Tuesday, 4 September 2007

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.



mathbfProposition:
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.



mathbfProof:
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 pm1, writing v=(x,t) for a decomposition with x in
the span (call it X+) of the basis vectors with B(ei,ei)=1,
and t in the span (call it X) of the basis vectors with B(ei,ei)=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 Yast says it is
maximal such that E(Yast,Y)>0. But since
Y=SY, it is also maximal such that E(SY,Y)equivB(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 partialY of Y,
the hyperplane Hy:=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 Hy=x:B(x,y)equivE(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 yinpartialY generate the closure of the cone Yast dual to Y
according to the Euclidean inner product E. I.e., Yast=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+yyzz---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.

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