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Consider two quadratic forms $Q$ and $P$ over a finite dimensional vector space $V$ over a quadratically closed (or perhaps Pythagorean) field $F$. If $V$ can be decomposed as $V = V_1 \oplus V_2 \oplus \dotsc \oplus V_k$ such that:

  • $V_i \perp V_j$ for all $i \neq j$ with respect to both $P$ and $Q$
  • No further decomposition of any $V_i$ is possible that respects this property

then is the set of pairs of quadratic subspaces $\{(Q|_{V_i}, P|_{V_i}) : i \in [k]\}$ unique up to isomorphism?

We justify the need for quadratic closure:

The finite fields are not quadratically closed. The conjecture is trivially false when we let $P = Q$ and we consider any finite field, like $\mathbb F_3$.

Let $P(x,y) = Q(x,y) = x^2 + y^2$ acting on $V := (\mathbb F_3)^2$. We get $V = (1,0)\mathbb F_3 \oplus (0,1)\mathbb F_3$ but also $V = (1,1)\mathbb F_3 \oplus (1,-1)\mathbb F_3$. Observe that $Q|_{(1,0)\mathbb F_3} \cong Q|_{(0,1)\mathbb F_3} \not \cong Q|_{(1,1)\mathbb F_3} \cong Q|_{(1,-1)\mathbb F_3}$.

Curiously, the basis $(1,1)$ and $(1,-1)$ is an orthogonal one -- but not a unit one -- with respect to $P$ (and $Q$). This suggests that the question would be more interesting for quadratically closed fields.

Improve this question
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  • $\begingroup$ If you allow $P=Q$, and $F$ is any field of characteristic $\neq 2$, then the claim is never true. In fact, any anisotropic vector (i.e., a $v \in V$ with $Q(v) \neq 0$) then has an orthogonal complement. In other words, any anisotropic $v \in V$ spans a $1$-dimensional subspace $V_1$ which is part of some decomposition $V_1 \oplus \dots \oplus V_k$ into $1$-dimensional pairwise orthogonal subspaces. $\endgroup$
    – Tom De Medts
    May 5 at 11:02
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    $\begingroup$ @TomDeMedts Sylvester's law of inertia shows that my claim is true for $\mathbb R$. The Takagi decomposition shows my claim is true for $\mathbb C$. [UPDATE] This is assuming $P = Q$ still. $\endgroup$
    – wlad
    May 5 at 11:08
  • $\begingroup$ Oh, I see that I have misinterpreted your question, my apologies. You do not ask uniqueness of the subspaces $V_i$, but only of the restrictions of the quadratic forms to those subspaces (up to isomorphism). $\endgroup$
    – Tom De Medts
    May 5 at 11:14

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