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Geometric realization of $B{\mathbb G}_{\mathfrak m}({\mathbb C})$ is ${\mathbb C}{\mathbb P}^\infty=\varinjlim_n~ {\mathbb C}{\mathbb P}^n_k$; what if one considers a separable field $k\neq {\overline k}$? Sheaf-theoretically, $B{\mathbb G}_{\mathfrak m}$ represents the simplicial sheaf $B{\mbox{hom}}_{k}(-,{\mathbb G}_{\mathfrak m})$ on the big etale site $Sch_k$; should I think of the analogue for ${\mathbb C}{\mathbb P}^\infty$ as $\varinjlim_n~ {\mbox{hom}}_{k}(-,{\mathbb P}^n_k)$?

Context: calculating etale cohomology of $B{\mathbb G}_{\mathfrak m}(k)$.

Note: For separably closed field $k$, theorem of Friedlander-Parshall says that $H^\ast_{et}(BG(k),{\mathbb Z}/m)\cong H^\ast_{et}(BG({\mathbb C}),{\mathbb Z}/m)$ for complex reductive group scheme $G({\mathbb C})$.

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In the blog post sbseminar.wordpress.com/2008/06/19/whats-a-stack , comment number 27 by "bb" begins with "There is a precise sense in which P^n approximates BG_m if one works in the A^1-homotopy category" and explains that since étale cohomology is $\mathbb{A}^1$-homotopy invariant, one may compute the cohomology of $B\mathbb{G}_m$ by taking a limit of cohomologies of projective spaces. –  S. Carnahan May 12 at 23:35
    
See also mathoverflow.net/questions/65728/… –  S. Carnahan May 12 at 23:37
    
Thanks for the references ... I am looking at them. –  sms1 May 13 at 3:06

1 Answer 1

up vote 4 down vote accepted

As far as I understand, yes, you can look at it this way. However, at least to me, this seems like a strange way of doing it, and there are a few problems with it:

  • Basically, you are saying that the cohomology of ${\mathbb P}^\infty$ and $B(\mathbb{Gm})$ coincide because the fiber (which is ${\mathbb A}^\infty-\{0\}$) is contractible. If you use something other than the `usual' cohomology, you need to justify the cohomological triviality of ${\mathbb A}^\infty-\{0\}$.

  • Before you even begin doing this, you must make sure that the cohomology theory you work with even allows ind-objects like ${\mathbb P}^\infty$ and ${\mathbb A}^\infty$.

  • When you are done, the argument will be tied to the specific group $\mathbb{Gm}$. What if you put a different reductive group there? Say, an exceptional one?

So it may be better to compute the cohomology of $B(\mathbb{Gm})$ directly, and view the claim about comparison with $\mathbb{P}^\infty$ as a bonus.

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