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Anton Petrunin

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Name Anton Petrunin
Member for 3 years
Seen 1 hour ago
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Location PSU
Age 45
from ratemyprofessors.com:

...Just stay away from this professor, he is useless and explains nothing ... He would just call you stupid if you asked a question ... his response, in russian english, was, "If you don't know this yet, maybe you perform suicide."...
2h
 comment Symmetric convex curve
Tell us which maps did you try and what are the bound which you can prove.
2h
 comment Connected sum in Alexandrov spaces
@Zimbrón, yes it should be this way (in CBB case), but there is no machinery is available.
2h
 accepted Local boundary symmetrisation of Riemannian metrics by coordinate changes
3h
 answered Local boundary symmetrisation of Riemannian metrics by coordinate changes
8h
 comment Symmetric convex curve
@djoke, it is not hard to perturb the map into a homeomorphism by making the Lipschitz constant little worse. On the other hand if you want to keep the constant then you can not do this --- square is an example.
1d
 revised Symmetric convex curve
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1d
 answered Symmetric convex curve
1d
 comment Connected sum in Alexandrov spaces
For general CBB-space, there is no known construction which keep an open set unchanged and change something outside. So your question seems hopeless...
2d
 comment Is a free alternative to MathSciNet possible?
Not for me --- I am not going to create Google+ account.
2d
 comment Is a free alternative to MathSciNet possible?
Filippo, "free" means "free content", it is not exactly about "free access".
2d
 awarded ● Good Answer
Jun
7
 accepted Riemann isometry vs Euclidean bi-Lipschitz mapping
Jun
7
 answered Riemann isometry vs Euclidean bi-Lipschitz mapping
Jun
5
 comment Delta-convex functions and inner products
It seems that for these functions the canonization does not depend on the choice of domain, am I right?
Jun
3
 comment Delta-convex functions and inner products
Yes, I thought about functions defined on a bounded domain. The canonization seems to depend on the choice of the domain and you may get more canonizations by taking weighted integrals. For the functions on $\mathbb R^n$, one might prove that there can be no canonical choice which invariant under translations. (It should be essentially the same argument as for non-existence of generalized limit for all {0,1}-sequences.)
Jun
1
 answered Delta-convex functions and inner products
May
27
 comment A notion of a ‘coarse’, parametrized dimension of an object, where the parameter determines how finely we can distinguish (say) a very thin rod from a line
Macroscopic dimension is exactly the one you decribing.
May
26
 comment Existence of particular embeddings in euclidean spaces for non compact manifolds
Oh, the Morse function should be the distance to 0, but not the coordinate.
May
26
 comment Why don’t more mathematicians improve Wikipedia articles?
There is positive part in this, you can improve your writing skills quite a bit. (If it would be more qualified contributors, it would be even better.)
May
26
 comment Why don’t more mathematicians improve Wikipedia articles?
@Mark M; WikiProject Mathematics has "eyes" of people who like to talk a lot. If you see how to make wikipedia better, simply do it and check what is the reaction of community. (I think it is called "Be Bold", it is a "rule" in wikipedia).
May
26
 comment Existence of particular embeddings in euclidean spaces for non compact manifolds
@Daniele, I think your idea works perfectly. Take the embedding as you describe. Note that in addition you can assume that between any pair of critical values there is an interval where the embedding is radial; i.e., it is swapped by segments in the radial direction. Now choose a function $f:\mathbb R_+\to \mathbb R_+$ such that it is locally constant away of the radial segments and "rescale by this function"; i.e., in the spherical coordinates you new embedding looks like $(f(\rho)\cdot \rho,\theta)$ if the old one was $(\rho,\theta)$. For right choice of $f$ you get the needed embedding.
May
20
 comment Closed geodesic loops around points in compact manifolds
@Igor, the manifold of broken geodesics is not complete. So you have to be bit more careful, to make sure that there is a critical point. In other words, read my answer.
May
20
 comment Closed geodesic loops around points in compact manifolds
@Misha, Yes, I do use minimax. I made an update, it should be more clear now.
May
20
 revised Closed geodesic loops around points in compact manifolds
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May
20
 revised Closed geodesic loops around points in compact manifolds
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May
20
 answered Closed geodesic loops around points in compact manifolds
May
18
 accepted Can one (block) diagonalize the curvature matrix of 2 forms on a Riemannian manifold?
