Impact
~1k
people reached
- 0 posts edited
- 0 helpful flags
- 18 votes cast
Nov
29 |
awarded | Popular Question |
May
16 |
awarded | Teacher |
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? |