Uncategorized | cozilikethinking

June 20, 2019

Day 2- Conference on Geometric Analysis

I am presenting my notes on today’s talks below. The talks were generally harder to follow than yesterday, not least because some of them were slide shows, and hence were prone to being thrown at the audience at brain-breaking speed.

Free Boundary Minimal Surfaces– The first talk today was by Lucas Ambrozio. Because it was a slide show talk, it sped by slightly fast, and I couldn’t take as many notes as I would have wanted. However, I will try to reproduce my impression of the talk.

Free Boundary Minimal Surfaces (we will call them FMBC’s from now on) are immersed hypersurfaces $(M,g)\hookrightarrow (N^{n+1},\overline{g})$ such that $\partial M\subset \partial N$ , and $M$ intersects $N$ orthogonally. Moreover, image $(M)\cap \partial N$ should not be contractible. As an example, one may imagine a circle along the boundary of a solid torus.

The first question of the day was whether out of all FMBC’s in a given manifold $N$ , there exists one that minimizes the $n-1$ dimensional volume.

As simple examples, we will consider FMBC’s in the unit ball $B_1^3(0)\subset R^3$ . One FMBC is any equatorial disc. Any other disc will not intersect the boundary orthogonally. Another example is the critical catenoid. Any catenoid is a minimal surface. However, only the critical catenoid intersects the boundary of the ball orthogonally. For any ball, there exists a unique critical catenoid.

The speaker then talked about the desingularization of intersecting minimal hypersurfaces. For instance, if we took the union of two intersecting equatorial discs in the unit ball, then we can smoothen out the area of the intersection by punching in some nice rounded holes (increasing the genus). We can do similar things for the intersection of the catenoid and an equatorial disc.

Let $\gamma$ be the genus and $r$ the number of boundary components of an FMBC. An open question is: Can we find an FMBC for any $(\gamma, r)$ ? This question seems like a hard, overarching question of the field.

We will talk briefly about Skeklov eigenvalues, which will play a role in the next section. Let $\phi$ be a function on an embedded hypersurface $\Sigma$ in $N$ . Let $\hat{\phi}$ be its harmonic extension beyond the hypersurface. Now consider the function $\partial\Sigma\to \frac{\partial \hat{\phi}}{\partial \nu}$ . We now construct a functional involving $\phi$ and $\frac{\partial \hat{\phi}}{\partial \nu}$ . The minimal value of this functional across all possible functions $\phi$ is the first Steklov eigenvalue.

The speaker answered some related questions to the open question discussed above. We can distinguish between various FMBCs of the unit ball by placing further constraints on the properties of the desired FMBCs. For example, if we require that the FMBC be totally geodesic, homeomorphic to a disc, of least possible area and of least possible Morse index (1), then the FMBC has to be the equatorial disc. Similarly, if we require that the FMBC be homeomorphic to an annulus along with either being immersed by Steklov eigenfunctions, being symmetric with respect to reflection across coordinate planes, or have a Morse index of $4$ , then such an FMBC has to be the critical catenoid. These theorems are important because all the FMBCs in the unit ball have not yet been classified.

The author turns this into a variational problems, given certain constraints. He proved the following: consider $M^2$ to be an FMBC in the unit ball in $R^3$ . If $\phi(x)=|x^{\perp}|^2A^2(x)\leq 2$ for all $x\in M$ , then $\phi$ can only take two values. Either $\phi=0$ , in which case the FMBC is an equatorial disc, or $\phi=2$ , in which case it is the critical catenoid. Note that here we’re talking about the nature of the immersion of $M^2$ inside $B_0^3(0)$ . $M^2$ might as well not be an FMBC at all. However, if $\phi(x)=|x^{\perp}|^2A^2(x)\leq 2$ , then it has to be an FMBC.

Ambrozio then talks about a quadratic form $Q(\phi,\phi)$ , whose exact form is not important for our current elementary discussion of the topic. Let index $(M)$ be the dimension largest subspace of $C^{\infty}$ where $Q(\phi,\phi)$ is negative definite (including $0$ of course). He then goes on to state the following result: if $\Omega\in R^{n+1}$ is smooth, and $n=2$ , then the FMBC $M$ satisfies the following condition: index( $M)\geq \frac{1}{3}(2\gamma+r-1)$ . If $n>2$ , there is a similar result with a more complicated right hand side.

He then goes on to state a result by Tran from 2016: an FBMS in the three dimensional unit ball that has index four must be star shaped with respect to the origin. Clearly, the critical catenoid satisfies these conditions. However, there might be other FMBC’s with index four too, as homeomorphism with annulus has not been assumed. However, if index $(M)\geq \frac{2}{3}(r-1)+3$ , then the critical catenoid are the only FMBC’s.

