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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.

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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".

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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

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