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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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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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