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There is a theorem that states that two quadratic forms over $\mathbb{Q}_p$ are equivalent iff they have the same rank, discriminant and the same $\epsilon$ invariant.

(The last is defined as follows: if $f = \sum_{i=1}^n a_ix_i^2$ then $\epsilon(f) = \prod_{1\le i < j \le n} (a_i,a_j)$, where $(\cdot,\cdot)$ is the Hilbert symbol over $\mathbb{Q}_p$.)

The proof (from Serre's Arithmetic) is based on a previous theorem, which relates these invariants to the elements the forms represent.

However, I do not know how I can show that two particular forms are equivalent (with the same invariants of course). So fix some prime $p$ and consider the following two forms over $\mathbb{Q}_p$:

\begin{equation} f(x) = x_1^2 + x_2^2 + x_3^2 + x_4^2 - (x_5^2 + x_6^2 + x_7^2) \nonumber \end{equation} and \begin{equation}
g(x) = -(x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 + x_6^2 + x_7^2) \nonumber \end{equation}

How can I show in particular that these two are equivalent (without using the classification theorem above)? Thanks.

  • 1
    Have you tried the book by O'Meara?2012-03-20
  • 0
    No, does it have such examples?2012-03-20
  • 0
    They are not equivalent when $p = \infty$2012-03-20
  • 0
    Yes of course not, but when $p$ is a "finite" prime they are (or should be)2012-03-20
  • 1
    The book "Introduction to Quadratic Forms " by Timothy O'Meara is considered a classic text. I hve not read it for some years, but it is very comprehensive.2012-03-20
  • 0
    O'Meara is, indeed, the standard reference. Too bad he had already been "promoted" to Provost before I found my way to Notre Dame. His students did carry the torch, but my interest got fixated on representation theory.2012-03-21

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