Ambrozio then goes on to state another theorem: let us take a three dimensional Riemannian manifold with non-negative Ricci curvature, and $\Omega$ a strictly convex domain in it. Then the set of FMBC’s in $\Omega$ with $\gamma,r$ bounded above by the same bound $C$ form a compact set. A corollary of this is the fact that the set of FMBC’s with Morse index bounded above by $C$ also forms a compact set. Somehow, Morse index captures information from both $\gamma$ and $r$ . The quadratic form used to define the Morse index probably somehow contains information about the boundary too.

We now generalize this to a manifold of dimension higher than three, which inevitably brings a constraint with it. Let $M^n$ be a Riemannian manifold with non-positive Ricci curvature, and $2\leq n\leq 6$ . Let $\Sigma$ be a compact, strictly convex and smooth domain of $M^n$ . Then the set of FMBC’s in $\Omega$ with both Morse index and area bounded above by $C$ , form a compact set.

Volume preserving stability of spherical space forms– The second lecture of the day was given by Celso Viana. The motivation for the talk was the isoperimetric problem: given the perimeter of a hypersurface (area $(\partial \Sigma)$ ), what is the maximum volume that it can enclose? This is a classical problem, and lots of advances in geometric analysis are made while attempting to solve this problem in various contexts. We shall assume that we are looking for solutions with constant mean curvature (or maybe the solution to the isoperimetric problem always happens to have constant mean curvature. I am not sure).

Let us consider an embedding $[X, (0,\epsilon)]\to M$ (essentially again reducing this to a homotopy-type argument). Note the change in notation: here $X$ is the hypersurface being immersed in $M$ . Then $\frac{d }{d t}|_{t=0} Area= -n\int Hfd_I$ , where $H$ is the mean curvature and $f=\langle \partial_t X,N\rangle$ . Note that $H$ is not assumed to be constant. One may ask that even area seems not to be constant here. That is true. However, the area given to us is achieved at $t=0$ . We are just looking for the rate of change of area at $t=0$ to study the properties of the volume enclosed by $X$ . If this rate of change satisfies certain properties, we will know that the volume enclosed by $X$ is maximal.

It turns out that $V'(0)=\int f$ . Hence, if $\int f=0$ , then the volume enclosed by $X$ is extremal. I would imagine that $f$ being identically $0$ would be a condition for the volume being extremal. They have weakened that condition, to just the integral being $0$ , here.

An interesting fact here is that $A''(0)\geq 0$ whenever volume is extremized. Hence, whenever volume is maximal, area is minimal, which would imply that if volume is held constant, this is the smallest perimeter that $X$ can have.

It is a famous theorem by Barbara Cormo(?)-Eschenburg that if $\Sigma$ is a stable constant mean curvature hypersurface in $R^n, S^n$ or $H^n$ , then $\Sigma$ has to be a geodesic sphere. By stability, I think they are referring to the stability of the area, as the second derivative being positive implies that the area there is a stable critical point.

In dimension 3, Rutone-Ros proved in 1992 that if $\Sigma$ is a stable constant mean curvature torus in $R^3/\Gamma, S^3/\Gamma$ or $H^3/\Gamma$ , then $\Sigma$ is flat. An example of this would be the Clifford torus embedded in $S^3$ . This example tells us that the Clifford torus is of maximal volume given the perimeter.

The speaker then goes on to talk about holomorphic forms on manifolds, but I did not understand the relevance of these topics to solving the isoperimetric problem. I will try and read his paper to find out more.

Wave Equations and Conformal Geometry– The third talk of the day was given by Sagun Chanillo. He talks about the functions $u$ on $R^2$ defined as $u:R\times R^2\to R^3$ satisfying the equation $-\partial_t^2u+\Delta u=2u_x\wedge u_y$ . As the solution is basically defined on two-dimensional space (the third coordinate being time), it looks like a two-dimensional space being embedded in $R^3$ .

Another function that we may consider is the following: $u: R\times S^2\to R$ satisfying the equation $\partial_t^2u-\Delta_g u=\alpha(\frac{e^{2u}}{\int_{S^2}r^{2u}}-1)$ . Here, $\alpha$ is a constant, and $g$ is the standard metric on the sphere.

We may want to impose extra conditions on the shape of $u$ , like making it the solution of an elliptic equation, and also a plateau, which would imply $|u_x|=|u_y|=1$ and $u_x\wedge u_y=0$ . As solutions to elliptic equations are constant over time, the wave equation becomes $\Delta u=u_x\wedge u_y$ . For the Liouville equation too, one might take the special case of $\alpha=1$ and $\int_{S^2} e^{2u}=1$ , which would make the equation $-\Delta u=e^{2u}$ . Interesting fact: If $u$ is indeed a solution to such a specialized Liouville equation, then $e^{2u}g$ is another metric on the sphere with constant curvature such that the area of the sphere is $4\pi$ .

