Proof that domains of positivity of symmetric nondegenerate bilinear forms are self-dual cones? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-18T11:14:46Z http://mathoverflow.net/feeds/question/16527 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16527/proof-that-domains-of-positivity-of-symmetric-nondegenerate-bilinear-forms-are-se Proof that domains of positivity of symmetric nondegenerate bilinear forms are self-dual cones? Howard Barnum 2010-02-26T16:39:04Z 2010-03-19T12:20:10Z <p>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). </p> <p>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 $B'$. </p> <p>To clarify the question: we will define a cone $K$ in a real vector space $X$ to be self-dual iff there <em>exists</em> an inner product----that is, a <em>positive definite</em> bilinear form $\langle . , . \rangle: X \times X \rightarrow \mathbb{R}$---such that <code>$K = K^*_{\langle . , . \rangle}$</code>. Here <code>$K^*_{\langle . , . \rangle}$</code> is the dual with respect to the inner product $\langle . , . \rangle$, that is <code>$K^*_{\langle . , . \rangle} := \{ y \in X: \forall x \in X ~\langle y, x \rangle &gt; 0 \}$</code>. 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, $B'$.</p> <p>Does anyone know, or can anyone come up with, a proof?</p> http://mathoverflow.net/questions/16527/proof-that-domains-of-positivity-of-symmetric-nondegenerate-bilinear-forms-are-se/17047#17047 Answer by Howard Barnum for Proof that domains of positivity of symmetric nondegenerate bilinear forms are self-dual cones? Howard Barnum 2010-03-04T02:51:45Z 2010-03-04T02:51:45Z <p>Here's what's true instead of the claim that domains of positivity are self-dual cones. </p> <p>$\mathbf{Proposition:}$ $Y$ 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. </p> <p>$\mathbf{Proof:}$ 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.<br> Also, note that for all $v,v'$, $B(Sv, Sv') = B(v,v')$, i.e. the form $B$ is reflection-symmetric. </p> <p>For "if": the definition of <code>$Y^\ast$</code> 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 $B$.</p> <p>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., $Y^\ast = SY$. $\diamond$</p> <p>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. </p> http://mathoverflow.net/questions/16527/proof-that-domains-of-positivity-of-symmetric-nondegenerate-bilinear-forms-are-se/18728#18728 Answer by Sergei Ivanov for Proof that domains of positivity of symmetric nondegenerate bilinear forms are self-dual cones? Sergei Ivanov 2010-03-19T12:20:10Z 2010-03-19T12:20:10Z <p>I believe that the statement you want is not true. In $X=\mathbb R^3$, begin with the standard cone <code>$x^2+y^2&lt;z^2$</code> 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. $-x^2+y^2+z^2$.</p> <p>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 <code>$X^*$</code>. A bilinear form defines a linear isomorphism between <code>$X^*$</code> 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.</p>