# Thursday, September 22

:::{.remark}
Today: continuing discussions of lattices over Dedekind domains $R$ with $k \da \ff(R)$.

- Smith and Hermite normal forms,
- The Frohlich invariant $\chi( \Lambda_1, \Lambda_2)$,
- Local-to-global principle for lattices.

Next time: endow a $K\dash$vector space $V$ with a bilinear form $\ip{\wait}{\wait}$ and see how it enriches the study of lattices.
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:::{.definition title="Normal forms"}
Let $V\in \Vect\slice k^{\dim = n}$ and let $\Lambda = \gens{y_1,\cdots, y_n}_R$.

1. (**Hermite normal form**) Let $\ts{y_i}_{i=1}^n$ be a $k\dash$basis for $V$ and let $M$ be a lattice in $V$.
  Then there exists $\ts{x_i}_{i=1}^n \subseteq V$ and $\ts{a_i} \subseteq k \da \ff(R)$ such that $M = \bigoplus a_i x_i$ where for all $1\leq i, j \leq n$ we have $x_j \in \gens{y_1,\cdots, y_j}_k$.

2. (**Smith normal form**) Let \( \Lambda_1, \Lambda_2 \in V \) be two lattices, then there exists a $k\dash$basis $\ts{x_i}$ of $V$ and $\ts{a_i}, \ts{b_i} \in \Frac(R)$ such that \( \Lambda_1 = \bigoplus a_i x_i \) and \( \Lambda_2 = \oplus b_i x_i \).
  For all $i$, set $\del_i \da a_i b_i\inv$. Then we may require that \( \del_1 \subseteq \cdots \del_n \) in which case the $\del_i$ are unique.
  Moreover $n_2 \leq n_1 \iff a_i \subseteq b_i$ for all $i \iff \del_i$ is an integral ideal for all $i$.

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:::{.remark}
We only need this locally. Try to deduce this from the usual SNF.
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:::{.definition title="Frohlich invariant"}
For \( \Lambda_1, \Lambda_2 \) lattices in $V$, define
\(
\chi( \Lambda_2 / \Lambda_1) \in \Frac(R)
\)
in the following way:

1. Suppose \( \Lambda_1 \subseteq \Lambda_2 \), then define $\chi(\Lambda_1 / \Lambda_2)$ to be the characteristic ideal (using that this is a finitely-generated torsion module, so take the product of primes from a JH series).
  \( \Lambda_2 / \Lambda_1 \) is finitely-generated since there exists a $d\in R\smz$ such that \( \Lambda_2 \subseteq {1\over d } \Lambda_1 \) and so \( {\Lambda_2\over \Lambda_1} \subseteq {{1\over d} \Lambda_1 \over \Lambda_1 } \) which is annihilated by $d$ and thus torsion.
  Note that this is an integral ideal $\chi( \Lambda_2, \Lambda_1) \normal R$.

2. If there is no containment, use fractional ideal and scale instead: choose $\alpha \in R\sm$ such that $\alpha \Lambda_1 \subseteq \Lambda_2$ and define
\[
\chi( \Lambda_2 / \Lambda_1) \da \chi\qty{\Lambda_1 \over \alpha \Lambda_1}\inv \chi\qty{ \Lambda_2 \over \alpha \Lambda_1} \da \gens{ \alpha}^{-n} \chi\qty{ \Lambda_2 \over \alpha \Lambda_1}
.\]
  I.e. scale the lattice, apply the first definition, see what scaling does in that case, and add in a correction factor to cancel it.


> See Sutherland's *module index*, or Serre's "Local Fields" where this goes unnamed.

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:::{.exercise title="?"}
Check that if \( \Lambda_2 \subseteq \Lambda_1 \), then the 2nd definition yields $\chi( \Lambda_2/ \alpha \Lambda_1)= \chi( \Lambda_2 / \Lambda_1)/ \chi( \Lambda_1 / \alpha\Lambda_1)$, and since \( \Lambda_1 / \alpha \Lambda_1 \cong (R / aR)^n \) and so $\chi( \Lambda_1/ \alpha \Lambda_1) = (\alpha)^n$.
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:::{.exercise title="?"}
Let $R=\ZZ$ with \( \Lambda_1, \Lambda_2 \subseteq \QQ^n \) with \( \Lambda_1 \subseteq \Lambda_2 \), then show that $\chi( \Lambda_2 / \Lambda_1) = \gens{\size( \Lambda_2/ \Lambda_1) } = \gens{[\Lambda_2: \Lambda_1]}$.
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:::{.exercise title="?"}
Show that