Coming back to the general wave equation without plateau or elliptic assumptions on $u$ , we define the conserved energy for “nice solutions” to be $\int_{R^2} \frac{1}{2} (|\partial_t u|^2 + |\nabla u|^2)+\frac{2}{3}u.(u_x\wedge u_y)$ . The first term on the right is some kind of energy term, and the second term is the volume term.

When does a global solution of the wave PDE fail to exist? Let $u_0$ be the part of $u$ that depends only on time coordinates, and $u_1$ the part that depends only on spatial coordinates. Then if $E(u_0,u_1)< E(W,0)$ and $\|\nabla u_0\|_{L^2}\leq\|\nabla W\|_{L^2}$ , then $u$ blows up in finite time. The existence of a global solution implies that $u$ is finite at all spacial points for all times. It seems that $W$ too only depends on the time coordinate, although the definition of $W$ was not given before. Maybe $W$ encodes some information about the energy bound for a system (beyond which the solution might blow up), and hence it makes sense that it only depends on time. This condition suggests that if the rate of increase of the energy of the solution is greater than the rate of increase of the upper bound of the energy, then despite being lower than the bound at first, within finite time, the solution will blow up. This makes intuitive sense. If $E(u_0,u_1)>E(W,0)$ , then the solution would not have existed even at $t=0$ .

The speaker then goes on to talk about self-similar blow up: assume that there exists a solution $u(x,y,t)=v(\frac{x}{t},\frac{y}{t})$ for some $v\in C^2(R^2)$ . Then $v$ is constant in the unit disc. This basically means that at any given point, there comes a time $t_0$ such that for $t>t_0$ , $u$ does not change. Moreover, every point comes to have the same value for $u$ as $t\to\infty$ . This is a much stronger situation than $u$ not blowing up. From now on, we shall consider only those $u$ for which there exists such a $v$ . In other words, we shall only consider solutions that do not blow up in finite time.

We shall now discuss Bourgain spaces $H^{s,b}$ . All functions that satisfy a particular integrability condition of their Fourier transforms are part of $H^{s,b}$ . A function $u$ belongs to $\mathcal{H}^{s,b}$ if $u\in H^{a,b}$ and $\partial_t u\in H^{s-1,b}$ . The reason we want to use $\mathcal{H}^{s,b}$ is that estimating energy of $u$ becomes very easy using the $\mathcal{H}^{s,b}$ norm.

The speaker then goes on to discuss the cases in which $u$ does exist, and proof of existence involves using the concept of Bourgain spaces, and also using randomization to overcome the difficulties faced whilst using traditional methods to prove the existence of $u$ .

Min-max construction for Constant Mean Curvature hypersurfaces– Much like yesterday, the last talk of the day provided the most intuitive and easy to understand results. The talk was given by Xin Zhou.

By now, we have had multiple talks on the fact that finding a hypersurface enclosing the largest volume, given a perimeter, can be phrased as a variational problem. The speaker here actually phrased the question of finding a hypersurface (we are going to refer to such a hypersurface as $\Sigma$ from now on) with constant mean curvature (CMC) as a variational problem. One’s intuition might suggest that a surface with CMC would extremize some property, like area of boundary, or volume enclosed inside. We are going to test this intuition and see how far it takes us.

The speaker then gets into the specifics. If a disc $D^2\subset R^2$ is embedded into a hypersurface in $R^3$ by the map $\phi$ , then its image has CMC if $\Delta \phi=C\phi_x\wedge \phi_y$ and $|\phi_x^2|-|\phi_y|^2=0=\phi_x.\phi_y$ . This is the same as the plateau condition in the previous lecture, aisde form the fact that $|\phi_k|$ do not necessarily have to be $1$ . Moreover, the following term is also minimized: $E(\phi)=\frac{1}{2}\int_D |\nabla\phi|^2$ (Energy term) $+ \frac{2}{3}c\int\phi.(\phi_x\wedge \phi_y)$ (volume term). Hence, although the volume or energy might not be individually minimized (our intuition would suggest that both are minimized at the same time), their sum is indeed minimized for a CMC. Intuition not being exactly right, but being right in a weak form, is a feature of much of variational calculus.

Methods to find CMC’s are:

1. Solving the isoperimetric problem- it would appear that $\Sigma$ that encloses the maximum volume given a perimeter has CMC. This agrees with the simple fact that a ball encloses the largest volume of air when it is fully inflated (of course we are assuming that the surface of the ball cannot be stretched to increase the area).

2. Perturbation method- Start with a hypersurface, and keep on perturbing it at various points until you get something with CMC.

