Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $\Lambda$ be an unimodular lattice with a quadratic form $(-,-)$ of signature $(m,n)$ , $m,n>0$.

I know that, fixed a base $e_1,\cdots,e_{m+n}$ for $\Lambda$, the matrix which has entries $a_{i,j}=(e_i,e_j)$ has determinant ugual to $\pm 1$.

I wonder if in this case it is always possibile to find an orthogonal base, i.e. a base $f_1,\cdots,f_{m+n}$ with $(f_i,f_i)=\pm 1$ and $(f_i,f_j)=0$.

I think yes, but i can't really prove it.

share|improve this question
3  
The hyperbolic lattice cannot be diagonalized with an integral basis. –  Atsushi Kanazawa Feb 27 '13 at 1:23

1 Answer 1

Given that $\Lambda$ is unimodular and indefinite, this can be done if and only if $\Lambda$ is odd (i.e. iff the diagonal entries $a_{i,i}$ are not all even). This follows from Milnor's classification. A couple of references where this is worked out are Serre's "A course in arithmetic" and Milnor and Husemoller's "Symmetric bilinear forms".

share|improve this answer
    
I'm interested in the case $\Lambda=3H \oplus -2E_8$, where $H$ is the hyperbolic plane, so i think this is the case you are talking about, right? –  rick Feb 27 '13 at 23:56
    
no, that one's an even lattice. so it doesn't have an orthogonal basis. –  Abhinav Kumar Feb 28 '13 at 0:19

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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