## Does the first singular cohomology of an ACM surface vanish?

Hi everybody, I am interested in the following:

Let $I\subset S=\mathbb{C}[x_0,\ldots ,x_n]$ be a graded ideal such that $\operatorname{depth}(S/I)\geq 3$, and let $X^h$ denote the analytic space associated to $X=\operatorname{Proj}(S/I)$.

Is it true that $H_{Sing}^1(X^h)= 0$?

The answer is yes if $X$ is smooth: In fact, in this case, if $H_{Sing}^1(X^h)\neq 0$, then the Hodge decomposition would give $H^1(X,O_X)=H_{S_+}^2(S/I)_0\neq 0$, a contradiction to the fact that $\operatorname{depth}(S/I)\geq 3$.

However, what can we say if $X$ is singular?

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Yes, by $H_{Sing}^1$ I mean singular cohomology over $\mathbb{C}$. I usually denote by $H^1(X^h,\mathbb{C})$ the sheaf cohomology of the constant sheaf associated to $\mathbb{C}$, however one can show that $H_{Sing}^1\cong H^1(X^h,\mathbb{C})$, so it does not matter. $H_{S_+}^i(S/I)$ means local cohomology with support in the ideal $S_+=(x_0,\ldots ,x_n)$. $H_{S_+}^i(S/I)_0$ means its degree $0$ part, and $\operatorname{depth}(S/I)\geq 3$ means $H_{S_+}^i(S/I)=0$ for all $i=0,1,2$. – Matteo Varbaro Dec 19 2011 at 10:05
I have a suggestion, but I'm not sure that it works. Take a smooth surface $S$ with $H^1>0$ in some projective space and project it generically to ${\mathbb P}^3$ to obtain a surface $X$. Intuitively, I would say that $H^1(X)\ge H^1(S)$, but I don't know how to prove this. (Recall that a generic projection is a birational map with finite fibers and the sigular locus of $X$ is a double curve with triple points that are triple also for the surface. So you can think of $X$ as being obtained form $S$ by identifying points on a curve with an involution). – rita Dec 21 2011 at 12:23
@Sandor: isn't a hypersurface ACM? I used to think it is, but I guess I got this wrong. – rita Dec 22 2011 at 6:31
To Rita and Sàndor: yes, a hypersurface is certainly ACM. – Angelo Dec 22 2011 at 7:18
I see, of course a hypersurface is ACM, I got stock with the smooth surface with the $H^1\neq 0$ which isn't... Sorry. – Sándor Kovács Dec 22 2011 at 10:03
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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 $Gr_F^0H^1(X,\mathbb C)=0$ 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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How does this imply vanishing of the first singular cohomology group? – Angelo Dec 22 2011 at 7:20
Oops, at some point I switched to thinking that the question was whether $H^1(X,\mathscr O_X)=0$.... This still says something, but certainly less. – Sándor Kovács Dec 22 2011 at 9:44
Concerning the (attempted) counterexample of Rita, I think that I can prove that the answer to my question is positive also if $X$ is a complete intersection. – Matteo Varbaro Dec 23 2011 at 16:42
Actually, I was interested in the question because I think I can show that a certain algebraic property of $I$ holds true if and only if the first singular cohomology of $X$ vanishes, under the assumption depth$(S/I)\geq 3$. Since I have not much feeling with singular cohomology, I wanted to be sure that the vanishing of this singular cohomology wasn't well known or easy, as it is under the smoothness assumption. In any case I stay very interested in the fact, and I am happy to have captured your attentions, thank you to think to the problem! – Matteo Varbaro Dec 23 2011 at 16:48

If you are willing to assume that $X$ is locally a complete intersection, then the result you want will follow from a theorem due to A.Ogus:

Theorem. Suppose $X\subset\mathbb{P}^n_\mathbb{C}$ is a local complete intersection of dimension $d=n-r$ with $d-r\geq1$. Then $\mathrm{H}^1(X,\mathbb{C})=0$.

In your question $X$ is an ACM hypersurface and the cone has depth $\geq3$. So the cone has dimension $\geq3$, which means $d:=\dim X\geq2$. But since $X$ is a hypersurface we see $d=n-1\geq2$, i.e., with notation of theorem, $r=1$ and $d-r=d-1=n-2\geq1$ which is the condition needed in the theorem.

The above theorem is Theorem 4.9, page 1106 in On the formal neighborhood of a subvariety of projective space, Amer. J. Math. 97 (1975), no.4, p.p. 1085-1107.

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 Is $r$ the codimension of $X$? In my question $X$ is not a hypersurface! As I said in one comment I can prove what I asked whenever $X$ is a complete intersection, so this is a particular case. However, if $X$ is a hypersurface, then it is automatically a locally complete intersection. In any case thank you for indicating me the paper of Ogus, I will read it! – Matteo Varbaro Dec 23 2011 at 18:35 Yes, $r$ is the codimension of $X$. I think the title of your question led me to think that $X$ was a hypersurface. – Mahdi Majidi-Zolbanin Dec 23 2011 at 18:52 I see. In the title I put ACM surface because it was shorter. This is the case in which dim$(S/I)$ $=$ depth$(S/I)=3$: I cannot answer the question neither in this special case. – Matteo Varbaro Dec 24 2011 at 9:19