$\newcommand{\R}{\mathbb{R}}$Suppose $\Omega$ is a bounded, convex domain in $\R^{m}$. Fix $x_1, x_2\in\Omega$ and an invertible matrix $A\in\mathrm{GL}^{+}(m)$ with positive determinant. Let $U\subset\Omega$ be an open convex neighbourhood containing $x_1, x_2$ and the Euclidean geodesic connecting them.

I want to construct a smooth diffeomorphism $\tau\in\mathrm{C}^{\infty}(\Omega;\Omega)$ of $\Omega$ such that $$ \begin{align} \tau(x_1)&=x_2 \tag{a}\\ \mathrm{d}\tau_{x_1} &= A \tag{b}\\ \tau|_{\Omega\setminus U} &\equiv \operatorname{Id} \tag{c} \end{align} $$

This seems (to me at least) like an underdetermined problem which should have a solution. How can one construct $\tau$? Am I completely mistaken and the boundedness of $\Omega$ is a topological obstruction to the existence of such a diffeomorphism?

I've tried the following:

  1. Let $r>0$ be so small that $\overline{B_r(x_1)}\subset U$ and consider a smooth bump function for $\overline{B_r(x_1)}$ supported in $U$, i.e. $\chi$ such that $0\leq\chi\leq1$, $\chi\equiv1$ on $\overline{B_r(x_1)}$ and $\chi\equiv0$ on $\Omega\setminus U$. Consider the affine transformation $T$, defined by $T(x)=x_2 + A(x-x_1)$. If need be let's make $r$ so small that the ellipsoid $x_2 + A(B_r(x_1))$ is contained in $U$ as well. Define $\tau$ as the convex combination $\chi \cdot T + (1-\chi) \cdot \operatorname{Id}$. This is a smooth map and fulfils $(a)-(c)$ but I believe it is not a diffeomorphism -- even in the case where all eigenvalues of $A$ are positive.
  2. Unsuccessfully tried finding a vector field by constructing flow lines between the points $x_1 + e_1, \ldots x_1 + e_m$ and $x_2 + Ae_1, \ldots x_2 + Ae_m$. I hoped that the (smoothly cutoff) flow would be the sought after diffeomorphism.
  3. Unsuccessfully tried finding an energy functional whose gradient flow induces the diffeomorphism.
  4. Might it be possible to write down an equation that is solved by applying the implicit function theorem (for Banach spaces, so $\tau\in\mathrm{C}^{k}(\Omega;\Omega)$)?
  5. I think I can reduce it to the case $x_1=x_2$ if that simplifies things.
  • 2
    $\begingroup$ Your "flow argument" (item 2) can be used to reduce to the case $x_1 = x_2$. Next take the singular value decomposition of $A = U\Sigma V^T$, where $U$ and $V$ rotations. Finally, try to write your diffeo as a composition of three diffeos: two doing local rotations around $A$ and one taking care of the $\Sigma$ term. These you can do fairly obviously and geometrically. $\endgroup$ Mar 20, 2023 at 17:01
  • $\begingroup$ Mentioning the SVD was very helpful @WillieWong! Turns out it's almost as easy as I originally thought. Thanks, I think I've fixed it now. $\endgroup$ Mar 20, 2023 at 20:20

1 Answer 1


As per @Willi Wong's comment, consider the singular value decomposition $A=U\Sigma V^T$. Because $A$ has positive determinant, we can make sure that $U$ and $V$ are both rotations without reflections, i.e. $U,V\in\mathrm{SO}(m)$. Furthermore, $\Sigma_{ii}>0$ for all $i$. Since $\mathrm{SO}(m)$ is both compact and connected, the exponential map is surjective and hence there are $X,Y\in\mathfrak{so}(m)$ such that $U = \exp(X)$ and $V=\exp(Y)$. (This even works if $\det(A)<0$ because then, say, $U$ can be chosen to be in the compact, connected component of $-\operatorname{Id}$ in $\mathrm{O}(m)$.)

