## dg.differential geometry – Does higher integrability of Jacobians hold between manifolds when the Jacobians are concentrated?

$$newcommand{M}{mathcal{M}}$$
$$newcommand{N}{mathcal{N}}$$

Let $$M,N$$ be two-dimensional smooth, compact, connected, oriented Riemannian manifolds.

Let $$f_n rightharpoonup f$$ in $$W^{1,2}(M,N)$$ with $$Jf_n > 0$$ a.e., and suppose that the volume $$V({x in M , | , Jf_n le r}) to 0$$ when $$n to infty$$, for some $$0. Is it true that $$Jf_n rightharpoonup Jf$$ in $$L^1(M)$$?

I am fine with assuming that $$f_n$$ are Lipschits and injective and that $$V(f_n(M)) to V(N)$$.

The “higher integrability property of determinants” implies that if $$M,N$$ are open Euclidean domains, then $$Jf_n rightharpoonup Jf$$ in $$L^1(K)$$ for any compact $$K subset subset M$$.

Without the assumption $$V(Jf_n le r) to 0$$, this clearly doesn’t hold, even when $$f_n$$ are conformal diffeomorphisms:

Take $$M=N=mathbb{S}^2$$. Let $$s: mathbb{S}^2 to mathbb{R}^2 cup {infty}$$ be the stereographic projection, and let $$g_k(x) = k x$$ for $$x in R^2$$ (and $$g_n(infty) = infty$$.).

Set $$f_n = s^{-1} circ g_n circ s$$. $$f_k$$ are conformal, orientation preserving, smooth diffeomorphisms
and thus $$int_{mathbb{S}^2 }Jf_n=V(mathbb{S}^2 )$$. By conformality $$int_{mathbb{S}^2 } |Df_n|^2 =2int_{mathbb{S}^2 }Jf_n$$ is uniformly bounded, so $$f_n$$ is bounded in $$W^{1,2}$$, and converges to a constant function. (asymptotically we squeeze bigger and bigger parts of the sphere to a small region around the pole).

So, we do not have weak convergence of $$Jf_n$$ to $$Jf=0$$. (the
Jacobians converge as measures to a Dirac mass at the pole.) The question is if by adding the non-degeneracy constraint $$V(Jf_n le r) to 0$$ we recover this ‘Jacobian Rigidity’ under weak convergence.

*(In my case of application $$r=frac{1}{4}$$ but I don’t think it matters).

## dg.differential geometry – Do smooth maps with nowhere-maximal rank have small image?

I’m trying to better understand the concept of “maps with small image” as used by Lipyanskiy in his construction of “geometric homology” in https://arxiv.org/abs/1409.1121. Lipyanskiy utilizes manifolds with corners, but for the purposes of this question I think it suffices to stick to ordinary manifolds, which we assume to be second countable.

By definition, a smooth map of manifolds $$f: Wto M$$ has small image if there is another smooth map of manifolds $$g: Tto M$$ such that $$dim(T) and $$f(W)subset g(T)$$. I’m interested in alternative formulations of this condition. In particular, it seems reasonable to conjecture that this condition is equivalent to the map $$f$$ having less than full rank at all points.

In fact, I’m pretty sure that having small image implies that $$f$$ is nowhere of full rank: otherwise $$f$$ will be an immersion at some point and so the image will have dimension at least $$dim(W)$$. Then I believe the argument about Hausdorff dimension from this question about space filling curves implies that $$f(W)$$ cannot be covered by a smooth map with domain of smaller dimension: Proof that no differentiable space-filling curve exists

So my question really comes down to the converse: if $$f$$ nowhere achieves maximal rank, does it have small image?

I would also be interested in any other equivalent conditions to having small image.

Thanks!

## dg.differential geometry – Does the continuous mapping space between topological manifolds always admit a Banach manifold structure?

Let $$M$$ and $$N$$ be smooth, i.e. $$C^infty$$, manifolds. Suppose that $$M$$ is compact. Then for every $$k geq 0$$ it is well known that $$C^k(M,N)$$ admits the structure of a smooth Banach manifold. I am interested in the continuous case, i.e. $$k = 0$$. I think that then one can drop the regularity of $$M$$ and $$N$$ and still get a Banach manifold $$C(M,N)$$ as one constructs the charts via an exponential map. My question: Is $$C(M,N)$$ a Banach manifold also in the case where $$M$$ and $$N$$ are merely topological manifolds, i.e. $$C^0$$ manifolds? I mean if $$M$$ and $$N$$ admit a $$C^1$$ structure or equivalently a $$C^infty$$ structure, then the statement is obvious, but this is not always the case.

