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Over number fields, two quadratic forms are equivalent iff they have the same dimension, signature, discriminant and Hasse invariant.

How is the situation like over finitely generated fields?

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2 Answers 2

By "Hasse invariant" I suppose you mean a 2-torsion class in the Brauer group (which can be further analyzed via local Brauer groups, but that step is specific to global fields). And by "signature" you're keeping track of the isomorphism class at the real places.

In general, if $k$ is any field and $(V,q)$ is a non-degenerate quadratic space over $k$ with $n := \dim V \ge 1$ then its isomorphism class corresponds exactly to an element of ${\rm{H}}^1(k, {\rm{O}}_n)$, so the question is how to describe this set. The case $n=1$ is elementary, so assume $n \ge 2$.

We have canonical short exact sequence $$1 \rightarrow \mu_2 \rightarrow {\rm{Pin}}_n \rightarrow {\rm{O}}_n \rightarrow 1$$ with central term on the left gives rise to a connecting map ${\rm{H}}^1(k, {\rm{O}}_n) \rightarrow {\rm{H}}^2(k, \mu_2) = {\rm{Br}}(k)[2]$. The image of the class of $(V,q)$ under this connecting map is the "Hasse invariant" of $q$.

Suppose $(V',q')$ is another $n$-dimensional quadratic space. The variety ${\rm{Isom}}(q',q)$ of isomorphisms of quadratic spaces is a left ${\rm{O}}(q)$-torsor, and the image in ${\rm{H}}^2(k,\mu_2) = {\rm{Br}}(k)[2]$ of its class in ${\rm{H}}^1(k,{\rm{O}}(q))$ under the connecting map arising from $$1 \rightarrow \mu_2 \rightarrow {\rm{Pin}}(q) \rightarrow {\rm{O}}(q) \rightarrow 1$$ is the ratio of the Hasse invariants of $q$ and $q'$ (since the above short exact sequences are central extensions related by "inner twisting"), so these latter invariants coincide precisely when the class of ${\rm{Isom}}(q',q)$ in ${\rm{H}}^1(k,{\rm{O}}(q))$ arises from ${\rm{H}}^1(k,{\rm{Pin}}(q))$.

Assuming such equality of Hasse invariants, we get a class in ${\rm{H}}^1(k,{\rm{Pin}}(q))$ well-defined up to the action of ${\rm{H}}^1(k,\mu_2)$. Using the exact sequence $$1 \rightarrow {\rm{Spin}}(q) \rightarrow {\rm{Pin}}(q) \rightarrow \mathbf{Z}/(2) \rightarrow 1$$ for even $n$ and the equality ${\rm{Pin}}(q) = \mu_2 \times {\rm{Spin}}(q)$ for odd $n$, the composite map ${\rm{Pin}}(q) \rightarrow \mathbf{Z}/(2)$ for even $n$ and ${\rm{Pin}}(q) \rightarrow \mu_2$ for odd $n$ kills $\mu_2 \subset {\rm{Pin}}(q)$ (as it lies inside ${\rm{Spin}}(q)$), so the class we have in ${\rm{H}}^1(k,{\rm{Pin}}(q))$ up to the action of ${\rm{H}}^1(k,\mu_2)$ yields a well-defined class in ${\rm{H}}^1(k,\mathbf{Z}/(2))$ for even $n$ and ${\rm{H}}^1(k,\mu_2)$ for odd $n$.

Now assume ${\rm{char}}(k) \ne 2$ (so $\mathbf{Z}/(2) = \mu_2$), so regardless of the parity of $n$ we have a well-defined class in ${\rm{H}}^1(k,\mu_2) = k^{\times}/(k^{\times})^2$. This is surely the ratio of the discriminants of $q$ and $q'$ if one unravels things.

So to summarize, if ${\rm{char}}(k) \ne 2$ and $(V,q)$ and $(V',q')$ have the same dimension, Hasse invariant in ${\rm{Br}}(k)[2]$, and discriminant (as a square class) then the obstruction to their equality is a class in ${\rm{H}}^1(k, {\rm{Spin}}(q))$. In the number field case, this encodes the discrepancy of signatures at real places of $k$. In general I suppose it is some kind of mystery, since the cohomology of simply connected groups over finitely generated fields is often a mysterious thing. And of course the Brauer group is a total mystery too beyond the case of global fields.

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up vote 3 down vote accepted

In general, quadratic forms are not determined by their Stiefel-Whitney invariants, see Scharlau, Winfried: Quadratische Formen und Galois-Cohomologie. In: Invent. Math., 4 (1967), 238–264, p. 251 f.

However, if $\dim{q} \leq 3$, $q$ is determined by dimension, discriminant and Hasse invariant, see Lam, Tsit-Yuen: Introduction to Quadratic Forms Over Fields. Graduate Studies in Mathematics 67, American Mathematical Society 2005, p. 120, Theorem 3.21.

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