Let $M$ be a $n$-dimensional Riemannian Manifold, fix $p\in M$, and $1<k<n$. Do we know if the following is true?

For any $k$-dimensional subspace $V$ of $T_p M$, there exists a minimal submanifold through $p$ and tangential to $V$ at $p$.

Or even better, do we have this?

There exists a diffeomorphism map $F$ from a ball $B\subset T_p M$ centered at 0 to $F(B)\subset M$ such that $F(0)=p$, and for any $k$-dimensional subspace $V$ of $T_p M$, $F(V\cap B)$ is an minimal submanifold.

Of course, if $k=1$, the exponential map will do.

Any comments or references are appreciated.

  • $\begingroup$ Not sure if this is relevant, but even when they exist, minimal surfaces are not necessarily unique (just think in the case of surfaces in $\mathbb{R}^3$; two distinct minimal surfaces can be tangent at one point). In terms of your motivation of getting a generalization of the exponential map, isn't this a problem? $\endgroup$ – Willie Wong Jun 22 '18 at 2:16
  • $\begingroup$ Oh I have considered this point. But I want to first check if this easier condidition is known. $\endgroup$ – A. Chu Jun 22 '18 at 2:24

Here's a sketch of an argument when $k=n-1$. It's possible it can be generalized to higher co-dimension, but I haven't thought about that.

Fix the open unit ball $B\subset \mathbb{R}^n$ and consider a smooth family of Riemannian metrics $g_t$, for $t\in [0,1]$ and with $g_0$ the standard euclidean metric.

Now for any unit vector $\mathbf{v}$ in $\mathbb{R}^n$ (i.e., so $\mathbf{v}\in \mathbb{S}^{n-1}$). There is a unique (oriented) minimal foliation $\mathcal{F}_{\mathbf{v}}(0)$ of $B$ where the leaves are minimal with respect to $g_0$ and the leaf $L_{\mathbf{v}}(0)$ of $\mathcal{F}_{\mathbf{v}}(0)$ that goes through $0$ is normal to $\mathbf{v}$ at $0$ (this orients the foliation). Indeed, just consider the intersection with $B$ of the foliation of $\mathbb{R}^n$ by planes normal to $\mathbf{v}$ and oriented by $\mathbf{v}$. Let $\mathcal{G}_{\mathbf{v}}$ be the foliation of $\partial B\backslash \{\mathbf{v},-\mathbf{v}\}$ given by intersecting the foliation of planes with $\partial B\backslash \{\mathbf{v},-\mathbf{v}\}$. That is for each $L\in \mathcal{F}_{\mathbf{v}}(0)$ one has $\partial L\in \mathcal{G}_{\mathbf{v}}$ and every leaf of $\mathcal{G}_{\mathbf{v}}$ arises this way. Write $\partial \mathcal{F}_{\mathbf{v}}(0)=\mathcal{G}_{\mathbf{v}}$.

One should be able to show (by an appropriate use of the inverse function theorem) that there is an $1>\epsilon>0$ small so that for each $t\in [0, \epsilon)$ there is a minimal folition of $B$, $\mathcal{F}_{\mathbf{v}}(t)$, with leaves minimal with respect to $g_t$ and with $\partial \mathcal{F}_{\mathbf{v}}(t)=\mathcal{G}_{\mathbf{v}}$. Furthermore one should have $\mathcal{F}_{\mathbf{v}}(t)$ continuous in $t$ (appropriately understood). and also continous in $\mathbf{v}$. One should be able to show that the constant $\epsilon$ depends only on $h=\frac{d}{dt}|_{t=0} g_t$.

Now for each $t$ let $L_{\mathbf{v}}(t)$ be the leaf of $\mathcal{F}_{\mathbf{v}}(t)$ that goes through $0$. Define a map $\phi_t:\mathbb{S}^{n-1}\to \mathbb{S}^{n-1}$ by $\phi_t(\mathbf{v})=\mathbf{w}$ where $\mathbf{w}$ is the unit normal to $L_{\mathbf{v}}(t)$ at $0$. This map is continouus and is also continuous in $t$. Moreover, by construction $\phi_0$ is the identity map and so has degree 1. Hence, $\phi_t$ has degree 1 for each $t\in [0, \epsilon)$ and so $\phi_t$ is surjective for such $t$. That is, for $g_t$ one can find a minimal surface through the origin with prescribed normal by choosing an appropriate $\mathbf{v}$.

Thus for any small perturbation of eucldiean space one can solve your problem. However, on any Riemannian manifold one can scale the metric on a small geodesic ball to get that the metric in normal coordinates is close to the euclidean one.

  • $\begingroup$ Just to be clear, you have addressed my first question, right? Also on paragraph 4 line 3, do you mean $\partial \mathcal{F}_{\mathbf{v}}(t)=\mathcal{G}_{\mathbf{v}}$? $\endgroup$ – A. Chu Jun 22 '18 at 17:09
  • $\begingroup$ I think for higher codimension n-k, we can just use n-k dimensional foliation. And we can consider the degree of the map $\phi_t$ between the Grassmanian manifold Gr(n,n-k). $\endgroup$ – A. Chu Jun 22 '18 at 17:35
  • $\begingroup$ @ᴊᴀsᴏɴ Yes, I am only addressing the first question, the second seems much harder. I have also fixed the typo. Regarding higher co-dimension, it might work as you say, they only part I'm unsure about is whether the perturbations are still foliations (it's easier to see this in the codimension one case as you can use the maximum principle). $\endgroup$ – Rbega Jun 22 '18 at 20:16
  • $\begingroup$ I don't quite see why we need the maximum principle here. Fixing a $\mathbf{v}$, let $f(t,x)$ as a function from $[0,1]\times B$ to $B$ such that $f(0, \cdot)$ is the identity and for each $t$, (1) $f(t, \cdot)=f(0, \cdot)$ on $\partial B$, and (2) for each slice $C\subset B$ (having its first coordinate being constant), $f(C)$ is a minimal submanifold. Since $\det \partial f(t,x)/\partial x\ne 0$ at $x=0$, and “everything is continuous” (I hope I am correct about this), we have $\det \partial f(t,x)/\partial x\ne 0$ for small $t$. $\endgroup$ – A. Chu Jun 23 '18 at 4:01
  • $\begingroup$ Hence by the inverse function theorem, $f(t,x)$ is a diffeomorphism for small $t$, so we get a foliation. $\endgroup$ – A. Chu Jun 23 '18 at 4:03

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