3. Glueing method- one may also glue together CMCs in a “nice way” to create larger CMCs.

An earth-shattering result of Zhou’s, from 2017, says the following: for $3\leq n+1\leq 7$ , for any given $c\in R$ m there exists a smooth, closed, almost embedded hypersurface $\Sigma\subset (M,g)$ such that $H_{\Sigma}\equiv c$ . An almost embedded hypersurface is one that may not be completely embedded, and may hence have self intersections. However, at the self intersection points, the hypersurface decomposes into two parts. Two spheres touching is an example, and the graph of $y^2=x^2(x+1)$ is not an example. Another condition is that the hypersurface is the boundary of an open set, and also has multiplicity $1$ .

Then the speaker talks about the method of getting such a hypersurface with CMC. $\Sigma$ has $H\equiv c\iff$ $\Sigma$ is a critical point of $A^c(\Omega)=$ Area $(\partial\Omega)-$ c vol $(\Omega)$ . Here again, our intuition is only correct in a weak form. We wold have expected both the area and volume to have been extremized at the same time for the existence of a CMC. However, it is only their difference that has to be extremized. Hang on. That’s not completely correct either.

$A^c$ doesn’t have to be extremized: the speaker said that $A^c$ can be varied with respect to two independent parameters (I don’t remember which). The saddle point of this variation is the one corresponding to the hypersurface with CMC; in other words we determine $L^c= \inf\max A^c$ . It is the hypersurface corresponding to this $L^c$ that gives us the CMC hypersurface.

Zhou then goes on to state an even more earth shattering result: given ANY function $h:M^n\to R$ , he can create a hypersurface $\Sigma^n$ in $M^{n+1}$ with the mean curvature being modeled by the function $h$ . This is a vast generalization from just the constant mean curvature case. He proves it by proving the following: $\Sigma^n\subset M^{n+1}$ has a prescribed mean curvature $\iff$ $\Sigma$ is a critical point of $A^n(\Omega)=$ Area $(\Omega)-\int_{\Omega} h$ dvol. He then probably proved the existence of this critical point by min-max considerations.

I will try to write notes for the talks tomorrow as well.

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June 20, 2019

Day 1 – Conference on Geometric Analysis

I am attending an NSF funded conference on Geometric Analysis in Princeton from 19 June to 22 June. The format is simple- 4 lectures everyday, interspersed with breaks, which see much fraternizing and sharing of ideas. I want to record my recollections of the four lectures that I attended today.

Antoine Song– Antoine is a graduating Princeton student who has been named Clay Fellow this year, and will be based at UC Berkeley come fall. One of the major reasons for the conferring of this great honor was his proof of Yau’s conjecture, which states that every 3-manifold has infinitely many minimal hypersurfaces. We are going go to through his talk briefly below.

Let us take a compact manifold $M^n$ , the superscript signifying the dimension of the manifold, and consider a minimal hypersurface $\Sigma$ in it. Let us assume that the dimension of the set of singularities of $\Sigma$ is $\leq n-7$ . Then there exists a sequence of hypersurfaces $\{\Sigma\}_i$ with unbounded index (for any $n\in\Bbb{N}$ , there exists a minimal hypersurface $\Sigma_k$ such that $index(\Sigma_k)\geq n$ ). The index here is the Morse index, which is the “number of directions in which one can go from a singularity to go down the manifold”. The speaker then explored the impact of unbounded genus on the intersection of hypersurfaces. Although the connection between index and genus seems unclear now, it will become clearer momentarily. As the genus of a sequence of minimal hypersurfaces increases to infinity, although each of the hypersurfaces in the sequence is smooth, the limit might actually be a non-smooth intersectin of hypersurfaces. The way this works is that although the genus (the number of holes) increases to infinity, we are assuming that the total measure of all the holes goes to $0$ .

Song then goes on to state that for $3\leq n\leq 7$ ,there are only a finite number of diffeomorphism classes of hypersurfaces with bounded are and index.

He then talks about the following result by Maximo: consider $\Sigma^2\subset M^3$ . Then if index is bounded ( $index\leq C)$ , then $genus\leq area$ . An obvious, and perhaps the more important implication, is that if area is also bounded, then the index being bounded makes the genus also bounded. This suggests an area being bounded $\implies$ $genus\leq constant\times index$ relation.

Note that for the rest of the talk after this, we assumed that the area of the hypersurfaces are bounded.

We have the following result: Let $2\leq n\leq 6$ . For areas of hypersurfaces being bounded, we have $\sum b^i(\Sigma)\leq C(A,g) (1+index(\Sigma))$ . Here $C(A,g)$ is a constant that depends only on the area and genus, and $\Sigma b^i(\Sigma)$ is the sum of all the Betti numbers of the hypersurface. This is exactly the result that we expected from Maximo’s result.