Let $\chi\in C^{\infty}(\Omega)$ be the smooth bump function from the question above. Let $\sigma_{max}$ be the maximal singular value of $B$. Then consider a smooth function $\psi\in C^{\infty}([0,1])$ such that $0\leq\psi\leq1$ and $\psi(0)=1$, $\psi(1)=0$ and $\psi^\prime(1)=0$ such that it satisfies the following differential inequality: $$ \begin{align} 1+ (\sigma_{max}-1)\psi(s) + s(\sigma_{max}-1)\psi^\prime(s) > 0. \tag{$\ast$} \end{align} $$ I'm not sure how to solve this explicitly for general $\sigma_{max}$ but one can mollify a piecewise affine function which satisfies this. This condition guarantees that the map $s\mapsto (\sigma_{max}\psi(s) + 1-\psi(s))s$ is bijective.

Consider the following three maps $M_1,M_2,M_3\in C^{\infty}(\Omega;\mathrm{GL}(m))$ given by $$ \begin{align} M_1(x) &= \exp(-\chi(x)\cdot Y), \\ M_2(x) &= \big(\psi\big(\tfrac{|x-x_1|}{r}\big)\cdot \sigma_{max} + 1-\psi\big(\tfrac{|x-x_1|}{r}\big)\big) \big(\chi(x)\cdot \tfrac{1}{\sigma_{max}}\Sigma + (1-\chi(x))\cdot \operatorname{Id}\big), \\ M_3(x) &= \exp( \chi(x)\cdot X). \end{align} $$ Thus, we have $M_1(x_1)=V^T$, $M_2(x_1)=\Sigma$ and $M_3(x_1)=U$. Indeed, $M_1$ and $M_3$ map into $\mathrm{GL}(m)$ because, e.g. $M_1(x)^{-1} = \exp(\chi(x)\cdot Y)$ can be given explicitly. The map $M_2$ also maps into $\mathrm{GL}(m)$ because we can bound each diagonal entry below by $$ \begin{align} (\psi \cdot \sigma_{max} + 1-\psi) (\chi \cdot \tfrac{\Sigma_{ii}}{\sigma_{max}} + (1-\chi)\cdot 1) \geq \min\{\Sigma_{ii},1\}>0 \end{align} $$ since $\Sigma_{ii}>0$.

Finally, define a smooth map $\tau\in C^{\infty}(\Omega;\Omega)$ by $$ \tau(x) = x_2 + M_3(x)M_2(x)M_1(x)\cdot (x-x_1). $$ This map is a diffeomorphism. Indeed, both $x\mapsto M_1(x)\cdot x$ and $x\mapsto M_3(x)\cdot x$ are diffeomorphism because they are merely rotations. The complicated scaling by $\tfrac{1}{\sigma_{max}}$ in the second factor in $M_2(x)$ and the differential inequality for $\psi$ which reverses this scaling guarantee that also $x\mapsto x_1 + M_2(x)\cdot (x-x_1)$ is a diffeomorphism.

Then by explicit calculation $\tau(x_1)=x_2$ and $\mathrm{d}\tau|_{x_1}=M_3(x_1)M_2(x_1)M_1(x_1) = U\Sigma V^T = A$. Further, by construction of $\chi$ the conditions $(a)-(c)$ are satisfied.

Edit: The original solution I posted was not quite correct. But it can be fixed by the above.

Here is an example of such a diffeomorphism for $x_1=x_2=0$, $r=1$ and $U=B_2(0)$. Enjoy! enter image description here

  • On the top left you see the bump function $\chi$ and on the top right the modulating function $\psi$.
  • Displayed at the bottom left is the domain of the diffeomorphism, color-coded by the level sets of $\chi$. Displayed at the bottom right is the image of the diffeomorphism.
  • I consider a line radiating outward from $x_1$ and look at its image under $\tau$, this is the red line that starts in one direction, then gets smoothly bent and eventually reaches the same direction of its preimage.
  • In green you can see multiple of these 'flow lines'. In particular, you can see the rotation and stretching by $A$.
  • The red circle gets mapped to an ellipsoid which emanates from $x_1$ with a certain velocity (the singular values of $A$), then gets slowed down until it reaches unit velocity (this is what $\psi$ and the scaling do). Subsequently it gets rotated and stretched until it reaches a multiple of the unit circle with unit velocity.

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