How can I find an approximation of a smooth function $R^2 \to R^4$ such that when restricted to $S^1$ it is an immersion?

Question

I am having trouble with the following exercise:

Given a smooth function $f: R^2 \to R^4$ and $S^{1}$ embedded in $R^2$, then $\forall \epsilon >0$ there exists a smooth function $f_{1}$ such that $sup|f(x)-f_{1}(x)|< \epsilon$, which is an immersion when restricted to $S^1$.

Answer 1

Here's a sketch of an elementary solution if you know the transversality theorem (e.g., see Guillemin & Pollack or "Parametric transversality theorem" here).

(1) Observe that you only need to perturb $f$ near $S^1$ to obtain the result. So let $\rho\colon \Bbb R^2\to\Bbb R$ be a smooth function that is $1$ in a small neighborhood of $S^1$ and $0$ outside a slightly larger neighborhood.

(2) When $A$ is a $4\times 2$ matrix, let $g_A(x) = f(x)+Ax$. Note that $h_A(x) = (dg_A)_x = df_x + A$. Since the matrices of rank $1$ in the space of $2\times 4$ matrices form a submanifold $Z$ of dimension $5$, if a map $h_A$ on $S^1$ is transverse to $Z$, it must be transverse by default and have image disjoint from $Z$. Likewise for the $0$ matrix. So, by the transversality theorem, $h_A(S^1)$ will be disjoint from $Z\cup \{0\}$ for almost every $A$, and this means that for such $A$, $g_A$ will be an immersion on $S^1$. Choose such an $A$ so that $|Ax|<\epsilon/2$ for all $x\in S^1$.

(3) Take $f_1 = f + \rho A$.

I leave it to you to work out the details (including a careful definition of the small and slightly larger neighborhoods). After all, this is an exercise for you :)

Answer 2

We can do this without invoking transversality. There is a standard theorem that specializes to your case:

Let $f : \mathbb R^n\to R^p$ be a smooth map with $2n \le p.$ Then for any $\epsilon > 0$ there exists a $p \times n$ matrix $A = (a_{ij} )$ such that $|a_{ij} | < \epsilon$ and the map $g : \mathbb R^n\to \mathbb R^p$ defined by $g (x) = f (x) + Ax$ is an immersion.

To prove this, first note that $M_k (p, n)\subseteq M (p,n) $, the set of all $p\times n$ matrices of rank $k$, is a submanifold of $M (p,n) $ of dimension $k (p + n − k).$

We need to find a matrix $A$ such that $\text{rank}\ dg(x)=n.$ Let $S=\{A_{p\times n}:|a_{ij}|<\epsilon|\}.$ Then, $m(S)=(2\epsilon)^{pn}>0.$ Consider $S_1 = \{A \in S : A = B − df (x);\ \text{rank}\ B < n\}.$ The result will follow if we can show that $m(S_1)=0.$

Define $F_k : M_k (p, n) \times \mathbb R^n \to M (p, n)$ by $F_k (B, x) = B − df(x)$ and $\Phi (k) = k (p + n − k) + n.$ Then $\Phi'(k) = −2k + p + n > 0$, so $\Phi$ is increasing. This implies that $k (p + n − k)< \text{dim}\ (M (p, n))$ and an application of Sard's theorem shows that $m\left(F_k(M_k (p, n) \times \mathbb R^n)\right)=0$, and so $m(S_1)=0$.