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Let $X$ be a smooth variety and let $Y\subset X$ be a smooth subvariety. Let $f:Y'\to Y$ be a (say, finite surjective) morphism. When $f$ is the identity, the cohomology group $H^0(Y',f^*N_Y)$ measures the first order deformations of $Y$ in $X$. My question is:

Does the group $H^0(Y',f^*N_Y)$ have a deformation-theoretic interpretation for other $f$?

Of course you could say that this group fits into an exact sequence involving $H^0(Y,f^*T_X|_Y)$ which measures deformations of the morphism $f$, but my question is really whether he above group measures something by its own.

I am mostly interested in the cases where $f$ is the Frobenius or the normalization map.

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  • $\begingroup$ You have the map $g:Y'\to X$ which is the composition of $f$ and of the inclusion. I think your $H^0$ measures first-order deformations of $g$. $\endgroup$ – user8268 Feb 29 '12 at 9:23
  • $\begingroup$ Wouldn't these lie in $H^0(Y',g^*T_X)$? $\endgroup$ – jco Feb 29 '12 at 9:34
  • $\begingroup$ ah sorry; my comment makes no sense $\endgroup$ – user8268 Feb 29 '12 at 18:25
  • $\begingroup$ Let $g: Y' \to X$ be the composition of $f$ followed by the inclusion map. I am not a 100% sure, but isn't the cohomology group you asked about the tangent space at $[g]$ to the space of deformations of the map $g: Y' \to X$ quotiented out by the group of reparametrizations of $Y'$? At least geometrically, this is what I expect it to be... $\endgroup$ – Malkoun Jul 27 '16 at 12:28

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