @CharlesRezk you can get any quotient of an affine space by a linear action of a finite group as the spectrum of such ring.
btw im not sure what happen if you force your Lie group to be semi-simple. I guess \$Spin_8\$ with the outer action of \$D_6\$ is a good place to look for example. 8:02 PM
@S.carmeli ah yes, my brain seemed to skip the semi--direct product. I'm not sure I understand the second part of your comment about the quotient of an affine space 8:16 PM
@NiallTaggart \$spec(A^H):=spec(A)//H\$ so the invariants of an action on a polynomial algebra has as spectrum a quotient of an affine space by a group action. If you specifically take a subgroup of \$SL_n(\mathbb{Z})\$ acting on a torus you get the quotient by a linear action.
the reason I talk about it geometrically is that for a linear action the algebra is a polynomial algebra iff the quotient is smooth if im not mistaken. 8:51 PM
@CharlesRezk Isn't it also in Simplicial Homotopy Theory by Goerss and Jardine?