## dg.differential geometry – Hodge decomposition seems to say harmonic and exact is zero (and possibly (harmonic and co-exact) and (exact and co-exact))

Part of Hodge Decomposition Theorem says that for a compact oriented Riemannian (smooth) $$m$$−manifold $$(M,g)$$ (I think M need not be connected, but you may assume connected if need be or you want) and for a smooth $$k$$-form $$omega$$, i.e. $$omega in Omega^k(M)$$

A. $$omega$$ decomposes into exact, co-exact and harmonic: $$omega =$$ (respectively) $$omega_d oplus omega_delta oplus omega_Delta in Omega^k(M) = B^kM bigoplus mathscr H^k(M,g) bigoplus image(delta_{k+1})$$

B. $$omega$$ is zero if $$omega$$ is at least 2 of the ff: harmonic, exact, co-exact. Specifically:

• B.1. ‘harmonic and exact is zero’

• B.2. ‘harmonic and co-exact is zero’

• B.3. ‘exact and co-exact is zero’

C. I think (B.1) is equivalent to the injectivity of the map $$phi: mathscr H^k(M,g) := ker(Delta_k) to H^k_{dR}M := frac{Z^kM}{B^kM}$$, which maps a harmonic form to its de rham cohomology class, $$phi(omega)=(omega):=omega + B^kM$$ (or ‘$$omegabmod B^kM$$‘).

Proof: $$phi$$ is the restriction of $$Phi:Z^kM to H^k_{dR}M$$ (from closed $$k$$-forms) to harmonic $$k$$-forms. Then $$ker(phi) = ker(Phi) cap Domain(phi) = B^kM cap mathscr H^k(M,g) = image(d_{k-1}) cap ker(Delta_k)$$. QED

• C.1. Note on notation: I personally would like to have $$phi$$ and $$Phi$$ to have subscripts $$k$$ since their domains and ranges depend on $$k$$, but I’m omitting the subscript $$k$$ because I’m following notation in the following powerpoint.

1. What exactly is going on in slide 46? This is part of proving that $$phi$$ is injective.
• 1.1. What I think: Here, Ryan Vaughn seems to conclude, for the decomposition of harmonic $$omega = omega_d oplus omega_delta oplus omega_Delta$$, that $$omega_{Delta}$$ is zero because $$omega_{Delta}$$ is exact and because of (B.1). However, I really don’t think decomposition is necessary because the other components of $$omega$$ will obviously be zero.

• 1.2. I assume that the decomposition itself (I mean without the ‘orthonormal’ part) doesn’t rely on (B.1) (or (B.2)-(B.3)), otherwise I think this would be circular. I’m guessing that (B.1) (or (B.2)-(B.3)) is used in the ‘ortho(normal)’ part of the decomposition theorem. Thus, it makes sense to talk about $$omega = omega_d oplus omega_delta oplus omega_Delta$$ even if we don’t know (B.1) (or (B.2)-(B.3)).

1. How does Ryan Vaughn prove (B.1), if Ryan Vaughn did? Otherwise, how does one prove (B.1) (of course without assuming the full Hodge Decomposition Theorem; soooo of course proving (B.1) is part of proving Hodge Decomposition Theorem)?

What I tried so far:

2.1 – I think the following is somehow relevant, if true:

a smooth $$k$$-form is harmonic if and only if closed and co-closed, i.e. $$ker (Delta_k) = ker(d_k) cap ker(delta_k) = Z^kM cap ker(delta_k)$$

2.2 – If true: It seems ‘harmonic and exact’ is equivalent to ‘co-closed and exact’. Please explain how to show ‘co-closed and exact is zero’.

## dg.differential geometry – Intersection of self-shrinkers

I have a problem regarding a statement in the paper Smooth compactness of self-shrinkers by Colding and Minicozzi.

In the article, they define a surface $$Sigma$$ in $$mathbb R^3$$ to be a self-shrinker if its mean curvature $$H$$ and the outer normal $$n$$ satisfy
$$H=frac{langle x,nrangle}2,$$
where $$x$$ is the position vector. In its corollary, it asserts, “Every self-shrinker must intersect the closed ball bounded by the spherical self-shrinker, which follows from the maximum principle since the associated MCF’s both disappear at the same point in space and time $$(0,0).$$

I think the spherical self-shrink is the sphere centered at the origin with radius $$2,$$ which is a self-shrinker by definition. However, I don’t understand how to use the maximum principle to derive that it intersects with any other self-shrinkers. Does this follows by something like avoidance principle for MCF? (But I think intersecting at the singularity doesn’t violate that principle…)

Any ideas or comments are appreciated!

## dg.differential geometry – On the orbit of a Frechet Lie group action

Suppose that $$G$$ is a Fréchet Lie group acting on a Fréchet manifold $$X$$.
Fix $$xin X$$ and let $$alpha(t)$$ be a smooth path in $$X$$ such that
$$begin{cases} alpha(0)=x\ alpha(t)in Gcdot x end{cases}.$$
Also denote $$rho_{x}:Grightarrow X:gmapsto gcdot x$$. Is it true that $$alpha'(0)in text{Im}(d_{e}rho_{x})$$?

In the finite dimensional setting, this is clearly the case: the orbit $$Gcdot x$$ is a weakly embedded submanifold of $$X$$, hence $$alpha(t)$$ is also smooth as a curve in $$Gcdot x$$. Consequently $$alpha'(0)in T_{x}(Gcdot x)=text{Im}(d_{e}rho_{x}).$$

In the case that is of interest to me, $$G=Diff(M)$$ is the space of diffeomorphisms of a compact manifold $$M$$, and $$X$$ is the Fréchet space of rank $$k$$ – distributions $$Gamma(Gr_{k}(M))$$.

