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6
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edited Dec 24 2011 at 9:57 |
Addendum I wrote this up thinking that the question was something different. As Angelo pointed out, this does not answer the actual question. I will leave this here just in case someone finds the computation useful. So this is a proof, that $H^1(X,\mathscr O_X)=0$. Not exactly what the question was, although it still implies that $Gr_F^0H^i(X,\mathbb Gr_F^0H^1(X,\mathbb C)=0$ where $F$ is Deligne's Hodge filtration. :( end of Addendum
Using the notations of the question, in addition let $Y=\mathrm{Spec}(S/I)$ be the affine cone over $X$, $P\in Y$ the vertex, and $U=Y\setminus \{P\}$.
Finally, let $\mathrm{depth}(S/I)=d\geq 3$.
First of all we have a long exact sequence:
$$
\dots \to H^i(Y,\mathscr O_Y) \to H^i(U,\mathscr O_U) \to H^{i+1}_P(Y,\mathscr O_Y) \to H^{i+1}(Y,\mathscr O_Y) \to \dots.
$$
Since $Y$ is affine, this implies that for $i>0$,
$$
H^i(U,\mathscr O_U) \simeq H^{i+1}_P(Y,\mathscr O_Y)
$$
and hence
$$
H^i(U,\mathscr O_U)=0 \tag{$\star$}
$$
for $0< i < d-1$.
Proposition
$\quad\
H^i(U,\mathscr O_U) \simeq \bigoplus_{n\in\mathbb Z} H^i(X, \mathscr O_X(n))
$
Proof $U$ is an $\mathbb A^1$-bundle over $X$. In fact, it is easy to see that $U\simeq \mathrm{Spec}_X ( \oplus _{n\in \mathbb Z} \mathscr O_X(n))$ with a projection $\pi:U\to X$. It follows that $\pi_*\mathscr O_U\simeq \oplus _{n\in \mathbb Z} \mathscr O_X(n)$ and $R^j\pi_*\mathscr O_U=0$ for $j>0$. Then the claimed isomorphism follows from the simple special case of the Leray spectral sequence when all $R^j$'s with $j>0$ are $0$.
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5
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edited Dec 22 2011 at 10:00 |
Addendum I wrote this up thinking that the question was something different. As Angelo pointed out, this does not answer the actual question. I will leave this here just in case someone finds the computation useful. So this is a proof, that $H^1(X,\mathscr O_X)=0$. Not exactly what the question was, although it still implies that $Gr_F^0H^i(X,\mathbb C)=0$ where $F$ is Deligne's Hodge filtration. :( end of Addendum
Using the notations of the question, in addition let $Y=\mathrm{Spec}(S/I)$ be the affine cone over $X$, $P\in Y$ the vertex, and $U=Y\setminus \{P\}$.
Finally, let $\mathrm{depth}(S/I)=d\geq 3$.
First of all we have a long exact sequence:
$$
\dots \to H^i(Y,\mathscr O_Y) \to H^i(U,\mathscr O_U) \to H^{i+1}_P(Y,\mathscr O_Y) \to H^{i+1}(Y,\mathscr O_Y) \to \dots.
$$
Since $Y$ is affine, this implies that for $i>0$,
$$
H^i(U,\mathscr O_U) \simeq H^{i+1}_P(Y,\mathscr O_Y)
$$
and hence
$$
H^i(U,\mathscr O_U)=0 \tag{$\star$}
$$
for $0< i < d-1$.
Proposition
$\quad\
H^i(U,\mathscr O_U) \simeq \bigoplus_{n\in\mathbb Z} H^i(X, \mathscr O_X(n))
$
Proof $U$ is an $\mathbb A^1$-bundle over $X$. In fact, it is easy to see that $U\simeq \mathrm{Spec}_X ( \oplus _{n\in \mathbb Z} \mathscr O_X(n))$ with a projection $\pi:U\to X$. It follows that $\pi_*\mathscr O_U\simeq \oplus _{n\in \mathbb Z} \mathscr O_X(n)$ and $R^j\pi_*\mathscr O_U=0$ for $j>0$. Then the claimed isomorphism follows from the simple special case of the Leray spectral sequence when all $R^j$'s with $j>0$ are $0$.
