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Let $M$ be a smooth $n$-dimensional manifold and $N\subset M$ be a closed embedded submanifold of codimension at least $2$. Furthermore, let $\mu$ be a volume form on $M$.

Question: Does there exists a vector field $X$ on an open neighborhood $U\subset M$ of $N$ (arbitrary small neighborhood of $N$, not fixed!) such that its divergence with respect to $\mu$ is $\equiv1$ (divergence is defined by $\mathcal{L}_{X}\mu = \text{div}(X)\mu$) and $X(p)=0$ for all $p\in N$?

I am not sure about the assumption on the codimension to be at least $2$. What is important for me is that the codimension is "big enough" (definitely it should not be $1$).

NEW QUESTION: Is the volume form $\mu$ on a sufficiently small neighborhood of $N$ exact, i.e. does there exists a $n-1$ form $\eta$ such that $d\eta = \mu$? And if so, is there any chance that this solves the original question? Under what conditions or further assumptions would this "NEW QUESTION" solve the original one?

Greetings, Stan

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    $\begingroup$ Is the open neighbourhood $U$ part of the data or are you asking whether there exists some $U$ where the vector field satisfies the conditions? $\endgroup$ May 13, 2020 at 8:48
  • $\begingroup$ actually the open set $U$ should be arbitrary small neighborhood on $N$. Hence not fixed. I will edit the question. Thanks for the advice. $\endgroup$
    – StanleyT
    May 13, 2020 at 8:55
  • $\begingroup$ Yes the volume form is exact. The de rham cohomology of a non compact manifold vanishes in top degree. $\endgroup$
    – Thomas Rot
    May 14, 2020 at 7:08
  • $\begingroup$ yes! what further assumptions are necessary so that exactness of $\mu$ solves the orginal question? $\endgroup$
    – StanleyT
    May 14, 2020 at 7:12
  • $\begingroup$ then since $\mu=d\eta$ and $\eta$ is a $n-1$ form there exists a vector field $X$ such that $\iota_{X}\mu = \eta$. Hence divergence of $X$ is $1$. But there is no guarantee that $X$ vanishes on $N$. Under what conditions does $X$ vanish on $N$? Or can it be choosen such that it vanishes? $\endgroup$
    – StanleyT
    May 14, 2020 at 7:26

2 Answers 2

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The answer is 'yes, there always is such a vector field $X$' and, in particular, the answer to your 'new question' is also 'yes'. (In fact, the first 'yes' implies the second 'yes', but the second 'yes' is used in the proof of the first 'yes'.)

Here is an outline of the proof (and it works also in the codimension $1$ case; I'm not sure why you want to exclude that):

First, since things are local around $N$, you can reduce to the case that $M$ is a vector bundle and $N$ is the zero section of that vector bundle (in which case, we can obviously identify $N$ with the base of the vector bundle). This is a standard result in differential topology, and one can find proofs in many places. Basically, you fix a Riemannian metric on $M$ and use the normal exponential map to show that a neighborhood of $N$ in $M$ is diffeomorphic to a neighborhood of the zero section of the normal bundle to $N$.

So suppose that $M^{n+k}$ is a vector bundle over $N^n$ and that $\mu$ is a volume form on $M$. Let $R$ be the vector field on $M$ whose time $t$ flow is scalar multiplication by $e^t$ in the fibers of $M\to N$ (which are vector spaces, so scalar multiplication is well-defined). Note that $R$ vanishes along $N$.

Then I claim that, for $k\ge 1$, there is a unique function $f$ on $M^n$ such that the vector field $X = fR$ has divergence $1$ with respect to $\mu$, i.e., that $\mathcal{L}_X\mu = \mathrm{d}\bigl(f\,\iota_{R}\mu\bigr) = \mu$, where $\iota_R\mu$ is the $(n{+}k{-}1)$-form that is the 'interior product' (aka 'left hook') of $R$ with $\mu$.

This is a claim about the unique solvability of a singular, linear first-order PDE: Let $\mathrm{d}\bigl(\iota_{R}\mu\bigr) = \kappa\,\mu$. Then $\kappa$ is a smooth function on $M$ that satisfies $\kappa(p) = k>0$ for all $p\in N$. I am claiming that there exists a unique smooth (and positive) function $f$ on $M$ that satisfies the linear inhomogeneous equation $$ \mathrm{d}f(R) + \kappa\,f = 1. $$ Since $R$ vanishes along $N$, this implies $f(p) = 1/k$ for all $p\in N$.

You can see what makes this a little delicate by examining what happens along a single line in a single fiber: If $v\in M_p$ is a nonzero element in the fiber over $p$, we can parametrize the line $\mathbb{R}v\subset M_p$ by $t\mapsto t{\cdot}v$ for $t\in \mathbb{R}$. In this case, the PDE to be solved becomes an ODE $$ tf'(t) + h(t)\,f(t) = 1 $$ where $h$ is a smooth function on $\mathbb{R}$ that satisfies $h(0) = k>0$. This is a regular, singular ODE, and while its (unique) smooth solvability is a classical fact, this seems not to be that well-known these days. The uniqueness on each line tells you that there is at most one smooth solution $f$ to the full equation on $M$, and it's not hard to use the uniqueness to show that, in fact, you get global smoothness for $f$ as well.

