1
$\begingroup$

Hi everybody.

I am trying to understand a proof of Kneser. the assertion is on a ''weak version'' of the local-global principle certain isometries: It is Satz (30.9) in kneser book ''Quadratische Formen'':

Let $L, M$ be lattices. An injective isometry $u : L \to M$ is called a presentation of $L$ (through $M$). Two presentations $u: L \to M, v:L \to N$ are called equivalent if there exists an isometric isomorphism $\phi : M \to N$ such that $v = \phi \circ u$. They are said to lie in the same genus if for every $p$ including $\infty$, there are isometric isomorphisms $\phi_p : M_p \to N_p$ such that $v_p = \phi_p \circ u_p$ where $X_p = X \otimes \mathbf{Z}_p$, $\mathbf{Z}_p = $ p-adic integers and $v_p(l \otimes x) = v(l) \otimes x$ and $u_p$ analogously. What Kneser proves is the following: Let $V$ be a fixed $n$-dimensional $\mathbb{Q}$-vector space together with a non-degenerate bilinear form $b$ and $u : L \to M \subset V$, a presentation of $L$ through $M$, $M$ being a complete (i.e. of rank $n$) lattice in $V$. For Simplicity, let us choose representatives for the genus of $M$ modulo isometric isomorphy, gen($M$) = $\{M_1, ..., M_n\}$, then $$|\{u : L \to V : \exists i : u(L) \subset M_i\} / \text{same genus}| = |\prod_{p} c(L_p, M_p)|$$ where $p$ runs through all primes including infinity and $c(L_p, M_p)$ is the set of presentations of $L_p$ through $M_p$ modulo equivalence, i.e. the number of genuses of presentations of $L$ through some lattice in the genus of $M$ (modulo the relation ''to be in the same genus'') is the same as the product over all local presentations modulo the relation ''to be in the same class''. Let us call the left set $X$ and the right set $Y$.

What kneser does in this proof is that he considers a map $\Phi : X \to Y$ that takes such a $u:L \to M'$, ($M'$ being -- more generally -- a lattice in $V$ that is in the same genus as $M$, i.e. there are isometric isomorphisms $\eta_p : M_p' \to M_p$) and maps that to the sequence $(O(M_p) \eta_p \circ u_p)_{p \in \mathbb{P} \cup \{\infty\}}$. One can check that this map is well defined and by definition it is injective. What i do not understand is the surjectivity: Given a set of local presentations $(O(M_p)u_p)_{p \in \mathbb{P} \cup \{ \infty\}}$, Kneser writes the following: for those finitely many primes dividing the determinant of $L$, one takes a $\tilde{U_p} \in O(V \otimes \mathbf{Q}_p)$ that satisfies $\tilde{U}_p(u_p(l_p)) = u_p(l_p)$ for all $l_p \in L_p$ and then puts $M_p' := \tilde{U}_p^{-1}(M_p)$. For the inifitely many remaining primes one just puts $M_p' := M_p$. Essentially, it does not matter what we do at these primes, because for $L_p$ unimodular, all presentations of $L_p$ through a fixed isomorphy class of a lattice like $M_p \cong M_p'$ are in the same class. There is a purely algebraic theorem (i.e. without the reference to any bilinear form) which states that one can take a $\mathbb{Z}$-lattice $M'$ in $V$ such that $M' \otimes \mathbf{Z}_p = M_p'$ for all $p$ including infinity. What he writes then confuses me: he writes that $M'$ is a preimage for the sequence of local presentations.

For applying the map $\Phi$ one needs a presentation of $L$ through $M'$ and i do not see where this should come from. Can somebody give a hint how to see that there is a presentation and how to show that this really is a preimage of the sequence of local presentations?

Thanks in advance,

Fabian Werner

$\endgroup$
1
  • $\begingroup$ Ok, thanks for the resources. I will check. FW $\endgroup$
    – user25160
    Nov 19, 2012 at 21:46

1 Answer 1

2
$\begingroup$

Would the last two pages of

http://wkchan.web.wesleyan.edu/qflecturenotes.pdf

help?

$\endgroup$
1
  • $\begingroup$ Ah, i applied the Witt extensin theorem in the wrong way, now i understand this. THANK YOU VERY MUCH!!! $\endgroup$ Nov 20, 2012 at 14:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.