# The work of Alice Chang

Sun-Yung Alice Chang is one of the pre-eminent scholars of modern geometry, and I wanted to understand the nature of her work. Although I’ve read papers by her before, this article written by her students Matthew Gursky and Yi Wang serves as a solid introduction to her work.

Given below are my notes as I try to understand the article. I first attach a screenshot from the article, and then below that write my notes and questions and such. I often make speculations which are either verified or proved wrong in the best next screenshot. I learned a lot from writing these notes. These can hopefully be useful to anyone working on PDE theory or Riemannian geometry.

Note: This article is unlikely to be helpful to anyone trying to learn about the topic. It is perhaps more useful as a record for myself of how I try and understand Mathematics (or anything at all). One of the reasons that I’ve started blogging less frequently on research papers is that I now try and write detailed notes on every paper before I even think of blogging about it, and that often leaves me too little time to devote to my day job.

• Why is it important to bound the norm of the function by that of its derivative? Why can’t we just measure both separately without having to compare both of them? I don’t know. It seems that we want to make the deduction that $W^{1,p}\subset L^{p*}$, where $1/p+1/p*=1/n$. Why would we want to make such a deduction though? Imagine that you’re trying to find out about an object. You see what it looks like. You hear what it sounds like. You taste what it tastes like. You throw it around and see how it flies in the air. It is the accumulation of all of these infinite observations that tells you all that object is and can be. Seeing if a function belongs to a certain $W^{k,p}$ or an $L^q$ is like that. We want to study the membership (or non-membership) of functions in all of these infinite sets to get a sense of what that function is like.
• Things like Sobolev’s inequality tell you that if a ball flies quickly through the air, it must be aerodynamically shaped, etc. Measuring one property tells you another. It gives you embedding theorems.
• Why do we need to consider weak derivatives? What is wrong with regular derivatives? Well it turns out that the mathematics of proving that smooth or differentiable functions indeed solve PDEs does not exist. Smooth functions do not form a nice enough algebra for which functional analysis can prove the existence of solutions. However, if we enlarge the set a little bit, functional analysis can prove that an element within this set satisfies this PDE. But in order to be able to access all elements of this set, we will need to weaken the question itself; ie weaken the definition of the PDE. It should now incorporate weak derivatives. After we have found the solution, we can prove by some limiting methods that the solution is indeed smooth. Hence, we have found a solution for the original question.
• Was this the only method possible? I am not sure. Maybe a new branch of mathematics would have rendered all of this obsolete. However, if we want to use functional analysis, we have to use this method.
• What is happening here? It turns out that the extremal function is not just one function. It is a family of functions. Moreover, by changing the parameters $a,b$, we can get a function almost completely concentrated at a point. Hence, the Sobolev constant is independent of the domain.
• We choose a cutoff function, and then let it remain fixed. Now we change $a,b$ such that as soon as the domain is inside the domain of the cutoff function, the Sobolev constant is realized.
• Why is the Sobolev constant not attained? I thought that it is attained for extremal functions. Let us walk back a little bit. In the second image, we are only considering compactly supported functions. We are able to reach the same upper bound with the much smaller set of compactly supported functions. However, the extremal function defined above is not compact supported, and it doesn’t become compactly supported for any values of $a,b$ (even though there is pointwise convergence). Hence, the Sobolev constant is not attained.
• Let us walk back a little bit. In the second image, we are only considering compactly supported functions. We are able to reach the same upper bound with the much smaller set of compactly supported functions. However, the extremal function defined above is not compact supported, and it doesn’t become compactly supported for any values of $a,b$ (even though there is pointwise convergence). Hence, the Sobolev constant is not attained.
• The assumption in the formula is that $n\neq p$. However, when $n=p$ we see the inequality shown above. This looks like the $\log$ formulae we generally see in extremal cases.
• Why are we restricting ourselves only to compactly supported functions? Probably because we don’t have the concept of an extremal function here. Hence, we want to better understand this space.
• What does the optimal value of $\beta$ mean? I think they mean an explicit value, which depends only on the dimension and not the domain.
