5
$\begingroup$

Can the Klein bottle be immersed in $\mathbb R^3$ so that the associated height function be of Morse-Bott type and have no centers?

That is, the height function would have only Bott-type extrema and saddle singularities. A Bott-type singularity is a non-degenerate singular circle: a circle where the derivative is zero with the function being quadratic on transverse curves. A center is a Morse-type local extremum: an isolated singularity around which the function is $\pm(x^2_1+x^2_2)$ in some local coordinates.

My intuition is that no. I think such function cannot have (Morse) singularities other than Bott-type extrema (because they would increase the genus), and I cannot see how to connect an (even) number of immersions of Bott-type extrema (circles) by tubes in a non-orientable way without additional singularities (this should follow from the Whitney–Graustein theorem).

For a torus, such an immersion (embedding) is a doughnut lying flat on the table. However, I can't see how this can be done for the Klein bottle. The answer here does not seem to do the trick because it also increases the genus.

$\endgroup$
6
$\begingroup$

I wanted to prove that this is impossible but instead proved that this is possible...

Unfortunately, it is a bit hard to draw the picture but I'll try to explain how this should look like.

Construction. In this construction the Klein bottle will be included between the planes $z=0$ and $z=1$. The curves $\{z=1\}\cap K$ and $\{z=0\}\cap K$ are both the eight figure curve (with rotation index $0$). And both are Bott circles. Let us call the first curve $S_1$ and the second $S_0$.

Now, the function $z$ restricted to $K\setminus S_0\cup S_1$ has no critical points. And $K\setminus S_0\cup S_1$ is the immersed image of two cylinders $C$ and $C'$, both propagating in $\mathbb R^3$ from the plane $z=0$ to the plane $z=1$. The intersection of $C$ and $C'$ with any plane $z=c$ (where $c\in [0,1]$) is a figure eight curve.

The last detail is two explain how $C$ and $C'$ look like. So we will take as $C$ just the direct product of a vertical interval with the figure eight curve. To construct $C'$ we need to do something a bit trickier. Namely to construct it we start from $S_0$ in $z=0$ and then start to rotate it so that it the plane $z=t$ it is the figure eight curve rotated by $\pi t$. Thus, for $t=1$ it will be rotated by $\pi$.

Now, one can easily check that if we rotate a figure eight by $\pi$, it changes its orientation! So if we glue $C$ with $C'$, we get the Klein bottle indeed.

To finish the construction one just needs to smoothen out the described surface at $S_0$ and $S_1$. But this is not hard to do.

$\endgroup$

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.