# Friday, April 16 :::{.remark} Last time: we defined the blowup $\Bl_0\CC^2$ as the closure of \[ \Bl_0 \CC^2 \da \cl \ts{ (x, y), [x:y] \st (x, y) \neq 0 } \subseteq \CC^2 \cross \CP^2 .\] This had the effect of adding in all limits of slopes as points approach $(0, 0) \in \CC^2$. We defined this using local holomorphic coordinate charts to $\CC^2$. Why is this a complex manifold? We can cover it with charts: given a point $(x, \mu)$ where $\mu = {y \over x}\in \PP^1$ is a slope, we can form a first chart by sending \[ (x, \mu) \mapsto \ts{ (x, x\mu), [1: \mu] } .\] This yields the first chart, as long as the slope is not infinite, so this applies to all finite slopes. The second chart will work for all nonzero slopes, where we take \[ (v, y)\in \CC^2 \mapsto \ts{ (yv, y), [v: 1] } .\] Note that restricting to $(x, y) = (0, 0)$, these give the standard $\CC\dash$charts on $\CP^2$. How do these two charts glue? When $\mu, \nu \neq 0$, we have well-defined transition functions $\mu = \nu ^{-1}$ and $x=y\nu$. ::: :::{.remark} Recall that for a complex curve $C \in \Mfd_\CC^2$, we have the blowup morphism $\pi: \Bl_pS\to S$ and we defined the **strict transform** $\hat{C} \da \cl \pi\inv(C\sm\ts{\pt})$. \begin{tikzpicture} \fontsize{42pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-16_14-02.pdf_tex} }; \end{tikzpicture} Here $E=\CP^1$ is the exceptional curve of the blowup, and intersects the curve twice. This has the effect of changing $D$ into an embedded curve. > Note that here $\pi^* D = \hat{D} + 2E$, where we'll define this next. ::: :::{.definition title="Pullback of a Curve"} The **pullback** of $C$, denoted $\pi^* C$, is constructed by writing $C = V(f)$ locally. We then set $\pi^* C \da V( \pi^* f)$. ::: :::{.example title="?"} Take $C \da \ts{ y=x } \subset \CC^2$ and consider $\Bl_0 \CC^2$. Then \[ \hat{C} \da \cl \ts{ \qty{ (x, x), [x:x]} \st x\neq 0 } = \cl \ts{ \qty{ (x, x), [1:1] } \st x\neq 0 } \subset \Bl_0 \CC^2 .\] By projecting onto the first component, $\pi:\hat{C} \mapsvia{\sim} C$ is an isomorphism. We can compute the pullback: we first have $\pi^* C = \pi^* V(y-x) = V( \pi^*(y-x))$, so consider $\pi^*(y-x)$ in the coordinate chart $(x, \mu)$. In this chart, $y=x\mu$, and so $\pi^*(y-x) = x\mu - x = x(\mu - 1)$, and so \[ V(\pi^*(y-x)) = V(x) + V(\mu -1) \implies \pi^* C = E + \hat{C} \text{ as a divisor} .\] ::: :::{.example title="A nodal curve"} Take the nodal curve $C = \ts{ y^2 - x^3 + x^2 }$: \begin{tikzpicture} \fontsize{44pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-16_14-15.pdf_tex} }; \end{tikzpicture} The pullback is then given by \[ \pi^* C &= V( \pi^* (y^2 - x^3 + x^2) ) \\ &= V(\mu^2x^2 -x^3 + x^2) \\ &= V(x^2) + V(\mu^2 - x + 1) \\ &= 2V(x) + V(\mu^2 -x + 1) .\] \begin{tikzpicture} \fontsize{44pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-16_14-17.pdf_tex} }; \end{tikzpicture} In the second coordinate chart, we have \[ \pi^* C = V(y^2 - y^4 \nu^3 +y^2 \nu^2) = 2V(y) + V(1-y\nu^3 + \nu^2 .\] \begin{tikzpicture} \fontsize{43pt}{1em} \node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Spring/FourManifolds/sections/figures}{2021-04-16_14-21.pdf_tex} }; \end{tikzpicture} Gluing along $\mu, \nu \neq 0$ we get the following picture for $\pi^* C$: \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}{2021-04-16_14-22.pdf_tex} }; \end{tikzpicture} Writing $C = \ts{ x=0 }$, note that $\hat{C}$ doesn't intersect the first coordinate chart. In the $\mu, x$ coordinate chart, for example, we can't get an infinite slop: \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}{2021-04-16_14-25.pdf_tex} }; \end{tikzpicture} ::: ## Change in Canonical Bundle Formula :::{.question} Given \( \Omega^2_S = K_S \to S \) the canonical line bundle, can we relate $K_{\Bl_p S}$ to $K_S$? ::: :::{.proposition title="Canonical of a blowup"} \[ K_{\Bl_p S} = \pi^* K_S \tensor \OO_S(E) .\] ::: :::{.proof title="?"} We'll abbreviate $\hat{S} \da \Bl_p(S)$. Let $\omega$ be a local section of $K_S$ near $p$, and in coordinate charts $(x, y)$, write $\omega = dx \wedge dy$. In the first coordinate chart on the blowup, we can write \[ \pi^* \omega = dx \wedge d(x\mu) = dx \wedge (\mu dx + x d\mu) = x\,dx\wedge d\mu .\] Note that $V(x) = E$, and that pulling back the canonical bundle yields something vanishing to order 1 (?). So $\pi^* K_S$ is isomorphic to the subsheaf of $K_{\hat S}$ whose sections vanish along $E$, which is isomorphic to $K_{\hat{S}} \tensor \OO(-E)$, since the latter are the functions which vanish along $E$. Tensoring both sides with $\OO(E)$ yields \[ K_{\hat{S}} = \pi^* K_S \tensor \OO_{\hat S}(E) \] as a line bundle, or in divisor notation $K_{\hat S} = \pi^* K_S + E$ where we take the divisor representing the line bundle instead. ::: :::{.remark} Using $\pi: \hat{S}\to S$, we get pullback maps \[ \pi^*: H^2(S; \ZZ) &\to H^2(\hat{S}; \ZZ) \\ \pi^*: \Div(S) &\to \Div(\hat{S}) .\] These are compatible in the sense that \[ [\pi^* C] = \pi^* [C] .\]. This can be seen by expressing $\OO_S(C) \cong \OO_S(A-b)$ for $A, B$ hyperplane section. We can assume $A, B$ avoid $p$ in their projective embeddings, making $[C] = [A] - [B]$ since $c_1(\OO_S(c)) = [C]$ is the fundamental class of $C$. So it suffices to prove the formula for curves *not* passing through $p$, but this is obvious! It follows from the fact that $\pi: \hat{S} \sm E \mapsvia{\sim} S\sm\ts{p}$ is an isomorphism. ::: :::{.remark} In fact, \[ H^2(\hat S; \ZZ) \cong \pi^* H^2(S, \ZZ) \oplus \ZZ[E] .\], which follows from Mayer-Vietoris. So this adds one to the rank. :::