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Let $\mathbb{T}^2=\mathbb{S}^1 \times \mathbb{S}^1$ be the flat $2$-dimensional torus, and let $0<\sigma_1 < \sigma_2$ satisfy $\sigma_1 \sigma_2=1$.

Does there exist an area-preserving diffeomorphism $f:\mathbb{T}^2 \to \mathbb{T}^2$ whose singular values are constant $\sigma_1 , \sigma_2$?

An immediate family of such diffeomorphisms which comes to mind are the affine (geodesic-preserving) maps which are induced by elements of $SL_2(\mathbb{Z})$. However, this family does not cover the entire range of pairs $\{ (\sigma_1,\frac{1}{\sigma_1}) \, | \, \sigma_1 \in (0,1) \}$, since it's countable. Furthermore the set of $\sigma_1$ which are admissible in this affine family is discrete away from zero, which is its only accumulation point.   Are there any non-affine examples?

Edit:

Robert Bryant has given an answer which shows that there is no non-affine $C^3$ example. I wonder what happens if we allow reduced regularity, say Lipschitz maps whose differential has a.e. the singular values $\sigma_1 , \sigma_2$.

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There are no non-affine examples with $f$ smooth (or even $C^3$). This follows from the fact that such an $f:\mathbb{T}^2\to\mathbb{T}^2$ would lift to a smooth (local) diffeomorphism $F:\mathbb{R}^2\to\mathbb{R}^2$ with constant singular values, and the argument I gave in my answer to this question implies that such a map $F$ would have to be affine.

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  • $\begingroup$ Thanks Robert. BTW, couldn't we get a lift which is an actual diffeomorphism? (not just a local diffeomorphism). Also, are you aware of any obstructions for such a "diffeomorphism with constant singular values" on other manifolds? (e.g. the sphere?). Do you perhaps have a guess regarding the case of a generic metric? (would it admit such a diffeomorphism?) $\endgroup$
    – Asaf Shachar
    Commented Nov 8, 2020 at 12:02
  • $\begingroup$ @AsafShachar: The lift $F$ already is a global diffeomorphism: $F:\mathbb{R}^2\to\mathbb{R}^2$ is clearly surjective, and, because it pulls back the complete metric on $\mathbb{R}^2$ to a comlete metric on $\mathbb{R}^2$, it's proper and a covering map, so it's one-to-one. I didn't need that, though, so I didn't mention it. $$\ $$ On other compact surfaces with nonzero Euler characteristic, there can't be such a map because it would split the tangent bundle into the sum of two line bundles. $$\ $$ There are plenty of metrics on the torus that admit such maps, though. $\endgroup$
    – Robert Bryant
    Commented Nov 9, 2020 at 13:56

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