Dimension of Affine Bundles on Projective Space - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-26T06:38:01Z http://mathoverflow.net/feeds/question/101817 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/101817/dimension-of-affine-bundles-on-projective-space Dimension of Affine Bundles on Projective Space Will Sawin 2012-07-10T01:03:45Z 2012-07-10T02:58:48Z <p>Let $X \to \mathbb P^n$ be a fiber bundle of algebraic varieties with $X$ an affine variety. What is the smallest dimension that $X$ can be?</p> <p>An obvious lower bound is $n+1$.</p> <p>An upper bound is $2n$, given by taking the complement of a generic effective divisor of bidegree $(1,1)$ in $\mathbb P^n \times \mathbb P^n$. (Alternatively, take the canonical "duality" $(1,1)$ divisor defined by viewing $\mathbb P^n$ as the Hilbert scheme of hyperplanes on $\mathbb P^n$.) The complement of an effective ample divisor is of course affine, and since the divisor is locally constant on the fibers its complement is as well.</p> <p>This construction is the construction in <a href="http://mathoverflow.net/questions/101801/projective-spaces-as-affine-varieties/101805#101805" rel="nofollow">this question</a>, which also inspired my question, with the second condition removed.</p> http://mathoverflow.net/questions/101817/dimension-of-affine-bundles-on-projective-space/101820#101820 Answer by Jason Starr for Dimension of Affine Bundles on Projective Space Jason Starr 2012-07-10T02:05:13Z 2012-07-10T02:58:48Z <p>Are you working over the complex numbers? Is $X$ smooth? If so, then I believe you can use the Leray spectral sequence and the Lefschetz hyperplane theorem to quickly deduce that $\text{dim}_{\mathbb{C}}(X) \geq 2n$. Indeed, denote by $m$ the relative dimension of $X$ over $\mathbb{C}P^n$, so that the (complex) fiber dimension is $m$. Let $p_0$ be the largest integer such that $H^{p_0}(Y,\mathbb{Q})$ is nonzero, where $Y$ is the fiber. Now consider the Leray spectral sequence computing the cohomology of the total space, $H^{r}(X,\mathbb{Q})$, starting with the page whose terms are $H^q(\mathbb{C}P^n,H^p(Y,\mathbb{Q}))$ -- I am using that $\mathbb{C}P^n$ is simply connected. In particular, for $q=2n$ and for $p=p_0$, the term is $$H^{2n}(\mathbb{C}P^n,\mathbb{Q})\otimes_{\mathbb{Q}}H^{p_0}(Y,\mathbb{Q}),$$ which is nonzero by hypothesis. In fact there is no nonzero term of higher $p$-degree or $q$-degree, hence there is no nonzero differential coming into or out of this term at any stage of the spectral sequence. Therefore $H^{2n+p_0}(X,\mathbb{Q})$ is nonzero. On the other hand, since $X$ is affine, we can use the Lefschetz hyperplane theorem to conclude that $H^d(X,\mathbb{Q})$ is zero for $d$ strictly larger than $\text{dim}_{\mathbb{C}}(X)$. Thus $n+m \geq 2n+p_0 \geq 2n$, so that $\text{dim}(X) \geq 2n$.</p> <p>EDIT: What I refer to as the "Lefschetz hyperplane theorem" is more properly the "Andreotti-Frankel" theorem.</p>