• The surprising thing about this theorem is that they were able to find an extremal function. But I thought extremals don’t exist…? But that’s only for $p\neq n$.
• Why should volume forms have anything to do with the metric? Why not just be completely coordinate dependent? I suppose we could have that. However, most manifolds don’t have a fixed coordinate system. Hence, we have a notion of integral that is invariant with respect to coordinate charts. That is why we include the metric in the definition of the volume form.
• I think that in the theorems given above, we were assuming that the dimension of $\Omega$ was the dimension of the space. Hence, when we have domains of lower dimensions, we need to modify the inequality, which is precisely what we have done on the sphere.
• Why should we care about these inequalities at all? I think we are making an observation about these functions. We are establishing a property of these functions.
• Why do we linearize differential operators? Probably because they are easier to solve, and help us figure out important properties of the actual operator. But the solutions of these linearized operators only give us a hint as to the nature of the actual solutions. A major fact in mathematics is that these linearized operators give us MUCH more information than we’d expect.
• Why must questions about geometry lead to PDE’s? Because the very basic notions of geometry, like curvature, are differential operators of the metric. Hence, differentiation is built into the very fabric of geometry.
• What is a variational approach? It is an approach in which you find the answer by minimizing some functional, because differentiation is easy! Hence, the hard part of this approach is finding the right functional.
• A conformal map of the sphere also changes the metric conformally. Why do we care about this? Because it turns out that these conformal maps generate an infinite family of extremal points for every extremal point that you can construct by yourself.
• What does it mean to say that the conformal group of the sphere is noncompact? Well it is a noncompact set in the space of all maps of the sphere to itself. But how does it matter? I don’t know. But it probably helps prove that some manifolds are not the sphere, or something.
• What does bubbling mean? Is there a bubble being formed somewhere? We are constructing a family of conformal maps on the sphere such that the conformal factors progressively accumulate at the north pole. Well….so what? $J_1$ is still minimized for all of these factors right? Yes, they are. However, this just shows us that $J_1$ doesn’t satisfy the Palais-Smale condition. But how does that matter? First of all, the round metric minimizes $J_1$. Hence, all of these metrics also minimize $J_1$. Who cares about the Palais-Smale condition? I think the point is that extremal points other than these would be difficult to find.
• Does bubbling happen only when we start with the round metric, and then construct the family described above? I think so. What is the point though? Why do we do it? Apparently it is the only thing that “messes up” the geometry of the sphere. We will probably have occasion to talk about it later.
• What does this theorem intend to say? I don’t think it is a generalization of the previous problem, because $\Delta_0 K\neq 0$. However, as long as there are at least two maximum points and all the saddle points are positive, there exists a metric that is conformal to the round metric such that its Gaussian curvature is given by this function.
• Note that we are not just concerned with any metric here. This metric here has to be conformal to the round metric. Hence, this is not like the Yamabe problem, where we start with a round metric and then try to get a conformal metric whose scalar curvature is constant.
• Is a saddle point also a critical point? Yes. Why do we want to find saddle points? Are those the points that satisfy the $\text{Gauss curvature}=K$ condition? I don’t understand why this has to be the case. How does $J_K$ come in? Well, we just insert the desired $K$ somewhere into the formula. Critical points will have exactly that Gaussian curvature. Why a saddle point though? Well maybe all of these critical points have to be saddle points. Critical point does not mean minima
• What is the conformal invariance of this problem? It just means that the pullback of the metric after a conformal change is also a critical point. Is this what conformal invariance means in general? Yes, conformal invariance means that a conformal change in the metric implies no change or very manageable change
• What does conformal invariance have to do with this? I think that through conformal invariance, we construct a saddle point of $J_K$ which is not even bounded in $W^{1,2}$. So what? Why is that a problem, as long as the $J_K$ value is finite? So remember that we are trying to find solutions of the original PDE, and hence the solution must lie inside $W^{1,2}$. Therefore, even though we may find a saddle point, it might not be the solution of the PDE. Therefore, we have not solved the problem we set out to solve. The key step in the proof is showing that at the saddle points where this might happen, $\Delta_0 K<0$. Hence, we don’t care about those solutions
• Why is the spectrum of the laplacian important? In some sense, it is the skeleton of the linear operator. If there is an eigenbasis, then we know exactly how the laplacian acts on the whole space. Note that this condition doesn’t come up naturally in Physics. We have equations like $\Delta \text{potential}=\text{charge}$ and $\Delta Q=\partial Q/\partial t$. The need for eigenvalues of the laplacian probably comes up somewhere else. Does it give us information about the metric or the manifold? It gives us (partial) information about the metric. Note that if the spectra of two linear operators (counter with multiplicities) are the same, then they are the same (provided they are of the same rank). However, this is not the same with metrics with the same spectra of the laplacian. Why did we have to choose the laplacian though? Why not a different linear transformation? That’s a good question. Maybe we can construct a linear transformation which satisfies this property? I’m not sure.
