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8h
comment Lower bound for the $C^*$-unitisation norm?
Unless I misunderstood the question, there is no such example: $|\lambda|\leq\|(a,\lambda)\|$ holds because the map $A^+\to\mathbb{C}$ is a $*$-homomorphism, and every $*$-homomorphism is norm-nonincreasing.
1d
comment Graph Isomorphism for Triangle Free graph
@Jim: please feel free to email me, it should be fun to take a look! But there also will be people out there who are much more qualified than me to give you helpful feedback. So you might instead want to ask somebody who has actually done research on graph isomorphism (which I haven't).
Apr
19
revised Graph Isomorphism for Triangle Free graph
added 128 characters in body
Apr
19
answered Graph Isomorphism for Triangle Free graph
Apr
19
comment Graph Isomorphism for Triangle Free graph
@Jim: alright, will do.
Apr
19
comment Graph Isomorphism for Triangle Free graph
@Jim: yes, that's what I mean. You check for isomorphism of the barycentric subdivisions, which are triangle-free. If these are not isomorphic, then your original graphs aren't isomorphic either; if the subdivisions are isomorphic, then so are the original graphs. The latter is what I gather from mathoverflow.net/questions/132408/…, but probably somebody else can say more about how isomorphism of the subdivisions implies isomorphism of the original graphs.
Apr
19
comment Graph Isomorphism for Triangle Free graph
@TonyHuynh: thanks, of course! I didn't see that this was just a special case of a familiar construction...
Apr
19
comment Graph Isomorphism for Triangle Free graph
What if you "blow up" a graph by replacing every edge by a path of length 2? This results in a triangle-free graph, and if all nodes of the original graphs have degree larger than 2, then two graphs are isomorphic if and only if their blow-ups are isomorphic. (BTW, does this construction have a name? It's definitely not blow-up, which already has a different meaning for graphs...) Hence graph isomorphism for triangle-free graphs should have complexity equal to graph isomorphism in general.
Apr
13
comment Flat coordinates of a Riemannian metric
If you can find a point $p$ for which you can compute all the geodesics that emanate from $p$, then you get the normal coordinates around $p$: en.wikipedia.org/wiki/Normal_coordinates Since you're assuming the metric in your chart to be isometric to some Euclidean $U\subseteq\mathbb{R}^n$, it follows that the normal coordinates establish an isometry between a neighbourhood of $p$ in $M$ and a neighbourhood of the origin in $T_pM$. Now you only need to choose a cartesian coordinate system in $T_pM$.
Apr
12
revised Estimate a Fourier Transform
improved formatting
Apr
12
reviewed Reviewed Estimate a Fourier Transform
Apr
12
comment Estimate a Fourier Transform
I've improved the formatting a bit; the question is still the same.
Apr
12
suggested approved edit on Estimate a Fourier Transform
Apr
11
revised Fast Fourier Transforms for non-trigonometric bases
improved language
Apr
11
awarded  Custodian
Apr
11
suggested approved edit on Fast Fourier Transforms for non-trigonometric bases
Apr
11
reviewed Reviewed Fast Fourier Transforms for non-trigonometric bases
Apr
3
awarded  Good Answer
Apr
2
awarded  Nice Answer
Apr
1
answered Examples of math hoaxes/interesting jokes published on April Fool's day?