21 reputation
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bio website winephysicssong.com
location Waterloo, Ontario
age
visits member for 4 years, 10 months
seen Mar 22 '10 at 3:25
I'm a theoretical physicist by Ph.D., but perhaps equally drawn to physics, mathematics, computer science, and philosophy. I've been working in quantum information theory and quantum computation since graduate school. I'm currently involved in a research program to understand quantum theory better by situating it in a broad framework, based on convex sets of states, measurement outcomes, and dynamical evolutions, for abstractly formulating candidate physical theories.

Sep
24
awarded  Autobiographer
Mar
7
comment Proof that domains of positivity of symmetric nondegenerate bilinear forms are self-dual cones?
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!
Mar
5
revised Proof that domains of positivity of symmetric nondegenerate bilinear forms are self-dual cones?
Pointed out that I now believe the statement to be proved is wrong, as discussed in my answer. Replaced "positive semidefinite" with "positive definite" in definition of inner product.
Mar
4
answered Proof that domains of positivity of symmetric nondegenerate bilinear forms are self-dual cones?
Mar
1
comment Proof that domains of positivity of symmetric nondegenerate bilinear forms are self-dual cones?
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.
Feb
28
comment Proof that domains of positivity of symmetric nondegenerate bilinear forms are self-dual cones?
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).
Feb
28
revised Proof that domains of positivity of symmetric nondegenerate bilinear forms are self-dual cones?
Added backticks around some LaTeX code for improved formatting
Feb
27
awarded  Supporter
Feb
26
awarded  Editor
Feb
26
comment Proof that domains of positivity of symmetric nondegenerate bilinear forms are self-dual cones?
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.
Feb
26
revised Proof that domains of positivity of symmetric nondegenerate bilinear forms are self-dual cones?
I've attempted to clarify the question by giving a more precise definition of self-dual cone, in response to the questions of Leonid Khovalev and another commenter. Thanks for prompting the clarifications.
Feb
26
awarded  Student
Feb
26
asked Proof that domains of positivity of symmetric nondegenerate bilinear forms are self-dual cones?