• What is the heat kernel?
• The heat kernel is not the solution to $\text{Heat PDE}=\delta$. It is the solution to $\text{Heat PDE}=0$, with $\text{temperature}=\delta(y)$. Hence, it is an actual solution to the heat equation, and not an attempt to form a general solution to $\text{Heat PDE}=f$
• Why do we need to form an equation over $\Bbb{R}^{d}\times \Bbb{R}^d\times \Bbb{R}_+$, instead of just $\Bbb{R}^d\times \Bbb{R}_+$? I don’t know. I would have just done the latter and said $\text{temperature}=\delta(a)$ at some $a\in \Bbb{R}^d$. I think that we want $y$ to be a variable because we want to be able to assess what happens when $y$ is varied. Hence, we might be able to solve a more general formulation of the problem.
• What does $\delta(x-y)$ mean? I interpret $y$ to be constant. However, if this were a function on $\Bbb{R}^{2d}$, this would be a function that is infinite on the diagonal. Is the integral of the $\delta$ function still $1$? I think so.
• I think that we are trying to form a solution for a generalized boundary condition, and not a generalized heat PDE.
• What does $\delta_x(y)$ mean? Well $\delta(x-y)$ is kind of symmetric. $\delta_x(y)$ makes it clear that $x$ is the independent variable and $y$ is fixed for the purposes of this computation.
• How is the spectrum of $\Delta$ related to the heat PDE? It turns out that the kernel of the heat PDE can be written in terms of the spectra and eigenbasis of the $\Delta$ operator. That is how the spectra is relevant to Physics. A lot of PDE theory is about reducing solving a differential equation problem to a Linear algebra problem. That is exactly what we’re trying to do here as well.
• What is the trace of the heat kernel? Assuming an orthonormal eigenbasis, maybe the trace is $\sum e^{-\lambda_n t}\phi^n(x)\phi_n(x)$? Yes, that would be the only way that this would make sense, as traces of functions don’t really exist.
• What did Brooks-Perry-Yang show? They showed that if all the isospectral metrics lay in the same conformal class, then this set would be compact. Alice showed that we don’t need to assume that all these metrics lie in the same conformal class. I think that the conformal class needs to contain a metric of negative curvature so that bubbling doesn’t occur (which destroys compactness)
• What does modulo conformal group mean? Well the action of the conformal group creates sequences that don’t have a limit inside the space. Thereby compactness is lost. But if all conformally equivalent metrics are considered to be one equivalence class, then the set is compact.
• Why do we define the laplacian to be $-\Delta$? Probably because $\Delta u +\lambda u=0$ implies $\text{laplacian }u=\lambda u$. We want those $\lambda$‘s to be our eigenvalues.
• Why do we have the $\zeta$ function for eigenvalues? Well we have them for $\{1,2,3,\dots\}$. Why not for eigenvalues? But that doesn’t really answer the question as to why this could be important. I think it’s possible that $\zeta(g_1)=\zeta(g_2)$ tells us something about the similarities between the metrics involves, although it is weaker than being isospectral.
