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Let $\left(\mathcal{M}^2,g_\mathcal{M};X\right)$ and $\left(\mathcal{N}^2,g_{\mathcal{N}};Y\right)$ be two smooth two-dimensional, simply connected Riemannian manifolds (with or without boundary), equipped with non-nonvanishing Killing fields $X$ and $Y$, respectively.

Does there exist a mapping $f:\mathcal{M}^2\rightarrow \mathcal{N}^2$ with constant, distinct singular values such that $\mathrm{d}f\left(X\right)=Y$ ?

In the affirmative case, how would one construct explicitly such a mapping, given the singular values ?

(It is well known that in general there are no local obstructions for mapping with constant singular values, but assuming that there exists a Killing field that is mapped to another Killing may set obstructions.)

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  • $\begingroup$ I curious whether you have looked at my answer. If you have questions about anything you don't understand, please ask them, and I'll try to answer. $\endgroup$ Commented Jan 27, 2022 at 12:50
  • $\begingroup$ @RobertBryant Thank you so much; my apologies as I have been into my thesis defense the last few days and could not read thoroughly the answer. $\endgroup$ Commented Feb 2, 2022 at 19:57

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To understand the local conditions, it's convenient to establish canonically associated local coordinate expressions for the quantities involved. Thus, let $(M^2,g,X)$ and $(N^2,h,Y)$ be as described and suppose that we want to test whether, for a given $p\in M$ and $q\in N$, there exists an open $p$-neighborhood $U\subset M$ and a local diffeomorphism $f:U\to N$ satisfying $f(p) = q$ with the desired properties. To simplify notation a little bit, let us fix orientations on $M$ and $N$ and require $f$ to be orientation-preserving. (We'll see what comes of this choice later.)

Then it is easy to show that there exist oriented, $p$-centered coordinates $(r,\theta):V\to (-\epsilon, \epsilon)\times (-\epsilon,\epsilon)$ on an open $p$-neighborhood $V\subset M$ and a smooth function $A:(-\epsilon,\epsilon)\to\mathbb{R}$ with $A(0)=0$ and $A'>0$ such that, on $V$, we have $g = \mathrm{d}r^2 + A'(r)^2\,\mathrm{d}\theta^2$ and $X = \partial/\partial\theta$. Similarly, there exist oriented, $q$-centered coordinates $(s,\phi):W\to (-\delta, \delta)\times (-\delta,\delta)$ on an open $q$-neighborhood $W\subset M$ and a smooth function $B:(-\delta,\delta)\to\mathbb{R}$ with $B(0)=0$ and $B'>0$ such that, on $W$, we have $h = \mathrm{d}s^2 + B'(s)^2\,\mathrm{d}\phi^2$ and $Y = \partial/\partial\phi$. Such adapted coordinates are locally unique.

[Note that, since $M$ is simply-connected (and, let's assume, connected, though the OP didn't include that condition), the functions $r$ and $\theta$ extend globally to $M$ uniquely so that $g = \mathrm{d}r^2 + a^2\,\mathrm{d}\theta^2$ and $X=\partial/\partial\theta$ where $a$ is a positive smooth function on $M$ satisfying $\mathrm{d}r\wedge\mathrm{d}a = 0$. The smooth mapping $(r,\theta):M\to\mathbb{R}^2$ is an immersion, but there is no reason to believe, under the given hypotheses, that it is an embedding, nor is it necessarily true that $a = A'(r)$ for some function $A:r(M)\to\mathbb{R}$. The situation with $(N,h,Y)$ is similar.]

