5
$\begingroup$

Gromov's waist inequality for unit n-sphere $\mathbb{S}^{n}$ says: For any continuous function $f: \mathbb{S}^{n} \rightarrow \mathbb{R}^{m} $, there is some $y \in \mathbb{R}^{m}$ s.t. $Vol_{n-m}(f^{-1}(y)) \geq Vol_{n-m}(\mathbb{S}^{n-m}) $.

I'm wondering if there is an averaged version of the inequality, comparing the averaged fiber volume and some averaged $\mathbb{S}^{n-m}$ volume. For example, is it true that for some constant $C(n, m)$ depending only on dimensions:

$ \int_{f(\mathbb{S}^{n})} Vol_{n-m}(f^{-1}(y)) \geq C(n, m) Vol_{m}(f(\mathbb{S}^{n})) Vol_{n-m}(\mathbb{S}^{n-m}) $

It is, of course, interesting to consider the tubular version:

$ \int_{f(\mathbb{S}^{n})} Vol_{n}(f^{-1}(y) + \epsilon ) \geq C(n, m) Vol_{m}(f(\mathbb{S}^{n})) Vol_{n}(\mathbb{S}^{n-m} + \epsilon ) $, as well as similar inequalities for ball, cube, etc. in place of sphere.

Improve this question
$\endgroup$
2
$\begingroup$

My guess would be that there are ways to manipulate the average -- say you take the unit sphere in $\mathbb{R}^3$, and a function like $f(x,y,z)=x^{100}$. Then the image is the interval $[-1,1]$, but the preimage of a random point is small. The coarea formula is one way around this -- it gives an exact formula for an integral of fiber volumes weighted by the Jacobian of $f$.

Improve this answer
$\endgroup$
2
$\begingroup$

It is more natural to estimate the total area of large fibres, but it also does not work:https://arxiv.org/abs/1402.2856

Improve this answer
$\endgroup$
  • $\begingroup$ Thanks for pointing out the literature! For the paper you cite, I'm wondering if there are some extra conditions we can impose on the map, so that estimating the total area of large fibres does work in higher dimensions. $\endgroup$ – random_shape Jan 18 '18 at 17:40

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.