• Why the $n/2$ part? A function is defined if we have finite values. An analytic function is defined if it’s equal to its Taylor series. We see that if $Re(s)>n/2$, we have convergence, as $\sum 1/n^k<\infty$ for $k>1$. The complex exponents don’t matter for convergence, which is all we care about right now. Why analytic though? How do we get powers of $s$ in the expansion? I don’t know. Maybe there is an extended proof for this.
• How do meromorphic functions make sense? They’re infinite! Well, they give us information on the nature of the poles, and that can help us determine important properties of the function in a neighborhood of the poles. Meromorphic functions are also important in Physics. Consider the electrostatic potential at the position of a charge. We still want to be able to study this potential, although it clearly has poles.
• The determinant of the laplacian is essentially the product of all eigenvalues. Why would this be an important property? Again, this is a weakening of the notion of isospectral matrices.
• What does a global invariant mean in this context? It doesn’t need to be measured at a point. “Niceness” is a local property of humans, because it needs to be measured within each human. The population of humans is not a local property. It’s a global property. In what sense is it an invariant? I’m not sure
• What does equation (11) say? It says that there’s an upper bound to the fraction on the right, if you start with the round metric on the circle, and that it is $0$ if the conformal metric is a pullback of a conformal map! Hence, not all conformal metrics are pullbacks of conformal maps. Therefore, because we have a $\log$ on the left, we have invariance under action of the conformal group of the determinant as long as we start with the round metric.
• The round metric only maximizes the determinant as long as we’re in the same conformal class.
• How does formula (10) tell you anything about compactness? Maybe the boundedness of the determinant tells us about the compactness of the metrics themselves? That’s possible. But wouldn’t isospectral metrics have the same determinant as well? Yes. I don’t understand how the proof works.
• What are the changes we make to the laplacian to make it conformally invariant? Well, we change the functional space we act on completely, and we also add a scalar curvature term. The good thing about this is what it matches the good old laplacian on Euclidean space, and hence it may be thought of as a “correct generalization” of the laplacian. As long as things “are the same” on Euclidean space, we’re fine.
• Why does the laplacian change the weights of the weight classes that it acts on? Doesn’t a laplacian generally preserve the rank of the differential forms it acts on? Well in this case, taking $\tilde{g}^{ij}$ brings along with it the factor $\frac{1}{t^2}$. This causes the weight to change.
• Aren’t we “cheating” when we act on weight classes instead of considering $\hat{g}=e^{2w}g$? Well, it’s like studying a line bundle on $\Bbb{R}$ instead of studying the space on all smooth functions on $\Bbb{R}$. We are saying that we can study $f:C^\infty (M)\to C^\infty (M)$ by just studying $\text{line bundle}\to \text{line bundle}$ because of the extraordinary homogeneity restrictions on the former map. Is that what is happening here? I think it is, because we are only consider the “extraordinarily homogeneous maps”- the conformally invariant maps.
• Why are Paneitz operators important? Because they are conformally invariant. But so what? There are several conformally invariant operators! Maybe this is a new operator that cannot be generated by previously known conformally invariant operators, that arose in a specific situation. Moreover, it seems to be the natural generalization of the laplacian to four dimensions. Wait. Why only four dimensions? Moreover, isn’t the correct generalization the conformal laplacian? Well the conformal laplacian is a generalization. We can include other terms as well that become $0$ in the Euclidean case, and get a valid generalization (of course, we need to have conformal invariance as well). If $X_{\hat{g}}=e^{nw}X_g$, is $X$ the correct generalization of the laplacian in $n$ dimensions? Shouldn’t it act on weighted classes? Shouldn’t it change the weight by $2$? Moreover, shouldn’t we look for a generalization that works in all dimensions (like the conformal laplacian)?
• Well first we generalize the notion of Gaussian curvature to $Q$, and then say that the Paneitz operator is the correct generalization of the laplacian. So a generalization of the laplacian doesn’t need to be equal to the usual laplcian in Euclidean space. It just needs to satisfy the same kind of PDE as the laplacian. Of course, equation (15) is only true in four dimensions.
