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Let $\mathbb{T}^2$ be the topological $2$-dimensional torus, and let $0<\sigma_1 < \sigma_2$ satisfy $\sigma_1 \sigma_2=1$. Let $g$ be an arbitrary smooth Riemannian metric on $\mathbb{T}^2$.

Does there exist an area-preserving diffeomorphism $f:(\mathbb{T}^2,g) \to (\mathbb{T}^2,g)$ whose singular values are constant $\sigma_1 , \sigma_2$ (Here the singular values of $df$ are w.r.t the metric $g$).

One naive approach is to consider the gradient flow of the functional $E:C^{ \infty}(\mathbb{T}^2,\mathbb{T}^2) \to \mathbb{R}$ defined by

$$ E(f)=\int_{\mathbb{T}^2} (\sigma_1-\sigma_1(df))^2+(\sigma_2-\sigma_2(df))^2. $$

Then starting with a given diffeomorphism $f \in \text{Diff}(\mathbb{T}^2)$ , we can try deforming it into a diffeomorphism with the required singular values.

For the flat torus there are only affine solutions ($SL_2(\mathbb{Z})$), so we have discretization in the admissible values of $\sigma_i$. I am interested to know more about what the answer could be for other metrics*.

This shows that the set of such special diffeomorphisms may be 'finite-dimensional'. (In general there is no reason that the set of such diffeomorphisms would form a group; however, it is always closed under taking inverses).

*I restricted the question to the torus due to a topological obstruction:

I want to consider oriented closed surfaces $M$. Since $\sigma_1 \neq \sigma_2$, one can choose the singular vectors in a smooth manner locally; this induces a splitting of the tangent $TM$ into a direct sum of line bundles, which implies $M$ has zero Euler characteristic. This leaves us only with the torus.

Improve this question
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    $\begingroup$ For a compact surface $M$, if such a map exists and is at least $C^1$, then it will induce a $C^0$-splitting of the tangent bundle into line bundles. This can only happen if the Euler characteristic of the surface is equal to $0$, i.e., for the torus and Klein bottle. In those two cases, such maps clearly do exist. For lower regularity, it is not clear what happens. $\endgroup$
    – Robert Bryant
    Nov 9, 2020 at 14:22
  • $\begingroup$ Thanks, I didn't think to consider this topological obstruction. Can you say if such maps exist for arbitrary metrics on the torus? (I wasn't sure if you were implying that such maps exist only for the flat "lattice-induced" metrics on the torus where we have affine maps, or for more general metrics as well.) $\endgroup$
    – Asaf Shachar
    Nov 9, 2020 at 16:49
  • $\begingroup$ @RobertBryant Actually, on a second thought, don't you need to assume $M$ is contractible for your splitting argument? The singular vectors correspond to orthogonal $1$-dimensional distributions on $TU$. If $U$ is contractible, then they are trivial, hence you can find global nonvanishing sections; but in general it seems to me that you would only be able to do that locally, no? (In general singular vectors can only be chosen smoothly locally: math.stackexchange.com/a/3163368/104576) $\endgroup$
    – Asaf Shachar
    Nov 11, 2020 at 7:48
  • $\begingroup$ @AsafShacar: I didn't assume that the line bundles are trivial; I didn't need to, all I need is existence: Since the singular values are distinct, the two line bundles exist and are continuous. After passing to a cover of degree at most 4, they become trivial, so the covering space has vanishing Euler characteristic. Since the Euler characteristic multiplies by degree under finite covers, the Euler characteristic of the original surface must have been zero. Hence the surface is either a Klein bottle or a torus. $\endgroup$
    – Robert Bryant
    Nov 11, 2020 at 9:44
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    $\begingroup$ Sure. This is a standard construction in algebraic topology. Suppose that $L\to M$ is a line bundle and, for convenience, assume that $L$ has an inner product. Let $\tilde M\subset L$ be the set of elements of $L$ of unit length. Then $\tilde M\to M$ is a double cover of $M$ (since $L$ is locally trivial). The pullback $\tilde L\to\tilde M$ of the line bundle $L\to M$ consists of the set of pairs $(u,ru)\in \tilde M\times L$ for $u\in \tilde M$ and $r\in\mathbb{R}$. Since $\sigma(u) = (u,u)$ is a nonvanishing section of $\tilde L$, it follows that $L$ is trivial. $\endgroup$
    – Robert Bryant
    Nov 19, 2020 at 13:22

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