Sergei Ivanov

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 Name Sergei Ivanov Member for 3 years Seen 2 hours ago Website Location Saint Petersburg, Russia Age 40
I am a research fellow at St.Petersburg Department of Steklov Math Institute
 16h accepted Lower bound on $L^2$ norm of mean curvature in general dimensions May16 answered Lower bound on $L^2$ norm of mean curvature in general dimensions May15 accepted A characterization of Hilbert spaces? May15 comment Lower bound on $L^2$ norm of mean curvature in general dimensionsThe exponent at $|\Sigma|$ should be $(n-2)/n$ for scale invariance. May14 comment Is every connected metrizable locally path connected space a length space?There is something wrong with the inequality on $f$. Take $t=0$ for example. May13 comment Isoperimetric inequality on a Riemannian sphereThe question came by association from a totally different problem. I am hoping that there might be a kind of rigidity result where one concludes that the metric is round (or not far from round) by looking at some rough measures of isoperimetric profile. If there is one, I could try to apply the technique of the proof in another context. May13 comment A characterization of Hilbert spaces?@Bill: this is plausible but I don't know for sure. In finite dimensions, you can differentiate the map at some point and get that a codimension 1 section is within a bounded distance from a Euclidean norm. I don't know how to carry this over to infinite dimensions. May12 comment Triangle area on surfaces of constant curvatureYou would need some normalization axiom, in order to distinguish between proportional measures. Do you have a specific one in mind? May12 answered A characterization of Hilbert spaces? May12 comment Isoperimetric inequality on a Riemannian sphere@katz: This is just an example of statement that might be true if dividing in half fails. It might as well be an integral inequality on the isoperimetric profile. In other words, I am more interested in a affirmative answer to a slightly different question than in a counter-example to the precise one. There is nothing magic about the constant $\pi$, I just want to avoid the trivial answer "look at a small neighborhood of a point where curvature is greater than 1". May10 comment Measuring the distance of a convex set from a ball (Nikodym distance)@alex: I think you are right. This simplification did not occur to me. May7 accepted Measuring the distance of a convex set from a ball (Nikodym distance) May6 revised Measuring the distance of a convex set from a ball (Nikodym distance)added an estimate of the Nikodym distance May6 comment Measuring the distance of a convex set from a ball (Nikodym distance)It is not immediately clear that the diameter is bounded. May6 answered Measuring the distance of a convex set from a ball (Nikodym distance) May6 awarded ● Nice Question May6 revised spectral radius monotonicitycorrected notice May6 comment spectral radius monotonicity@Hans: yes, sure. I misread the question as "there exist $a$ and $b$ such that...". I am not sure about monotonicity. And actually I now see a flaw in the argument: $S'T$ is not symmetric, so its spectral radius is not equal to the norm. Sorry about this confusion. May6 comment spectral radius monotonicity@Hans: sorry, I meant "positive definite", not "positive elementwise". The inequality follows by diagonalization. The transition from $b\to\infty$ to a large $b$ is basically the definition of limit. May5 answered spectral radius monotonicity May5 asked Isoperimetric inequality on a Riemannian sphere May5 comment Optimal paintbrush geodesicsHow about $w=\pi$ and arbitrarily short $\gamma$? I think in this case $\gamma(w)$ covers the sphere. May2 awarded ● Popular Question Apr10 accepted Monotonicity of Loewner ellipsoid? Apr10 answered Monotonicity of Loewner ellipsoid? Apr8 awarded ● Nice Answer Apr7 answered The sum of same powers of all matrices modulo p Apr7 answered Maximal cross sections of the Cartesian product of two planar domains Apr7 comment Maximal cross sections of the Cartesian product of two planar domainsThe function can be a constant and constants seem to be unimodal by your definition. Apr5 comment What kinds of manifolds