• What is the $Q$ curvature? It is a geometric quantity whose integral is a topological invariant in four dimensions. Is this the only geometric quantity with this property? It is possible that it is not. However, it comes up naturally in the generalization of Gauss-Bonnet, which is called Chern-Gauss-Bonnet. But was this really needed? What does this tell us about the manifold? I think it mostly tells us what is not possible, given the topology of the manifold, because we can have really wild metrics on the manifold. Of course all of this is independent of the embedding.
• Both $K$ and $Q$ are completely determined by the metric.
• What is the point of this generalization though? It essentially tells us what a $4$-dimensional manifold given a certain topology can “look like”. It restricts the class of metrics, and hence the ways that this manifold can be embedded inside Euclidean space isometrically. Generally, topology tells us very little about what a manifold can look like. This takes us towards that goal, but eliminating what a manifold cannot look like.
• Why is scale invariance important? Well it’s sort of like conformal invariance, but weaker. Hence, when trying to construct conformally invariant quantities, we may mod out by scaling stuff, and then tweak the right things to get conformal invariance. I think of it as a stepping stone.
• What is happening here? It is interesting that for all kinds of operators $A_g$, $F_A$ can be represented as a sum of three functionals with different coefficients. We sort of have a vector space with a basis thing going on here.
• How do we associate a Lagrangian to a geometric problem? Well the functional that we try to find the extremal point of is generally the lagrangian.
• They’re not saying that $I$ is the Euler-Lagrange equation given on the side. They’re just discussing the nature of critical points and the corresponding Euler-Lagrange equation on the right. Hence, if you want to make $|W_g|^2, Q_g$ or $\Delta_g R_g$ constant, you’ll have to extremize the functionals I, II or III
• What is the theorem that they proved? They proved that the round metric extremizes every normalized functional of the form $F_A$ on $S^4$. This is a pretty strong result. I think that we don’t have the tools to study $S^{2n}$ in general.
• So we are specifically considering $A_g=L_g$
• How is the dimension being $4$ relevant here? I think we are using $Q$ curvature and Paneitz operator here. Note that II has $Q_g=c$ its its Euler-Lagrange equation. It is possible that we extremize all three functionals I, II and III, and we get the Paneitz operator from II.
• I think that $R$ affects the definition of $L$
• How might the proof go? The positive scalar curvature might make $L$ “nice”, leading to a relatively simple fourth order PDE that we’re able to solve. The $\chi\leq 0$ probably makes the PDE even easier to solve. Hence, we’re affecting both the metric and $L$ to make things “easier”. Why do we need to be four dimensions though? Well we have to prove that II can be extremized. It’s probably difficult to do in other dimensions, but gives us a solvable 4th order PDE in four dimensions.
• I find it difficult to understand theorems like these. How are these conditions intuitive? How do you know how many to apply and when to stop? Are these the weakest conditions that can be imposed? Let’s take an analogy. We know that compact sets are closed. OK. But is this the weakest condition possible? No. There are lots of non-compact sets that are closed. However, this is still useful information. There are probably some important metrics that satisfy the conditions given above. And it is good to know that this large class of metrics satisfy this property. We don’t always want to construct the largest class of objects that satisfy a given property. We mostly want to understand individual objects, or classes of objects. And sometimes specifying this awkward class of objects takes awkward wording and seemingly arbitrary conditions, like those stated above. These conditions may in fact encode very easy to see geometric properties that we are missing. Much like the formal definition of compactness hides the intuitive notion of “closed and bounded”. But how can we be sure that we are including all “nice” metrics that we want to study? I think that we are definitely including a large subclass. $R>0$ and $\chi\leq 0$ aren’t that hard to visualize.
• How do we have a parallel for surfaces with high genus and four dimensional manifolds with $\chi\leq 0$? Well a high genus implies $\chi\leq 0$. The only difference is that we can discard the $R>0$ assumption in surfaces. I think the article suggests that we may be able to remove that assumption in four dimensional manifolds as well.
