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5 votes
1 answer
275 views

Equivalence of quadratic forms over $p$-adic integers vs over localisation at $p$

To discern whether two integral quadratic forms are equivalent over the $p$-adic integers, one can compute a Jordan decomposition at $p$ and read off some invariants. Restricting to $p\ne2$ for ...
a196884's user avatar
  • 323
23 votes
1 answer
3k views

A list of proofs of the Hasse–Minkowski theorem

I am currently doing a project in which I intend to include the most insightful possible proof of the Hasse–Minkowski theorem (also known as the Hasse principle for quadratic forms, among other names) ...
6 votes
1 answer
263 views

Hasse invariant and the Clifford algbera

Let $$q = a_1 x_1^2 + \cdots + a_n x_n^2$$ be a quadratic form over some $p$-adic field $\mathbb{Q}_p$. We thus have its Hasse invariant $$\mathcal{h}(q) = \prod_{1 \leq i < j \leq n} (a_i,a_j)_p \...
Rita's user avatar
  • 103
2 votes
2 answers
230 views

The stabiliser group of an isotropic quadratic form over $\mathbb{Q}_p$ is non-compact?

Let $\mathbb{Q}_p$ denote the $p$-adic integers. Let $V$ be a $\mathbb{Q}_p$-vector space and $Q : V \rightarrow \mathbb{Q}_p$ be a non-degenerate integral quadratic form. We say that the pair $(Q,V)$ ...
Andrew Musso's user avatar
0 votes
1 answer
432 views

$p$-adic orthogonal groups in four variables

Let $p>2$ be prime. By the classification of quadratic forms, there are $8$ pairwise non-equivalent isotropic orthogonal groups in $4$ variables. Is there a concrete classification of orthogonal ...
Keivan Karai's user avatar
  • 6,224