Elementary computation of direct image sheaves. - MathOverflow

Elementary computation of direct image sheaves.

2012-09-03T02:22:07Z

I am a physicist and would like to understand the section 1 of this math paper, which explains how the SYZ conjecture implies topological mirror symmetry. I have some technical problem and would appreciate your help. My question is purely mathematical and nothing scary to most of you, I think.

The setting is the following; let $f:X\rightarrow B$ be a 3-torus $T^3$-fibration of a Calabi-Yau threefold (which has singular fibers). Since $X$ is a real sixfold, $B$ is a real threefold. Under certain technical assumption, using Leray spectral sequence, the author shows that the dual torus fibration gives another Calabi-Yau threefold with mirrored Hodge diamond.

My questions concern the Leray spectral sequence associated with $f$. First I don't quite understand why $E_2$ of the Leray spectral sequence looks like the diagram in the paper; how can we conclude that $$ H^{i}(B,R^{j}f_{*}\mathbb{R})=\mathbb{R} \ (i=0,3), \ \ 0 \ (i=1,2) $$ for $j=0,3$? My understanding is that roughly the direct image sheaves $R^{i}f_{*}\mathbb{R}$ are local system, so on a small open subset we have $R^{i}f_{*}\mathbb{R}\cong H^{i}(T^{3},\mathbb{R})$. How should we understand the computation above?

Secondly why $\pi_{1}(X)=0$ (or rather $H^{1}(X,\mathbb{R})=H^{5}(X,\mathbb{R})=0$) implies that $$ H^{0}(B,R^{1}f_{*}\mathbb{R})=H^{3}(B,R^{2}f_{*}\mathbb{R})=0? $$ I know that the problem is caused by my poor understanding of direct image sheaves. I hope to make my understanding clearer with this example.

I am afraid that my questions are elementary and this may not be a place to ask such a question. Any help, comments and reference suggestions will be appreciated.

Thank you,

Answer by Tim Perutz for Elementary computation of direct image sheaves.

2012-09-03T13:45:15Z

Edited following algori's comment.

In the context of the paper I think one can make these deductions on rather formal grounds - deep understanding of higher direct images is not needed.

Gross assumes that the base $B$ is a simply connected closed 3-manifold (i.e., it's $S^3$). Let $i\colon B_0\to B$ be the inclusion of the (open and dense) regular locus, and $f_0\colon f^{-1}(B_0)\to B_0$ the restricted fibration. Gross also tells us that $f$ is "simple", meaning that $i_*(R^jf_{0\ast} \mathbb{R})\cong R^j f_\ast \mathbb{R}$ for each $j$. In degree $j=0$, simplicity is automatic: the fibers of $f_0$ are connected, and since disconnectedness of a fiber is an open property on $B$, the fibers of $f$ are also connected. So $f_{0\ast}\mathbb{R}=\mathbb{R}$ and $f_{\ast}\mathbb{R}=\mathbb{R}$.

The fibers of $f_0$ are also oriented. Moreover, after choosing orientations for $X$ and $B$ we have a consistent notion of a "positive" fiber-orientation: one which, after wedging with the pullback of a positive orientation form on $B$, agrees with the orientation of $X$. Hence $R^3f_{0\ast}\mathbb{R}=\mathbb{R}$. By simplicity, $R^3 f_{\ast}\mathbb{R}\cong \mathbb{R}$. We can therefore complete the $0$th and $3$rd rows of $E_2$ - each is the de Rham cohomology of $S^3$: $$ \mathbb{R}\quad 0\quad 0 \quad \mathbb{R}.$$

The spectral sequence converges to $H^\ast(X;\mathbb{R})$, and we know that $E_2^{2,0}=0$. We must have $E_2^{0,1}=0$ because otherwise it would survive to $E_\infty$ and we would have $H^1(X;\mathbb{R})\neq 0$. Similarly, since $E_2^{1,3}=0$ and $H^5(X;\mathbb{R})=0$, we must have $E_2^{3,2}=0$.