• When you minimize II, you get $Q_g=c$. How do we get $c=0$? It has something to do with $P_g$ having a trivial kernel. Maybe $c$ is the dimension about the kernel, assuming $P_g\geq 0$? I don’t know.
• Note that all our minimization and extremization is occuring in the same confomal class.
• Why couldn’t we do this for the round metric? Well we know that the round metric maximizes everything in its conformal class, and also has constant $Q$ curvature. Alice has studied this for all other conformal classes.
• We no longer start with a functional that is the $\log$ of determinants. We play around with coefficients $\gamma_i$ such that we get an “easy” Euler-Lagrange equation.
• If we find critical points of $F_A$ with positive scalar curvature, then the Ricci curvature will also be positive. So? Well we want to construct metrics with positive Ricci curvature, and this is hard in general. However, playing around with values of $\gamma_2,\gamma_3$ so that we get the “right” second order PDE is easy. And that is what we do here
• What does it mean to say that the conformal change of the curvature tensor is determined by the Schouten tensor? Well if the Schouten tensor is conformally invariant (which it is not), the curvature tensor will also be conformally invariant. However, if the Schouten tensor has a different transformation law, that will determine the transformation law of the curvature tensor.
• Keeping $\text{volume}=1$ is equivalent to normalizing the relevant functionals so as to have scale invariance.
• Why are $\sigma_k$ curvatures important? They are an attempt to generalize the $\int R$ functional. We want to minimize this functional, so as to get $\sigma_k=c$ manifolds.
• Why don’t we want to attain manifolds with $K=c$ or $Q_g=c$? Why $\sigma_k=c$? I think we want to solve both problems. Of course the former is dependent on the embedding, while the latter isn’t.
• Note the following interesting fact: $\int K$ when $n=2$ and $\int Q$ when $n=4$ are topological invariants. However, we can solve PDE’s so as to find $K=c$ or $Q=c$ solutions. For $\int \sigma_1$ when $n=2$ and $\int \sigma_2$ when $n=4$ on the other hand, we have conformal invariance (at least for $n=4$) instead of topological invariance. There are no PDE’s here, and obviously no minimization. But we should still be able to somehow make $\sigma_2=c$ for $n=4$ right, even though the integral doesn’t change? I’m not sure.
• It seems that $Q_g$ and $\sigma_2$ are related. Are $K$ and $R$ related in the same way? I’m not sure. Maybe they are in two dimensions
• I don’t understand why the conformal invariance of $\int \sigma_2$ precludes the possibility that $\sigma_2=c$
• Note that $K,Q$ are found to be constant only in dimensions $2,4$ respectively. On the other hand, $\sigma_1,\sigma_2$ are found to be constant in all dimensions except $2,4$
• We assume local conformal flatness. Do we have the same assumption in the Yamabe problem? No. I suppose it’s needed for higher $k$. Yes, that’s exactly what is stated.
• What is happening with $\sigma_1$? I think that we can find it to be constant in all dimensions. Hence, the problem starts only with $k\geq 2$. Therefore, the analogy with $K,Q$ isn’t exactly clear. $\int R$ may not even be a conformal invariant. In fact it clearly isn’t. We minimize it to find the solution of $R=c$
• So I was right. In $4$ dimensions, we can still have $\sigma_2=c$. Of course, the assumption is that $\sigma_1>0$. I don’t think we had to assume that $K>0$ to prove $Q=c$.
• What does it mean to say that the conditions are conformally invariant? Is $R>0$ a conformally invariant condition? I’m not sure.
• What are quermassintegral inequalities? They’re easier to describe in convex geometry, and compare volumes of the form $\int \sigma_k(L)d\mu$, where $L$ is the second fundamental form. These inequalities are not expected to hold in nonconvex domains. However, Alice and her collaborators could prove such an inequality for $(k+1)$-convex domains.
• But $(k+1)$-convex domains are also convex right? Where does the non-convex part come in? I’m not sure of this point. I think the implication is that the non-convex domains that lie inside this $(k+1)$-convex domain satisfy the inequality. Again, I’m not sure.