a. $\chi( \Lambda_2/ \Lambda_1)$ is independent of the choice of \( \alpha \).
b. Suppose the Frohlich invariant is an integral ideal, so $\chi( \Lambda_2 / \Lambda_1) \normal R$.
c. Does it follow that \( \Lambda_1 \subseteq \Lambda_2 \).
  Check $R = \ZZ, V = \QQ^2, \Lambda_1 = \ZZ \oplus \ZZ$ and $\Lambda = p \ZZ \oplus {1\over p} \ZZ$.
  Scale \( \Lambda_1 \) to be in \( \Lambda_2 \) to get 
  \[ 
  \chi( \Lambda_2 / \Lambda_1) = p^{-2} \chi\qty{p\ZZ \oplus {1\over p} \ZZ \over p\ZZ \oplus p\ZZ} =p^{-2} p^2 = 1 
  ,\]
  why is this an example or counterexample?
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:::{.exercise title="?"}
Given $I, J \in \Frac(R)$ which are lattices in $k$, then
\[
\chi(I/J) = J I\inv
.\]
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:::{.exercise title="Very important, why have SNF for Dedekind domains instead of just PIDs"}
Let \( \Lambda_1, \Lambda_2 \) be lattices in $V$, and use SNF to get $\ts{x_i}\in V$ and $\ts{a_i}, \ts{b_i} \in \Frac(R)$ with \( \Lambda_1 = \oplus a_i x_i, \Lambda_2 = \oplus b_i x_i \) and $\del_i = a_i b_i\inv$.
Then
\[
\chi( \Lambda_1 / \Lambda_2 )= \prod \del_i\inv \text{ or } \qty{\prod \del_i\inv}\inv
.\]

> What an unfortunate formula. See Sutherland's notes for more details and to sort whether it's this thing or its inverse!

:::

:::{.exercise title="?"}
Let \( \Lambda_i, i=1,2,3 \) be lattices in $V$ and show
\[
\chi( \Lambda_3 / \Lambda_1) = \chi( \Lambda_3 / \Lambda_2) \chi( \Lambda_2/ \Lambda_1)
\]
and
\[
\chi( \Lambda_1 / \Lambda_2)\inv = \chi ( \Lambda_2 / \Lambda_1)
.\]
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:::{.proposition title="?"}
Let $M \in \Aut_k(V)$ and \( \Lambda \) a lattice in $V$.
Then 
\[
\chi( \Lambda, M \Lambda) = \gens{ \det M}
.\]
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:::{.remark}
Note that if $M: \ZZ^n\to \ZZ^n$ whose image is a sublattice, then $[\ZZ^n: M(\ZZ^n)] = \abs{\det M}$, a hugely useful result.
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:::{.proof title="?"}
Both sides are fractional ideals and thus can be computed locally, so without loss of generality, assume $R \in \DVR$. Then \( \Lambda \) is free; let $\ts{x_i}$ be an $R\dash$basis for $\Lambda$, then the corresponding $k\dash$basis for $V$ yields an identification $V \iso k^n$.
View $M\in \GL_n(k)$, then by SNF for matrices, there exist unimodular matrices $P, Q\in \GL_n(R)$ (so $\det P, \det Q\in R\units$) such that $PMQ = D \da \diag(d_1,\cdots, d_n)$ with $d_i\in k\units$.
Choose a scaling factor $\alpha \in R\smz$ such that $\alpha d_i \in R$ for all $i$, then $\alpha M \Lambda$ has basis $\ts{ \alpha d_i x_i}$.
We then have
\[
\Lambda/ \alpha M \Lambda\cong \oplus R/ \alpha d_i R \implies \chi\qty{ \Lambda/ \alpha M \Lambda} = \prod \gens{ \alpha d_i } = \gens{ \alpha^n}\gens{\prod d_i}
,\]
and by definition we have 
\[
\chi( \Lambda/ M \Lambda) 
&= \gens{ \alpha}^{-n} \chi( \Lambda/ \alpha M \Lambda) \\
&= \gens{\prod d_i} \\
&= \gens{\det PMQ} \\
&= \gens{\det P} \gens{\det M} \gens{\det Q} \\
&= \gens{1} \gens{\det M} \gens{1} \\
&= \gens{\det M}
.\]

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