I have heard the claim: "Any closed manifold admits a Morse function which has one local minimum and one local maximum" often used in talks without a reference.

This does not seem to be very easy to prove "hands on". Trying to perturb the minima/maxima away locally will create new local minima/maxima, so I don't believe this will work.

One idea I had is to embed the manifold in $\mathbb{R}^n$, using Whitney, taking the height function, and turning on some flow in one fixed direction (I see this as stretching a balloon). I could not get the details right at all though...

Thus my question is: Is there an elementary proof of this fact? What are the standard references?


1 Answer 1


Here's a direct proof. Let $M$ be a smooth $n$-manifold and $f$ a Morse function on $M$. Let $p_i$ be the local minima, $q_j$ be the index-$1$ critical points, $f(p_i)<f(q_j)$ for all $i,j$, and $\gamma_{ij}$ the gradient flow line from $p_i$ to $q_j$. Connect each $p_i$ to $p_0$ by a direct path of $\gamma_{ij}$s. Note in particular the graph we have defined is contractible. Homotope $f$ so the $q_j$ which are not on any paths have a greater value than the $q_j$ which are on the path, say (by scaling and translating) $f^{-1}(1)$ separates the greater $q_j$ and the $q_j$ on the path. Now consider $B = f^{-1}(\infty,1]$. By construction, this is diffeomorphic to an $n$-ball. In $B$, replace $f$ by the radial distance function (possibly homotoped to match smoothly with $f$ on $\partial B$). This new $f$, call it $\widetilde{f}$, has exactly one index $0$ critical point.

Repeat the procedure for $-\widetilde{f}$ to get exactly one index $n$ critical point.

Intuitively, think of a level surface flow. It starts with a bunch of dots, the index $0$ critical points, expanding. Then they send out tendrils which join together at the index $1$ critical points. Controlling the Morse function is tantamount to controlling the level surface flow. We slow down the level surfaces near some of the index $1$ points so that the level surfaces all join into a giant sphere. This is our motivation; inside the sphere, we modify the function so that the sphere has expanded from a single dot, instead of many dots.

Here's another proof using handlebody decompositions.

By duality between Morse theory and handlebody theory, "Every manifold admits a Morse function with exactly one local maximum and exactly one local minimum" is equivalent to saying that every closed manifold admits a handlebody decomposition with exactly one $0$-handle and exactly one $n$-handle. To see this, take a smooth handlebody decomposition of $M$ ("smooth" so that the attaching maps are all smooth maps). Since this is done by gluing successive handles, we may focus our attention on the $0$-handles $M^0$ and the manifold obtained by gluing $1$-handles, $M^1$.

Since $M$ is connected and the only handle with disconnected attaching spheres are $1$-handles, $M^1$ is connected (that is, the gluing of $1$-handles kills all elements of $\pi_0$). Therefore, we may pick a single $0$-handle and connect it to each other $0$-handle by a path of $1$-handles (possibly through other $0$-handles). This union of $0$- and $1$-handles is homeomorphic to a ball, so we may replace it with a single $0$-handle and regard the remaining $1$-handles as attaching to the single $0$-handle.

Turn the handlebody decomposition upside down and repeat to obtain a single $n$-handle.

References to look into: Milnor's Morse Theory, Milnor's "Killing paper", Ranicki's book on surgery.

  • $\begingroup$ Thanks. I had the "c" right, though! $\endgroup$
    – Neal
    Jun 27, 2012 at 17:43
  • $\begingroup$ This is great, thank you very much, I think the image is clear now. Just to be sure: the graph of index 0 and 1 points and its gradient flowlines is connected because the CW-complex generated by the critical points must be connected, and this cannot happen by attaching 2-cells or higher right? $\endgroup$
    – Thomas Rot
    Jun 29, 2012 at 7:40
  • $\begingroup$ Two more minor points: The sentence $f(p_i)<f(q_i)$ for all $i,j$ is not correct right (because these are local minima, not global ones, so the index 1 points can be lower than some of the index 0 points)? And there can be more than one flow lines from $q_i$ than $p_i$ as well? $\endgroup$
    – Thomas Rot
    Jun 29, 2012 at 7:42
  • 1
    $\begingroup$ Let me think about why the graph is connected (although I'm sure it is). For $f(p_i)<f(q_j)\forall i,j$, you can always homotope $f$ in a neighborhood of an index $0$ point to be arbitrarily low by adding a negative bump function supported on the neighborhood. $\endgroup$
    – Neal
    Jun 29, 2012 at 10:44
  • $\begingroup$ Revisiting this: I believe your reason for the graph being connected is correct. $\endgroup$
    – Neal
    Aug 3, 2012 at 13:00

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