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Given a closed manifold $M$ and an integer $k\geq 0$, let $G_k(M)$ denote the space of smooth functions $f:M\to\mathbb R$ with at most $k$ critical points.

To what extend has the topology of the spaces $G_k(M)$ been studied?

Obviously, a complete understanding of the topology of $G_k(M)$ in general seems hopeless, but that does not mean there aren't interesting things to be said.

Some things I know already:

  • The smallest $k$ for which $G_k(M)$ is non-empty is called the Lusternik--Schnirelman category of $M$. Obviously, constructing functions with few critical points is important in topology, e.g. for proving the h-cobordism theorem.
  • The space of functions on $N\times[0,1]$ which have no critical points and which are standard near the boundary is a main object of study in the theory of pseudo-isotopies. (This space isn't exactly of the form $G_k(M)$, but seems close enough to include on this list.)
  • The spaces $G_{2n}(S^1)$ and $G_{2n+1}(S^1)$ are homotopy equivalent to $S^{2n-1}$ (proved using the heat equation, which is apparently useless once $\dim M\geq 2$).
  • There is a natural map $S^n\to G_2(S^n)$ which associates to a given vector $\mathbf v$ the function $\mathbf w\mapsto\mathbf w\cdot\mathbf v$ on the unit sphere $S^n$ in $\mathbb R^{n+1}$. This element of $\pi_n(G_2(S^n))$ becomes trivial in $G_{2n+2}(S^n)$, and I would conjecture this is sharp (i.e. it is non-trivial in $G_{2n+1}(S^n)$). This conjecture is proven for $n\leq 3$ in [1], and I believe the methods used there can also be used to treat the case $n=4$.
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    $\begingroup$ One other small observation: $G_k(\mathbb R^n)$ is contractible, using an Alexander trick / straight line homotopy argument. $\endgroup$
    – Ryan Budney
    Commented Jul 11, 2017 at 15:40
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    $\begingroup$ @RyanBudney: False. For example, $G_0(\mathbb R^n)$ is homotopy equivalent to $S^{n-1}$ (consider the "slope at the origin"). $\endgroup$
    – John Pardon
    Commented Jul 24, 2017 at 11:21
  • $\begingroup$ Well, I should have been a little more detailed. In the k=0 case (which I did not think you were interested in) the Alexander argument gives you a deformation-retract to the linear subspace. That's a once-punctured dual space , homotopy-equivalent to a sphere as you say. When $k>0$ this Alexander argument gives a deformation-retract to the subspace with at most one critical point at the origin. Perhaps that has the homotopy type of $\Omega^n S^{n-1}$, or something like that. $\endgroup$
    – Ryan Budney
    Commented Jul 24, 2017 at 18:31
  • $\begingroup$ Didn't Vassiliev study a question similar to this? Perhaps it was Morse functions with restricted numbers and types of critical points. $\endgroup$
    – Ryan Budney
    Commented Jul 24, 2017 at 18:33

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