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The demonstration of the theorem 3.2 in the book Morse theory by Milnor

THEOREM $\mathbf{3.2.}$ Let $f:M\to\bf R$ be a smooth function, and let $p$ be a non-degenerate critical point with index $\lambda$. Setting $f(p)=c$; suppose that $f^{-1}[c-\epsilon,c+\epsilon]$ is compact, and contains no critical point of $f$ other then $p$, for some $\epsilon\gt0$. Then, for all sufficiently small $\epsilon$, the set $M^{C+\epsilon}$ has the homotopy type of $M^{C-\epsilon}$ with a $\lambda$-cell attached.

is given in the special case whene the manifold is the Torus ,

My question is : can i prove it in the case where the manifold is a manifold with dimension 1 ? enter image description here enter image description here this is a part of proof , in dimension 1 we chose only $(u^1)$ as a coordinate systeme, the index of the critical point is 0 or 1 ,so $f=c±(u^1)^2$

my first probleme is when i must define $e^{\lambda}$ ,I can't say that is the set of points in $U$ with : $u^1)^2\leq \varepsilon$ ,and $u^1 = 0$ !

so how to define $e^{\lambda}$

and in the book ,they applicated this to the torus , how to applicated it on a manifold with dimension 1 ?

Please

Thank you

@Yvoz

enter image description here

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  • $\begingroup$ IIRC, this theorem is stated without any restrictions on the dimension of the manifold. So of course this holds, unless you are looking for a way to prove it without invoking to the general theorem, in which case the classification of manifolds of dimension $1$ might help. Hope this is of any use to you. Regards. $\endgroup$ – awllower May 3 '13 at 9:34
  • $\begingroup$ I will try to demonstrate and tell you if I succeed $\endgroup$ – Vrouvrou May 3 '13 at 13:16
  • $\begingroup$ Then I shall look forward to the good news. :D $\endgroup$ – awllower May 3 '13 at 13:33
  • $\begingroup$ I edited my question $\endgroup$ – Vrouvrou May 3 '13 at 14:42
  • $\begingroup$ Could you formulate your problem more explicitly? I think that the index coincides with the CW-complex that you should supply, but there is no index in dimension $1$, so that I am confused by your question. Thanks in advance. $\endgroup$ – awllower May 3 '13 at 15:22
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I just give an idea to prove the result when $M$ is a polygonal line embedded in $\mathbb{R}^3$ and $\,f\,\colon\, M \to \mathbb{R}$ is the height function defined by $\,f(x,y,z) = z$.

The critical points of $\,f$ of index $0$ (resp. 1) are just its local minima (resp. maxima). Since $\,f$ is linear, the set of its critical points is a subset of the vertices of $M$; since $\,f$ is Morse, there can't be consecutive vertices at the same height (i.e. no edge of $M$ is parallel to the $xy$-plane).

In the hypothesis of the theorem, consider the set $S = \,f^{-1}[c-\varepsilon,c+\varepsilon] =M \cap \{c-\varepsilon \leq z \leq c+ \varepsilon\}$. Choosing sufficiently small $\varepsilon$ (and slightly perturbing $M$ if necessary) we can assume that $\,p$ is the only vertex contained in $S$. Now, it's easy to see that $\,S = P \sqcup Q$, where $P$ is made up of disjoint straight segments joining $\{z = c-\varepsilon\}$ with $\{z = c+\varepsilon\}$ and $Q$ is composed of two edges with common vertex in $\,p$ and the remaining two vertices in $\{z = c+\varepsilon\}$ (resp. $\{z = c-\varepsilon\}$) if $\lambda = 0$ (resp. $\lambda = 1$). In particular $Q$ is a one-dimensional cell.

Now recall that $\,M^{c - \varepsilon} = M \cap \{z \leq c - \varepsilon\}$ and $\,M^{c + \varepsilon} = M^{c - \varepsilon} \cup S$. By retracting the straight lines on the lower base, we get that $M^{c-\varepsilon} \cup P$ has the same homotopy type of $M^{c-\varepsilon}$ and so $M^{c+\varepsilon} \cong M^{c-\varepsilon} \cup Q$ where

  • $M^{c-\varepsilon} \cap Q = \varnothing$ if $\lambda = 0$
  • $M^{c-\varepsilon} \cap Q = \,\,$two points if $\lambda = 1$

So $Q$ is attached to $M^{c - \varepsilon}$ precisely as a $\lambda$-cell.


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  • $\begingroup$ thank you for the answer , you don't use the methode of milnor , i don't understand this part:"Since f is linear, the set of its critical points is a subset of the vertices of M; since f is Morse, there can't be consecutive vertices at the same height (i.e. no edge of M is parallel to the xy-plane).", can you explain me ,or if you can draw on a sheet, and scan it please ,please it's very important to me $\endgroup$ – Vrouvrou May 10 '13 at 19:11
  • $\begingroup$ @Vrouvrou If $\,L$ is a linear function defined on a segment $[PQ]$, it's easy to see that $\,L$ attains its extremal values on $P$ and on $Q$. So if $\,f$ is linear and defined on a polygonal line $M$ (i.e. $M$ is made up of consecutive segments), then extremal values of $\,f$ are attained on the endpoints of the segments which $M$ is made up of. $\endgroup$ – Ivo May 11 '13 at 7:37
  • $\begingroup$ Note that it is improper to say that $\,f$ is Morse on a polygonal line (which is not a differentiable manifold); but you know that critical points of Morse functions are isolated. Now, if two consecutive vertices $P$ and $Q$ of $M$ have the same height, then all points in the segment $[PQ]$ are critical. P.S. This observation can be avoided assuming that the vertices are all at different heights. $\endgroup$ – Ivo May 11 '13 at 7:50
  • $\begingroup$ i think that $Q$ is composed of two edges with common vertex in $p$ and the remaining two vertices in {z=c−ε} (resp. {z=c+ε}) if λ=1 (resp. λ=0) no ? $\endgroup$ – Vrouvrou May 12 '13 at 6:50
  • $\begingroup$ please why $Q$ is a 1-cell ? , 1-cell is $e^1=\lbrace x\in \mathbb{R}, ||x||<1\rbrace$ $\endgroup$ – Vrouvrou May 12 '13 at 7:04

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