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4
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edited Dec 22 2011 at 9:52 |
Addendum I wrote this up thinking that the question was something different. As Angelo pointed out, this does not answer the actual question. I will leave this here just in case someone finds the computation useful. So this is a proof, that $H^1(X,\mathscr O_X)=0$. Not exactly what the question was. :( end of Addendum
Using the notations of the question, in addition let $Y=\mathrm{Spec}(S/I)$ be the affine cone over $X$, $P\in Y$ the vertex, and $U=Y\setminus \{P\}$.
Finally, let $\mathrm{depth}(S/I)=d\geq 3$.
First of all we have a long exact sequence:
$$
\dots \to H^i(Y,\mathscr O_Y) \to H^i(U,\mathscr O_U) \to H^{i+1}_P(Y,\mathscr O_Y) \to H^{i+1}(Y,\mathscr O_Y) \to \dots.
$$
Since $Y$ is affine, this implies that for $i>0$,
$$
H^i(U,\mathscr O_U) \simeq H^{i+1}_P(Y,\mathscr O_Y)
$$
and hence
$$
H^i(U,\mathscr O_U)=0 \tag{$\star$}
$$
for $0< i < d-1$.
Proposition
$\quad\
H^i(U,\mathscr O_U) \simeq \bigoplus_{n\in\mathbb Z} H^i(X, \mathscr O_X(n))
$
Proof $U$ is an $\mathbb A^1$-bundle over $X$. In fact, it is easy to see that $U\simeq \mathrm{Spec}_X ( \oplus _{n\in \mathbb Z} \mathscr O_X(n))$ with a projection $\pi:U\to X$. It follows that $\pi_*\mathscr O_U\simeq \oplus _{n\in \mathbb Z} \mathscr O_X(n)$ and $R^j\pi_*\mathscr O_U=0$ for $j>0$. Then the claimed isomorphism follows from the simple special case of the Leray spectral sequence when all $R^j$'s with $j>0$ are $0$.
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3
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edited Dec 22 2011 at 9:41 |
Yes, it is true pretty much for the reason you're saying, but you don't need Hodge theory or even to work over $\mathbb C$. Using the notations of the question, in addition let $Y=\mathrm{Spec}(S/I)$ be the affine cone over $X$, $P\in Y$ the vertex, and $U=Y\setminus \{P\}$.
Finally, let $\mathrm{depth}(S/I)=d\geq 3$.
First of all we have a long exact sequence:
$$
\dots \to H^i(Y,\mathscr O_Y) \to H^i(U,\mathscr O_U) \to H^{i+1}_P(Y,\mathscr O_Y) \to H^{i+1}(Y,\mathscr O_Y) \to \dots.
$$
Since $Y$ is affine, this implies that for $i>0$,
$$
H^i(U,\mathscr O_U) \simeq H^{i+1}_P(Y,\mathscr O_Y)
$$
and hence
$$
H^i(U,\mathscr O_U)=0 \tag{$\star$}
$$
for $0< i < d-1$.
Now the desired vanishing (and more) follows from the following:
Proposition
$\quad\
H^i(U,\mathscr O_U) \simeq \bigoplus_{n\in\mathbb Z} H^i(X, \mathscr O_X(n))
$
Proof $U$ is an $\mathbb A^1$-bundle over $X$. In fact, it is easy to see that $U\simeq \mathrm{Spec}_X ( \oplus _{n\in \mathbb Z} \mathscr O_X(n))$ with a projection $\pi:U\to X$. It follows that $\pi_*\mathscr O_U\simeq \oplus _{n\in \mathbb Z} \mathscr O_X(n)$ and $R^j\pi_*\mathscr O_U=0$ for $j>0$. Then the claimed isomorphism follows from the simple special case of the Leray spectral sequence when all $R^j$'s with $j>0$ are $0$.
