Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $f : X \to Y$ be a [fill in the blank] morphism of [fill in the blank] complex varieties. Then we have the pushforward $f_! : K(X) \to K(Y)$ which is defined by $f_!(E) = \sum_i (-1)^i [R^i f_\ast E]$, the alternating sum of the higher direct images. Here we take $K(X)$ to mean the $K$-group of coherent sheaves.

On the other hand we can also define $K(X)$ as the $K$-group of $C^\infty$ complex vector bundles on $X$ considered as a real manifold. Then we can define a Gysin map $f_!$ using the Thom isomorphism theorem for $K$-theory.

  1. Which adjectives do I need to fill in the blanks with to make the two notions of $K(X)$ agree?

  2. Which adjectives do I need to fill in the blanks with to make the two notions of $f_!$ agree?

If $X$ is smooth and projective, then any coherent sheaf has a finite resolution by locally free sheaves, so we have a map $K^{alg}(X) \to K^{top}(X)$. On the other hand, I don't think it's true that any $C^\infty$ complex vector bundle has a holomorphic structure, so I don't think there is a map $K^{top}(X) \to K^{alg}(X)$ ...

share|improve this question
1  
I know next to nothing but I think that the answer is that this is very rarely an isomorphism... it doesn't seem like the left side is two periodic in any way. If you invert the Bott element than you get closer. Here is are some papers that describes a lot of intricate relationships related to this question: archive.numdam.org/ARCHIVE/ASENS/ASENS_1985_4_18_3/… math.unl.edu/~mwalker5/papers/finite.pdf –  Daniel Pomerleano Oct 11 '11 at 23:24
    
Well, I'm not looking at graded anything. That is, I'm only looking at $K^0$. –  Kevin H. Lin Oct 12 '11 at 0:02
5  
Re 1, elaborating on Daniel's comment: If you rationalize on both sides, this is asking to compare rational Chow theory with rational even-dimensional cohomology. In general, that won't go well (e.g., for $X$ a positive genus curve). It will be true only under some very restrictive hypotheses on $X$ (e.g., if $X$ is "cellular" in the sense of a locally closed decomposition by affine spaces). The result of Thomason is saying that things in fact get better if you look at higher and higher K-theory and look at torsion(/complete) instead of rationalizing. –  Anatoly Preygel Oct 12 '11 at 0:50
    
And Remark 4.17 of op.cit., and the text around it, discusses the compatibility of that relationship with pushforwards. –  Anatoly Preygel Oct 12 '11 at 0:55
    
It's false already for elliptic curves. –  Tom Goodwillie Oct 12 '11 at 0:57
show 1 more comment

2 Answers

Well, let me say something really straightforward. First, the map from $K^{alg}(X)$ to $K^{top}(X)$ is defined even if $X$ is not necessarily projective. Second, $f_!$ is defined only if $f$ is proper, and in that case I think it always agrees with the topological push-forward.

share|improve this answer
add comment

I found a paper which I think answers #2:

Riemann-Roch and topological K-theory for singular varieties, by Baum, Fulton, MacPherson

http://www.springerlink.com/content/k284857584wp9032/

They prove that the algebraic $f_!$ and the topological $f_!$ agree in the case of proper morphisms of (not necessarily smooth) quasi-projective varieties.

In other words we have a commutative square

K^{alg}(X) ---> K^{alg}(Y)
   |               |
   V               V
K^{top}(X) ---> K^{top}(Y)
share|improve this answer
    
btw, how can I do commutative diagrams here on MO? –  Kevin H. Lin Oct 12 '11 at 7:48
    
@Kevin: You can do commutative diagrams the same way Knuth does them in TeXbook. –  Dmitri Pavlov Oct 13 '11 at 13:34
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.