May
18
 comment Can one (block) diagonalize the curvature matrix of 2 forms on a Riemannian manifold?
@Robert, You are right, but note that I state "curvature tensor equals to a curvature of product"; it does not mean that space is a product.
May
17
 comment Reference request: affine transforms + circle inversion?
Do you know if $\cal{T}$ is finite-imensional?
May
17
 comment Can one (block) diagonalize the curvature matrix of 2 forms on a Riemannian manifold?
P.S. For example, if sectional curvature has definite sign then the curvature tensor can not be block-diagonalized. Further, the rank of generic curvature tensor is $n\cdot(n-1)/2$, but any block-diagonalized has rank at most $\lfloor\tfrac n2\rfloor$
May
17
 comment On a version of gradient descent
@Robinson1. I would better leave it as an exercise :) BTW, if it would be wrong then so is the original statement.
May
17
 comment Is there any proof that you feel you do not “understand”?
@Steven, this proof is cheating. The Pythagorean theorem is way simpler than existence of the area functional which is used in the proof (and many proofs of existence of the area use Pythagorean theorem).
May
17
 answered Can one (block) diagonalize the curvature matrix of 2 forms on a Riemannian manifold?
May
16
 comment Triangle area on surfaces of constant curvature
@Alexandre, Euclid (and Kiselev) did not prove the existence, essentially they add the existence as an axiom, but they did not say that it is an "axiom". This axiom follows from the rest of axioms, but it takes 20 pages at least. Instead of unit square you have to use other normalization (which essentially defines curvature). A rigourous way to introduce area given in "Elementary Geometry From An Advanced Standpoint" by Moise 35 pages Euclidean plane onlyy,and in "Geometry: A Metric Approach with Models" by Millman and Parker 40 pages neutral plane and contains a gap.
May
16
 answered On a version of gradient descent
May
16
 comment Triangle area on surfaces of constant curvature
@BS, check this question mathoverflow.net/questions/119953/… I would be very happy if you know a better answer.
May
15
 comment Triangle area on surfaces of constant curvature
It use the properties of the area which (if you look carefully) already include the original statement inside.
May
15
 comment Triangle area on surfaces of constant curvature
@Alexandre, I do not see what exactly you disagree with. A rigorous intro to area from the axioms takes 20-40 pages, and once it is done the formula is already proved. So these sort of "proofs" confuse poorly educated students and they prove nothing to those who know what area is.
May
13
 comment Triangle area on surfaces of constant curvature
Chapter 5 is "Affine-projective relationship" did you really mean this?
May
13
 comment Triangle area on surfaces of constant curvature
P.S. surface of constant curvature κ are spheres plane or Lobachevsky plane. All these things are "elementary" for me.
May
13
 comment Triangle area on surfaces of constant curvature
@Kofi, you ask for an elementary derivation. For me "Riemannian metric" and "integral" are not elementary and the geometry as it was used to be covered in the school (but not any more) is elementary.
May
13
 revised Triangle area on surfaces of constant curvature
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May
13
 comment Triangle area on surfaces of constant curvature
@Sergei, yes sure, all I wanted to say is that if one knows what is area and curvature then there is nothing to prove.
May
12
 revised Triangle area on surfaces of constant curvature
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May
12
 answered Triangle area on surfaces of constant curvature
May
12
 comment The Isoperimetric problem for domains constrained to lie between two parallel planes
You say "Since the problem is completely symmetric in the y,z directions, the solution can be represented as the rotation of the graph of a function r(x) around the x axis." This is true and if you want to prove it you may use Schwarz symmetrization in the directions of yz-plane. $$ $$ Once you get to this point the remaining part is ODE. It is a simple problem, but I can not help since you do not specify how you want to count the area where your set touches $x=\pm a$ planes.
May
10
 comment Euclid with Birkhoff
Yes, Moise is quite good.
May
9
 awarded ● Nice Answer
May
8
 answered Converse to Milnor’s theorem on manifolds with nonnegative Ricci curvature.
May
6
 revised Measuring the distance of a convex set from a ball (Nikodym distance)
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