Suppose I have a smooth manifold $M$, and an embedded $C^2$ submanifold $N \subset M$, of codimension at least 3. Does there exist, for every point $x \in N$, a smooth ($C^\infty$) hypersurface in a ball in $M$, which contains the portion of $N$ in that ball?



C^2 submanifolds contained in a hypersurface
The answer is negative.$\newcommand{\RR}{\mathbb{R}}$$\newcommand{\abs}[1]{\lvert #1 \rvert}$$\newcommand{\set}[1]{\left\lbrace #1 \right\rbrace}$$\newcommand{\ceiling}[1]{\left\lceil #1 \right\rceil}$


More precisely, $f:\RR\to\RR^n$ defined by
$$ f(t) = \bigl( t,\abs{t}^{3+\frac{1}{2}},\abs{t}^{3+\frac{1}{3}},\ldots,\abs{t}^{3+\frac{1}{n}} \bigr) $$
gives an embedded $C^2$ curve in $\RR^n$ such that no neighbourhood of zero in the curve is contained in a smooth (or even $C^4$) hypersurface in $\RR^n$. Taking products of $f$ with any $\RR^l$ produces examples of embedded $C^2$ submanifolds of $\RR^{n+l}$ of dimension $l+1$ for which there exists no neighbourhood of zero in the submanifold which is contained in a smooth hypersurface in $\RR^n$.


The argument below is divided into two parts: the first involves only the local description of submanifolds of $\RR^n$; the second part involves an estimate using Taylor expansions. The proofs are elementary and will be provided in some detail.


**Smoothly independent functions**


For convenience, let us introduce a little bit of terminology.


**Definitions:**


1. Let $f:\RR\to\RR$, $g:\RR\to\RR^n$ be functions $\RR\to\RR$. Say that $f$ *depends smoothly* on $g$ at $t\in\RR$ if there exists a $C^\infty$ function $h:\RR^n\to\RR$ such that $f = h\circ g$ on some neighbourhood of $t$.
2. Call a function $f:\RR\to\RR^n$ *smoothly dependent* at $t\in\RR$ if $f\_i$ depends smoothly on $(f\_1,\ldots,f\_{i-1},f\_{i+1},\ldots,f\_n)$ at $t$, for some $1\leq i\leq n$. Otherwise, say $f$ is *smoothly independent* at $t$.


Let $f:\RR\to\RR^n$ be continuous, and denote by $L$ the image of $f$ as a subspace of $\RR^n$.


**Claim 1:**
Let $S$ be a $C^\infty$ submanifold of $\RR^n$ of codimension one (i.e. a smooth hypersurface) which contains some neighbourhood of $f(0)$ in $L$. Then $f$ is smoothly dependent at zero.


*Proof*:

Since $S$ is a smooth hypersurface in $\RR^n$, it is locally the graph of a smooth function. More precisely, there exists $i\in\set{1,\ldots,n}$ such that some neighbourhood $U$ of $f(0)$ in $S$ is parametrized as
$$
U = \set{ (x\_1,\ldots,x\_{i-1},g(x),x\_i,\ldots,x\_{n-1}) \mid x \in V }
$$
where $V$ is an open in $\RR^{n-1}$ and $g:\RR^{n-1}\to\RR$ is a smooth function. Since $S$ contains some neighbourhood of $f(0)$ in $L$, we conclude that
$$
f\_i(t) = g\bigl(f\_1(t),\ldots,f\_{i-1}(t),f\_{i+1}(t),\ldots,f\_{n-1}(t)\bigr)
$$
for $t$ sufficiently close to zero. ■


**Examples of smoothly independent functions**


In view of claim 1, it remains to construct a $C^2$ embedding $f:\RR\to\RR^{n+1}$ which is smoothly independent at zero, for each $n>0$.


**Claim 2:**
Take any finite sequence $a\_0=1,a\_1,\ldots,a\_n$ of positive real numbers such that none of them can be written as a linear combination of the others with non-negative integer coefficients. Then the function
$$
f(t) = (t,\abs{t}^{a\_1},\ldots,\abs{t}^{a\_n})
$$
is smoothly independent at zero.


Here is a fairly explicit instance of claim 2 which recovers the example given at the beginning of the answer.


**Example:**
Pick some positive integer $l$ and a sequence of pairwise distinct real numbers $b\_1,\ldots,b\_n \in (0,1)$. Take $a\_i = l+b\_i$ for $i>0$ (and $a\_0 = 1$). Then the condition in claim 2 above is satisfied, hence the corresponding $f$ is smoothly independent. Moreover, this choice of $f$ is a $C^{l-1}$ embedding.


Claim 2 is a direct consequence of the next lemma.


**Lemma:** Let $a\_0,\ldots,a\_n$ be positive real numbers such that $a\_0$ is not a linear combination of $a\_1,\ldots,a\_n$ with non-negative integer coefficients. There exists no smooth function $h:\RR^n \to \RR$ such that $t^{a\_0} = h(t^{a\_1},\ldots,t^{a\_n})$ for all $t>0$ in some neighbourhood of zero.


*Proof*:
Assume that such a smooth function $h$ does exist. Take the Taylor polynomial for $h$ of a sufficiently large degree $k$ (it actually suffices to take $k = \ceiling{ \frac{a\_0}{ \min\set{a\_1,a\_2,\ldots,a\_n} } }$), together with the corresponding [Peano remainder term](https://en.wikipedia.org/wiki/Taylor%27s_theorem#Taylor.27s_theorem_for_multivariate_functions). Since $h$ is $C^k$, we have (in little-o notation)
$$
h(x) = \sum\_{\abs{I}\leq k} a\_I x^I + o\left(\sum\_{\abs{I} = k} \abs{x^I}\right)
$$
where $I=(i\_1,\ldots,i\_n)$ ranges over $n$-tuples of non-negative integers, $\abs{I}=i\_1+\ldots+i\_n$, and $x^I = (x\_1)^{i\_1} \cdots (x\_n)^{i\_n}$. The little-o symbol above is considered in the limit $x\to 0$.


Replace the preceding equality into $t^{a\_0} = h(t^{a\_1},\ldots,t^{a\_n})$ to produce:
$$
t^{a\_0} = \sum\_{\abs{I}\leq k} a\_I t^{i\_1 a\_1 + \cdots + i\_n a\_n} + o(t^{a\_0})
$$
Given that $a\_0$ is not a linear combination of $a\_1,\ldots,a\_n$ with non-negative integer coefficients, all the terms $a\_I t^{i\_1 a\_1 + \cdots + i\_n a\_n}$ appearing above have degree different from $a\_0$. Therefore, after grouping together monomials of the same degree, and moving the monomials of degree greater than $a\_0$ into the little-o symbol, we get
$$
t^{a\_0} = \sum\_{0\leq r<a\_0} C\_r t^r + o(t^{a\_0})
$$
Here $r$ varies over the real numbers, but only finitely many coefficients $C\_r$ are non-zero. In fact, all the coefficients $C\_r$ are zero: if the lowest degree non-zero term on the right hand-side is $C\_s t^s$ (with $0\leq s < a\_0$), then
$$
C\_s t^s = t^{a\_0} - \sum\_{s<r<a\_0} C\_r t^r - o(t^{a\_0}) = o(t^s)
$$
and hence $C\_s=0$. In conclusion, all the terms $C\_r t^r$ are zero, and thus $t^{a\_0} = o(t^{a\_0})$, which is impossible. ■


