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I'm working through the details of Deligne and Mumford's 69' paper, "The Irreducibility of the Space of Curves of Given Genus", and I had a few quick questions:

1) On p. 77, they claim that for $x$ a node, $\pi:C'\rightarrow C$ the blowing-up of $x$ and $x_1,x_2\in C'$ the two points of $\pi^{-1}(x)$. Then supposedly it's "easy" to show that for an invertible sheaf $\mathcal L$ on $C$, we have $$\rm{Hom}(\mathfrak m_x,\mathcal L)\cong H^0(C',\pi^*\mathcal L),\rm{Hom}(\mathfrak m_x^2,\mathcal L)\cong H^0(C',\pi^*\mathcal L(x_1+x_2)),$$ but I'm having trouble seeing it. I've tried to show it locally by assuming the curve looks like $xy=0$ in $\mathbb A^2$, but then I'm finding that on an affine open $U$ which trivializes $\mathcal L$ and doesn't contain $x$, then the restriction of such a homomorphism is equivalent to giving a section of $\mathcal L$ over $U$, and thus a section of $\pi^*\mathcal L$ over the isomorphic preimage of this open. But around $x$ I'm finding that such a homomorphism is equivalent to giving two functions, one on each branch, which vanish at $x_1,x_2$, which seems to be giving an isomorphism with $H^0(C',\pi^*\mathcal L(-x_1-x_2))$. I find the same thing for the square. Any help? I'm sure this is easy for experts...

On p. 78 2) At the beginning of \textit{Case 3} they claim, and I agree, that the restriction of $\omega_C^{-n}$ has degree at most -2 on each component $E$ of $C$. Shouldn't this automatically imply it has no sections on the preimage of that component in the blow-up? Why do they go on to speak about when $n=2$, the degree of $\omega_C$ is 1 on that component, and two poles are allowed? Furthermore, if the whole point is when $n\geq 3$, why is this necessary?

3) Later after the corollary on that page, why can we assume that the projective bundle $\mathbb P(\pi_*(\omega^n_{C/S}))$ is trivializable so we can actually get a morphism from $S$ to the Hilbert Scheme?

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Re 1): $Hom(\mathcal{O}_x, \mathcal{L})$ injects into $Hom(\mathfrak{m}_x, \mathcal{L})$ so it is clear that what you compute cannot be correct. –  ulrich Nov 21 '12 at 4:20
    
I was pretty sure it was incorrect since it wasn't providing the correct answer :). Any thoughts on how to show it correctly? –  HNuer Nov 21 '12 at 17:00
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Let $A'$ be the ring $k[x] \times k[y]$, $A$ the subring consisting of pairs of polynomials having the same value at $0$ and $m$ the maximal ideal of $A$ consisting of pairs of polynomials vanishing at $0$. Given an element $a$ of $A'$, multiplication by $a$ gives an $A'$-linear map from $m$ to $A$. Conversely, given such a map $\phi$ , $\phi(x,0)$ is an element of $A'$ which must be of the form $(xf(x),0)$ for some element $f \in k[x]$. Similarly, $\phi(0,y) = (0,yg(y))$ for some $g \in k[y]$. So $\phi \mapsto (f,g) \in A'$ gives the desired inverse. –  ulrich Nov 25 '12 at 10:15
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The reason for considering $n=2$ is Serre duality: to show $H^0(C,\omega^3)$ is constant one needs to show that $H^1(C,\omega^3)$ vanishes and this is equivalent to showing that $H^0(C, \omega^{-2})$ vanishes. –  ulrich Nov 25 '12 at 10:32
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For 3), it is not claimed that the projective bundle is trivializable for all families of stable curves. –  ulrich Nov 25 '12 at 13:52