If you are doubtful, I can supply the proof. While this is a classical ODE/PDE fact, I don't, off-hand, remember a good source, just a proof (which must be the standard one).

Added remark about ODE/PDE fact: First, consider the ODE on the real line $\mathbb{R}$ $$ t\,f'(t) + h(t) f(t) = g(t) $$ where $h$ and $g$ are given smooth functions on $\mathbb{R}$ and $h(0) = k>0$. We want to show that there is a smooth solution $f$ and that it is unique. Write $h(t) = k - t m'(t)$ for some smooth function $m$ on the real line. When $g = 0$, the only solutions are $f(t) = c\,t^{-k}\mathrm{e}^{m(t)}$, where $c$ is a constant, and hence only when $c=0$ is the solution smooth. For general $g$, we use variation of parameters and look for a solution of the form $f(t) = c(t) t^{-k} \mathrm{e}^{m(t)}$ for some function $c(t)$ that vanishes to order at least $k$ at $t=0$. Substituting this into the above equation, we find that $c'(t) = t^{k-1}g(t)\mathrm{e}^{-m(t)}$, so, since we want $c(0)=0$, we have $$ c(t) = \int_0^t \tau^{k-1}g(\tau)\mathrm{e}^{-m(\tau)}\,d\tau. $$ Since $k>0$, this integral vanishes to order $k$ at $t=0$. Thus, $$ f(t) = \mathrm{e}^{m(t)}t^{-k}\int_0^t \tau^{k-1}g(\tau)\mathrm{e}^{-m(\tau)}\,d\tau $$ is a smooth solution. It is unique because we have already shown uniqueness when $g=0$. For use below, note that, if $g$ vanishes to order $l>0$ at $t=0$, then so does $f$, and we always have $f(0) = g(0)/k$.

Now, returning to the general case of $\mathrm{d}f(R) + \kappa f = 1$, we see, by applying the above argument to each line in $M_p\subset M$ for $p\in N$, that there is a unique function $f$ on $M$ that satisfies this equation and that $f$ is smooth on $M$ except possibly along the zero section $N\subset M$ itself. It is clearly smooth on every line through $0_p\subset M_p$, but one might worry that it is not smooth on a neighborhood of $N$ in $M$.

However, the following argument shows that this is not the case: Let $F_k(M,N)$ denote the space of smooth functions on $M$ that vanish to order at least $k$ along $N$. This is a decreasing filtration of $C^\infty(M) = F_0(M,N)\supset F_1(M,N)\supset\cdots$. One easily sees that the linear operator $D(f) = \mathrm{d}f(R) + \kappa\,f$ maps $F_i(M,N)$ into $F_i(M,N)$ and hence induces a linear operator $D_i:G_i(M,N)\to G_i(M,N)$ on the associated graded $$ G_i(M,N) = F_i(M,N)/F_{i+1}(M,N)\simeq S^i(M^*) $$ (bearing in mind that $M$ is a vector bundle over $N$). Since $R$ is the Euler (radial) vector field on each fiber $M_p$ and since $\kappa(p)=k$ for all $p\in N$, it follows that $D_i$ is simply multiplication by $i{+}k$ for $i\ge0$ and hence is an isomorphism of $G_i(M,N)$ with itself for $i\ge0$.

Consequently, for every $i\ge0$, there exists a smooth $f_i$ on $M$ such that $\mathrm{d}f_i(R) + \kappa\,f_i = 1 - h_i$ where $h_i\in F_{i+1}(M,N)$. Using the above 'integration construction on each line', we can find a function $u_i$ on $M$ that is smooth away from $N$, vanishes to order $i{+}1$ along $N$ and satisfies $\mathrm{d}u_i(R) + \kappa\,u_i = h_i$. It follows that $f = f_i + u_i$. Thus, for every $i\ge0$, $f$ can be written as the sum of a function $f_i$ that is smooth on $M$ and a function $u_i$ that vanishes to order $i{+}1$ along $N$ and is smooth away from $N$. Consequently, $f$ is differentiable to all orders along $N$ and hence is smooth on all of $M$ as desired.

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@Robert Bryant:

thank you very much for the answer. Unfortunately I have not been able to register completely and my account details have been deleted. therefore i cannot mark the question as "answered". the one I'm writing about is a new accout. I am very sorry. I also had a similar idea about the existence of the solution, but I didn't manage to finish. I would be very happy about details uniqueness and smoothness of the solutions.

PS. with this new accout I am not able to write this as a comment.

Greetings, Stan

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  • $\begingroup$ I believe that you can ask the moderators to merge the two accounts. (These kinds of mistakes happen from time to time, and the only way to fix them is to ask the moderators to do it.) $\endgroup$ May 18, 2020 at 9:23
  • $\begingroup$ StanleyT, please go here: mathoverflow.net/help/merging-accounts $\endgroup$
    – S. Carnahan
    May 18, 2020 at 11:56

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