# 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.