Max Koecher (for example, in The Minnesota Notes on Jordan Algebras and Their Applications; new edition: Springer Lecture Notes in Mathematics, number 1710, 1999), defined a domain of positivity for a symmetric nondegenerate bilinear form $B: X \times X \rightarrow \mathbb{R}$ on a finite dimensional real vector space $X$, to be an open set $Y \subseteq X$ such that $B(x,y) > 0$ for all $x,y \in Y$, and such that if $B(x,y) > 0$ for all $y \in Y$, then
$x \in Y$. (More succinctly, perhaps, we could say it's a maximal set $Y \subseteq X$ such that $B(Y,Y) > 0$.) Aloys Krieger and Sebastian Walcher, in their notes to chapter 1 of this book, state that "In the language used today, a domain of positivity is a self-dual open proper convex cone." [I now believe this is wrong; see my answer below for what I think is true instead.] It's quite easy to prove that it's an open proper convex cone. (Proper means it contains no nonzero linear subspace of $X$, i.e. that its closure is pointed.) But, although I have a vague recollection of having encountered a proof once in a paper on homogeneous self-dual cones, I haven't succeeded in finding it again, or in supplying it myself. I'm pretty sure Krieger and Walcher's claim is correct—for example, the 1958 paper by Koecher that is generally cited (along with a 1960 paper by Vin'berg) for the proof of the celebrated result that the (closed) finite-dimensional homogeneous self-dual cones are precisely the cones of squares in finite dimensional formally real Jordan algebras, is titled "The Geodesics of Domains of Positivity" (but in German).
The most natural way to prove this would be to find a positive semidefinite nondegenerate $B'$, such that the cone is a domain of positivity for $B'$ as well. In principle, $B'$ might depend on the domain $Y$. (While maximal in the subset ordering, domains of positivity for a given form $B$ are not unique.) But a tempting possibility, independent of $Y$, is to transform to a basis for $X$ in which $B$ is diagonal, with diagonal elements $\pm 1$, change the minus signs to plus signs, and transform back to obtain $B'$.
To clarify the question: we will define a cone $K$ in a real vector space $X$ to be self-dual iff there exists an inner product—that is, a positive definite bilinear form $\langle . , . \rangle: X \times X \rightarrow \mathbb{R}$—such that $K = K^*_{\langle . , . \rangle}$. Here $K^*_{\langle . , . \rangle}$ is the dual with respect to the inner product $\langle . , . \rangle$, that is $K^*_{\langle . , . \rangle} := \{ y \in X: \forall x \in X ~\langle y, x \rangle > 0 \}$. So in asking for a proof that a domain of positivity is a self-dual cone, we are asking whether some inner product $\langle . , . \rangle$ with respect to which $K$ is self-dual exists. Above, I considered the case $K=Y$, and called the inner product I was looking for, $B'$.
Does anyone know, or can anyone come up with, a proof?
Best Answer
I believe that the statement you want is not true. In $X=\mathbb R^3$, begin with the standard cone $x^2+y^2<z^2$ and perturb it so that the resulting cone $K$ is symmetric to its Euclidean dual through the $yz$-plane and has no affine symmetries (that is, no nontrivial linear maps that map it to itself). As your argument shows, this cone is self-dual w.r.t. $-x^2+y^2+z^2$.
I claim that this is a unique non-degenerate form which makes $K$ self-dual. Indeed, the dual cone is naturally (canonically) defined in the dual space $X^*$. A bilinear form defines a linear isomorphism between $X^*$ and $X$, and the dual cone in $X$ is the image of the canonical dual cone under this isomorphism. Since $K$ has no affine symmetries, there is only one linear map from $X^*$ to $X$ that sends the canonical dual cone to $K$. Therefore there is only one non-degenerate bilinear form that makes $K$ self-dual. And it is not positive.