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This is a cross-post.

Let $(M,g)$ be a two-dimensional compact surface, endowed with a Riemannian metric.

Fix $s>0$, and suppose that for any two geodesic triangles $A,B$ having area $s$, there exists an affine onto map $f:A \to B$, where I say $f$ is affine if $\nabla df=0$. (equivalently, $f$ maps parametrized geodesics to parametrized geodesics. Here $\nabla=\nabla^{T^*M} \otimes \nabla^{f^*TM}$).

I assume $s<<\text{Area}(M)$ is very small, so there are a lot of triangles of area $s$.

What can we say about the metric $g$? Does it have to be flat? Are there any restrictions on its curvature?

I do not require $f$ to be the restriction of an affine map $M \to M$; (I think this is a stronger requirement than the existence of "local" or piece-wise affine maps. e.g. for the flat torus, globally we only have $SL_2(\mathbb{Z})$.)

I believe that the assumption means that that there a lot of affine maps locally $M \to M$; perhaps we can translate this into showing $M$ is flat.

If $\nabla^{T^*M} \otimes \nabla^{f^*TM}$ has zero curvature, then $M$ is flat. And 'many affine maps' means roughly 'many parallel sections of $T^*M \otimes TM$ '-- although not exactly, since for every $f$, $df$ is a section of a vector bundle which depends on $f$, i.e. $T^*M \otimes f^*TM$.

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  • $\begingroup$ You know a lot of advanced mathematics so it would be nice if you could contribute more answers to other's questions. If you ask many questions expecting people to spend their time on your problems, you should give something in return. $\endgroup$ May 25, 2021 at 15:03
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    $\begingroup$ Well, first, I wish I would also share your sentiment that "I know a lot of advanced mathematics":) Seriously, I really do not feel that way. It is very rare that I see any question here to which I can contribute an answer, or even a comment. My impression from this site, is that I am using very (very) elementary machinery and tools compared to most participants here (even in "my" areas of research). Nevertheless, perhaps your comment will encourage me to try more! Regarding your comments on expectations, you made me wonder a bit. $\endgroup$ May 25, 2021 at 15:36
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    $\begingroup$ I will say that I certainly not expect from anyone to invest time on a problem, if it isn't interesting for them. I do get the impression that asking well-phrased and interesting questions, do contribute value to the site, even if one does not answer too much. In fact this is also the opinion of the Stack Exchange management. I do try to put here questions which might interest other people as well, $\endgroup$ May 25, 2021 at 15:37
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    $\begingroup$ and invest a lot of time of phrasing the questions in a suitable generality. (so the question won't be too specialised for "my purposes" only, but will have a wider scope/interest/value). Although, I must admit that sometimes I fail doing it -- and your comment made me think that perhaps I should try to be more selective in what I post here - out of respect to the community here, and not to insert "garbage" to the site. Again, you certainly encouraged me to try to participate more in the "helping side" as well. Maybe I can do more than I believe... $\endgroup$ May 25, 2021 at 15:38
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    $\begingroup$ @PiotrHajlasz: the conventional view is that question asking and question answering are equally important to the success of MathOverflow. $\endgroup$ May 26, 2021 at 20:26

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Using the structure equations, it is not difficult to show that, if $f:(M,g)\to(N,h)$ is a diffeomorphism of (not necessarily complete) connected surfaces that is affine in the OP's sense, i.e., $\nabla(\mathrm{d}f)=0$, then $f$ has constant singular values and $L\bigl(f(p)\bigr) = K(p)/|\det(f)|$ for any $p\in M$, where $K:M\to\mathbb{R}$ and $L:N\to\mathbb{R}$ are the Gauss curvatures of $g$ and $h$ respectively, and $\det(f)$ is the product of the singular values of $f$ (and thus is constant).

Moreover, if we restrict attention to the open sets $M^*$ and $N^* = f(M^*)$ where the respective Gauss curvatures are nonvanishing, then $f:M^*\to N^*$ is a homothety, i.e., an isometry up to a (constant) scale factor.

Thus, the only situation in which the OP's desired flexibility holds is for a flat surface.

