# Tuesday, February 21

:::{.remark}
Last time: $M$ a compact manifold, $\dim_\RR M = 2n$, there is a perfect pairing 
\[
H^k(M;\ZZ)/\tors \tensor_\ZZ H^k(M; \ZZ)/\tors &\to \ZZ \\
\alpha \tensor \beta &\mapsto \int_M \alpha\wedgeprod \beta
.\]
Interpret $\alpha . \beta$ as $[N_1] . [N_2] = \sum_{p\in N_1 \transverse N_2} \pm 1$ where the $N_i$ are Poincaré duals.
Satisfies \( \alpha . \beta= (-1)^{n} \beta \alpha \) for \( \alpha, \beta\in H^{n}(X; \ZZ)/\tors \) where $n\da \dim_\CC M = {1\over 2}\dim_\RR M$.
:::

:::{.definition title="Lattices"}
An (orthogonal) lattice is a free abelian group \( \Lambda \cong \ZZ^k \) of finite rank, together with an integral symmetric bilinear form 
\[
\cdot: \Lambda\tensor \Lambda\to \ZZ 
.\]

- $\Lambda$ is **symplectic** if $\cdot$ is alternating, so $\alpha.\beta = -\beta . \alpha$.
- $\Lambda$ is **unimodular** if for all primitive nonzero vectors $x\in \Lambda$, $\exists y\in \Lambda$ such that $x.y = 1$, where $x$ is primitive if $x\neq \lambda z$ for any $z\in \Lambda$
- $\Lambda$ is nondegenerate if $\forall x\in \Lambda$, $\exists y \in \Lambda$ with $x.y \neq 0$.

- The **Gram matrix** of a basis $\ts{e_i}$ for a lattice \( (\Lambda, \cdot) \) is $M_{ij} \da e_i . e_j$.
  $M$ is symmetric for orthogonal lattices and skew-symmetric for symplectic lattices.
  One can write $v.w = v^t M w$.
- If \( \Lambda_i \) are lattices, so is \( \bigoplus_{i} \Lambda_i \).
  
:::

:::{.example title="?"}
The standard example: \( \Lambda \in \ZZ^2 \) with \( \tv{x_1, y_1}.\tv{x_2, y_2} \da x_1 x_2 + y_1 y_2 \) is nondegenerate and unimodular.
Here $\tv{2,2} = 2\tv{1,1}$ is not primitive, but $\tv{2,1}$ is primitive.
Proving unimodularity: if $\gcd(m, n) = 1$, one just needs to solve $mx + ny = 1$ for $\tv{x,y}\in \ZZ^2$.
This has Gram matrix $\matt 1 0 0 1$.
:::

:::{.example title="?"}
A degenerate lattice: $(\ZZ^2, \vector x . \vector y \da x_1 x_2)$.
This is symmetric, but $\vector x . \tv{0, 1} = 0$ for every $\vector x \in \Lambda$.
The Gram matrix is $\matt 1 0 0 0$.
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:::{.example title="?"}
A symplectic lattice: $(\ZZ^2, \vector x . \vector y \da x_1 y_2 - x_2 y_1)$, which has Gram matrix $\matt 0 1 {-1} 0$.
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:::{.example title="?"}
There is a $2g\dash$dimensional symplectic lattice for every $g\geq 0$ given by $\ZZ^{2g}_\mathrm{symp} \da \bigoplus_{i=1}^{g} (\ZZ^2, \cdot_\mathrm{symp})$, which has Gram matrix comprised of $g$ diagonal blocks of $\matt 0 1 {-1} 0$.
This is the only unimodular symplectic lattice up to isomorphism, and is the intersection form on $H^1(\Sigma_g; \ZZ)$.
The vectors here can be represented by fundamental classes of 1-dimensional submanifolds, i.e. real curves:

\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2023/Spring/k3surfaces/sections/figures}{2023-02-21_10-03.pdf_tex} };
\end{tikzpicture}

More generally, if $\dim_\RR M = 4k+2$, then $H^{2k+1}(M; \ZZ) \cong \ZZ^{2g}_\mathrm{symp}$ for some $g$.
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:::{.definition title="Orthogonal complements"}
For $M \subseteq \Lambda$, define $M^\perp \da \ts{x\in \Lambda\st x.m =0 \, \forall m\in M}$.
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:::{.example title="?"}
For $\ZZ^2_\mathrm{symp} = \ZZ \alpha \oplus \ZZ \beta$, we have $\ZZ\alpha^\perp = \ZZ\alpha$.
For $\ZZ^2_\std = \ZZ \alpha \oplus \ZZ \beta$, one instead has $\ZZ\alpha^\perp = \ZZ\beta$.
:::

