# Monday, March 29 :::{.remark} Last time: the Hodge decomposition theorem. Let $(X, g) \in \Mfd_\CC^\cpt(\Kah)$, then the space of harmonic $k\dash$forms \( \mathcal{H}^k(X) \tensor_\RR \CC \) decomposes as \( \bigoplus_{p+q = k} \mathcal{H}^{p, q}(X) \). There is also a symmetry \( \bar{\mathcal{H}^{p, q}(X) } = \mathcal{H}^{q, p}(X) \). We have an isomorphism to the de Rham cohomology \( \mathcal{H}^k(X) \tensor_\RR \CC \cong H^k_\dR (X; \CC) \). We know the constituent pieces as well, as well as several relationships: \[ \mathcal{H}^{p, q}(X) &= \ker (\Delta_d: A^{p, q}(X) \selfmap) \\ \Delta_{\delbar} &= \delbar \delbar^\dagger + \delbar^\dagger \delbar \\ \Delta_d &= 2 \Delta_{\delbar} .\] There was a proposition that \( \ker(\Delta_d) = \ker(d) \intersect \ker(d^\dagger) \), and the same proposition holds for \( \Delta_{\delbar} \). In this case we have \( \ker(\Delta_{\delbar}) = \ker(\delbar) \intersect \ker( \delbar^\dagger) \) on $A^{p, q}(X)$, and this is isomorphic to \( \ker(\delbar) / \im(\delbar) \). Recall that we resolved the sheaf \( \Omega^p \) of holomorphic $p\dash$forms by taking the Dolbeault resolution \[ 0 \to \Omega^p \to A^{p, 0} \mapsvia{\delbar} A^{p, 1} \mapsvia{\delbar} A^{p, 2} \to \cdots .\] Thus we can identify \( \ker(\delbar)/\im(\delbar) \cong \Hsh( X; \Omega^p) \) as sheaf cohomology. We defined \( h^{p, q}(X) \da \dim_\CC H^{p, q}(X) \). ::: :::{.corollary title="Homology is independent of the choice of Kähler form"} $h^{p,q }(X)$ is independent of the Kähler form, noting that the isomorphism to sheaf cohomology doesn't involve taking adjoints, and $\dim_\CC \Hsh^q(X; \Omega^p)$ doesn't depend on the complex structure. ::: :::{.remark} A priori, one could vary the Kähler form and have some $h^{p, q}$ jump or drop dimension. It also turns out that varying the complex structure will also not change these dimensions. ::: :::{.remark} Whenever the Hodge-de Rham spectral sequence degenerates, one generally gets $\sum_{p+q} h^{p,q } = h^k$. Note that there is a resolution: \[ 0 \to \constantsheaf{\CC} \to \OO \mapsvia{d} \Omega^1 \mapsvia{d} \Omega^2 \mapsvia{d} \cdots ,\] which is not acyclic and thus has homology. In general, the spectral sequence is \[ E^1_{p,q} = \Hsh^q(X; \Omega^p) \abuts \Hsh^{p+q}(X; \constantsheaf{\CC}) .\] ::: :::{.fact} A fact about the cohomology of vector bundles: given a family of Kähler manifolds $X_t$, one can consider \( H^q(X_t; \bundle{E}_t \) where \( \bundle{E}_t \) is a family of holomorphic vector bundles. This can only jump upward in dimension, i.e. $\dim_\CC H^q(X_t; \bundle{E}_t)$ is **lower semicontinuous**. ::: :::{.example title="?"