Due to Cerf, there's exist a certain homotopy between two Morse functions. It is said that this homotopy is "generic". What is a precise definition of the property to be "generic" in this case?
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1$\begingroup$ Usually in differential topology the term "generic" is used to describe a property of smooth maps which defines an open, dense subset of the space of smooth maps, e.g "Immersions are generic amongst all maps $f\colon\thinspace M\to N$ if $2m<n$". Without more context, it's difficult to give a more precise answer than that. See p 141 of "Stable Mappings and their Singularities" by Golubitsky and Guillemin. $\endgroup$– Mark GrantCommented Apr 25, 2013 at 13:33
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2 Answers
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Intuitively, for the homotopy to be generic has the same meaning as for the Morse function itself to be generic: the set of points where it fails to be a submersion is as simple as possible. Practically, this means that for all but a finite number of times (values of the $t$ parameter) the function $H(x,t)$ is a Morse function of the variable $x$ and the Morse singularities trace out smooth curves transverse to the $x$ direction; and for each of those exceptional times $t$ the function $H(x,t)$ is Morse at all values of $x$ except for a single value $x$ (I'm assuming compactness here) where it undergoes one of a certain class of very special singularities obtained by collapsing two Morse singularities, called "birth-death" singularities. Just as with Morse singularities themselves, which are described locally by specific functions of $x$ in some coordinate system, birth-death singularities are described locally by specific functions of $x,t$ in some coordinate system.
The function $H(x,t) = x^3 - tx$ at $t=0$ is an example of a birth-death singularity in one dimension: for $t>0$ there are no critical points; and for $t<0$ there is one maximum and one minimum.
answered Apr 25, 2013 at 13:38
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1$\begingroup$ For your example, do you mean $H(x,t) = x^3 - tx$ at time $t = 0$? I'm pretty sure the critical points of $x^2-t$ don't change with $t$, but it's certainly possible I'm not understanding what $H$ is supposed to be. $\endgroup$– dvitekCommented Mar 26, 2016 at 6:48
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$\begingroup$ Oops, that's a good catch in an old answer. Thanks. $\endgroup$ Commented Mar 26, 2016 at 13:32
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"Generic" usually refers to open and dense.
Assume $M$ is a closed smooth manifold. Let $$C^\infty(M)$$ be the space of all smooth real valued functions. Topologize this with respect to the Whitney $C^\infty$ topology (see: http://ncatlab.org/nlab/show/C-infinity+topology)
Let $$H(M) \subset C^\infty(M)$$ be the subspace of functions whose singularities are either Morse or birth-death (in particular, $H(M)$ contains the space of Morse functions). Then $H(M)$ is a generic subspace (it's open and dense). It is also path connected. Even more is true: the complement $\cal D$ of $H(M)$ inside $C^\infty(M)$ has codimension two, in the sense that any smooth map $D^1 \to C^\infty(M)$ whose endpoints are in $H(M)$ can be infinitesimally perturbed, relative to its boundary $S^0$, to a smooth map that has image in $H(M)$. Here smooth means that the adjoint map $M\times D^1\to \Bbb R$ is a smooth map.
This is discussed in beautiful detail in the book:
In any case, note that $C^\infty(M)$ is a contractible space (in fact it's affine). This means that if we are given a smooth map $S^0 \to H(M)$, then it admits an extension to a map $D^1 \to C^\infty(M)$. By the above discussion, this map can be perturbed slightly, relative to $S^0$, so that its image lies in $H(M)$.