:::{.remark}
Over $\RR$, symmetric bilinear forms are classified by their signature: they can all be diagonalized to $\diag(1,\cdots, 1, -1,\cdots,-1,0,\cdots, 0)$ for some multiplicities $n_+, n_-, n_0$ where $n_+ + n_- + n_0 = \rank_\ZZ \Lambda$.
Any lattice can be extended via $\Lambda_\RR \da \Lambda \tensor_\ZZ \RR$, so define $\sgn \Lambda \da (n_+, n_-, n_0)$.
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:::{.example title="?"}
Let \( \Lambda = \ZZ\adjoin{\sqrt{5}} \) and let $u.v = \Re(x\bar {y})$, so 
\[
\norm{u} = u.u = (a+b\sqrt{5})(a-b\sqrt{5}) = a^2 - 5b^2
.\]
Note that \( \Lambda_\RR \cong \RR^{1,1} \), which in the standard basis has Gram matrix $\matt 1 0 0 {-1}$.
This can be visualized as a 2-dimensional subspace of $\RR^2$ spanned by $1, \sqrt{5}$.
Note that $\norm{\sqrt 5} = -5$.
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:::{.remark}
Define the *hyperbolic signature* as $(1, n)$ for any $n\geq 0$.
One can visualize positive/negative norm vectors using the light cone: for $\RR^{1, n}$, solving $v.v=0$ to get $x_1^2 = x_2^2 + \cdots + x_{n_1}^2$.
This is a cone over $S^1$ at height 1 in $\RR^3$:

\begin{tikzpicture}
\fontsize{45pt}{1em} 
\node (node_one) at (0,0) { \import{/home/zack/SparkleShare/github.com/Notes/Class_Notes/2023/Spring/k3surfaces/sections/figures}{2023-02-21_10-27.pdf_tex} };
\end{tikzpicture}

Note that $\ts{v.v > 0}$ has two connected components.
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:::{.definition title="Hyperbolic planes"}
A *hyperbolic cell/plane* is the lattice $H$ defined as $\ZZ^2$ with pairing given by the Gram matrix $\matt 0 1 1 0$ and signature $(1, 1)$.
This admits an orthonormal basis ${e_1 + e_2\over \sqrt 2}, {e_1 - e_2 \over \sqrt 2}$, and $H_\RR \cong \RR^{1, 1}$.
:::

:::{.remark}
Recall the ADE Dynkin diagrams: 

![](figures/2023-02-21_10-32-44.png)

One can build a root lattice out of each diagram: 

- Take one basis vector $e_i$ for each node, 
- $e_i^2 = -2$,
- $e_i.e_j = 1 \iff$ nodes $i, j$ are connected, and zero otherwise.

The lattices will be negative definite, i.e. of signature $(0, n)$ with $n$ the number of nodes.
The only unimodular such lattice corresponds to $E_8$.
:::

:::{.example title="?"}
$A_1$ corresponds to the matrix $[-2]$, and thus $\ZZ$ with bilinear form $-2n^2$.
$A_2$ yields $\matt{-2}{1}{1}{-2}$, and $A_3$ yields $\mattt {-2} 1 0 1 {-2} 1 0 1 {-2}$.
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:::{.question}
How does one check that a lattice is unimodular?
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:::{.definition title="?"}
Let \( (\Lambda, \cdot) \) be a nondegenerate lattice, so \( \Lambda\injects \Lambda_\RR \cong \RR^{a, b} \).
Define 
\[
\Lambda\dual \da \ts{ y\in \Lambda_\RR \st x.y\in \ZZ \, \forall x\in \Lambda}
.\]
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:::{.example title="?"}
Consider $\Lambda \da \sqrt{2}\ZZ \injects \RR^1$ with the standard pairing, then $1\over \sqrt{2}\in \Lambda\dual$.
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:::{.remark}
One can always find a basis of $\Lambda\dual$ given by $e_i\dual$ where $e_i\dual e_j = \delta_{ij}$.
Since $e_i\dual M e_j = \delta_{ij}$ for $M$ the Gram matrix of a form, one finds that $e_i\dual$ is the $i$th row of $M\inv$.
Why: letting $N$ be the matrix with rows $e_i\dual$, one has $NM = I$.
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:::{.proposition title="?"}
$\Lambda$ is unimodular iff \( \Lambda\dual = \Lambda \).
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:::{.proof title="?"}
Note $x^2\in \ZZ$ by definition, so $\Lambda \subseteq \Lambda\dual$.
If $v\in \Lambda\dual\sm \Lambda$, then one can show that the minimal $n$ such that $nv\in \Lambda$ yields a primitive element of $\Lambda$.
Since $Nv.w \in n\ZZ$ for all $w$, so can't pair to 1.
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:::{.remark}
So $\Lambda\dual \da \bigoplus \ZZ e_i\dual \subseteq \Lambda \implies e_i\dual \in \Lambda \implies M\inv \in \GL_n(\ZZ)$, and applying the same argument to duals yields $\det M = \pm 1$.
In general, $\det M = \size( \Lambda\dual / \Lambda)^2$ is the covolume.
So 

- $\vol(\RR^n / \Lambda) = \det M$
- $\vol(\RR^n / \Lambda\dual ) = \det M\inv$, which is the Gram matrix of \( \Lambda\dual \).
- $\size(\Lambda\dual / \Lambda)^2 = \covol( \Lambda)^2/ \covol( \Lambda\dual)^2 = \det(M)^2$.

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