## dg.differential geometry – On the perturbation of one vertex of an n-simplex so that point the given point on one of its face gets into its interior

I have a closed cube $$Q_{0}=(0,1)^l$$ and half-plane $$H={(z_1,z_2,…,z_l) in mathbb{R}^l : z_1 + z_2 +…+ z_l > alpha }$$ in Euclidean space $$mathbb{R}^l$$ with $$0. Consider the open convex set

$$C={(z_1,z_2,…,z_l) in mathbb{R}^l : z_1 + z_2 +…+ z_l > alpha, 0 in $$mathbb{R}^l$$.

Also, suppose I have $$l+1$$ points, $$vec{z_1},vec{z_2},…,vec{z_l},vec{z_{l+1}}$$ so that each point is in the intersection of a different face of $$Q_0$$ with $$H$$ so that they do not lie on the same affine hyperplane (meaning these $$l+1$$ pants form $$l$$-simplex in $$mathbb{R}^l$$).

Now, consider point $$vec{z} in C$$ which also lie on the open line segment $$(vec{z_1}, vec{z_2})$$.

My question is to provide some argument which guarantees that I can perturb $$vec{z_1}$$ (or all $$vec{z_{i}}, i=1,2,…,l$$) on the face containing it to, say, $$vec{z_{1, epsilon}}$$ so that the given point $$vec{z}$$ now belong to the open convex hull of $${ vec{z_{1, epsilon}},vec{z_2},…,vec{z_l},vec{z_{l+1}} }$$.

## dg.differential geometry – question about the book “Holomorphic Morse Inequalities” by Marinescu-Ma

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## dg.differential geometry – What would be a good introductory reference for learning jet-bundle theory?

I am interested in learning the theory of Jet bundles, and am aware of the standard reference “The geometry of jet bundles” by D. J. Saunders. However this appears to be a detailed book, suitable for those who wish to specialise in this area. Can somebody recommend a relatively more introductory book (for a reader who knows the necessary differential geometric pre-requisites for learning this subject, but has never encountered Jet bundles) ? Thanks so much !

## dg.differential geometry – Extensions of minimal hypersurfaces

Let $$B subset mathbf{R}^{n+1}$$ be the unit ball, and $$M subset B$$ be a minimal hypersurface. By this we mean that $$M$$ is an embedded $$n$$-dimensional submanifold with vanishing mean curvature. We allow for the closure $$overline{M}$$ of $$M$$ in $$B$$ to not be embedded, write $$mathrm{sing} , M = overline{M} setminus M$$ and call this the singular set of $$M$$. However this is assumed to be small enough for $$M$$ to be stationary in $$B$$: compactly supported deformations $$X in C_c^1(B;mathbf{R}^{n+1})$$ do not change the area of $$M$$ up to first order. For example one might take $$n geq 2$$ and consider a surface $$M$$ embedded outside the origin with $$mathrm{sing} , M = { 0 }$$.

Question 1. Are there conditions that allow the extension of $$M$$ to a globally defined immersed minimal hypersurface in $$mathbf{R}^{n+1}$$? That is, when is there a minimal hypersurface $$tilde{M}$$ in $$mathbf{R}^{n+1}$$ (immersed away from a small singular set) with $$tilde{M} cap B = M$$?

Let me make some remarks summarising my own conclusions.

• The Cauchy–Kovalevskaya theorem could be relevant, but I am not sure whether this can be used to construct a globally defined extension. Moreover, one would have to worry about pieces coming together and meeting tangentially.
• This is not a purely PDE-theoretic question. If one considers the case where $$M$$ is the graph of a smooth function $$u$$ defined on $$mathbf{R}^n cap B$$—this function satisfies the (quasi-linear) minimal surface equation—then it is not hard to see that $$u$$ can in general not be extended to a globally defined function $$tilde{u}: mathbf{R}^n to mathbf{R}$$. The Bernstein theorem is one way to see this, but also simple examples can be constructed using a suitable portion of the catenoid.
• One can use the unique continuation property of minimal surfaces against the question, by taking $$M$$ to be a portion of a known surface. For example, by taking $$M$$ to be an embedded portion of an immersed minimal surface $$tilde{M}$$ one can see that one cannot hope for a globally defined and embedded minimal extension. Moreover, if one chooses $$M$$ to be portion of a singly-periodic Scherk surface one sees that $$tilde{M}$$ may have unbounded area growth: $$lim_{R to infty} mathcal{H}^n(tilde{M} cap B_R)/R^n = infty$$.

I am especially interested in the case where $$M$$ is one of the surfaces constructed by Caffarelli–Hardt–Simon. These are defined in $$B$$, embedded outside the origin, where they are prescribed to be tangent to a given minimal cone $$mathbf{C}$$.

Question 2. How does the answer change if $$M$$ is one of those surfaces? Is there $$tilde{M}$$ extending $$M$$, perhaps even with bounded area growth, that is with a constant $$C > 0$$ so that $$mathcal{H}^n(tilde{M} cap B_R) leq C R^n$$ for all radii $$R > 1$$?