Timeline for Deformations of hypersurfaces
Current License: CC BY-SA 2.5
6 events
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| Aug 20, 2010 at 13:17 | comment | added | Barbara | The contribution of the eigenspaces, not just for a hypersurface but in general when the variety you're working with is (say) smooth, is that they can be used to defined so-called "natural" deformations to which the group action doesn't extend, at least when $G$ is abelian. See Pardini's paper Abelian Covers (Crelle early nineties) for the definition of natural deformation. | |
| Aug 19, 2010 at 12:44 | history | edited | Francesco Polizzi | CC BY-SA 2.5 |
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| Aug 18, 2010 at 19:59 | comment | added | Francesco Polizzi | Once you know that the invariant part is $8$-dimensional, the fact that also all the other eigenspaces are $8$-dimensional comes almost immediately by symmetry considerations. I never tried the computation you are suggesting, anyway probably the action on the latter group should be not too difficult to understand: write a basis for $H^0(P^3, T_{P^3})$ by using Euler sequence and restrict it to $X$ (here you use the particular form of the equation). | |
| Aug 18, 2010 at 19:26 | comment | added | bellini | That is an interesting example, as it both shows that the action may be nontrivial, and gives an interpretation of the fixed part. What would you say is the best way to 'check' the decomposition of $H^1(X,T_X)$ as a $G$-representation? In the given case, I would try and write that as the quotient of $H^0(X,\mathcal{O}(5))$ by $H^0(X,T_{\mathbb{P}^3}$, but this does no seem to depend on the given equation (except for its degree), and I wouldn't know the action on the latter. | |
| Aug 18, 2010 at 18:44 | history | edited | Francesco Polizzi | CC BY-SA 2.5 |
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| Aug 18, 2010 at 18:35 | history | answered | Francesco Polizzi | CC BY-SA 2.5 |