linear algebra - Proof that domains of positivity of symmetric nondegenerate bilinear forms are self-dual cones?

Here's what's true instead of the claim that domains of positivity
are self-dual cones.

$mathbf{Proposition:}$
$Y$ is a domain of positivity for a nondegenerate
symmetric bilinear form $B$ if and only if it is an open cone whose dual,
according to the Euclidean inner product $E$ associated with a basis
orthonormalizing the form, is its image under reflection of $X_-$
through $X_+$, the ``negative and positive eigenspaces'' associated
with the form in this basis.

$mathbf{Proof:}$
We'll write $v,v'$ for vectors in $X$. We'll use an orthonormal basis
as described above, in which the form is diagonal with diagonal
elements $pm 1$, writing $v = (x,t)$ for a decomposition with $x$ in
the span (call it $X_+$) of the basis vectors with $B(e_i, e_i) = 1$,
and $t$ in the span (call it $X_-$) of the basis vectors with $B(e_i,
e_i) = -1$. Let $S$ be the linear map $(x,t) mapsto (x, -t)$,
i.e. reflection of the subspace $X_-$ through the subspace $X_+$.
Note that $E(x,y) := B(x,Sy)$ is a positive semidefinite
symmetric nondegenerate bilinear form.
Also, note that for all $v,v'$, $B(Sv, Sv') = B(v,v')$, i.e. the form $B$
is reflection-symmetric.

For "if": the definition of $Y^ast$ says it is
maximal such that $E(Y^ast,Y) > 0$. But since
$Y=SY$, it is also maximal such that $E(SY,Y) equiv B(Y,Y) > 0$,
i.e., it is a domain of positivity of $B$.

For ``only if'': let $Y$ be a domain of positivity for $B$. For
every $y$ in the boundary $partial Y$ of $Y$,
the hyperplane $H_y := {x: B(x,y) = 0}$ is
a supporting hyperplane for the cone $Y$, and these are all the
supporting hyperplanes. But it's standard convex geometry that the
supporting hyperplanes of a proper convex cone $Y$ are the precisely
the zero-sets of the linear functionals that constitute the boundary
of $Y$'s dual cone. We have $H_y = {x: B(x,y) equiv E(x,Sy) = 0}$;
that is, this hyperplane is just the plane normal to $Sy$ according to
the Euclidean inner product. That is to say, the vectors $Sy$, for $y
in partial Y$ generate the closure of the cone $Y^ast$ dual to $Y$
according to the Euclidean inner product $E$. I.e., $Y^ast = SY$.
$diamond$

Offline (or rather, off-math-overflow) correspondence with Will Jagy helped
stimulate this solution. He gave
another example---which I'd come up with a few weeks ago, but forgotten about---of a DOP for $xx' + yy' - zz'$---namely, the positive orthant generated by $(0, 1, 0)$, $(1, 0, 1)$
and $(1, 0, -1)$ (or in his dual description, defined by inequalities $x > z$, $x > -z$, $y > 0$), which is of course not isomorphic to an ice-cream cone, but is symmetric under reflection through the xy plane. The hypothesis that the DOPs were precisely the self-dual cones symmetric under reflection suggested itself to me, and attempts to prove the hypothesis ended up providing the proof of the proposition above.