Constructing Diffeomorphism with prescribed derivative

Question

I am in the following situation: I am working in $S^1\times\mathbb{R}^2$ and have two different frames for $T(S^1\times\mathbb{R}^2)$, given by $X_1,X_2, X_3$ and $Y_1, Y_2=\partial_x, Y_3=\partial_y$, both of which are defined only on $S^1$ ($\partial_\theta$ is a non-zero vector field which is a positive basis for $TS^1$). Furthermore all the vector fields are nowhere tangent to $S^1\times\{0\}$. I now want to find a diffeomorphism $f$ from an open neighbourhood of $S^1\times\{0\}$ to an open neighbourhood of $S^1\times\{0\}$ such that

• $f$ is given by the Identity on $S^1\times\{0\}$
• for any $p\in S^1\times\{0\}$ we have $T_p f(X_i) = Y_i$.

Does anyone know of a theorem which would guarantee the existence of $f$, so far I have only found Whiteney's Extension Theorem and the extension lemma (Lemma 2.26) in Lee's Introduction to Smooth Manifolds, both of which do not cover my case since there are restrictions on either the domain or range of $f$.

Answer 1

This is not always possible. To see this, suppose $X_1$ is actually tangent to $S^1\times \{0\}$ and $Y_1 = -X_1$. Then there cannot be an $f$ meeting your criteria.

The point is that since $f$ is the identity on $S^1\times \{0\}$, $T_pf$ must be the identity on $T_p(S^1\times \{0\}) \subseteq T_p (S^1\times \mathbb{R}^2)$ for any $p\in S^1\times \{0\}$.

I'm not sure what kind of conditions one should impose on how $\{X_i\}$ is related to $\{Y_i\}$ in order to guarantee such an $f$ exists.

Answer 2

This map won't generally exist; there are all kinds of obstructions to consider:

• Let $t(X)\subset S^1$ denote the set of points for which $X$ is tangent to $S^1\times\{0\}$. Since this property is preserved by any diffeomorphism $f$ which preserves $S^1\times\{0\}$, we must have $t(X_i)=t(Y_i)$.

• Every vector field on $S^1\times\{0\}$ nowhere tangent to $S^1\times\{0\}$ can be identified with a loop in $\mathbb{R}^2\setminus\{0\}$, and thus an element of $\pi_1(\mathbb{R}^2\setminus\{0\})\cong \mathbb{Z}$, allowing us to compute the winding number $w(X)\in\mathbb{Z}$ of any such vector field $X$. For each $f$ there is an integer $w(f)$ such that $w(Tf(X))=w(X)+w(f)$, so if $X_i,X_j$ are nowhere tangent to $S^1\times\{0\}$, the same must be true for $Y_i,Y_j$, and $w(X_i)-w(X_j)=w(Y_i)-w(Y_j)$.

• Let $\partial_\theta$ be the canonical tangent vector field of $S^1\times\{0\}$, and $\omega$ be a choice of orientation of $S^1\times\mathbb{R}^2$. For any $f$ there is a fixed $c\in C^\infty(S^1\times\{0\})$ such that $\omega(V_1,V_2,V_3)=c\omega(Tf(V_1),Tf(V_2),Tf(V_3))$ for any vector fields $V_i$ on $S^1\times\{0\}$. This along with $Tf(\partial_\theta)=\partial_\theta$ results in several restrictions.

Post Edit: In the edited form of the question, the first condition completely rules out the existence of $f$, since $Y_1=\partial_\theta$ is tangent to $S^1\times\{0\}$ while $X_1$ are nowhere tangent.