It is well-known that a Morse function on $T^2$ has at least $4$ critical points, but also that there exist functions $f\colon T^2\to\mathbb R$ with only 3 critical points (the least possible number by Lusternik-Schnirelmann theory): a minimum, a maximum, and a degenerate saddle.

It is not hard to describe these functions by means of their levelsets, but it seems difficult to produce immersions of $T^2$ into $\mathbb R^3$ with a height function that does the job. According to Banchoff and Takens, there are no smooth embeddings with such height functions, only immersions. I was trying to look for pictures of such immersions and came across the following beautiful image by Cassidy Curtis:

Strange 2-torus immersed in R^3

I expected that these would be somehow easier to produce -- but probably I am wrong... still:

Is there a "simpler" immersion of $T^2$ into $\mathbb R^3$ whose image has a height function with exactly 3 critical points? Or is the above example "optimal" in some sense?

The only necessary conditions I see are that the saddle point $p$ must be degenerate and that there are 3 arcs with both endpoints at $p$ that lie in the same levelset as $p$. I find it hard to believe that these conditions are not sufficient, and that there are no easier immersions; but I have not been able to prove this or find any such immersions.

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    $\begingroup$ I suppose the first thing you need to do is settle on a notion of complexity. On the analytic end, you could talk about something like the elastic bending energy of the immersion. Perhaps more directly amenable to computation would be the total number of double-points created and destroyed in these level set pictures. I suppose this would be the same as the number of local maxima and minima on the co-dimension two strata of the immersion. $\endgroup$ Commented Feb 2, 2016 at 5:57
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    $\begingroup$ Those are lovely hand-sketches by-the-way. Curtis must have read the Topological Picturebook. Looks almost like Francis drew them. $\endgroup$ Commented Feb 2, 2016 at 8:33
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    $\begingroup$ Another construction of such a function $f$: construct the torus as the quotient of ${\bf R}^2$ by the lattice $L = A_2$, and consider the Green's function with charges of $+1$ and $-1$ at the two nontrivial points of $A_2^* \, / \, A_2^{\phantom.}$. The degenerate saddle is at the origin. The logarithmic singularities can be capped artificially, or smoothed by applying a heat kernel. I don't know whether this helps construct an immersion with height function $f$. $\endgroup$ Commented Feb 3, 2016 at 1:46
  • $\begingroup$ A nice explicit construction of such a function $f$ is the following. Construct the torus as $\mathbb{R}^2/\pi\mathbb{Z}^2$, then let $f = \sin(x)\sin(y)\sin(x+y)$. The three critical points are the origin, $(\frac{\pi}{3},\frac{\pi}{3})$, and $(\frac{2\pi}{3},\frac{2\pi}{3})$. $\endgroup$
    – jwsiegel
    Commented May 30, 2019 at 13:32

3 Answers 3


Warning: this is not an immersion (it has twelve Whitney-umbrella-like pinch points)

Here is a relatively simple explicit realization: the $z$ coordinate for the parametric surface$$(x,y,z)=(\sin(2u),\sin(2v),\sin(u)\sin(v)\sin(u-v));$$after an affine shift leaving $z$ unchanged the graph looks like this:

enter image description here

It is thus similar to the Steiner's Roman surface except that the latter has three double lines and this one has six.

I've tried to cut it to make the central monkey saddle more visible:

enter image description here

The implicit equation, in slightly different coordinates, is$$(X+Y+Z)(X+Y-Z)(X-Y+Z)(-X+Y+Z)=(XYZ)^2,$$

enter image description here

with function $X+Y+Z$:

enter image description here

Mathematica codes:


 {u, 0, \[Pi]}, {v, 0, \[Pi]},
 BoxRatios -> {1, 1, 1}, Mesh -> None,
 RegionFunction -> (Abs[#3]<.05 \[Or] .2<Abs[#3]<.4 \[Or] .6<Abs[#3]<.8 \[Or] Abs[#3]>1 &),
 PlotPoints -> 250,
 PlotStyle -> FaceForm[Red, Cyan],
 BoundaryStyle -> Black,
 Boxed -> False,
 Axes -> False,
 SphericalRegion -> True

  (U-V-W) (U+V-W) (U-V+W) (U+V+W)+(U V W)^2==0,
  • $\begingroup$ Is it obvious that this surface has genus 1? $\endgroup$
    – j.c.
    Commented Jul 27, 2018 at 7:19
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    $\begingroup$ @j.c. It's defined as a map from $\Bbb R^2/2\pi \Bbb Z^2$. $\endgroup$
    – mme
    Commented Jul 27, 2018 at 8:29
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    $\begingroup$ @MikeMiller Thanks, yes. In facte even $\pi$, instead of $2\pi$ will work. I mean, it is a map from $f$ from $[0,\pi]\times[0,\pi]$ to $\mathbb R^3$ which agrees (together with all derivatives) on the boundary in a way to be extendable to the torus, i. e. $f(u,0)=f(u,\pi)$ and $f(0,v)=f(\pi,0)$ $\endgroup$ Commented Jul 27, 2018 at 9:06
  • $\begingroup$ @მამუკაჯიბლაძე I was paranoid about the $z$-coordinate and just did not check whether it was $\pi$-invariant. :D $\endgroup$
    – mme
    Commented Jul 27, 2018 at 9:11
  • $\begingroup$ Sorry for misprints, $f(0,v)=f(\pi,v)$ it should be $\endgroup$ Commented Jul 27, 2018 at 9:17

I would recommend to look at the paper (here is a free original in russian)

  • Elena Kudryavtseva, Realization of smooth functions on surfaces as height functions. (Russian) Mat. Sb. 190 (1999), no. 3, 29--88; translation in Sb. Math. 190 (1999), no. 3-4, 349–405

where the structures of immersions realizing given function as a height function are described.

Let $M$ be a closed surface and $f:M\to\mathbb{R}$ be a smooth (not necessarily Morse) function having exactly $N$ critical points. Say that $f$ can be realized as a height function if there exists an immersion $j:M \to\mathbb{R}^3$ and an orthogonal projection $p:\mathbb{R}^3 \to\mathbb{R}$ to some line $l$ such that $f = p\circ j$.

Notice that in this case at each critical point $z\in\mathbb{R}^3$ the normal vector to the surface is parallel to the line $l$. Say that two immersions with the same height functions are normally equivalent if the directions of the normals to these immersed surfaces at all critical points of the height functions are the same.

Elena Kudryavtseva proved (see Theorem 1 of the paper above) that $f$ can be realized as a height function in the following cases:

1) $M$ is either a sphere or a torus;

2) $M$ is any orientable surface and $f$ is Morse, and in this case there are $C^{N/2}_{N} = \frac{N!}{(N/2)! (N/2)!}$ normally non-equivalent immersions;

3) $M$ is non-orientable, and $f$ is not necessarily Morse but has only finitely many critical points, and in this case there are $2^{N}$ normally non-equivalent immersions.


The complexity of surface immersions in 3-space has been studied in a series of papers by Nowik and collaborators. See

Nowik, Tahl. Higher-order invariants of immersions of surfaces into 3-space. Pacific J. Math. 223 (2006), no. 2, 333–347

as well as

Nowik, Tahl. Order one invariants of immersions of surfaces into 3-space. Math. Ann. 328 (2004), no. 1-2, 261–283.


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