3
$\begingroup$

Consider the quadratic form $5x^2+6y^2$. This has Conway Sloan $2$-adic symbol $[1^{-1}2^{-1}]_0$. After a sign walk from $1$ to $2$ the symbol becomes $[1^{+1}2^{+1}]_4$. However, there doesn't exist a form with this symbol as the oddities of each summand are limited to $1$ and $7$ and there is no solution modulo $8$ to the equation. This seems to contradict the statement on page 384 of Conway-Sloan about existence that the existence condition for compartments is that the total oddity have the same parity as the total dimension.

It is still possible to ensure any compartment in a train not the first one has an even number of minus signs in it, and all such expressions will have the same oddities and sequence of dimensions for each compartment. But it is not clear the compartments will be isomorphic given the above failure to carry out intracompartment sign-walking, and so I don't know if normalizing the compartments will produce a normal form.

My question then: if two trains are isomorphic, and both trains have been normalized so that any compartment that is not the first compartment (or if the train starts with a Type II form, all compartments) has an even number of minus signs, are the compartments necessarily isomorphic over the $2$-adic integers?

$\endgroup$
1
  • 3
    $\begingroup$ Page 384 of what? $\endgroup$ – Yemon Choi Nov 10 '17 at 1:52
0
$\begingroup$

The answer is yes. There is a sign walk entirely within each compartment that removes the minus signs and produces the same normalized symbol for each train. Therefore the compartments of the two different trains are isomorphic as they have the same normalized symbol, even if that normalized symbol isn't the symbol of an actual quadratic form.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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