Sometimes I see questions answered on MathOverflow in such a way that I don't really understand the answers. Sometimes I work out what they mean, and other times I can't. I'd like to ask for more clarification, but the answers are often from years ago, and the fact that the things that confuse me are themselves supposed to be answers to someone's question indicates that that further questions are not at a sufficient level for MathOverflow.

So I thought I might try to ask one such question here, just to see what happens.

At https://mathoverflow.net/questions/21651/cohomology-of-flag-varieties, Allen Knutson says the following of a compact Lie group $K$ with maximal torus $T$:

$$H^*_T(K/T) = H^*_{T \times T}(K) = H^*_{T×K×T}(K\times K) = H^∗_{K}(K/T×K/T) = H^*_K(K/T) \otimes_{H^*(BK)} H^*_K(K/T).$$

The equivariant cohomology $H^*_G(X)$ can be thought of, and is sometimes defined as, the singular cohomology of the homotopy quotient $X_G = (EG \times X)/G$, where $X$ is a $G$-space, $EG$ is the total space of a universal principal $G$-bundle, and the action we are quotienting by is the diagonal action $g\cdot(e,x) = (eg^{-1},gx)$.

If the action of $G$ on $X$ is free, the homotopy quotient is homotopy equivalent to the regular quotient: $X_G \simeq X/G$. Given these definitions, how can I see Knutson's isomorphism chain?

The first isomorphism seems to be given by letting the new $T$ factor act on $K$ by right multiplication, so that the total action is $(t_1,t_2)\cdot k = t_1kt_2^{-1}$, and then using the homotopy equivalence above.

The second one seems to be given by interposing a new $K$ factor which is to be quotiented out by a new $K$-action: $(t_1,k,t_2)\cdot (k_1,k_2) = (t_1k_1t_2^{-1},kk_2)$.

The last isomorphism comes from an equivariant Künneth theorem, and implies, to my eyes, that the action of $K$ on $K/T \times K/T$ is given by $k\cdot(k_1T,k_2T) = (kk_1T,kk_2T)$.

On the other hand, the penultimate homotopy equivalence would naively seem be given by quotienting out each factor by a $T$, so that the action of $T \times K \times T$ on $K \times K$ should be given by $(t_1,k,t_2)\cdot(k_1,k_2) = (kk_1t_1^{-1},kk_2t_2^{-1})$.

But this doesn't agree with the other action I'd guessed on $K \times K$. Perhaps what I'm asking is this:

Is there an obvious self-homeomorphism of $K \times K$ that is $(T \times K \times T)$-equivariant in the sense that it takes the one action to the other? Failing that, what is some other homotopy equivalence between the quotients?

I tried to ask this with the apposite "equivariant-cohomology" tag, but I lack the reputation necessary to create it.


1 Answer 1


1) When I stuck the $K$ in the middle it was by seeing $K \cong (K\times K)/K$, where the group divided by acts "in the middle". So your formula $(t_1,k,t_2)\cdot (k_1,k_2) = (t_1 k_1 t_2^{-1}, t_2 k_2)$ is slightly off; it should be $(t_1 k_1 k^{-1}, k k_2 t_2^{-1})$.

2) I then pulled a cheap trick and inverted the first factor in $K\times K$, just so I could have $K$ acting on the left on both factors. This wasn't really necessary, I just thought the notation would be a little nicer.

Put those together and you'll get the action you correctly guessed in your paragraph with "penultimate".

  • $\begingroup$ Just to check that I'm getting this, when we pass from $(K \times K)_{T \times K \times T}$ (with your (in retrospect, natural) action to $(K/T \times K/T)_K$, flipping the actions on the first $K$-coordinate on the way, we're changing the first $T$-action on $EK$ too, right? That is, I think we have $(t_1,k,t_2)\cdot (e,k_1,k_2) = (et_1^{-1},kk_1t_1^{-1},kk_2t_2^{-1})$, but since it's a free action, we replace it by the action $(t_1,k,t_2)\cdot (e,k_1,k_2) = (e,kk_1t_1^{-1},kk_2t_2^{-1})$ to get an equivalent quotient, then contract the $EK$ factor. Is this right? $\endgroup$
    – jdc
    Mar 14, 2014 at 23:52
  • 1
    $\begingroup$ I think it's a mistake to explicitly include $EG$ factors in one of these deeper calculations. Accept that $H_G(X) = H(X/G)$ for free actions and then black-box that away. Partly because, when you move beyond $H^*()$ to $K()$ you don't use the Borel construction any more. $\endgroup$ Mar 15, 2014 at 13:03
  • $\begingroup$ And since I somehow didn't say this earlier, thanks so much for coming here to tell me! $\endgroup$
    – jdc
    Mar 15, 2014 at 16:52
  • $\begingroup$ Allen, it seems like there isn't anything special about $T$ in this construction, so that for other subgroups $S$ and $H$ of $K$, one would find that $H^*_S(K/H) = H^*_S \otimes_{H^*_K} H^*_H$. Is this so? $\endgroup$
    – jdc
    Jun 11, 2014 at 1:17
  • $\begingroup$ Well, the key step is equivariant Kunneth, which I'm happy to apply if e.g. $H^*_S$ and $H^*_H$ are concentrated in even degree, e.g. if $S,H$ are connected. Without that I become squeamish about equivariant Kunneth. $\endgroup$ Jun 11, 2014 at 5:36

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