Timeline for Proof that domains of positivity of symmetric nondegenerate bilinear forms are self-dual cones?
Current License: CC BY-SA 3.0
12 events
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| May 16, 2015 at 15:52 | history | edited | Ricardo Andrade | CC BY-SA 3.0 |
replaced tags with more specific or (apparently) more relevant ones; feel free to re-add 'mp.mathematical-physics' if you feel it is appropriate
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| May 16, 2015 at 14:54 | answer | added | Ettore Minguzzi | timeline score: 3 | |
| Mar 19, 2010 at 12:20 | answer | added | Sergei Ivanov | timeline score: 3 | |
| Mar 5, 2010 at 1:25 | history | edited | Howard Barnum | CC BY-SA 2.5 |
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.
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| Mar 4, 2010 at 2:51 | answer | added | Howard Barnum | timeline score: 1 | |
| Mar 1, 2010 at 22:33 | comment | added | Will Jagy | 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? | |
| Mar 1, 2010 at 21:14 | comment | added | Howard Barnum | 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, 2010 at 1:14 | comment | added | Howard Barnum | 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, 2010 at 1:03 | history | edited | Howard Barnum | CC BY-SA 2.5 |
Added backticks around some LaTeX code for improved formatting
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| Feb 26, 2010 at 23:48 | comment | added | Howard Barnum | 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, 2010 at 23:38 | history | edited | Howard Barnum | CC BY-SA 2.5 |
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.
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| Feb 26, 2010 at 16:39 | history | asked | Howard Barnum | CC BY-SA 2.5 |