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Jan
29 |
comment |
What is the mathematical significance of the IHES logo?
How do you see this? Does it matter how you connect the ends (at oo)? JV |
Jan
29 |
comment |
What is the mathematical significance of the IHES logo?
I saw this webpage, but I couldn't discern specific content. There are several knots shown on this page: at the top, there are the Mobius figure 8 knot as well as a tubular (3,5) torus knot. These are different knots, and it is not claimed that the IHES knot is either of these. Can you help to clarify why this website answers my question? JV |
Jan
28 |
asked | What is the mathematical significance of the IHES logo? |
Jan
27 |
revised |
Average of Fourier coefficients of a cusp form of half integral weight
added 10 characters in body |
Jan
27 |
awarded | Revival |
Jan
26 |
answered | Average of Fourier coefficients of a cusp form of half integral weight |
Jan
2 |
awarded | Self-Learner |
Jan
2 |
awarded | Yearling |
Dec
23 |
revised |
Quaternion orders such that every proper ideal is invertible
added 17 characters in body |
Dec
23 |
answered | Quaternion orders such that every proper ideal is invertible |
Dec
23 |
asked | Quaternion orders such that every proper ideal is invertible |
Dec
4 |
answered | Does $S$ being a free rank-$n$ $R$-algebra imply that $S/R$ is free rank $n-1$? |
Dec
3 |
revised |
Reference request for $R$-index
edited body |
Dec
2 |
asked | Reference request for $R$-index |
Oct
26 |
comment |
Locally square implies square
Yes, it is a difference; the matter of whether or not this is a "big" difference depends on the context. Personally, I found it helpful to see the example and how it was used, YMMV. That being said, I also do not know if there is an irreducible counterexample to the original question. |
Oct
25 |
answered | Locally square implies square |
Sep
25 |
accepted | Noncommutative group of invertible ideals of a ring |
Aug
1 |
comment |
Segments of Voronoi Diagrams on smooth manifolds. Are they geodesics?
Nevermind: on a small enough neighborhood around each point of the bisector, the geodesics from the point cover all possible tangent directions. In particular, if the bisector contains all geodesics between points, then it is totally geodesic. |
Aug
1 |
comment |
Segments of Voronoi Diagrams on smooth manifolds. Are they geodesics?
This paper only says that bisectors in $M$ are totally geodesic if and only if $M$ has constant curvature. This makes no difference for the original poster's question which was one-dimensional, but isn't it true that a bisector $B$ can be geodesic (the geodesic between any two distinct points in $B$ is contained in $B$) without being totally geodesic? I'm thinking, for example, of products. |
Jul
29 |
awarded | Revival |