Let $A$ be an algebra over $k$, $\operatorname{tr_A}(x, y):=\operatorname{tr}(m_{xy})$ be a trace form on $A$, and $V_A$ be its restriction on the orthogonal complement to $1$. I wonder why a map $A \mapsto V_A$ gives a bijection $$\left\{ \begin{array}{cc} \text{central simple algebras over } k \\ \text{ of dimension } 4 \end{array} \right\} \leftrightarrow \left\{ \begin{array}{cc} \text{quadratic forms of rank } 3 \\ \text{ with discriminant } 4^3 \end{array} \right\}.$$

Any help or reference is welcome!


Everything is in Lam's book Introduction to Quadratic Forms over Fields. Theorem III 5.1 says:

All central simple algebra $A$ of dimension $4$ is quaternion. That is $A \cong \left(\frac{a,b}{k}\right)$.

Note that as quadratic space, $A = \langle 1, -a, -b, ab\rangle$ and $V_A = \langle -a, -b, ab\rangle$.

Then Theorem III 2.5 says:

Two quaternion algebras $A, A'$ are isomorphic as $k$-algebra iff $V_A, V_{A'}$ are isomorphic as quadratic spaces.

On the other hand, all quadratic forms of rank three with $d=1$ ($=4^3$ in $k^*/(k^*)^2)$) are of the form $\langle c, d, cd\rangle$. Thus the one-to-one correspondence.


In the formulation, presumably on the right side what is intended are 3-dimensional non-degenerate quadratic spaces (up to isomorphism), with discriminant 1 (same as $4^3$ mod squares as John Ma notes). But to make this work also in characteristic 2, it is better to proceed with a different point of view: that of conformal isometry of quadratic spaces (i.e., isomorphisms $T:V \simeq V'$ such that $q' \circ T = \lambda q$ for some $\lambda \in k^{\times}$). More specifically, we claim that away from characteristic 2, every 3-dimensional non-degenerate quadratic space is conformal to a unique one with discriminant 1. Thus, by working with conformal isometry classes we will be able to work in a fully characteristic-free manner.

To see what is going on, recall that the set of isomorphism classes of central simple algebras of dimension 4 is ${\rm{H}}^1(k, {\rm{PGL}}_2)$, and the set of conformal isometry classes of dimension 3 is ${\rm{H}}^1(k, {\rm{PGO}}_3)$. But ${\rm{GO}}_{2m+1} = {\rm{GL}}_1 \times {\rm{SO}}_{2m+1}$, so ${\rm{PGO}}_{2m+1} = {\rm{SO}}_{2m+1}$. Hence, ${\rm{PGO}}_3 = {\rm{SO}}_3$. Since ${\rm{SO}}_3 \simeq {\rm{PGL}}_2$ through the representation of ${\rm{PGL}}_2$ via conjugation on the 3-dimensional space of traceless $2 \times 2$ matrices equipped with the determinant as the standard split non-degenerate quadratic form $xy - z^2$ (preserved by that conjugation action!), that answers the entire question at the level of isomorphism classes of objects. (The link to ${\rm{SO}}_3$ encodes the link to discriminant 1.)

But we can do better than keep track of isomorphism classes: we can also keep track of isomorphisms, as explained below. This is a refinement of John Ma's answer, as well as that of Matthias Wendt (which appeared at almost exactly the same time as this answer first appeared, so I didn't see it until this one was done).

The following notation will permit considering finite fields on equal footing with all other fields. For a finite-dimensional central simple algebra $A$ over an arbitrary field $k$, let ${\rm{Trd}}:A \rightarrow k$ be its "reduced trace" and ${\rm{Nrd}}:A \rightarrow k$ be its "reduced norm". These are really most appropriately viewed as "polynomial maps" in the evident sense. That is, if $\underline{A}$ is the "ring scheme" over $k$ representing the functor $R \rightsquigarrow A \otimes_k R$ (i.e., an affine space over $k$ equipped with polynomial maps expressing the $k$-algebra structure relative to a choice of $k$-basis) then we have $k$-morphisms ${\rm{Trd}}:\underline{A} \rightarrow \mathbf{A}^1_k$ and ${\rm{Nrd}}:\underline{A} \rightarrow \mathbf{A}^1_k$.

