Timeline for Is the Torelli map an immersion?
Current License: CC BY-SA 2.5
12 events
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| Jul 24, 2017 at 20:26 | comment | added | inkspot | You are right: at the non-hyperelliptic points, the map fails to be an immersion in the same way that $Spec\ k\to B(\mathbb Z/2)$ fails to be an immersion. To get an immersion, first pass to the quotient $A_g\to A_g/[-1]$, where $A_g/[-1]$ has the same objects as $A_g$ but $Mor(X,Y)$ is replaced by $Mor(X,Y)/[-1]$, and then $M_g\to A_g/[-1]$ is an immersion along the non-hyperelliptic locus. To aid intuition (?) note that over $\mathbb C$ we have $A_g=\mathfrak H_g/Sp_{2g}(\mathbb Z)$ while $A_g/[-1]=\mathfrak H_g/PSp_{2g}(\mathbb Z)$. | |
| Aug 9, 2016 at 3:46 | comment | added | Aaron Landesman | Also, does the fact that it is an immersion (and injective) outside of the hyperelliptic locus imply that the image of any non-hyperelliptic point in M_g is smooth in the image of M_g? | |
| Aug 9, 2016 at 3:38 | comment | added | Aaron Landesman | Can someone clarify the definition of immersion/infinitessimal torelli map being used in this answer? Am I correct that it means the map is an isomorphism on tangent spaces at every point, or is immersion being used to denote something more, like a locally closed immersion? It would seem that the Torelli map is 2:1 on the locus of non-hyperelliptic curves because the Jacobian has an extra multiplication by -1 automorphism that the curve does not have. Hence, it wouldn't be a locally closed immerion away from the hyperelliptic locus. | |
| Dec 19, 2009 at 18:15 | history | edited | VA. | CC BY-SA 2.5 |
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| Dec 19, 2009 at 16:50 | comment | added | Tony Pantev | Excellent point! I was been too quick and did my infinitesimal calculation away from the stacky points. So of course I missed this important point. Thanks VA! | |
| Dec 19, 2009 at 13:55 | history | edited | VA. | CC BY-SA 2.5 |
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| Dec 19, 2009 at 12:28 | comment | added | VA. | I mean the space of the first-order deformations of the jacobian J=J(C), and indeed it is the tangent space to the moduli STACK $A_g$. Dimension of $H^1(O_C)=H^1(O_J)$ is g, and dimension of $Sym^2(H^1(O_J))$ is g(g+1)/2. | |
| Dec 19, 2009 at 8:14 | vote | accept | David Zureick-Brown | ||
| Dec 19, 2009 at 6:51 | comment | added | shenghao | can you explain a little bit about what you mean by "deformation space" and why it's Sym^2(H^1(O_J))? I thought you mean the tangent space of the moduli space A_g at the point J=Jac(C), but then the dimension of Sym^2(H^1) is not g(g+1)/2, and I got confused... | |
| Dec 19, 2009 at 4:42 | history | edited | VA. | CC BY-SA 2.5 |
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| Dec 19, 2009 at 3:27 | history | edited | VA. | CC BY-SA 2.5 |
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| Dec 19, 2009 at 2:19 | history | answered | VA. | CC BY-SA 2.5 |