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Let $ f: \mathbb{P}^n \dashrightarrow (\mathbb{P}^1)^n , [x_0:\dots :x_n] \mapsto ([x_0,x_1],[x_0,x_2], \dots ,[x_0,x_n])$ a birational map. Is $f$ a Crepant birational map?

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Kontsevich proved in '95 that the hodge polynomial is independent of crepant resolution over a field of characteristic zero. He created the field of motivic integration to prove this fact. Here are the respective hodge polynomials of the two spaces in question where I repress the fact that we are working over the field $k$ just for notational simplicity:

$$H([\mathbb{P}^n], u,v) = \sum_{i=0}^{n} H([\mathbb{A}^i], u,v) =\sum_{i=0}^{n} (uv)^i $$

and

$$H([\prod_{i=1}^{n}\mathbb{P}^{1}], u,v) = \prod_{i=1}^{n} (H([\mathbb{A}^1], u,v)+1) = \prod_{i=1}^{n} (uv+1) = (uv+1)^n = \sum_{i=0}^{n} {n \choose i} (uv)^i $$

where we denote by $[X]$ the equivalence class of the variety $X$ in the grothendieck ring of varieties. Therefore, there are no crepant birational maps between $\mathbb{P}^n$ and $\prod_{i=1}^{n} \mathbb{P}^1$.

Note, actually, here $[X]$ lives in the image of the Grothendieck ring of varieties localized at $[\mathbb{A}^1]$ under the completion homomorphism given by the dimension filtration. But, the hodge polynomial ring homomorphism `'$$H(X,u,v) = \sum_{i,j} (-1)^{i+j} dim(H^i(X, \Omega_{X}^{j}))u^iv^j $$' factors through the ring homomorphism $X \mapsto [X]$.

I am only sure that this argument works over a field of characteristic zero for two reasons: 1) there is not currently a resolution of singularities over fields of positive characteristic, and 2) I don't know how to define the hodge polynomial in positive characteristic.

Note that this result is sometimes known as Batyrev's Theorem because he originally proved a similar result in '95 involving the beta numbers via p-adic integration and the weil conjectures.

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Sorry, i make a mistake in the statement. I will edit my post. Obviously there are no crepant birational maps between $\mathbb{P}^n$ and $(\mathbb{P}^1)^n$ because they have a distinct number of generators of their Picard Groups. –  Joaquín Moraga Mar 6 '13 at 6:30
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That is annoying that you completely changed the question. –  Andrew Stout Mar 6 '13 at 6:39
    
anyway, this argument I posted still works. $f$ cannot be a birational map or else the hodge polynomials would be the same (by the same work of Kontsevich mentioned above). next time you should either add to your question or start a new question instead of asking a completely different question after someone has answered you. –  Andrew Stout Mar 6 '13 at 6:48
    
Really sorry, i get confused and wrote the wrong question cause both had the birational morphism f. –  Joaquín Moraga Mar 6 '13 at 6:49
    
I will edit it or will get very confusing. –  Joaquín Moraga Mar 6 '13 at 6:50

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