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I don't have a clue with this problem. Thank you very much for your help & guidance.

(a) Suppose $F(x,t): X\times I \rightarrow R$ is a homotopy of Morse functions. That is, $f_t: X \rightarrow R$ is Morse for every $t$. Show that the set $C = \{(x,t)\in X\times I : d(f_t)_x = 0\}$ forms a closed, smooth submanifold of dimension one of $X\times I$. Assume the homotopy is constant near the ends of I and use an open interval.

(b) Let $\pi: X \times I \rightarrow I$. Show that $d(\pi|_C)_{(x,t)}: TC_{(x,t)} \rightarrow TI_t$ is surjective.

(c) Show that if $X$ is compact, there is no homotopy of Morse functions between two Morse functions with different numbers of critical points.

1LiterTears
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1 Answers1

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(a) Whenever you are asked to show that something defined by an equation is a smooth manifold, the Implicit Function theorem should come to mind. Let $n$ be the dimension of manifold $X$. Since being a submanifold is a local property, we can work with a coordinate patch: that is, consider $X$ as being $\mathbb R^n$. Then $d(f_t)$ is a smooth map into $\mathbb R^n$, let's denote it by $F:\mathbb R^n\times (0,1)\to \mathbb R^n$. We must prove that the derivative of $F$ has maximal rank at the points where $F=0$. This is where you will use the Morse condition.

(b) Between one-dimensional manifolds, "surjective derivative" is the same as "nonzero derivative". The implicit function theorem can tell you why it's nonzero.

(c) Use parts (a) and (b) to show that for every integer $k$, the set of values $t$ for which $f_t$ has exactly $k$ critical points, is open.

ˈjuː.zɚ79365
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  • Thank you very much~ For (a), I use when $dF = 0, d^2 F \neq 0$ to show that when $df = 0, d(df) \neq 0$ to show $df$ is non-singular and therefore a smooth manifold, right?

    For (c), sorry I couldn't follow. Could you point out what happens when the set is open? Thanks!

     – 1LiterTears May 30 '13 at 19:00
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    @MathSnail You have $(0,1)=\bigcup_k {t: f_t \text{ has } k \text{ critical points}}$. An interval is connected; therefore it cannot be written as a disjoint union of open sets, unless all but one of them are empty. – ˈjuː.zɚ79365 May 30 '13 at 23:17
  • Hi @user79365 - thank you so much for your help. But I am very unfamiliar with the notation $\cup_k$, and I don't really understand what it represents. So I could not understand why the union equals (0,1) and how it relates to the disjoint union of open sets. Would you mind giving me further guidance? Thank you very much! – 1LiterTears May 31 '13 at 01:38
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    @MathSnail $\bigcup_k$ is fairly common notation for union of sets, indexed by $k$. What I mean is: for every $k=0,1,2,3,\dots$ consider the set of numbers $t$ such that $f_t$ has exactly $k$ critical points. These sets are disjoint. Their union is the interval. If you can show that these sets are open, the connectedness of the interval implies that all but one of them are empty. – ˈjuː.zɚ79365 May 31 '13 at 03:10
  • Oh I got it! Thank you so much @user79365!! – 1LiterTears May 31 '13 at 03:50