Tobias Fritz
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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?