2 Answers 2

The arithmetic genus $g$ of a connected curve $C$ is defined as $g = h^{1}(C,\mathcal{O}_{C})$. Suppose that $C$ has at most nodal singularities. Let $C = \bigcup_{i=1}^{\gamma}C_{i}$ be the irreducible components decomposition of $C$, and set $\delta :=\sharp Sing(C)$. Let $$\nu:\overline{C} = \bigsqcup_{i=1}^{\gamma}\overline{C_{i}}\rightarrow C$$ be the normalization of $C$. The associated morphism $\mathcal{O}_{C}\hookrightarrow\mathcal{O}_{\overline{C}}$ on the structure sheaves yield the following sequence in cohomology $$0\mapsto H^{0}(C,\mathcal{O}_{C})\rightarrow H^{0}(\overline{C},\mathcal{O}_{\overline{C}})\rightarrow \mathbb{C}^{\delta}\rightarrow H^{1}(C,\mathcal{O}_{C})\rightarrow H^{1}(\overline{C},\mathcal{O}_{\overline{C}})\mapsto 0.$$ We get a formula for the arithmetic genus $g$ of $C$ $$g = h^{1}(\overline{C},\mathcal{O}_{\overline{C}})+\delta-\gamma + 1 = \sum_{i=1}^{\gamma}g_{i} + \delta-\gamma + 1$$ where $g_{i} = h^{1}(\overline{C_{i}},\mathcal{O}_{\overline{C_{i}}})$ is the geometric genus of $C_{i}$.

Now, let us assume $C$ to be Deligne-Mumford stable. That is $C$ and hence any component $C_i$ of $C$ (endowed with its special points) have finitely many automorphisms. Let $D = \omega_{C}^{\otimes 3}$. We can considet the irreducible components of $C$. I will denote the component by $C$.

If $C$ is smooth of genus $g\geq 2$ then $\deg(D) = 6g-6\geq 2g+1$ and $D$ is very ample.

If $g(C) = 1$ then there is at least one special point on $C$ and $deg(D_{|C}) \geq 3$ implies that $D$ is very ample.

If $g(C) = 0$ then there are at least three special points on $C$ and again $deg(D_{|C}) \geq 3$.

If the irreducible component $C$ is singular the worst case is when it is of arithmetic genus one. Then there is at least another special point which is smooth. In this case $deg(D_{|C}) \geq 3(-2+2+1) = 3$, where $-2$ come from the canonical of $\mathbb{P}^1$, $+2$ is the contribute of the node, and $+1$ is the contribute of the special point.

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up vote 1 down vote accepted

Since no one has written an official answer to this question and the bounty is ending in a few hours, I'll provide an answer of my own to the question, though see ulrich's comments for another approach.

1) The crucial observation here is that $\mathfrak m_x=\pi_*(\mathcal O_{C'}(-x_1-x_2))$ where $x\in C$ is a node, $\pi:C'\rightarrow C$ is the partial normalization at $x$ and $x_1,x_2$ are the points in $C'$ lying over $x$. This is fairly clear from the description of $\mathcal O_C(V)$ as the set of $ f\in \mathcal O_{C'}(\pi^{-1}(V))$ such that $f(x_1)=f(x_2)$ for any affine $V$ such that $x$ is its only singular point. Then the claim is seen as follows:

Writing $\mathcal L=(\mathcal L\otimes \omega_C^{-1})\otimes \omega_C$, we get that $$\text{Hom}(\mathfrak m_x,\mathcal L)\cong \text{Hom}(\mathfrak m_x (\mathcal L^{-1}\otimes \omega_C),\omega_C)\cong H^1(C,\mathfrak m_x(\mathcal L^{-1}\otimes \omega_C))^{\vee}.$$ Then using the crucial observation above, the projection formula, and the fact that $\pi$ is finite so the higher direct images vanish, we see that

$H^1(C,\mathfrak m_x (\mathcal L^{-1}\otimes \omega_C))^{\vee}\cong H^1(C',\mathcal O_{C'}(-x_1-x_2)\otimes\pi^*\mathcal L^{-1}\otimes \pi^*\omega_C)\cong H^0(C',\pi^*\mathcal L),$

where I've used the fact that $\pi^*\omega_C\cong \omega_{C'}(x_1+x_2)$ and Serre duality on $C'$. The proof of the $\mathfrak m_x^2$ case is almost identical.

2) They choose $n\geq 2$ since when they used Serre duality an exponent of $1-n$ appeared for $\omega_C$, and $n$ was originally $n\geq 3$, so they WLOG assume now $-n$ is the exponent of $\omega_C$ but $n\geq 2$. For the rest, a possibly clearer proof of this result is given as Lemma 10.6.1 in "Geometry of Algebraic Curves, Vol. II" by ACG.

3) The projective bundle is not necessarily trivial, and the Hilbert scheme constructed there parametrizes only those families for which a trivialization exists. However, any family can be pulled back to the principal $PGL$ bundle associated to the projective bundle, and there it does become trivial. Thus there is a map from the associated $PGL$-bundle to the Hilbert scheme.

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