For $n>6$ , we have a similar result: $\mathcal{H(\Sigma)}^{n-7}\leq C_1(1+index(\Sigma))^{7/n}$ . This again is an area being bounded $\implies$ $genus\leq constant\times index$ relation.

He then takes the above two results to prove that $genus(\Sigma)\cong index(\Sigma)$ . I am not clear on how he proves the converse inequality: that $index\leq constant\times genus$ .

He then also speaks of a small generalization: that if area is not bounded, then $b_1(\Sigma)\leq constant\times index$ . $b_1$ is “part“ of the genus of the manifold, and hence this is a result of the same flavor as in the case with bounded area.

Song then goes on to describe in brief the method used to arrive upon the above result; the method is called Ros’ method. We have to find an intermediate term, say $H$ , such that $b_1(\Sigma)\leq H\leq index(\Sigma)$ . This $H$ turns out to be the set of harmonic one forms defined on the manifold. These harmonic forms help us construct a vector field (one example would be choosing the a vector field from the kernel of the harmonic form). These vector fields would obviously have to be chosen such that one goes “down” the manifold while traveling along any of these directions. How exactly this is done was not made clear.

The speaker then also goes on to talk about the proof of Geometric covering. For all $x\in \Sigma$ , consider $r=$ supremum of radii such that $B(x,2r)$ is stable (just touches a singularity). The folding number, which is the maximum number of such (disjoint) balls on the hypersurface, is less than or equal to $1+index(\Sigma)$ . Maybe I don’t understand this result well…I’d expect that $2 index/geq$ folding number. Still have to think about this.

The speaker then goes on to talk about Cech cohomology for a bit (considering we are talking about covering a space with balls, this is a natural generalization). However, I will not be going into this.

Pingfei Guan– Guan is a geometric analyst based at McGill University. He talked about whether immersions of manifolds into other manifolds of one higher dimension, under certain constraints, was possible, and what we could say about the regularity of those immersions. Consider $(M^n, g)\hookrightarrow{} (N^{n+1}, \overline{g})$ . Under the constraint that $K_g>0$ , what can we say about the regularity of the immersion? Inevitably, this becomes a question of homotopy.

Consider $(S^2, g_t)$ , where $g_t$ is a slightly deformed metric of $g$ (I am assuming that $g_0=g$ , but I could be wrong). For what $t$ can we embed $(S^2,g_t)$ smoothly into $R^{n+1}$ ? As smoothness is an open condition (slight deformations do not make the immersion non-smooth), this set of $t\in [0,1]$ is likely to be an open set.

Then speaker then talks about regularity estimates by Nirenberg: Step 1: Control the mean curvature (so that it is positive, or heeds whatever other constraint you place on the immersion). Step 2: Get the $C^{2,\alpha}$ estimate. This condition, I suppose, helps us check the regularity of the immersion beyond the second derivative.

Now consider $(M^n, g)\hookrightarrow (N^{n+1},\overline{g})$ . If $\{h_{ij}\}$ is the second fundamental form, then $\sigma_2(\{h_{ij}\})=\sum h_{ii}h_{jj}-h_{ij}^2$ is the symmetric polynomial of degree $2$ (although it clearly is not a fundamental symmetric polynomial as defined by Newton). We have the following result: $\sigma_2(\{h_{ij}\})=\frac{R_g-R_{\overline{g}}}{2}+Ric_{\overline{g}}(\nu,\nu)$ . Note that we do not know if the right hand side is positive, as the left hand side does not have to be positive. The Evans-Krylov Theorem states that if an immersion satisfying the given constraints on curvature exists, and is also $C^{2,\alpha}$ , then $\frac{R_g-R_{\overline{g}}}{2}+Ric_{\overline{g}}(\nu,\nu)\geq 0$ .

Although the speaker then goes on to talk about other things, including maixmizing the mean curvature $H$ amongst all embeddings, I will not discuss that here.

Matthew Gursky– This was a slide show talk, so I couldn’t talk as many notes in my notebook. He talks about using the Chern-Gauss-Bonnet formula for singular Yamabe metrics in dimension $3+1=4$ . A result of Nirenberg’s from 1974 is the following: Let $\Omega\subset R^{n+1}$ be a smooth bounded domain ( $n\geq 2$ ). Then there exists $\hat{g}=u^{-2}ds^2$ such that $R_{\hat{g}}=-n(n+1)$ . This essentially solves the Yamabe problem for subsets of Euclidean space. Aviles- McOwen then generalized this for non-Euclidean domains too: if we have a compact manifold with boundary, then the same kind of metric $\hat{g}$ exists with the same $R_{\hat{g}}$ .

In general, the existence of such a metric is not obvious at all. However, a variational reformulation of this problem can be helpful. We shall consider this shortly. But before that, we shall talk about renormalization, so that we can motivate the definition of the renormalization coefficient $V$ .

