Construct a diffeomorphism $\phi:M\to M$ such that $\phi(p)=q$ and also $d\phi(X_p)=Y_q$

Question

PROBLEM: Construct a diffeomorphism $\phi:M\to M$ such that $\phi(p)=q$ and also $d\phi(X_p)=Y_q$, where $M$ is a connected smooth manifold and $p,q \in M$ , $X_p \in T_pM$ and $Y_q \in T_qM$

I know how to construct a diffeomorphism such that $\phi(p)=q$. One can see Guillemin and Pollack's book Differential Topology,page142, "Isotopy Lemma".

Alternatively, we can consider a compactly supported vector field $X$ with $\text{supp} X \subset U$ for some coordinate ball centered at $p$. WLOG, may assume $q$ lies in $U$ such that $q=(c,0,...,0)$ and $X= \partial_1$ on some $V \subset \subset U$. Now, since $X$ is compactly supported, the flow $\theta$ of $X$ is complete and we pick $\theta_c :M\to M$. Then,note that $\theta_c$ is a diffeomorphism and given by $$(u^1,u^2,...,u^n) \to (u^1+c,u^2,...,u^n)$$ on some neighborhood of $p\in U$; hence, it is the desired diffeomorphism such that $\theta_c(p)=q$

However, with an additional requirement $d\phi(X_p)=Y_q$, I don't know how to continue. I try to adjust the above argument but fail to have any useful idea.

Answer 1

You know how to get $p$ to $q$, so the question really amount to, "construct a diffeomorphism $M\to M$ which takes $X$ to $Y$, for $X,Y\in T_pM$."

The basic idea you outlined is to pass to local coordinates and find a vector field whose flow is yields the desired diffeomorphism. The same idea's going to work for moving tangent vectors around, too. Pass to local coordinates, say, so that $p$ goes to $0$. Find a nonsingular linear map $A$ taking $X\mapsto Y$ and let $A_t$ be a smooth path of invertible maps with $A_0 = I$ and $A_1=A$.

Our vector field is going to be built from $\hat{V} = \frac{d}{dt}\big|_{t=0}A_t$. The point of the construction is the flow of $\hat{V}$ is $A_t$, so that after flowing along $\hat{V}$ for one unit of time, we have taken $X$ to $Y$.

Now we just have to localize in the standard manner. Let $\phi$ be a bump function identically equal to $1$ near the origin and identically zero a little further away. The final vector field is $V = \phi\hat{V}$. The bump function allows us to restrict the support of $V$ to the coordinate neighborhood so it can be pulled back to $M$. The support of $V$ is compact, so its flow is defined for all time, and by construction after flowing for unit time we have taken $X$ to $Y$.

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I think it's worth mentioning this proves something slightly stronger than the OP's question: on a connected manifold, any two tangent vectors can be related by a diffeomorphism isotopic to the identity.

Answer 2

(The idea behind this answer is essentially outlined in Neal's answer).

One, of course, assume that $X_p,Y_q\neq 0$, or there is nothing extra to prove.

First of all, let $T : M\to M$ be a diffeomorphism so that $T(p) = q$. Let $\psi: U \to \mathbb R^n$ be local coordinate at $p$. Then $T^{-1}\circ \psi : T(U)\to \mathbb R^n$ is a local coordinate of $q$, and $T =id : \psi(U)\to \psi(U)$ under these two coordinate. We assume that $T(U)$ is the unit ball $B_1(0)$ in $\mathbb R^n$ and $\psi(p) = 0$.

Under this identification, $X_p, Y_q$ are both vectors in $\mathbb R^n$ (rooted at $0 = \psi(p)$). Let $A : \mathbb R^n \to \mathbb R^n$ be an invertible linear transformation (with positive determinant) so that $A(X_p) = Y_q$. Let $r$ be small (depending on $A$), so that $A(B_r(0)) \subset B_1(0)$.

Let $B\in gl(n, \mathbb R)$ so that $e^B = A$. Define a vector fields on $B_r(0)$ given by $Z(x) = Bx$. Now let $\rho : B_r(0)\to \mathbb R$ be a smooth cutoff function which is $1$ on $B_{r/2}(0)$. Then define a vector fields on $X$ given by

$$Q(x) = \rho(x) Z(x).$$

This vector fields can be treated as vector fields on $M$. Let $\alpha$ be the time one diffeomorphism corresponding to $Q$. As $Q(p) = 0$, we have $\alpha(p) = p$. Lastly, let $\phi = T\circ \alpha$. Then $\phi$ is a diffeomorphism and $\phi(p) = q$. To see $d\phi (X_p) =Y_q$, it suffices to check that in local coordinate $\psi: U \to \mathbb R^n$, $T^{-1}\circ \psi : T(U)\to \mathbb R^n$.

Under these coordinate, note that as $\rho$ is locally one, we have $\alpha_t (x) = e^{tB}x$ (locally on $B_{r/2}(0)$). In particular, in this coordinate,

$$\phi(x) = \alpha (x) = e^B x = Ax\Rightarrow d\phi_p = A\Rightarrow d\phi_p(X_p) = Y_q. $$