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
$$\mathcal{h}(q) = \prod\_{1 \leq i < j \leq n} (a\_i,a\_j)\_p \in \{\pm 1\},$$
where $(a\_i,a\_j)\_p$ is the usual Hilbert symbol.


Let $\mathcal{C}(q)$ be the Clifford algebra associated to $q$. If $n$ is even, this is a central simple algebra; in fact, it is a tensor product of quaternion algebras. It thus gives a $2$-torsion element in the Brauer group $\mathbb{Q}/\mathbb{Z}$ of $\mathbb{Q}\_p$, i.e. an element $\mathcal{h}'(q) \in \{\pm 1\}$.


**Question**: How is $\mathcal{h}'(q)$ related to $\mathcal{h}(q)$? We should be able to express $\mathcal{h}'(q)$ in terms of $\mathcal{h}(q)$ along with $n$ and the discriminant, but I don't quite understand all the constructions well enough to do this.



Hasse invariant and the Clifford algbera
You can find some information about this in Lam's book "Introduction to quadratic forms over fields," particularly in the 3rd chapter. I'll give the answer in a "field agnostic" way, not specifically for $p$-adic fields.


In general, the Hasse invariant is a slight modification of either the Brauer class of the Clifford algebra (if the form has even dimension), or the Brauer class of the even part of the Clifford algebra (if the form has odd dimension). If $h'$ denotes the class in the Brauer group of the Clifford algebra, or even Clifford algebra in the odd case (in either case, a central simple algebra), the relationship looks like this (taken from Lam, prop 3.20, and depending on the residue of $n$ modulo $8$):


if $n \equiv\_8 1, 2$ then $h = h'$,


if $n \equiv\_8 3, 4$ then $h = h' + \delta$,


if $n \equiv\_8 5, 6$ then $h = h' + \gamma$,


if $n \equiv\_8 7, 8$ then $h = h' + \delta'$,


where $\gamma, \delta, \delta'$ are quaternion algebras as follows: $\gamma = (-1, -1)$ is Hamilton's quaternion algebra, $\delta = (-1, -d)$ and $\delta' = (-1, d)$, and $d$ is the discriminant of the form $q$.


