770 reputation
717
bio website perimeterinstitute.ca/…
location
age
visits member for 2 years, 11 months
seen 13 hours ago

2d
comment Can we drop commutativity assumption?
By construction, the multiplication map $A^{\otimes n}\otimes A^{\otimes n}\to A^{\otimes n}$ is an intertwiner for the permutation representation of the symmetric group $S_n$. Since $S^nA$ is the invariant subspace of this representation, this intertwiner restricts to $S^n A\otimes S^n A\to S^n A$, independently of whether the original multiplication is commutative or not. On the other hand, the resulting map $\bigwedge^n A\otimes\bigwedge^n A\to A^{\otimes n}$ also lands in $S^n A$, so that $\bigwedge^n A$ is closed under multiplication only in the degenerate cases $char(k)=2$ or $dim(A)<n$.
Aug
26
awarded  Nice Answer
Aug
21
revised Algebras for probability monad
added 43 characters in body
Aug
21
answered Algebras for probability monad
Aug
21
comment Vector Fields in a Riemannian Manifold
Actually I'm not sure that my statement on "isospectralities" makes sense: the principal symbol of the Laplace operator is the metric itself. So doesn't this imply that every vector field that commutes with the Laplacian must be Killing?
Aug
21
comment Vector Fields in a Riemannian Manifold
Clearly Killing vector fields have this property. But in general, I guess that one can think of such vector fields as generators of "isospectralities" (instead of isometries). What would be a nice example of a manifold with such a vector field that is not Killing?
Aug
17
answered General additive function of probability
Aug
17
awarded  Informed
Aug
16
comment Looking for the name of a mathematical symbol that looks remotely like 1 (answer: indicator function)
Another commonly used term for this concept is "indicator function".
Jul
31
comment SO$(4)$ (& SO$(n)$) characterization?
For the case of SO(4), see this MO discussion, of which the current question actually seems to be a duplicate: mathoverflow.net/questions/37136/…
Jul
31
comment SO$(4)$ (& SO$(n)$) characterization?
Some trivial comments. First, in order to embed a finite group $G$ into $O(n)$, it is enough to embed it into $GL(n,\mathbb{R})$: if you have the latter embedding, start with any scalar product on $\mathbb{R}^n$, and use averaging over $G$ to turn it into an invariant scalar product, and you have an embedding into $O(n)\subseteq SO(n+1)$. Second, $G$ always embeds into $S_{|G|}$ by Cayley's theorem, and hence into $SO(|G|+1)$. These observations suggest to me that a general answer to your question is probably too much to hope for...
Jul
16
revised Reference request for translating from Top to C*-alg
improved formatting a bit
Jul
16
comment Reference request for translating from Top to C*-alg
I've improved the answer's formatting a bit. FWIW, 'covariant' in the Pedersen quote should be 'contravariant'.
Jul
16
suggested approved edit on Reference request for translating from Top to C*-alg
Jul
10
comment Why is the identity element of a group denoted by $e$?
For German speakers: am I missing something, or does Burkhardt's Encyklopädie really define a "group" as a cancellative semigroup, and then claim that the existence of the unit and inverses follow?
Jun
24
comment A probability distribution in n dimensional space which its projection on any line is a uniform distribution?
Addendum to my comments: now I understand that Will achieves something equivalent by considering a random projection. Clever!
Jun
24
comment A probability distribution in n dimensional space which its projection on any line is a uniform distribution?
Hence we work wlog with a measure of the form $f(r)\, d\Omega\, dr$, where $d\Omega$ is the volume element of the sphere (and $f$ may be distributional), i.e. an integral over the uniform measures on spheres for varying radius of the sphere. For these measures, it should be a straightforward calculation to determine whether there exists some $f$ for which $f(r) \, d\Omega\, dr$ satisfies the OP's requirement.
Jun
24
comment A probability distribution in n dimensional space which its projection on any line is a uniform distribution?
Using Will's observation on mean and covariance, one can also assume the measure to be symmetric under rotations without loss of generality. The reason is that having projection uniform on $[-1,1]$ is a property which is invariant under taking mixtures of measures. Hence if some measure has this property, then so does its convolution with a random rotation (chosen from the Haar measure on $O(n)$).
Jun
24
comment A probability distribution in n dimensional space which its projection on any line is a uniform distribution?
@Salavati: do you know the answer for the two-dimensional case? A positive answer in three dimensions would imply a positive answer in two dimensions by projecting the measure onto the plane. So I think that it will help to consider the two-dimensional case first.
May
29
comment Reference for an unbiased definition of a symmetric monoidal category
I had asked this question on the nForum about a year ago: nforum.ncatlab.org/discussion/3101/symmetric-monoidal-category/…