Does the first singular cohomology of an ACM surface vanish? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-20T09:01:05Z http://mathoverflow.net/feeds/question/83809 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/83809/does-the-first-singular-cohomology-of-an-acm-surface-vanish Does the first singular cohomology of an ACM surface vanish? Matteo Varbaro 2011-12-18T17:33:46Z 2011-12-24T09:57:38Z <p>Hi everybody, I am interested in the following:</p> <p>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 $X=\operatorname{Proj}(S/I)$. </p> <p>Is it true that $H_{Sing}^1(X^h)= 0$?</p> <p>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 $\operatorname{depth}(S/I)\geq 3$.</p> <p>However, what can we say if $X$ is singular?</p> http://mathoverflow.net/questions/83809/does-the-first-singular-cohomology-of-an-acm-surface-vanish/84051#84051 Answer by Sándor Kovács for Does the first singular cohomology of an ACM surface vanish? Sándor Kovács 2011-12-21T22:31:27Z 2011-12-24T09:57:38Z <p><strong>Addendum</strong> 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. :( <strong>end of Addendum</strong></p> <p>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:</p> <p>$$\dots \to H^i(Y,\mathscr O_Y) \to H^i(U,\mathscr O_U) \to H^{i+1}_P(Y,\mathscr O_Y) \to H^{i+1}(Y,\mathscr O_Y) \to \dots.$$ 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 $0&lt; i &lt; d-1$.</p> <blockquote> <p><strong>Proposition</strong> $\quad\<br> H^i(U,\mathscr O_U) \simeq \bigoplus_{n\in\mathbb Z} H^i(X, \mathscr O_X(n))$ </p> </blockquote> <p><strong>Proof</strong> $U$ 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 $0$.</p> http://mathoverflow.net/questions/83809/does-the-first-singular-cohomology-of-an-acm-surface-vanish/84174#84174 Answer by Mahdi Majidi-Zolbanin for Does the first singular cohomology of an ACM surface vanish? Mahdi Majidi-Zolbanin 2011-12-23T17:42:26Z 2011-12-23T17:42:26Z <p>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:</p> <p><strong>Theorem.</strong> Suppose $X\subset\mathbb{P}^n_\mathbb{C}$ is a local complete intersection of dimension $d=n-r$ with $d-r\geq1$. Then $\mathrm{H}^1(X,\mathbb{C})=0$.</p> <p>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.</p> <p>The above theorem is Theorem 4.9, page 1106 in <em>On the formal neighborhood of a subvariety of projective space</em>, Amer. J. Math. <strong>97</strong> (1975), no.4, p.p. 1085-1107.</p>