9
$\begingroup$

It can be proved that all $\mathbb{R^2}\rightarrow\mathbb{R^2}$ mappings with constant singular values are affine. In three dimensions, however, there are non-trivial examples, like

$$ \begin{align} x'&=\lambda_2 x -\frac{1}{\lambda_2}\sqrt{\frac{\lambda_2^2-\lambda_1^2}{\lambda_3^2-\lambda_2^2}}\int_0^{z}\sin f(\xi)\,\mathrm{d}\xi \\ y'&=\lambda_2 y +\frac{1}{\lambda_2}\sqrt{\frac{\lambda_2^2-\lambda_1^2}{\lambda_3^2-\lambda_2^2}}\int_0^{z}\cos f(\xi)\,\mathrm{d}\xi \\ z'&=\frac{\lambda_1\lambda_3}{\lambda_2}z \end{align} $$

for an arbitrary differentiable $f$ and $\lambda_3>\lambda_2>\lambda_1$, whose differential has constant singular values $\lambda_3\,,\lambda_2\,,\lambda_1$. So I wonder if one can classify or say something generic about such $\mathbb{R^3}\rightarrow\mathbb{R^3}$ maps, as is the the case in two dimensions.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.