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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.

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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 $f$.

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

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  • $\begingroup$ 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

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