# Friday, February 05

:::{.remark}
Recall that a lattice is **unimodular** if the map $L\to L\dual \da \Hom(L, \ZZ)$ is an isomorphism, where $\ell \mapsto \ell \cdot (\wait)$.
To check this, it suffices to check if the Gram matrix $M$ of a basis $\ts{e_i}$ satisfies $\abs{ \det M } = 1$.
:::

:::{.example title="Determinant 1 Integer Matrices"}
The matrices $[1]$ and $[-1]$ correspond to the lattice $\ZZ e$ where either $e^2 \da e\cdot e = 1$ or $e^2 = -1$.
If $M_1, M_2$ both have absolute determinant $1$, then so does
\[
\begin{bmatrix}
M_1 & 0 
\\
0 & M_2
\end{bmatrix}
.\]

So if $L_1, L_2$ are unimodular, then taking an orthogonal sum $L_1 \oplus L_2$ also yields a unimodular lattice.
So this yields diagonal matrices with $p$ copies of $+1$ and $q$ copies of $-1$.
This is referred to as $rm{1}_{p, q}$, and is an *odd* unimodular lattice of signature $(p, q)$ (after passing to $\RR$).
Here *odd* means that there exists a $v\in L$ such that $v^2$ is odd.
:::

:::{.example title="Even unimodular lattices"}
An even lattice must have no vectors of odd norm, so all of the diagonal elements are in $2\ZZ$.
This is because $(\sum n_i e_i)^2 = \sum_i n_i^2 e_i^2 + \sum_{i<j} 2 n_i, n_j e_i \cdot e_j$.
Note that the matrix must be symmetric, and one example that works is 
\[
\begin{bmatrix}
0 & 1 
\\
1 & 0
\end{bmatrix}
.\]
We'll refer to this lattice as $H$, sometimes referred to as the *hyperbolic cell* or *hyperbolic plane*.
:::

:::{.example title="A harder even unimodular lattice"}
This is built from the $E_8$ Dynkin diagram:

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\begin{tikzcd}
	&&&& {\bullet_{e_8}} \\
	{\bullet_{e_7}} & {\bullet_{e_6}} & {\bullet_{e_5}} & {\bullet_{e_4}} & {\bullet_{e_3}} & {\bullet_{e_2}} & {\bullet_{e_1}}
	\arrow[no head, from=2-1, to=2-2]
	\arrow[no head, from=2-2, to=2-3]
	\arrow[no head, from=2-3, to=2-4]
	\arrow[no head, from=2-4, to=2-5]
	\arrow[no head, from=2-5, to=1-5]
	\arrow[no head, from=2-5, to=2-6]
	\arrow[no head, from=2-6, to=2-7]
\end{tikzcd}

The rule here is
\[
e_i \cdot e_j = 
\begin{cases}
-2  & i =  j
\\
1 & e_i \to e_j \\
0 & \text{if not connected}.
\end{cases}
\]
So for example, $e_2 \cdot e_6 = 0, e_1 \cdot e_3 = 1, e_2^2 = -2$.
You can check that $\det(e_i \cdot e_j) = 1$, and this is referred to as the $E_8$ lattice.
This is of signature $(0, 8)$, and it's negative definite if and only if $v^2 < 0$ for all $v\neq 0$.
One can also negate the intersection form to define $-E_8$.
Note that any simply-laced Dynkin diagram yields some lattice.
For example, $E_{10}$ is unimodular of signature $(1, 9)$, and it turns out that $E_{10} \cong E_8 \oplus H$.
:::

:::{.definition title="Unimodular lattice II"}
Take 
\[ 
\mathbf{II}_{a, a+8b} \da \bigoplus_{i=1}^a H \oplus \bigoplus_{j=1}^b E_8 
,\]
which is an even unimodular lattice since the diagonal entries are all $-2$, and using the fact that the signature is additive, is of signature $(a, a+8b)$.
Similarly, 
\[ 
\mathbf{II}_{a+8b, a} \da \bigoplus_{i=1}^a H \oplus \bigoplus_{j=1}^b (-E_8) 
,\]
which is again even and unimodular.
:::

:::{.remark}
Thus 

- $\mathbf{I}_{p, q}$ is odd, unimodular, of signature $(p, q)$.
- $\mathbf{II}_{p, q}$ is even, unimodular, of signature $(p, q)$ only for $p \equiv q \mod 8$.