For $A$ of dimension 4 we set $\underline{V}_A$ to be the kernel of ${\rm{Trd}}:\underline{A} \rightarrow \mathbf{A}^1_k$; speaking in terms of kernel of ${\rm{Trd}}$ is a bit nicer than speaking in terms of orthogonal complements so that one doesn't need to separately consider characteristic 2 (where the relationship between quadratic forms and symmetric bilinear forms breaks down). This $\underline{V}_A$ is an affine space of dimension 3 over $k$ on which ${\rm{Nrd}}$ is a non-degenerate quadratic form $q_A$ (i.e., zero locus is a smooth conic in the projective plane $\mathbf{P}(V_A^{\ast})$, where $V_A := \underline{V}_A(k)$): indeed, these assertions are "geometric" in nature, so it suffices to check them over $k_s$, where $A$ becomes a matrix algebra and we can verify everything by inspection.

We will show that the natural map of affine varieties $$\underline{{\rm{Isom}}}(\underline{A}, \underline{A}') \simeq \underline{{\rm{CIsom}}}((\underline{V}_A, q_A), (\underline{V}_{A'}, q_{A'}))/\mathbf{G}_m$$ from the "isomorphism variety" to the "variety of conformal isometries mod unit-scaling" is an isomorphism; once that is shown, by Hilbert 90 we could pass to $k$-points to conclude that isomorphisms among such $A$'s correspond exactly to conformal isometries among such $(V_A, q_A)$'s up to unit scaling. It suffices to check this isomorphism assertion for varieties over $k_s$, where it becomes the assertion that the natural map $$\underline{{\rm{Aut}}}_{{\rm{Mat}}_2/k} \rightarrow {\rm{CAut}}_{({\rm{Mat}}_2^{{\rm{Tr}}=0}, \det)/k}/\mathbf{G}_m$$ from the Aut-scheme to the scheme of conformal isometries up to unit-scaling is an isomorphism. But this is precisely the natural map $${\rm{PGL}}_2 \rightarrow {\rm{PGO}}(-z^2-xy) = {\rm{PGO}}(xy+z^2) = {\rm{SO}}(xy+z^2) = {\rm{SO}}_3$$ between smooth affine $k$-groups that is classically known to be an isomorphism over any field $k$ (can check bijectivity on geometric points and the isomorphism property on tangent spaces at the identity points).

Finally, we want to show that every 3-dimensional non-degenerate quadratic space $(V, q)$ is conformal to $(V_A, q_A)$ for some $A$. Note that if such an $A$ exists then it is unique up to unique isomorphism in the sense that if $A$ and $A$ are two such equipped with conformal isometries $(V_A, q_A) \simeq (V, q) \simeq (V_{A'}, q_{A'})$ then this composite conformal isometry arises from a unique isomorphism $A \simeq A'$ of $k$-algebras. Hence, by Galois descent (!) it suffices to check existence over $k_s$! But over a separably closed field the smooth projective conic has a rational point, so $(V, q)_{k_s}$ contains a hyperbolic plane and thus is isometric to $xy + \lambda z^2$ on $k_s^3$ for some $\lambda \in k_s^{\times}$. This is conformal to $(-1/\lambda)q_{k_s}$, but $(1/\lambda)xy - z^2$ and $-x'y' - z^2$ are clearly isometric, and the latter is ${\rm{Mat}}_2^{{\rm{Tr}}=0}$ equipped with the restriction of det.


On a conceptual level, I would like to see this as an instance of a sporadic isomorphism of algebraic groups. Take the conjugation action of $GL_2$ on the space of $2\times 2$-matrices with trace 0, equipped with the trace form. One can show that this induces an isomorphism $PGL_2\cong SO(3)$. This is discussed e.g. in this MO-question. From the isomorphism of algebraic groups, you get a bijection $$ H^1(k,PGL_2)\cong H^1(k,SO(3)). $$ These are étale/Galois cohomology groups classifying étale locally trivial torsors over $\operatorname{Spec} k$ with the appropriate structure group. Now identify $PGL_2$-torsors with central simple algebras of dimension $4$, this is done e.g. in the book of Gille-Szamuely "Central simple algebras and Galois cohomology". On the other side, identify $SO(3)$-torsors with quadratic form of trivial discriminant. This should give the bijection in the question.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.