• Why is this space called conformally compact Einstein? Well it is conformally compact because it is conformally equivalent to an almost compact manifold, which is the disc. Note that the Poincare disc is not the same as the Euclidean disc. It is the Euclidean disc that is almost compact. The Poincare disc has to have its metric conformally changed to make it the Euclidean disc. It is conformally Einstein because $\text{Ric}(g_+)=-ng_+$. It is actually Einstein, and no conformal factors are needed to make it Einstein
• I think this is where the Einstein part of Poincare-Einstein comes in. $g_+$ doesn’t even have to be in normal form for the Einstein condition to hold.
• I think $g_+$ is always fixed. Moreover, the conformal class $[g]$ on the boundary is also fixed.
• Why do we need a conformal class on the boundary at all? We want to construct quantities that are conformally invariant, so that we may use them in Physical laws.
• I think that the volume is in terms of the metric $g_+$, and not $\rho^2 g_+$
• What does $o(1)$ mean? It refers to a quantity $a(n)$ which satisfies $\lim\limits_{n\to\infty} a(n)=0$. Why do we need to use this notation? Why should we compare functions to powers of $n$? Moreover, why should we care about behaviour as $n\to\infty$? Computational costs depend only on $n\to\infty$. But why? This is because $n$ often does get very large. And we want a good feeling for the computational costs involved. $O(n^3)$ might be manageable, but $O(n^{3.1})$ might become completely impossible. We could also have had things like $O(2^n)$ right? Well those problems are completely intractable, with impossible computational costs. We want to restrain ourselves to polynomials and monomials. I still don’t understand why knowing something is $O(n^3)$ instead of $O(n^4)$ would help. Well, suppose we want a result for $n=100$ in finite time. The former will take $10^6$ seconds, which is approximately 9 days, while the latter will take $900$ days, which is about three years. Hence, if we are already aware of the value of $n$ needed in our computation, and it is high, then knowing the exponent of $n$ is pivotal.
• The coefficients are not conformal invariants, and depend upon curvature terms of the boundary metric. Hence, a defining function has already been defined.
• I’m always surprised by inequalities leading to homeomorphism or diffeomorphism theorems. How do they work? Can’t we make the volume arbitrarily small? Not when we have a blowup near the boundary.
• How is this volume renormalized? We are not talking about the whole volume expansion. We are only talking about the conformally invariant constant term $V$
• What was the Anderson formula? It relates $V$ to $\chi$, although it need to say $V>\frac{1}{2}\frac{4\pi^2}{3}\chi$ (which implies homeomorphism to $B^4$)
• It turns that this is true only when $n=3$. There is a generalization of this to higher dimensions, which involves a conformally invariant integral.
• Have I constructed a fractional power laplacian in my paper? I have proven its existence. However, I haven’t factorized a fractional power laplacian. I have constructed GJMS operators anyway whole symbols would be fractional power laplacians.
• Is the Paneitz operator a GJMS operator for $k=2$? Yes. That is correct.
• How can Fourier analysis be used to construct fractional power laplacians? Well differential operators are analogous to multiplication by a variable. Hence, fractional differential operators are probably multiplication with fractional powers of polynomials, which are not that bad. We can then take the inverse Fourier transform.
• What is an elliptic extension problem? I think it is just an elliptic PDE with a certain boundary condition.
• How can the Cafarelli-Silvestre construction be interpreted as acting on the hyperbolic half space? Maybe we modify the elliptic PDE on Euclidean space to become the scattering operator on hyperbolic PDE? Yes, that is possible.
• How does making the space a hyperbolic half plane give us the extended interval $\gamma\in (0,n/2)$? This is because of the usual ambient restrictions. But what if $n$ is odd, which I think is the assumption here? I’m not sure. However, what I learned is that $w=-\frac{n}{2}+k$ is always negative. As $f\in \xi[w]\implies \tilde{f}=t^{-w/2}f$, this implies that $t$ will be raised to positive powers.