:::

:::{.theorem title="Serre"}
Every unimodular lattice which is not positive or negative definite is isomorphic to either $\mathbf{I}_{p, q}$ or $\mathbf{II}_{p, q}$ with $8\divides p-q$.
:::

:::{.remark}
So there are obstructions to the existence of even unimodular lattices.
Other than that, the number of (say) positive definite even unimodular lattices is 

| Dimension | Number of Lattices |
|-----------|--------------------|
| 8         | 1: $E_8$               |
| 16        | 2: $E_8^{\oplus 2}, D_{16}^+$ |
| 24        | 24: The Neimeir lattices (e.g. the Leech lattice) |
| 32        | >$8\times 10^{16}$!!!!   |

Note that the signature of a definite lattice must be divisible by 8.
:::

:::{.remark}
There is an isometry: $f: E_8 \to E_8$ where $f\in O(E_8)$, the linear maps preserving the intersection form (i.e. the Weyl group $W(E_8)$, given by $v\mapsto v + (v, e_i) e_i$.
The Leech lattice also shows up in the sphere packing problems for dimensions $2,4,8,24$.
See Hale's theorem / Kepler conjecture for dimension 3!
This uses an identification of $L$ as a subset of $\RR^n$, namely $L \tensor_\ZZ \RR \cong \RR^{24}$ for example, and the map $L \injects (\RR^{24}, \cdot)$ is an isometric embedding into $\RR^n$ with the standard form.
Connection to classification of Lie groups: root lattices.
:::

:::{.remark}
If $M^4$ is a compact oriented 4-manifold and if the intersection form on $H^2(M; \ZZ)$ is indefinite, then the only invariants we can extract from that associated lattice are

- Whether it's even or odd, and
- Its signature

If the lattice is even, then the signature satisfies $8\divides p-q$.
So Poincaré duality forces unimodularity, and then there are further number-theoretic restrictions.
E.g. this prohibits $\beta_2 =7$, since then the signature couldn't possibly be 8 if the intersection form is even.
:::

## Characteristic Classes

:::{.definition title="Classifying space"}
Let $G$ be a topological group, then a **classifying space** $EG$ is a contractible topological space admitting a free continuous \(G\dash\)action with a "nice" quotient.
:::

:::{.remark}
Thus there is a map $EG \to BG \da EG/G$ which has the structure of a principal \(G\dash\)bundle.

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Here we use a point $p$ depending on $U$ in an orbit to identify orbits $g\cdot p$ with $g$, and we want to take transverse slices to get local trivializations of $U\in BG$.
It suffices to know where $\pi ^{-1} (U) \cong U \cross G$, and it suffices to consider $U \cross \ts{e}$.
Moreover, $EG\to BG$ is a universal principal $G\dash$bundle in the sense that if $P\to X$ is a universal $G\dash$bundle, there is an $f:X\to BG$.

\begin{tikzcd}
	P && EG \\
	\\
	X && BG
	\arrow[from=1-1, to=3-1]
	\arrow["f", from=3-1, to=3-3]
	\arrow[from=1-3, to=3-3]
	\arrow[dashed, from=1-1, to=1-3]
\end{tikzcd}

> [Link to Diagram](https://q.uiver.app/?q=WzAsNCxbMCwyLCJYIl0sWzIsMCwiRUciXSxbMiwyLCJCRyJdLFswLDAsIlAiXSxbMywwXSxbMCwyLCJmIl0sWzEsMl0sWzMsMSwiIiwyLHsic3R5bGUiOnsiYm9keSI6eyJuYW1lIjoiZGFzaGVkIn19fV1d)


Here bundles will be classified by homotopy classes of $f$, so 
\[
\ts{\text{Principal $G\dash$bundles} {}_{/ X} } \mapstofrom [X, BG]
.\]
:::

:::{.warnings}
This only works for paracompact Hausdorff spaces!
The line $\RR$ with the doubled origin is a counterexample, consider complex line bundles.
:::


\todo[inline]{Revisit this last section, had to clarify a few things for myself!}