# Immersion in $\mathbb R^3$ of a Klein bottle with Morse-Bott height function without centers

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.

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.