Timeline for Quotients of curves of genus $4$ by a free $\mathbb{Z}/ 3 \mathbb{Z}$-action
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 1, 2016 at 14:46 | vote | accept | Francesco Polizzi | ||
| Feb 1, 2016 at 14:09 | comment | added | Francesco Polizzi | @abx: I checked the computations and everything seems ok, thank you again for your help. Just to be pedantic: if $Q$ is the invariant quadric, one obtains an invariant cubic of the form $V_3$ by taking a general element in the invariant subspace of $\mathsf{S}^3 H^0(C_4, \, K)$ and adding suitable multiples of $x_0Q$ and $x_1Q$ in order to get rid of the monomials $x_0x_1x_3$, $x_0x_2x_3$. | |
| Jan 29, 2016 at 20:01 | comment | added | Holonomia | @Francesco Polizzi: That's right I agree. | |
| Jan 29, 2016 at 19:55 | comment | added | Francesco Polizzi | @Holonomia: well, I see that the existence of an invariant cubic in the ideal of $C_4$ easily follows using David Speyer's averaging argument. | |
| Jan 29, 2016 at 19:40 | comment | added | Holonomia | @Francesco Polizzi: OK, I understand. But since $Z_3$ is finite, abelian and acts linearly you have a simultaneous diagonalization hence fixed cubics, isn't it?. | |
| Jan 29, 2016 at 19:01 | comment | added | David E Speyer | Of course, this invariant cubic is not unique; I am just saying there is one, well defined modulo the invariants in $H^0(\mathbb{P}^3, \mathcal{O}(1)) Q$ | |
| Jan 29, 2016 at 19:01 | comment | added | David E Speyer | Letting $Q$ be the invariant quadric, and $C$ the space of cubics, we have a short exact sequence $0 \to H^0(\mathbb{P}^3, \mathcal{O}(1)) Q \to C \to \mathbb{C} \to 0$ of $\mathbb{Z}/3$ representations, where $\mathbb{Z}/3$ acts trivially on the final $\mathbb{C}$. Such a sequence must split, the image of the splitting $\mathbb{C} \to C$ is an invariant cubic. (More explicitly, take any cubic $f$ not in $H^0(\mathbb{P}^3, \mathcal{O}(1)) Q$, then $(1/3) (f+\sigma f + \sigma^2 f)$ is an invariant cubic.) | |
| Jan 29, 2016 at 18:49 | comment | added | Francesco Polizzi | @Holonomia: the quadric through $C_4$ is unique (so necessarily invariant), but the cubic is not: there is a $5$-dimensional vector space of them. | |
| Jan 29, 2016 at 18:08 | comment | added | abx | The quadric is in the kernel of $\mathsf{S}^2H^0(C_4,K)\rightarrow H^0(C_4,2K)$. The invariant subspace in $\mathsf{S}^2$ has dimension 4; in $H^0(C_4,2K)$, it is $H^0(C_2,2K)$, which has dimension 3. Similarly, the kernel of $\mathsf{S}^3H^0(C_4,K)\rightarrow H^0(C_4,3K)$ has dimension 5, with eigenvalues $(1,1,1,\xi ,\xi ^2)$. The eigenvalues $(1,1,\xi ,\xi ^2)$ correspond to the multiples of the quadric, it remains one invariant cubic. | |
| Jan 29, 2016 at 17:56 | comment | added | Holonomia | The invariance of both the quadric and the cubic did not follow from the uniqueness ? | |
| Jan 29, 2016 at 17:42 | comment | added | Francesco Polizzi | (2) How can I detect the space of invariant cubics through the canonical curve? Using Koszul resolution I can compute $$h^0(\mathscr{I}_{C_4})=5,$$ and it seems to me that there are $8$ invariant monomials, namely $$x_0^3, \, x_0^2x_1, \, x_0x_1^2, \, x_1^3, \, x_0x_2x_3, \, x_1x_2x_3, x_2^3, \, x_3^3.$$ Why you say Writing down the invariant quadric and cubic(s) containing $C_4$ leads to your formulas? How can I check that there exist an invariant cubic containing $C_4$ and whose equation does not contain the invariant monomials $x_0x_2x_3, \, x_1x_2x_3$? | |
| Jan 29, 2016 at 17:37 | comment | added | Francesco Polizzi | Thank you for the answer, it is really helpful. I have a couple of questions. (1) Your computations show that the curve $C_4$ embeds in $\mathbb P^4$ and must be invariant by the action $$[x_0: x_1 : x_2 : x_3] \mapsto [x_0: x_1 : \xi x_2 : \xi^2 x_3].$$ On the other hand, it is a canonical curve, so it is intersection of a (unique) quadric and a cubic. But why both the quadric and the cubic must be invariant, too? | |
| Jan 29, 2016 at 11:46 | history | answered | abx | CC BY-SA 3.0 |