bio | website | perimeterinstitute.ca/… |
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location | ||
age | ||
visits | member for | 2 years, 7 months |
seen | 3 hours ago | |
stats | profile views | 362 |
Mar 24 |
revised |
Dissolution of Tensors
fixed tag, typo and latex |
Mar 24 |
suggested | approved edit on Dissolution of Tensors |
Mar 9 |
comment |
Numbers greater than Skewes's whose existence can be found in number theoretic proofs
This seems close to the "eventual counterexamples" question: mathoverflow.net/questions/15444/… |
Mar 6 |
comment |
Abstract connectedness
@MikeShulman: yes, I strongly agree about the minimality of the separoid axioms! It feels like they're missing something important... |
Mar 6 |
comment |
Abstract connectedness
The notion of "separoid" may be related: en.wikipedia.org/wiki/Separoid "Given a topological space, we can define a separoid saying that two subsets are separated if there exist two disjoint open sets which contains them (one for each of them)." But possibly it captures something slightly different from what you're looking for; I haven't thought about it. |
Mar 3 |
comment |
Base of a cone in a vector space: can one always choose a convex base?
@WillieWong: sure, done! |
Mar 3 |
revised |
Base of a cone in a vector space: can one always choose a convex base?
added example due to request in the comments |
Mar 1 |
revised |
Base of a cone in a vector space: can one always choose a convex base?
small improvements |
Feb 28 |
revised |
Base of a cone in a vector space: can one always choose a convex base?
yet another correction |
Feb 28 |
revised |
Base of a cone in a vector space: can one always choose a convex base?
this is final, sorry for the many edits ;) |
Feb 28 |
answered | Base of a cone in a vector space: can one always choose a convex base? |
Feb 27 |
comment |
Base of a cone in a vector space: can one always choose a convex base?
@AlexDegtyarev: that's a good question, but actually the answer is negative. Here's a simple example of a pointed convex cone which does not have a convex base. In $\mathbb{R}^2$, define $(x,y)$ to be in $C$ if $x>0$, or if $x=0$ and $y\geq 0$. (This cone is a simple counterexample also to many other elementary conjectures that one might have about convex cones, by the way.) |
Feb 27 |
comment |
Base of a cone in a vector space: can one always choose a convex base?
@WillieWong: thanks, that's exactly right. I'm not sure whether a convex base always exists; if it does, then the question is uninteresting. So I assume that the OP knows examples in which a convex base does not exist. In these cases, my argument shows that a non-convex base still exists, for trivial reasons. This shows that what the OP is trying to do is not possible. |
Feb 27 |
comment |
Base of a cone in a vector space: can one always choose a convex base?
Every cone has a base in your sense, and this follows from the axiom of choice. In $C\setminus\{0\}$, put $x\sim y$ whenever $x$ and $y$ are positive scalar multiples of each other. This is an equivalence relation, and the axiom of choice lets you pick one representative of each equivalence class. These representatives form a base in your sense. |
Feb 26 |
comment |
Symplectic (contact) structure on $M_{n}(\mathbb{R})$
Concerning the second sentence: the derivative of the Hamiltonian indeed vanishes in certain directions, but how do you conclude that the associated vector field vanishes as well? (A generic Hamiltonian has vanishing directional derivatives in a subspace of codimension 1 of the tangent space, but that doesn't mean that the flow is zero!) |
Feb 21 |
comment |
Are there only countably many compact topological manifolds?
You also want to assume connectedness, since otherwise any disjoint union of closed manifolds would again be a closed manifold, and these disjoint unions can be arbitrarily big. |
Feb 7 |
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Which group algebras in analysis are “true group algebras”?
It might be worthwhile mentioning that the "suitable forgetful functor" is the one which assigns to every C*-algebra $A$ its unitary group $\mathcal{U}(A)$, considered as a discrete group. |
Feb 7 |
comment |
Which group algebras in analysis are “true group algebras”?
At least the maximal group C*-algebra $C^*(G)$ of a discrete group $G$ does have a well-known universal property: just as $\mathbb{C}[G]$ classifies ordinary representations of $G$, $C^*(G)$ classifies unitary representations, in the sense that every group homomorphism $\pi:G\to \mathcal{U}(B)$ to the unitary group $\mathcal{U}(B)$ to another C*-algebra $B$ uniquely extends to a $*$-homomorphism $C^*(G)\to B$. More concisely speaking, the functor $G\mapsto C^*(G)$ which assigns to every discrete group its maximal group C*-algebra is right adjoint to the unitary group functor. |
Feb 3 |
revised |
graphs with independence number = Shannon capacity
corrected (following comment) |
Feb 3 |
comment |
graphs with independence number = Shannon capacity
@Wolfgang: whoops, I hadn't realized that you were pointing out a typo! Thanks, corrected. |