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2
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edited Dec 22 2011 at 6:07 |
Yes, it is true pretty much for the reason you're saying, but you don't need Hodge theory or even to work over $\mathbb C$.
Using the notations of the question, in addition let $Y=\mathrm{Spec}(S/I)$ be the affine cone over $X$, $P\in Y$ the vertex, and $U=Y\setminus \{P\}$.
Instead of setting it to $3$, let's say that Finally, let $\mathrm{depth}(S/I)=d$.
\mathrm{depth}(S/I)=d\geq 3$.
First of all we have a long exact sequence:
$$
\dots \to H^i(Y,\mathscr O_Y) \to H^i(U,\mathscr O_U) \to H^{i+1}_P(Y,\mathscr O_Y) \to H^{i+1}(Y,\mathscr O_Y) \to \dots.
$$
Since $Y$ is affine, this implies that for $i>0$,
$$
H^i(U,\mathscr O_U) \simeq H^{i+1}_P(Y,\mathscr O_Y)
$$
and hence
$$
H^i(U,\mathscr O_U)=0
$$
for $0< i < d-1$.
Now the desired vanishing (and more) follows from the following:
Proposition
$\quad\
H^i(U,\mathscr O_U) \simeq \bigoplus_{n\in\mathbb Z} H^i(X, \mathscr O_X(n))
$
Proof $U$ is an $\mathbb A^1$-bundle over $X$. In fact, it is easy to see that $U\simeq \mathrm{Spec}_X ( \oplus _{n\in \mathbb Z} \mathscr O_X(n))$ with a projection $\pi:U\to X$. It follows that $\pi_*\mathscr O_U\simeq \oplus _{n\in \mathbb Z} \mathscr O_X(n)$ and $R^j\pi_*\mathscr O_U=0$ for $j>0$. Then the claimed isomorphism follows from the simple special case of the Leray spectral sequence when all $R^j$'s with $j>0$ are $0$.
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1
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answered Dec 21 2011 at 22:31 |
Yes, it is true pretty much for the reason you're saying, but you don't need Hodge theory or even to work over $\mathbb C$.
Using the notations of the question, in addition let $Y=\mathrm{Spec}(S/I)$ the affine cone over $X$, $P\in Y$ the vertex, and $U=Y\setminus \{P\}$.
Instead of setting it to $3$, let's say that $\mathrm{depth}(S/I)=d$.
First of all we have a long exact sequence:
$$
\dots \to H^i(Y,\mathscr O_Y) \to H^i(U,\mathscr O_U) \to H^{i+1}_P(Y,\mathscr O_Y) \to H^{i+1}(Y,\mathscr O_Y) \to \dots.
$$
Since $Y$ is affine, this implies that for $i>0$,
$$
H^i(U,\mathscr O_U) \simeq H^{i+1}_P(Y,\mathscr O_Y)
$$
and hence
$$
H^i(U,\mathscr O_U)=0
$$
for $0< i < d-1$.
Now the desired vanishing (and more) follows from the following:
Proposition
$\quad\
H^i(U,\mathscr O_U) \simeq \bigoplus_{n\in\mathbb Z} H^i(X, \mathscr O_X(n))
$
Proof $U$ is an $\mathbb A^1$-bundle over $X$. In fact, it is easy to see that $U\simeq \mathrm{Spec}_X ( \oplus _{n\in \mathbb Z} \mathscr O_X(n))$ with a projection $\pi:U\to X$. It follows that $\pi_*\mathscr O_U\simeq \oplus _{n\in \mathbb Z} \mathscr O_X(n)$ and $R^j\pi_*\mathscr O_U=0$ for $j>0$. Then the claimed isomorphism follows from the simple special case of the Leray spectral sequence when all $R^j$'s with $j>0$ are $0$.
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