Does the first singular cohomology of an ACM surface vanish? - MathOverflow

Does the first singular cohomology of an ACM surface vanish?

2011-12-18T17:33:46Z

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?

Answer by Sándor Kovács for Does the first singular cohomology of an ACM surface vanish?

2011-12-21T22:31:27Z

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 $0$.

Answer by Mahdi Majidi-Zolbanin for Does the first singular cohomology of an ACM surface vanish?

2011-12-23T17:42:26Z

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.