Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Max Koecher (in, for example, 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 Ch. 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 $+/- 1$, 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?

share|improve this question
    
Thanks for the comments, Leonid and Will; I have edited the post to attempt to clarify. Briefly, I want to prove that the cone is self-dual in the sense that there exists a positive semidefinite bilinear form (i.e., an inner product) with respect to which it is self-dual. It's not obvious that that's the same thing as the existence of a symmetric nondegenerate bilinear form with respect to which it's self dual; the question, essentially, is whether these two are in fact the same thing. –  Howard Barnum Feb 26 '10 at 23:48
    
Will, in $$\mathbb{R}^2$$, every pointed open cone is self-dual (and in fact, isomorphic (as a cone) to $$\mathbb{R}^2_+$$ (the strictly positive quadrant). So you're certainly right there. The way I like to visualize things in $$\mathbb{R}^3$$ is to consider the "diagonalized" bilinear forms $$tt' - xx' - zz'$$ and $$-tt' + xx' + zz'$$. (The question is trivial for the other signatures.) For $$+,-,-$$ it's easy: the positive and negative "light cones" are the only DOPs; while for $$-,+,+$$, I conjecture many nonisomorphic ones, in the complement of these light cones (the "conic doughnut). –  Howard Barnum Feb 28 '10 at 1:14
    
This sounds correct. You've built the Lorentz (alias quadratic, alias second-order, alias ice-cream) cone with central axis $(1,0,0)$, in $\mathbb{R}^3$. Its interior is one domain (of many) of positivity of the bilinear form $B$ in question ($xx' + yy' - zz'$), as well as of the Euclidean inner product. Orthogonality according to $B$ is not the same thing as according to the Euclidean inner product, except when $z=0$, but that's okay. The set of vectors $B$-orthogonal to a given boundary vector $x$ is still a supporting hyperplane, just not opposite $x$; these hyperplanes bound the cone. –  Howard Barnum Mar 1 '10 at 21:14
    
So the question becomes, in this standardized situation, are there any other DOP's in the conic doughnut except the rotations of my ice-cream cone around the z-axis? –  Will Jagy Mar 1 '10 at 22:33

2 Answers 2

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.

share|improve this answer

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.

share|improve this answer
    
Hi Howard, there is some problem with msri updating software, I was not able to print out your answer today, and for that matter ssh is disabled. Good work, anyway. Will –  Will Jagy Mar 5 '10 at 1:29
    
I should also point out that this question arose in ongoing work with Alex Wilce and Ross Duncan---and Alex's insistent unwillingness to to quote and rely on the editors' chapter-end notes without seeing a proof turned out to be well-founded, and crucial motivation for investigating the question! –  Howard Barnum Mar 7 '10 at 16:28

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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