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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?
 

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