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Mike Miller
Mike Miller
6:36 PM
If someone wants me I'll check this occasionally.
 
 
2 hours later…
loch
loch
8:42 PM
@MikeMiller Consider EU(n) -> BU(n). Let T be a macimal torus. Then we have an action T on EU(n), which then gives us BT-> BU(n). I’m not sure how to see that the induced map on cohomology is injective ! (Eventually I want to say that the image is precisely the Sn invariant ring of H*BT)
 
loch
loch
9:12 PM
I suspect the key word here is Leray Hirsch
 
Mike Miller
Mike Miller
@loch What do we know (if anything?)
 
loch
loch
We know that BT -> BU(n) is a fibre bundle with fibres U(n)/T, which is the (complete) flag variety
 
Mike Miller
Mike Miller
Yeah, so I think there is a desirable inductive argument here
It suffices to show that BS^1 x BU(n-1) -> BU(n) is injective on cohomology
Now the fiber is a complex projective space
which is easier to understand the cohomology of (and apply Leray-Hirsch)
I remember an argument in Burt Totaro's book on group cohomology and algebraic varieties that says that whenever you have a projective space bundle, the induced map on cohomology (chow groups, to him) is injective - and I suspect his argument is sufficiently formal that it would apply in most settings
 
loch
loch
Yeah the key step is to find a class on the total space which generates the cohomology of each fiber ---- when you're looking at a vector bundle E-> B and its associated projective bundle PE-> B, there is a tautological line bundle on PE which restricts to the tautological bundle on each fiber which does the trick (since, of course, its first chern class is a generator of the cohomology of P^n)

In this case I think taking the universal bundle on BU(n-1) (or what's it called - the associated bundle?) does the same trick too
 
Mike Miller
Mike Miller
9:27 PM
That sounds likely... I wasn't sure how much one had already about chern classes etc
if you haven't looked at Burt's book before I think you'll really like it
 
loch
loch
Oh yeah I did mention first chern class - but that's fine (BU(1) is easy!)
I'll check it out! It combines something I'm slightly familiar with (varieties) and something I am significantly less familiar with (group cohomology) --- so that sounds like a good way for me to learn the latter
 
Mike Miller
Mike Miller
I have to remember the actual nem
Group Cohomology and Algebraic Cycles
The setup is "algebraic group cohomology", which is essentially the Chow groups of (X x EG)/G - to define this, you take fin dim approximations to EG
 
loch
loch
Yeah I think that's how equivariant chow groups are done
 
Mike Miller
Mike Miller
Lemma 2.3 in his book is relevant but no proof
which he quotes from Fulton
@loch I have an alternate proof idea which might be easier
This should essentially be an instance of the "splitting theorem", which says that for any space $X$ equipped with bundle $E$, one may find a space $X'$ and a map $f: X' \to X$ for which 1) $f^*E$ splits as a direct sum of line bundles, and 2) $f^*$ is injective on cohomology
Hmm and now I am struggling to carry that out
The argument I wanted for this is as follows: let $X'$ be the unit sphere bundle of $E$. Argue by the Gysin sequence that the projection is an injection on cohomology. (This is the part I am skeptical of!) In any case, $f^*E$ naturally splits off a line bundle: the generator of this line bundle is given by the element of $S(E)$ you live at.
So you just induct upwards
 
loch
loch
9:45 PM
I am familiar (well I was once familiar) with Fulton's proof on the chow ring of projective bundles! It relies quite crucially on the fact that for chow groups you have the following exact sequence A(Z) -> A(X) -> A(U) -> 0, where Z is a closed subscheme of X and U is its complement, the first map being proper pushforward of cycles and the second map being restriction (but no spectral sequence arguments etc).
So now that I think of it I'm not sure if the proof idea goes through in cohomology -- but perhaps one can say something about cycle maps to go from Chow to cohomology --- but I guess the upshot is that it might not work here (at least not by literally translating the proof) since of course BU(n) etc. are not algebraic
Hmm I'm having office hours soon but I'll think about what you said
 
Mike Miller
Mike Miller
Gotcha
One can make sense of fd approximations, replacing U(n) by GLn(C), but cycles and cohomology are just different
I also have to do some work
 

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