Here is a little bit more detail: It's a local calculation, so choose a $g$-orthonormal coframing $g={\omega_1}^2+{\omega_2}^2$ on $U\subset M$ and an $h$-orthonormal coframing $h={\eta_1}^2+{\eta_2}^2$ on $V = f(U)\subset N$. Let $\omega_{12}$ on $U$ and $\eta_{12}$ on $V$ satisfy $$ \mathrm{d}\omega_1=-\omega_{12}\wedge\omega_2\quad \mathrm{d}\omega_2=+\omega_{12}\wedge\omega_1\quad \mathrm{d}\omega_{12}=-L\,\omega_1\wedge\omega_2 $$ and $$ \mathrm{d}\eta_1=-\eta_{12}\wedge\eta_2\quad \mathrm{d}\eta_2=+\eta_{12}\wedge\eta_1\quad \mathrm{d}\eta_{12}=-K\,\eta_1\wedge\eta_2\,. $$ There will exist functions $a_{ij}$ on $U$ such that $f^*(\eta_i) = a_{ij}\,\omega_j$. The condition $\nabla(\mathrm{d}f) = 0$ then translates into the equations $$ \mathrm{d} \begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix} = \begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix} \begin{pmatrix}0&\omega_{12}\\-\omega_{12}&0\end{pmatrix} -\begin{pmatrix}0&\bar\eta_{12}\\-\bar\eta_{12}&0\end{pmatrix} \begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix} $$ where $\bar\eta_{12} = f^*\eta_{12}$. This immediatly implies the equations $$ \mathrm{d}(a_{11}a_{22}-a_{12}a_{21}) = \mathrm{d}(a_{11}^2+a_{12}^2+a_{21}^2+a_{22}^2) = 0, $$ so the singular values of the matrix $a$ are constant.

If the singular values are equal, it follows that $f$ is either the constant map or else $f$ is a homothety. If they are not equal, we can choose our coframings so that $0<a_{11}<a_{22}$ and $a_{12}=a_{21}=0$. Now, the constancy of the $a_{ij}$ together with the above equations implies that we must have $$ a_{11}\omega_{12}-a_{22}\bar\eta_{12} = a_{22}\omega_{12}-a_{11}\bar\eta_{12} = 0. $$ In other words $\omega_{12} \equiv \bar\eta_{12} \equiv 0$. Now the above structure equations imply that $K \equiv L \equiv 0$.

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  • $\begingroup$ Dear Robert, thank you for this interesting answer. Just to be sure: Are you saying that every affine map between surfaces with nonvanishing curvatures is a homothety? $\endgroup$ May 26, 2021 at 20:25
  • $\begingroup$ @AsafShachar: Yes, in fact, if there is any point where the curvature is nonzero, then the affine map is necessarily a homothety. $\endgroup$ May 26, 2021 at 20:59
  • $\begingroup$ Thanks. I think I understand why one point of non-zero curvature suffices - since the singular values are constant, they are determined by a single point. I also understand why the singular values of any affine map should be constant . Could you please elaborate on the analysis using the structure equation, which leads you to $L\bigl(f(p)) = K(p)/\det(f)^2$ and to the conclusion regarding the homothetic nature of $f$? $\endgroup$ May 29, 2021 at 16:16
  • $\begingroup$ I understand that this should be slightly similar to your analysis here: mathoverflow.net/a/351550/46290, although in the current situation we might have non-zero curvature, and stronger assumption on the map $f$. BTW, I think that $L\bigl(f(p)) = K(p)/\det(f)$ instead of $L\bigl(f(p)) = K(p)/\det(f)^2$ $\endgroup$ May 29, 2021 at 16:19
  • $\begingroup$ @AsafShachar: You are right about there not being a square on the $\det(f)$. That was caused by my mis-translating the formulae that I got into the language of the answer. Instead of $\det(f)^2$, it should be $|\det(f)\,|$. As for why it has to be homothety unless $K\equiv L\equiv 0$, if the two singular values are equal (and nonzero constant), it's a homothety by definition. If the two singular values are not equal, you can adapt coframes on domain and target to make $f$ diagonal with distinct, constant eigenvalues. Then the structure equations force $K\equiv L\equiv 0$. $\endgroup$ May 29, 2021 at 17:05

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