bio | website | perimeterinstitute.ca/… |
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location | ||
age | ||
visits | member for | 1 year, 6 months |
seen | 13 hours ago | |
stats | profile views | 298 |
Apr 1 |
comment |
Numbers of distinct products obtained by permuting the factors
What about generalizing the problem: given any equivalence relation on $S_n$, can one find $G$ and the $g_i$ such that the map $S_n\to G,\: \sigma \mapsto g_{\sigma(1)}\cdot\ldots\cdot g_{\sigma(n)}$ identifies two permutations if and only if they are equivalent? Or is this trivially false? |
Mar 25 |
awarded | Suffrage |
Feb 28 |
comment |
Does every separated measurable space embed into a power of $\{0,1\}$?
That's precisely what I had tried initially (see the original version of my question). If you know about Gelfand duality and Stone duality, it seems like a natural thing to try and indeed the resulting injective function is measurable (with respect to the Baire, and hence also the Borel $\sigma$-algebra on $\{0,1\}^\Sigma$). However, I don't see why this function should also take measurable sets to measurable sets, and I think this fails in general. In fact, I suspect that its image may not even be measurable, but I don't have a concrete example of that. |
Feb 28 |
accepted | Does every separated measurable space embed into a power of $\{0,1\}$? |
Feb 28 |
comment |
Does every separated measurable space embed into a power of $\{0,1\}$?
Neat answer! I can't accept both, so I will go with Joseph's which covers the Borel case, since it also contains a sufficient condition which is close to his necessary one. |
Feb 27 |
comment |
Does every separated measurable space embed into a power of $\{0,1\}$?
@Joseph: I updated the question to reflect your correction. A more detailed answer would be great! |
Feb 27 |
revised |
Does every separated measurable space embed into a power of $\{0,1\}$?
added 838 characters in body |
Feb 27 |
comment |
Does every separated measurable space embed into a power of $\{0,1\}$?
@Joseph: you're right. I will have to think about it a bit more. Sorry for the confusion... |
Feb 27 |
revised |
Does every separated measurable space embed into a power of $\{0,1\}$?
clarified the question after a request in the comments |
Feb 27 |
comment |
Does every separated measurable space embed into a power of $\{0,1\}$?
@Joel: yes, that's exactly what I mean. I'll clarify the question accordingly. |
Feb 27 |
revised |
Does every separated measurable space embed into a power of $\{0,1\}$?
deleted 132 characters in body |
Feb 27 |
asked | Does every separated measurable space embed into a power of $\{0,1\}$? |
Feb 14 |
awarded | Scholar |
Feb 14 |
accepted | graphs with independence number = Shannon capacity |
Feb 14 |
answered | graphs with independence number = Shannon capacity |
Jan 29 |
comment |
Resource for learning quantum mechanics from the viewpoint of representation theory
Maybe this should be Community Wiki? |
Jan 29 |
answered | Resource for learning quantum mechanics from the viewpoint of representation theory |
Jan 11 |
comment |
graphs with independence number = Shannon capacity
@Graphth and everybody else: we have finally disproved all these conjectures and are currently in the process of revising our manuscript accordingly. The proof is somewhat indirect and actually does make use of the relationship to quantum information theory that we develop. I will compose a more detailed answer as soon as the updated manuscript is on the arXiv. |
Nov 20 |
revised |
Replacing commutative C*-algebras by simple ones
added missing tag |
Nov 19 |
suggested | suggested edit on Replacing commutative C*-algebras by simple ones |