Timeline for Equivariant Hilbert schemes of points
Current License: CC BY-SA 3.0
8 events
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| Mar 16, 2017 at 20:09 | history | edited | Xudong | CC BY-SA 3.0 |
posted a follow-up question
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| Mar 14, 2017 at 0:02 | comment | added | Jason Starr | Please note: even if $G$ is not in $\textbf{SL}_2(\mathbb{C})$, you can still restrict the $2$-form from $\text{Hilb}^{rg}(\mathbb{A}^2)$ to the $G$-invariant locus. However, "averaging over an orbit" will sum up over powers of a non-identity root of unity, and that sum will equal zero. So when $G$ is not contained in $\textbf{SL}_2(\mathbb{C})$, the restricted $2$-form will be identically zero. | |
| Mar 13, 2017 at 23:59 | comment | added | Jason Starr | "Can I say ..." Yes, for a finite quotient stack such as $[\mathbb{A}^2/G]$, you can construct $\beta_r$ in that way (up to dividing through by an integer). For a smooth, $2$-dimensional Deligne-Mumford stack with Gorenstein coarse moduli space that is not necessarily a finite quotient stack, it is better to use traces of $p$-forms as in my article with de Jong about rational curves on cubic threefolds. | |
| Mar 13, 2017 at 22:46 | comment | added | Xudong | @JasonStarr: Can I say that the form $\beta_r$ is restricted from the one on $\mathrm{Hilb}^{rg}(\mathbb{A}^2)$ onto this $G$-invariant smooth subvariety? | |
| Mar 13, 2017 at 17:46 | comment | added | Jason Starr | "But I am not comfortable to say that $\beta = dx \wedge dy$ is $G$-invariant, since it is only so up to a scalar multiple." If $G$ is a subgroup of $SL_2(\mathbb{C})$, then $G$ preserves $\beta$. | |
| Mar 13, 2017 at 17:11 | comment | added | Xudong | @JasonStarr: Thanks! This is what I had in mind. But I am not comfortable to say that $\beta = dx \wedge dy$ is $G$-invariant, since it is only so up to a scalar multiple. | |
| Mar 12, 2017 at 23:24 | comment | added | Jason Starr | Assuming that the characteristic of $k$ does not divide the order of $G$, the $G$-invariant element $\beta = dx\wedge dy$ in $H^0(\mathbb{A}^2,\omega_{\mathbb{A}^2/k})$ induces an everywhere nondegenerate element $\beta_n$ on the Hilbert scheme $\text{Hilb}^{P_r}_{[\mathbb{A}^2/G],k}$, where $P$ is the Hilbert polynomial of $r$ times the $G$-representation $k[G]$. Because $\text{Hilb}^{P_r}_{[\mathbb{A}^2/G],k}$ is smooth over $k$, it suffices to check that $\beta_r$ is nondegenerate at every codimension $1$ point, and this is the same computation as nondegeneracy of $\beta_1$. | |
| Mar 12, 2017 at 23:02 | history | asked | Xudong | CC BY-SA 3.0 |