\chapter{Lattice basics}
We want to abstract the notion of $\bZ^2 < \bR^2$.
From such humble origins comes a very deep and central theory.
\section{Definition}
Given a field $F$ with a ring $R$ inside it
(for instance, $\bZ < \bR$, or the ring of integers of a number field),
one defines a lattice $\Lambda$ in a vector space $V$ (of dimension $n$)
as a free $R$-module of rank $n$ whose $F$-span is $V$.
The reason for the technicalities is to rule out cases like $\bZ +
\sqrt{2} \bZ < \bR$, which is free over $\bZ$ but not $\bR$.
Note that we've defined a lattice of maximal rank; one can also define
lattices of lower rank, by insisting that the dimension of their span
equals their rank (for the same technical reason).
\subsection{Other uses of ``lattice''}
One also calls discrete subgroups of Lie groups (often required
to be maximal rank) ``lattices''; this is a non-linear generalization.
There is also a rather different algebraic/logic notion of lattice,
which abstracts the powerset, union and intersection: $\cP(X),\cup,\cap$;
alternatively, it abstracts logical AND and OR.
\subsection{Stabilizer of a lattice: $\GL(\Lambda)=\GL_n(R)$}
The stabilizer of a lattice (the linear operators $T$ s.t. $T(\Lambda)
= \Lambda$) is a subgroup of $\GL(V)$; with respect to a basis of the
lattice, they are the matrices with entries in $R$ that are invertible
over $R$. A matrix with entries in $R$ that is not invertible over $R$
will take the lattice \emph{into} a sublattice of itself.
Consider for instance $\psmallmatrix{2}$, acting on $\bZ < \bR$.
\subsection{Dual lattice}
Given a lattice $\Lambda < V$, we can define the dual lattice
$\Lambda^* < V^*$ as ``those elements of the dual that pair to
`integers'''. Formally,
$$\Lambda^* := \set{w \in V^* \st \forall v \in \Lambda, \left \in R}$$
More concretely, one can take a basis for the lattice (which is then a
basis for the vector space), and let the dual lattice be the $R$-span
of the dual basis. This works (the choice of basis doesn't matter)
because $\GL_n(R)^*=\GL_n(R)$: the transpose of an invertible matrix
is invertible, so a different choice of basis yields a different
choice of dual basis, but still the same lattice.
\subsection{Lattices with respect to a bilinear form}
On a bilinear space (a vector space with a bilinear form $B$;
equivalently, with a map $T\from V \to V^*$),
one can ask for a lattice to be \Def{integral},
i.e., evaluate to integers on itself,
meaning $B(u,v)\in R$ for all $u,v \in \Lambda$.
If the form is non-degenerate, one has an internal dual
and an \Def{integral lattice} is exactly one such that $\Lambda^* < \Lambda$.
\subsubsection{Old ``twos in'' convention}
Previously some people used the ``twos in'' convention,
which said that the matrix representing the form (with respect
to a basis for the lattice) had to have integer coefficients,
but the intrinsic ``twos out'' convention (which we've been using)
is superior and won out.
\section{Structure}
\subsection{Moduli space of lattices}
XXXXXXXXXXXXXXXXXX
$\GL_n(K)/\GL_n(R)$
\subsection{Unimodular Lattices}
There are 2 very important exceptional lattices,
E_8 and the Leech lattice.
For indefinite lattices there's only 1, except in dim 8k where there's
another
in $\bR^n$ (and $\bR^{p+q}$)}
Unimodular integral lattices in $\bR^{p+q}$ (i.e., w/r/t the quadratic form
of signature $(p,q)$) are aka indefinite lattices
unique odd one
unique even one in dim $p-q$ divisible by 8
$(8,0)$ is the E_8 lattice
For positive definite lattices, it's the same up to dim 8,
then grows
$E_8$ and the Leech lattice (
\section{Applications}
\subsection{Abelian varieties}
Given an abelian variety, which is $V/\Lambda$,
one can construct the dual variety by taking the dual space and dual
lattice (and quotienting).
A fundamental example is that the Albanese variety is the dual to the
(connected component of the identity of the) Picard variety.
A principally polarized abelian variety has a distinguished choice of
basis, hence it is identified with its own dual (in the usual way for
vector spaces).
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in general given a free R module,
can make a space that it's a lattice in,
and get a torus via quotienting
\subsection{Middle cohomology of an oriented manifold}
Poincar\'e duality gives a perfect pairing
$H_k(M;\bZ) \times H_{n-k}(M;\bZ) \to \bZ$
XXXXXXXXXXXXXXXXXXXXXXXX
it's unimodular!
in dim 4
simply connected, oriented
even unimodular: 1 topologically
odd unimodular: 2 topologically
btw, even unimodular forms are the ones that come form a bilinear form?
a priori the sym -> quad -> sym
and quad -> sym -> quad
(which are both *2)
look symmetric, like scalar -> trace -> scalar
...but actually, sym -> quad is iso,
but quad -> sym is *2
(I think...)