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Can you fill a Klein bottle $K \subseteq \mathbb R^4$, embedded in four dimensional Euclidean space $\mathbb R^4$, with hyperraindroplets (hyperballs $B_\varepsilon$) from the outside? Will it enclose a four-dimensional region, similar to how the three-dimenaional torus $T^2 \subseteq \mathbb R^3$ partitions three-dimensional space $\mathbb R^3$ into an inside and an outside? In the case of the three-dimensional torus embedded into three-dimensional space, three-dimensional droplets cannot "fill" the inside from the outside without intersecting with the torus itself (there is no path for a ball which does not also lead to crossing the torus).

Or will it be analogous to the Möbius band, which doesn't enclose an inside? "Filling" a Möbius strip with water doesn't really make sense since it's just a surface and doesn't enclose any three-dimensional volume.

What throws me off is that it seems unreasonable to think that a surface can enclose a four-dimensional volume, since analogously a curve in three dimensional space cannot enclose a three-dimensional volume.

I have no real intuition behind this.

I asked this somewhere else and it turns out that if $\mathcal N$ is a closed submanifold of connected $\mathcal M$ with dimension $\dim\left(\mathcal N\right) < \dim\left(\mathcal M\right) - 1$, then $\mathcal M\setminus \mathcal N$ is (path) connected. Does this also mean that a Klein bottle does not partition $\mathbb R^4$ into an inside and an outside, since $\mathcal M\setminus \mathcal N$ is path-connected? Here: we take $K = \mathcal N \subseteq \mathcal M = \mathbb R^4$.

Even if it is path-connnected, can a hyperraindroplet freely pass through $\mathcal M\setminus \mathcal N$? That is, can you reach every point of $\mathcal M\setminus \mathcal N$ by translated balls $B_\varepsilon(x)$ if you squeeze $\varepsilon$ small enough?

Markus Klyver
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    "it seems unreasonable to think that a surface can enclose a four-dimensional volume, since analogously a curve in three dimensional space cannot enclose a three-dimensional volume" - this is correct. Your droplets can just go around the embedded manifold in the same way that 3D droplets can go around a smooth 1D curve. – Karl Feb 03 '24 at 17:24
  • @Karl If we don't care about smoothness, there is maybe a way to make a space-filling curve enclose a three-dimensional region. – Markus Klyver Feb 03 '24 at 17:27
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    Hmm, yeah. It's the topology of the filled region that determines what it encloses, not the topology of the domain of the curve. – Karl Feb 03 '24 at 17:31
  • How are you defining the outside of a non-orientable surface? There is a Mobius strip in there. – CyclotomicField Feb 03 '24 at 19:42
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    @CyclotomicField Regardless of whether the surface is orientable or non-orientable, a surface embedded in $\Bbb R^4$ has no "inside" or "outside." – Ted Shifrin Feb 03 '24 at 20:51
  • Just as $S^1 = \partial D^2 \subset \mathbb{R}^3$ and $\mathbb{R}^3\setminus S^1$ is path-connected, one can show that $K^2 = \partial M^3 \subset \mathbb{R}^4$ and $\mathbb{R}^4\setminus K^2$ is path-connected. (Here I am denoting the Klein bottle by $K^2$ to indicate dimension.) – Michael Albanese Feb 04 '24 at 14:12
  • @MarkusKlyver You can't do this even in the topological category: Space-filling curves and other such structures are not topological embeddings. – Kajelad Feb 05 '24 at 19:28

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