Consider $\Omega$ with the metric $u^{-1}ds^2$ . As this metric blows up near the boundary ( $u=0$ at the boundary; one might think of it as the distance function), the volume element $d vol_{\hat{g}}$ does too. Hence, so does the integral $\int{R_{\hat{g}} dvol_{\hat{g}}}$ , as the curvature is constant everywhere (as $\hat{g}$ is the solution to the Yamabe problem). To somehow scrape together an integral that does not diverge, we need to find a finite volume element. This can be found in the following way: let $V_{\epsilon}$ be the volume of all points that are at least $\epsilon$ away from the boundary of the manifold. Then $vol_{\hat{g}}=$ some terms with negative powers of $\epsilon$ (as volume decreases with an increase in $\epsilon +$ Covariant energy $\log \epsilon+V\epsilon+O(1)$ . The $V$ here is finite. It is this that we substitute for $dvol_{\hat{g}}$ .

Now back to the variational problem corresponding to finding the Yamabe metric: turns out that $8\pi\chi(M)\leq$ an expression involving a non-local tensor, a traceless tensor, and $V$ . Moreover, if we have equality, then the Yamabe metric does exist. The author proved that in a special case (where the boundary is umbilic, amongst other conditions), we indeed have equality, which proves that the Yamabe metric exists.

Robin Neumayer– This was perhaps my favorite talk of the evening, if only for the fact that it was easy to follow and well illustrated. Consider a hollow manifold, say a sphere, and a hypersurface on it, say a latitude $M$ on it. The latitude $M$ divides the sphere into two parts- say $S_1$ and $S_2$ . If $\inf \frac{|M|}{|S_1|,|S_2|}$ exists as we run through all hypersurfaces $M$ on the sphere, then this infimum is known as the Cheeger constant $h(S)$ . By $|X|$ , we mean the volume of $X$ (of the right dimension). There are loads of applications of the Cheeger constant, including in designing landscapes (if the landscape has a Cheeger constant above a threshold value, then it will not be prone to landslides).

Let the sets minimizing the Cheeger constant be called Cheeger sets. Do they exist, and are they unique? They do exist, but may not be unique. Examples of multiple Cheeger sets in bowties with narrow necks were given. The concept of a narrow neck will play an important part soon. Note that the union of Cheeger sets is also a Cheeger set. Hence, the concept of a maximal Cheeger set does exist.

Now we come to a special case: consider a convex set $\Omega\subset R^2$ . The convexity, and the fact that $\Omega$ is two-dimensional, are important assumptions here. In this case, a unique Cheeger set exists, and is given by $\Omega^r\oplus B(x,r)$ , where $x\in \Omega^r$ . Here $\Omega^r$ is the set of all points that are at least $r$ away from the boundary. Moreover, in the case of this Cheeger set, $r$ turns out to be $\frac{1}{h(\Omega)}$ . Note that $h(\Omega)$ is an independent quantity, and is not defined based on this construction. Moreover, the area of the Cheeger set turns out to be $\pi r^2$ .

This is a particularly simple Cheeger set. We’d like to know what other kinds of manifolds have unique Cheeger sets that can be constructed in the same way. Remember that we had assumed convexity for $\Omega$ . We want to see if we can somewhat weaken that assumption.

The following is the impressive result that the speaker obtained: consider a domain $X\subset R^2$ such that it has no necks of radius $r$ , which just means that a ball of radius $r$ can travel throughout the inside of the compact manifold without any stoppages are bottlenecks. If $X$ is also a Jordan domain and the boundary $\partial X$ has measure $0$ , then it has a Cheeger set $\Omega^2\oplus B(x,r)$ given by the above construction.

Another formulation of this fact is the following: if no necks of all $r$ such that $\frac{inr(\Omega)}{2}\leq r\leq \frac{1}{2}(\frac{|\Omega|}{A})^{1/2}$ exist, then $\Omega$ has a Cheeger set defined by the above construction. Note that the right hand side is just another way of writing $\frac{1}{h(\Omega)}$ , as the area of the Cheeger set has to be $\pi(\frac{1}{h(\Omega)})^2$ .

This is all for the talks today. I shall try to type up notes from other talks too.

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May 13, 2019

Effective Altruism- May

I turned *way too old* earlier this month. Hence, on my birthday month, I would like to record the donation I made to Effective Altruism:

Screen Shot 2019-05-13 at 5.54.44 PM

I have also started reading on poverty in India. The first paper that I perused (very) partially is this.

