Timeline for Are spaces of holomorphic maps manifolds?
Current License: CC BY-SA 2.5
5 events
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| Jun 1, 2010 at 10:15 | comment | added | Dmitri Panov | BCnrd, thanks a lot for the suggestion of how to prove that the degree 1 End is an Auto! I wonder if the other thing that I said has any chances to be true -- for a smooth compact Y the spaces Hom(Y,Y) of degree>0 are smooth? | |
| Jun 1, 2010 at 2:31 | comment | added | BCnrd | Dmitri, I suppose $Y$ is connected (via def'n of "variety"; always confuses me). Quasi-finite maps between smooth connected proj. varieties (over any field) of the same dimension are finite flat of constant degree, and finite flat maps with constant degree 1 between noetherian schemes are isom (pass to affine case and use Nakayama to reduce to considering fiber algebras of degree 1 over fields). To make a subfunctor of endomorphism functor out of the condition "finite flat of degree 1", we check it is an open condition on the Hom-scheme (e.g., by usual flatness, properness, fiber-dim. stuff). | |
| May 31, 2010 at 17:19 | comment | added | Dmitri Panov | I have assumed, that Y is compact. In this case any self-map of degree one is an automorphism. Indeed, if we had a holomorphic map $Y\to Y$ of degree one that is not an automorphism, its differential would degenerate at one point and hence on a divisor. This would give a contradicton if we compaire the canoncial class of $Y$ with itself (sorry for been a bit clumsy). So these degree one maps form a group. | |
| May 31, 2010 at 16:03 | comment | added | Oblomov | In fact, I meant \mathrm{Hom}(X,Y) in the last par of my question (I have edited it now), but looking at self-maps is fine. However, you claim that \mathrm{Hom}_1(Y,Y) is Lie group (for composition, I guess). Why is it so? (Sorry to be stupid, but I don't see why it's a group and why it's a manifold). | |
| May 31, 2010 at 13:54 | history | answered | Dmitri Panov | CC BY-SA 2.5 |