EDIT: The answer is trivially positive; the question arose from my misunderstanding of the figure below.

Can a non-orientable closed surface of odd genus 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 $\pm(x_1^2+x_2^2)$ in some local coordinates.

My intuition is that no. (I've asked this question on math.SE but did not get any answer.)

Consider the projective plane $\mathbb RP^2$ as the Boy surface (left) [EDIT: wrong, both figures are not immersions, and the left figure is not the Boy surface] and the Klein bottle $K^2$ (right):

|Fig. 1

(image from the book). The 8-shaped level sets are immersions (i.e., not self-intersecting) except for where singular points are shown. The vertical line in the right-side figure is a homologically non-trivial cycle.

For even genera $g$ (except $g=2$, which is a different story), it is easy to do: e.g., connect the top and bottom of $K^2$ (right) by a tube (as if you drill a wormhole along the vertical axis), which will form a surface of genus $g=4$ immersed [EDIT: wrong, this is not an immersion] with two Bott-type extrema (circles) and two Morse-type saddles. (You can get any even genus $g\ge4$ by adding more handles.)

However, adding such a handle to the $\mathbb RP^2$ (left) seems not possible. Suppose you add such a handle connecting the bottom to the top of the figure (left). There must be a singularity on that handle. Indeed, consider the evolution of the level sets from the bottom to the top along this handle. The level sets at its endpoints are circles $S^1$ immersed in the plane: O-shaped at the bottom and 8-shaped at the top, which are not regularly homotopic by the Whitney–Graustein theorem. Therefore, there must be a singularity in between.

My intuition is that the singularity will be similar to the saddle shown on the picture (roughly speaking, not preserving the parity of the total turning number below and above). Though the singular contour can be more complicated (e.g., connecting more handles), it will effectively convert the left-side picture into the right-side one: it would cause an additional cycle passing through the first singularity (like the cycle between the two singularities shown on the right), thus making the genus $g$ even.

I think this argument would generalize to a surface with more handles, as soon as any cycle exists between the "bottom" and "top" of the singular level of the type shown in the figure (left): in the absence of centers, for each saddle "that changes the orientation," along the evolution of the level sets (passing transparently the Bott-type extrema along the handle) there will occur another similar saddle adding a second cycle.

Unfortunately, I lack the skill to convert this into a formal proof, and even if I could do it for this particular type of immersion [EDIT: wrong, this is not an immersion] of $\mathbb RP^2$ (Boy surface), it would not prove the claim in the general case. Could you provide such a proof, or point to sources where a proof can be found? Detailed explanations would be greatly appreciated, since I am not an expert.


This is not correct, and here is why.

Take an immersion of $\mathbb RP^2$ in $\mathbb R^3$, for example, https://en.wikipedia.org/wiki/Boy%27s_surface .

After a little perturbation, we can assume that the height function is a Morse function. After a further perturbation we can assume that this function has an even number of local maxima and minima. Indeed, we can always create additional pairs of critical points, one local max and one saddle. Moreover for any large enough $n$ there will be a perturbation of the original immersion with the total $2n$ local maxima and minima.

Assume that these maxima minima are on different level sets. Let us pair them together in any way and join them by segments in $\mathbb R^3$ along which the height function has no critical points. It is not hard to see then, that we can attach a cylinder to the surface that will go along the segment. We might need to create up to two Bott circles on this cylinder so that the condition on the absence of centres is satisfied.

Remark. Note that both surfaces depicted on the figure are not immersed surfaces. The first one has one singularity at the top, the second has two singularities.

  • $\begingroup$ Than you! Yes, centers are what you say (I've edited the question). Could you please clarify how to make the cylinder connecting the two centers an immersion? The figure (left) is precisely what you say: a Morse height function on an immersion of $\mathbb RP^2$ with two centers, but I can't see how to connect them by an immersed cylinder without a second saddle on it. $\endgroup$ – Alexander Gelbukh Oct 12 '19 at 15:00
  • $\begingroup$ (Sorry, I meant "Thank you!") $\endgroup$ – Alexander Gelbukh Oct 12 '19 at 16:27
  • 1
    $\begingroup$ The figure on the left is not an immersion of $\mathbb RP^2$ to $\mathbb R^3$. Namely, there is a problem with the very top point. Close to this point the surface doesn't look infinitesimally like an embedded disk. By the way, for the same reason, the surface on the right is not an immersion at two points. An immersed surface doesn't have this type of singularities, en.wikipedia.org/wiki/Boy%27s_surface Does this answer your question? $\endgroup$ – Dmitri Panov Oct 13 '19 at 8:05
  • $\begingroup$ Yes, now I see: it was a stupid question. I mistakenly thought the figure showed an immersion. So your answer is that on any surface there can be defined a simple Morse function, any such function is the height function of a suitable immersion (dfgm.math.msu.su/files/bf-engl.pdf, page 76), and any Morse center can be turned into a Bott-type circle by adding a handle. Obvious. Sorry to bother you with a stupid question :( $\endgroup$ – Alexander Gelbukh Oct 13 '19 at 20:37
  • $\begingroup$ Well, the singularities of both surfaces that are depicted on your figure are quite famous objects. They are called Whitney umbrellas. en.wikipedia.org/wiki/Whitney_umbrella If you think of a surface in $\mathbb R^3$ as an image of a smooth map, then it turns out that you can not get rid of such a Whitney umbrella if you slightly perturb the map. So you need to change the map quite significantly to get an immersion. To be honest I don't understand well enough how this Boy surface looks like... $\endgroup$ – Dmitri Panov Oct 13 '19 at 21:21

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