# Monday, November 08

:::{.remark}
Today: Poincaré duality, relates to *smooth* loci, and rational (i.e. $\QQ$) smoothness.
Take all varieties to be quasiprojective subvarieties of a flag variety $G/B$.

From algebraic topology, there is a relative cup product in singular cohomology:
\[
\cupprod: H^i(X, U; \ZZ) \tensor_\ZZ H^j(X, U; \ZZ) \to H^{i+j}(X, U_1\union U_2)
.\]
Even better, we have a pairing with Borel-Moore homology for $Y \leq X$ a closed subvariety:
\[
\capprod: H^j(X, X\sm Y) \tensor_\ZZ \bar{H}_j(X) &\to \bar{H}_{j-i}(Y)
.\]
This yields
\[
\capprod: H^j(X, X\sm Y) \tensor H^{2n-j}(G/P, (G/P)\sm X) \to H^{2n-j+i}(G/P, (G/P) \sm Y)
.\]
Think of $H^j(A, B)$ as chains in $A$ vanishing along $B$:

\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/FlagVarieties/sections/figures}{2021-11-08_14-01.pdf_tex} };
\end{tikzpicture}

:::

:::{.proposition title="Poincaré duality"}
For $X$ smooth and irreducible, capping against the fundamental class induces an isomorphism
\[
H^i(X) &\iso H_{2n-i}(X) \\
\alpha &\mapsto \alpha \capprod [X]
,\]
which is induced by
\[
H^i(X) \times \bar{H}_{2n}(X) &\to \bar{H}_{2n-i}(X) \\
(\alpha, [X]) &\mapsto \alpha \capprod [X]
.\]
:::

:::{.remark}
Recall that there is an affine stratification $G/P_Y = \Disjoint_{w\in W^Y} BwP_Y/P_Y$, and
\[
\bar{H}_{2k} (G/P) = \bigoplus_{\substack{ w\in W^Y \\ \ell(w) = k }} \ZZ[X_w^Y]
.\]
Pulling back along the isomorphism there is some element such that $d_{X_w^Y} \capprod [G/P] = [X_w^Y]$, so we often identify $d_{X_w^Y} = [X_w^Y]$.
:::

:::{.remark}
An alternative perspective on Chern classes: compose the maps

\begin{tikzcd}
	{\K(X)} && {A(X)} && {\bar{H}_*(X)} && {H^*(X)} \\
	\\
	\eps && {c_1(\eps)} && {c_1(\eps)} && {c_1(\eps)} \\
	&& {[Z(s)]} && {[Z(s)]} && {[Z(s)]}
	\arrow[from=1-1, to=1-3]
	\arrow[from=1-3, to=1-5]
	\arrow[from=1-5, to=1-7]
	\arrow[from=3-1, to=3-3]
	\arrow[from=3-3, to=3-5]
	\arrow[from=3-5, to=3-7]
	\arrow[from=4-3, to=4-5]
	\arrow[from=4-5, to=4-7]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsMTEsWzAsMCwiXFxLKFgpIl0sWzIsMCwiQShYKSJdLFs0LDAsIlxcYmFye0h9XyooWCkiXSxbNiwwLCJIXiooWCkiXSxbMCwyLCJcXGVwcyJdLFsyLDIsImNfMShcXGVwcykiXSxbNCwyLCJjXzEoXFxlcHMpIl0sWzYsMiwiY18xKFxcZXBzKSJdLFsyLDMsIltaKHMpXSJdLFs0LDMsIltaKHMpXSJdLFs2LDMsIltaKHMpXSJdLFswLDFdLFsxLDJdLFsyLDNdLFs0LDVdLFs1LDZdLFs2LDddLFs4LDldLFs5LDEwXV0=)

Here $Z(s)$ is the zero divisor of a section $s$ coming from the class of a bundle in $\K(X)$.
For a line bundle $\mcl$, we have $c_1(\mcl) \in A^1(X) \iso \bar{H}_{2n-2} \iso H^2(X)$.
:::

:::{.theorem title="On nilpotent orbits, Borho-MacPherson"}
TFAE:

- $H^i(X, X\sm\ts{x} ) = \QQ[2n]$
- $\RR\globsec{X, \IC_X} = \QQ[0]$

:::

:::{.remark}
Mentioned by Geordie: $\IC_X \cong \QQ_X$, the constant sheaf.
:::

:::{.example title="?"}
Let 
\[
f(x,y,z) = x^3 + y^3 - xyz
.\]
Let $X\da V(f) \subseteq \PP^2\slice \CC$, and define
\[
\xi: \PP^1\slice \CC &\to X \\
[a: b] &\mapsto [ab^2: a^2b : a^3 + b^3]
.\]

Check that this is well-defined:
\[
(ab^2)^3 + (a^2b)^3 - a^3b^3 (a^3 + b^3)
.\]
Note $\xi$ is projective and thus proper, and finite since it is quasifinite (finite fibers).
One can check
\[
\xi\inv[0:0:1] &= \ts{ [0:1], [1:0] } \\
\xi\inv[x:y:z] &= \ts{\pt}
.\]

:::{.exercise title="?"}
Check that $\xi$ is birational.
:::

Thus $\xi$ is the normalization of $X$, but isn't an isomorphism, so smoothness must fail.

:::{.question}
Is $X$ *rationally* smooth?
:::
Since $X$ is compact, $\bar{H}_k(X) \iso H_k(X)$.
Since $X$ is connected we get $H^0 = \QQ$, and by duality $H^2(X) \cong H_2(X) \cong \bar{H}_2(X) \cong \QQ$, we have $H^k(X) = \QQ[0] \oplus \QQ[1] \oplus  \QQ[2]$.
Note that the Poincaré polynomial $p(q) = 1 + q + q^2$ has symmetric coefficients.
What this morphism looks like:

\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2021/Fall/FlagVarieties/sections/figures}{2021-11-08_14-25.pdf_tex} };
\end{tikzpicture}

:::{.claim}
$X$ is not rationally smooth.
:::

:::{.proof title="?"}
By the projection formula,
\[
\xi_*(\xi^* \alpha\capprod \beta) = \alpha\capprod \xi_* \beta
.\]
Let $\alpha \in H^1(X)$ be nonzero, then
\[
\alpha\capprod [X] = a \capprod \xi_* [ \PP^1] = \xi_*(\xi^* \alpha \capprod [\PP^1] )
.\]
Since $\xi$ is birational, $\xi_* [\PP^1] = [X]$ and $H^k(\PP^1) = \QQ[0] \oplus \QQ[2]$.
Rationally smooth implies PD, and since PD doesn't hold here we can't have rational smoothness.
:::

:::