Supposing that an $f$ exists with all the specified properties, we can, by shrinking $\epsilon$, assume that $f(V)\subset W$ and hence, using the fact that $f_\ast(X) = Y$, conclude that $$ f^*(s) = s\circ f = R(r)\quad\text{and}\quad f^*(\mathrm{d}\phi) = \mathrm{d}\theta + M(r)\,\mathrm{d}r.\tag1 $$ for some functions $R$ and $M$ on $(-\epsilon,\epsilon)$ with $R(0)=0$ and $R'>0$. This implies that, relative to the orthonormal coframings, we must have $$ f^*\begin{pmatrix}\mathrm{d}s\\ B'(s)\,\mathrm{d}\phi\end{pmatrix} = \begin{pmatrix}R'(r) & 0\\ B'(R(r))M(r) & B'(R(r))/A'(r)\end{pmatrix} \begin{pmatrix}\mathrm{d}r\\ A'(r)\,\mathrm{d}\theta\end{pmatrix}\tag2 $$

Now, the constancy of the singular values implies that, in particular, the determinant of the above coefficient matrix must be constant, i.e., that there must be a constant $c_2>0$ such that $$ R'(r)B'(R(r))/A'(r) = c_2\,.\tag3 $$ Since $B(0) = R(0) = A(0) = 0$, we then integrate to get $B(R(r)) = c_2\,A(r)$. In particular, since $B$ is invertible, $R(r) = B^{-1}\bigl(c_2\,A(r)\bigr)$ for some positive constant $c_2$.

Now, the sum of the squares of the singular values of the coefficient matrix must be another constant $c_1 > 2c_2$ (so that the two constant singular values will be distinct) such that $$ R'(r)^2 + B'(R(r))^2\,M(r)^2 + B'(R(r))^2/A'(r)^2 = c_1\,.\tag4 $$ Using the above formula for $R'(0) = c_2 A'(0)/B'(0)$, we see that, by taking $c_1$ sufficiently large, we can guarantee that the above equation for $M(r)$ has (two) real solutions on a neighborhood of $r=0$.

Conversely, for $c_2>0$ and $c_1 > 2c_2$ sufficiently large and $\epsilon>0$ sufficiently small, there will be functions $R(r)$ and $M(r)$ that satisfy the above equations (3) and (4) and the initial condition $R(0)=0$, and hence, via (1) and the initial condition $f(p)=q$, they will determine a unique mapping $f$ with the desired properties.

Thus, local mappings $f$ with constant, distinct singular values always exist carrying any desired point to any other. Moreover, it is clear that there is a 2-parameter family of such local mappings carrying any given point in the domain to any given point in the range.

The existence of a global such mapping $f:M\to N$ depends on the growth properties of the functions $A$ and $B$ and the validity of their domains. Little more can be said about this without more information or hypotheses about the functions $A$ and $B$.

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  • $\begingroup$ Thank you so much for such a great answer. I wish to clarify just a couple of details. In constructing $ f^*\left( d \phi\right)$ we use the fact that $ f^*\left( d \phi\right)\left(\frac{\boldsymbol{\partial}\:\:}{\boldsymbol{\partial}\theta} \right)=d \phi \left(d f\left(\frac{\boldsymbol{\partial}\:\:}{\boldsymbol{\partial}\theta} \right) \right)=1$, then $f^*\left(d \phi\right)=M(r,\theta)d r+d \theta$, that is shouldn't be $M$ in general a function of $r,\theta$ ? Also, why is it guaranteed that $c_2A(r)$ is in the image of $B(r)$, when solving for $R(r)$ ? Thank you. $\endgroup$ Commented Feb 2, 2022 at 19:51
  • $\begingroup$ I guess that there should be a typing error, so that in the first paragraph it should read $(s,\phi):W$ ... on an open $q$-neighborhood $W\subset N$, isn't it ? $\endgroup$ Commented Feb 2, 2022 at 19:54
  • $\begingroup$ @DanielCastro: Since $f^*(\mathrm{d}\phi)$ and $\mathrm{d}\theta$ are closed, we have $\mathrm{d}(M\,\mathrm{d}r) = 0$, so $M$ is a function of $r$. Since $A(0)=B(0)=0$ while $B'(0)>0$, by shrinking $\epsilon$, we can guarantee that $c_2\,A(r)$ is in $\bigl(B(-\delta),B(\delta)\bigr)$ when $|r|<\epsilon$. Yes, that $V$ should be a $W$. I'll fix those points. $\endgroup$ Commented Feb 3, 2022 at 10:21

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