It is a paper written by two Indian PhD students at Columbia University, who talk about the fact that there are basically two poverty lines used by the government of India. The latest one is in fact a harsher scale of poverty, and according to both such lines poverty in India has been steadily decreasing, especially in the 2004-2005 and the 2009-2010 period. The authors do not use the 5-year studies on poverty as a basis for their conclusions, but the annual expenditure survey done by the government of India. The basis for their choice is the following: people who spend more are probably earning more, and vice-versa. Hence, whether people are above or below the poverty line can be easily approximated by how much they’re spending.

I didn’t complete reading the paper for the following reasons: it seems motivated at the very outset to show that India is “shining”, it is more an instance of statistical jugglery than a commentary on the causes of poverty, and bases its conclusions on poverty lines that I don’t take to be credible. Every government is motivated to suppress data on poverty, or introduce measures of poverty that suggests that there are less poor people in their country than there really are. And I find the two poverty lines to not be a good measure for poverty in India.

I then found this book written up by people at the World Bank.

I will try to peruse relevant sections of this book and complete this article by (hopefully) this weekend

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May 9, 2019

Putnam 1994, Question 1

Just want to record a solution to a relatively simple Putnam problem here. The problem is the following:

Show that a sequence $a_1,a_2,a_2,\dots$ satisfies the condition $0<a_n\leq a_{2n}+a_{2n+1}$ . Prove that $\sum a_i$ diverges.

I tried attacking it with the usual methods of showing that a sequence diverges, but nothing seemed to work. However, every good Olympiad worth its salt asks for some experimentation, which will then yield important observations/patterns. I did the same here.

Without loss of generality, let $a_1=1$ . Then $a_2+a_3\geq a_1=1$ . Hence, $a_2+a_3\geq 1$ . Similary, $a_4+a_5\geq a_2$ and $a_6+a_7\geq a_3$ . Therefore, $a_4+a_5+a_6+a_7\geq a_2+a_3\geq 1$ . Continuing this pattern, we observe that for all $n\in\Bbb{N}$ , $a_{2^n}+\dots+a_{2^{n+1}-1}\geq 1$ . Hence $\sum a_i$ , which can be thought of as $\sum\limits_{n=1}^{\infty}(a_{2^n}+\dots+a_{2^{n+1}-1})$ , diverges.

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April 16, 2019

Effective Altruism- April

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Recording my donation here for April. This was an eventful month, as I was part of a team that organized a massive fundraising event in State College. Minor glitches notwithstanding, it was an enriching experience.

This post will be updated soon with some views on child malnutrition in India.

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March 29, 2019

A cute combinatorial identity

Just wanted to record a cute combinatorial argument. Prove that $\sum\limits_{i=0}^n {n\choose i} {i\choose m}= {n\choose m} 2^{n-m}$

I read this formula in the article “Pascal functions” in “The American Mathematical Monthly”.

They prove it using the new machinery of Pascal functions that they develop in the article. I tried proving it using an algebraic argument, but couldn’t find the right formulation. I could finally prove it using a combinatorial argument, which I have explained below:

The left hand side counts the number of ways of first selecting $i$ items from $n$ items, and then selecting $m$ items from those already-selected $i$ items. Obviously, for $i<m$ , this does not make sense. Hence, ${i\choose m}=0$ in those cases. For $i\geq m$ , the above interpretation is valid.

We can interpret the right hand side as the number of ways of selecting $m$ items from $n$ items, and then going to each of the left-over items (there are $n-m$ of them) and deciding if they had been pre-selected or not. Here pre-selection implies being selected amongst the $i$ items, out of which the final $m$ items have been chose.

Hence, the right hand side equals the left hand side.

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March 15, 2019

Some solutions

I solved two math problems today. The solutions to both were uniquely disappointing.

The first problem was the first problem from IMO 1986:

Let $d$ be a number that is not $2,5$ or $13$ . Prove that out of the set $\{2,5,13,d\}$ , we can select two different numbers $a,b$ such that $ab-1$ is not a square.

A quick check would show that $2.5-1, 2.13-2$ and $5.13-1$ are all squares. This shows that of the two numbers that we select from $\{2,5,13,d\}$ , one of them has to be $d$ . We essentially need to prove that for $d\neq 2,5,13$ , $2d-1, 5d-1$ and $13d-1$ are not all squares (a maximum of two of them might be. But all three cannot).

I tried figuring out some of the implications of all of them being squares, but couldn’t come up with anything concrete, as I didn’t write anything at all, and just tried to devise a strategy mentally. Ultimately, I came up with possibly the most mundane proof of all time.

Proof: Consider squares $\mod 16$ (I couldn’t come up with any contradiction up until $\mod 8$ ). The possible residues for squares $\mod 16$ are $0,1,4,9$ . Now my strategy was to plug in all possible residues $\mod 16$ in place of $d$ , and checking that in every such instance, at least one of $2d-1, 5d-1$ and $16d-1$ gave a residue that was not $0,1,4,9\mod 16$ . This I could successfully verify. Hence proved.