} Consider \[ X_t \da \ts{ x^3 + y^3 + z^3 + txyz = 0 } \subseteq \CP^2 ,\] where $t$ varies in $\CC$. These all admit a line bundle \( \bundle{L}_t \da \ro{ \OO(1) } {X_t} \), the anti-tautological line bundle on $\PP^2$. \begin{tikzpicture} \fontsize{45pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-03-29_14-14.pdf_tex} }; \end{tikzpicture} The real points of this vanishing locus form an elliptic curve, and each $X_t$ is a Riemann surface of genus 1. Note that $h^{0, 1}$ can jump on closed sets, but $H^1$ is constant since Riemann-Roch involves genus and degree. What is $\deg \ro {\OO(1)}{X_t}$? Take a section $s \in H^0(\PP^2; \OO(1))$ which vanishes on a line in $\PP^2$. How many points lie in a line intersected with $X_t$? Looking at fundamental classes, we have $[X_t] = 3\ell$, and by Bezout $3\ell \cdot \ell = 3$. The point is that $H^q(X_t; \Omega^p)$ can only possibly increase at special values of $t$. Assuming the $X_t$ are all diffeomorphic, then $h^k(X_t)$ is constant and $h^{p, q}(X_t)$ can't jump. So the $h^{p, q}$ are invariants of families. ::: :::{.definition title="Hodge Diamond"} The **Hodge Diamond** of $X \in \Mfd(\Kahler)$ (which won't depend on the choice of Kähler form) is given by \begin{tikzcd} &&& {h^{n, n}} \\ && {h^{n-1, n}} && {h^{n, n-1}} \\ & \ddots &&&& \ddots \\ \ddots &&& \vdots &&& \ddots \\ & {h^{2, 0}} && {h^{1, 1}} && {h^{0, 2}} \\ && {h^{1, 0}} && {h^{0, 1}} \\ &&& {h^{0, 0}} \arrow["\hodgestar"{pos=0}, dotted, tail reversed, from=6-3, to=2-5] \arrow["{z\mapsto \bar{z}}"', dotted, tail reversed, from=6-5, to=2-5] \arrow["{z\mapsto \bar{z}}", dotted, tail reversed, from=6-3, to=2-3] \arrow["\hodgestar"{description, pos=0.1}, dotted, tail reversed, from=6-5, to=2-3] \end{tikzcd} > [Link to Diagram](https://q.uiver.app/?q=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) Note that there are symmetries, e.g. $\hodgestar$ takes $h^{1, 0} = h^{n-1, n}$ and $\bar{h^{p, q}} = h^{q, p}$. ::: :::{.proposition title="CYs have extra Hodge diamond symmetry"} If $X$ is **Calabi-Yau**, so $K_X = \OO_X$ (i.e the canonical bundle is trivial), then the Hodge diamond has an orientation preserving $(\ZZ/2)^2$ symmetry, i.e. there is a rotation by $\pi/2$. > Note: this isn't extra symmetry! Just a proof of the symmetry in this case. ::: :::{.proof title="?"} Let \( \Omega^k_X \) be the sheaf of holomorphic $k\dash$forms, then there is a map \[ \Omega_X^k \tensor \Omega_X^{n-k} &\to \Omega_X^n \da K_X \\ \alpha \tensor \beta &\mapsto \alpha \wedge \beta .\] Fiberwise, this is a perfect pairing. If one takes $\alpha \da e_{i_1} \wedge \cdots e_{i_k} \in \Wedge^k T_x\dual X$, there is a unique basis wedge $\beta \da e_{j_1} \wedge \cdots \wedge e_{j_n - k}$ then \( \alpha\wedge \beta \) is a basis wedge \( e_1 \wedge \cdots \wedge e_n \). So \( \Omega_X^k \cong ( \Omega_X^{n-k} )\dual \) if $X$ is Calabi-Yau. By Serre duality, \[ \Hsh ^p(X; \Omega_X^q)\dual \cong \Hsh^{n-p}(X; (\Omega_X^q)\dual \tensor K_X ) .\] ::: :::{.example title="?"} In dimension 3, take \[ X \da \ts{ x_0^5 + \cdots + x_4^5 = 0 } \subseteq \PP^4 \in \Mfd^3(\CC) .\] See Hodge diamond. ::: :::{.remark} Note that $K3$s are special CYs. An example is $\CC^2 / \Lambda$ for \( \Lambda \) a rank 4 lattice. This is diffeomorphic to $(S^1)^4$, for example $E\cross E$. :::