admit concave boundary?What do you mean by convexity of the boundary? It has no meaning without additional structure (e.g. a Riemannian metric or an embedding into $\mathbb R^n$). Apr2 comment Invariance of the l.h.s. of Euler-Lagrange equationThank you. It seems that this description gets better when translated to $T^*M$ by Legendre transform. If $s(t)\in T^*M$ is the Legendre transform of $\gamma'(t)$, the analog of $\beta$ is a 1-form $\omega(s'(t),\cdot)+dH(\cdot)\in T^*_{s(t)}T*M$, where $\omega$ is the symplectic form and $H$ is the Hamiltonian. And it projects down to $M$ because vanishes on the fiber of $T^*M$. Apr2 answered Infimum of a finite number of distances in the plane Apr2 asked Invariance of the l.h.s. of Euler-Lagrange equation Mar31 answered Iterates converging to a continuous map Mar31 comment Iterates converging to a continuous mapThere is a tag 'reference-request'. Mar31 comment Iterates converging to a continuous mapYour wording suggests that you have a proof and only seek a reference. It this correct? Mar31 comment Iterates converging to a continuous mapWhy $\varphi_\infty=0$? What is $\varphi(x)=x$? Mar26 comment On Lipschitz embeddability of certain compact metric spaces into $\mathbb{R}^n$I think it is true for $n=1$ and false for $n=2$. For $n=2$ you can do a similar construction using circles and points inside them (but the resulting metric is no longer intrinsic). For $n=2$ and an intrinsic metric on a disc, it is an open problem although not a very popular one. For $n=1$, it seems that one can can construct a Lipschitz embedding using the linear order on the real line: divide the interval containing the set in half, then in 4 pieces, etc, and choose images of the division points carefully. Mar26 accepted On Lipschitz embeddability of certain compact metric spaces into $\mathbb{R}^n$ Mar25 revised A question of compactness in the geometry of numbersfixed constants Mar25 accepted A question of compactness in the geometry of numbers Mar25 answered A question of compactness in the geometry of numbers Mar24 comment Average weighted value of a linear functional over increasing bounded subsets of Z^n$\mu^m(f)$ involves a sum and a limit. The sum equals a Riemannian sum. therefore the limit of sums equals the integral. Taking to the power $1/m$ yields that $\mu^m(f)=\hat\mu^m(f)$. And $\hat\mu(f)$ is by definition the $L^m$ norm of the function $f(x)/\|x\|$ restricted on the ball. That's all I'm saying. Mar24 comment Average weighted value of a linear functional over increasing bounded subsets of Z^nFor $f(x)/\|x\|$ the technical details are more technical because the function is discontinuous at 0. But this singularity is not too bad and one can check that the Riemannian integral sums still converge to the integral. Not sure which 'two integrals' you mean, the integrals of $f(x)^m$ and $(f(x)/\|x\|)^m$ are obviously different. Mar23 comment positive sectional curvature of submanifold in $R^n$?@Ralph, yes, probably you read the question better than I did. I reacted to the phrase 'best lower bound', which I read as 'the minimum sectional curvature at a point'. Mar23 awarded ● Enlightened Mar23 awarded ● Nice Answer Mar23 comment Diameter estimate of distance sphere of positive curved manifoldThe sphere may be disconnected even for $r>\pi/2$, isn't it? Mar23 comment positive sectional curvature of submanifold in $R^n$?No, the spheres estimate principal curvatures. In $\mathbb R^3$ the sectional curvature equals Gauss curvature, which is a single number at a given point, but there are two principal curvatures which correspond to the spheres. Mar23 comment Average weighted value of a linear functional over increasing bounded subsets of Z^n@Mike: Sorry, I misread the denominator in your formula. In fact, $\mu^m(f)$ is the normalized $L^m$ norm of the function $f(x)/\|x\|$ over the unit ball. In order to get $\hat\mu^m$, one has to replace the denomitator $\|e\|$ by $r$. To see why the sum approaches the integral, rescale the lattice by the $(1/r)$-homothety. Then the sum over the lattice becomes a Riemannian integral sum for the integral over the norm's unit ball. Since the function and the ball are nice, the integral sums converge to the intergal.