The solution that I could find in the IMO compendium was much smarter, but involved some algebra that would obviously involve putting pen to paper.

The second question that I solved today was proving the Koszul formula, which states that

$2g(\nabla_X Y,Z)=D_Xg(Y,Z)+D_Yg(X,Z)-D_Zg(X,Y)-g([Y,X],Z)-g([X,Z],Y)+g([Z,Y],X)$

I’d seen proofs before in textbooks, that would involve writing down expansions, and then combining all of them in some clever way. The way that I came up with was pretty algorithmic, and perhaps even more elementary. It is simply the following: Take $g(\nabla_X Y,Z)$ . Use the torsion free condition to write it as $g(\nabla_Y X,Z)+g([X,Y],Z)$ . Now use the formula for connection acting on a tensor to write $g(\nabla_Y X,Z)$ as $D_Y g(X,Z)-g(X,\nabla_Y Z)$ . Performing the two operations, we have $D_Yg(X,Z)-g(X,\nabla_Y Z)+g([X,Y],Z)$ . Performing this operation of first using the torsion free condition and then the formula for a connection on a tensor three times, we get the Koszul formula.

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March 8, 2019

Effective Altruism – March

I’m recording the payment I made to Effective Altruism in the month of March, to keep my pledge of donating 10% of my lifetime income to it.

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I had said that I’ll soon write an article about causes that concern me/that I want to make monetary contributions to. Some articles that I read in the recent past are the wiki article on poverty in India, sexual injustice in rural India, and also Tolstoy’s views on charity and religion (which for sure have changed the world. For starters, Gandhi pretty much took all of his views and practices from Tolstoy, almost verbatim, which of course decided the course for Indian self-determination). However, I feel that I have not read enough to write anything original/meaningful. Hopefully this will change by next month.

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March 4, 2019

Here’s a slightly badly written proof to a competitive math problem. I guess I could expand it slightly if readers find it unreadable.

The following is a question from the International Mathematics Competition, 1994.

Prove that in any set $S$ of $2n-1$ different irrational numbers, there exist $n$ irrational numbers $\{s_1\dots,s_n\}$ such that for any $\{a_1,\dots,a_n\}\in \Bbb{Q}$ such that

(1) $a_i\geq 0,\forall i\in\{1,\dots,n\}$ and (2) $\sum\limits_k a_k>0$ ,
$\sum\limits_{i=1}^n a_is_i$ is also irrational.

Proof: We will prove this by induction. This is obviously true for $n=1$ . Now let us assume that this is also true for $n$ . We shall now prove the case for $n+1$ : that amongst any $2(n+1)-1$ different irrational numbers, there exist $n+1$ numbers such that the given condition is satisfied. We will prove this by contradiction. Let us call this $2(n+1)-1$ element set $A_{n+1}$ .
Remove any two elements from $A_{n+1}$ . We get $2n-1$ elements. By the inductive hypothesis, there exist $n$ elements $\{s_1,\dots,s_n\}$ such that for any $\{a_1,\dots,a_n\}$ non-negative rational numbers amongst which at least one is strictly positive, $\sum\limits_{i=1}^n a_is_i$ is irrational. Let us call this set of elements $S_n$ . Now one by one, add each element of $A_{n+1}\setminus S_n$ , to $S_n$ to form $n+1$ element sets. If each of these $n+1$ element sets can be put into some linear combination to give us a rational number, then we can subtract the right rational linear combination of $A_{n+1}\setminus S_n$ (which by our assumption should also be rational) to give us $S_n$ back, which should again be a rational number. But that is a contradiction. Hence, there has to exist an $n+1$ tuple of elements in $A_{n+1}$ too such that no linear combination with non-negative rational coefficients, such that at least one coefficient is positive, can give us a rational number.

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February 7, 2019

Effective Altruism

The reason that I’m writing this non-mathematical post here is that I want to help spread the word about this fantastic organization. I also heard about it in Vipul Naik’s blog.

I decided to start donating to Effective Altruism December last year. It is an organization that does research on the most effective ways to donate money, and then makes suggestions based on where one might want to donate. Typically, it asks its members to pledge 10% of their lifetime income to such causes.

I was initially introduced to Effective Altruism by some of my math heroes like Vipul Naik. Some causes that facilitated my joining this organization were that I now have a disposable income through a monthly stipend, not all of which I spend, and that I’ve gotten increasingly involved in fundraising at my university through various organizations, and find my involvement to be meaningful and fulfilling.

Below is the screenshot of my donation for this month. I shall soon update this blog with details of the specific organizations linked with EA that are doing amazing work, especially organizations that are fighting malaria/other preventable diseases in Africa and other parts of the world.

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