3
$\begingroup$

I asked this question on http://math.stackexchange.com but no unswers!

I have this paragraph from K.C. Chang Infinite dimensional Morse theory

In comparison with degree theory, which has proved very useful in nonlinear analysis in proving existence and in estimating the number of solutions to an operator equation, Morse theory has a great advantage if the equation is variational. Relative homology groups and critical groups are series of groups that provide both a finer structure and better estimate of the number of solutions than does the degree, which is only an integer. The relationship between the Leray-Schauder index and critical groups is established.

And I don't Understand how to see that Morse theory is better then degree theory,

how to see that Relative homology groups and critical groups provide both a finer structure and better estimate of the number of solutions than does the degree ?

Please help me

Thank you.

Improve this question
$\endgroup$
10
  • 1
    $\begingroup$ It might help you get answers if you mention what you already know about critical groups, etc. For instance, it is clear that the topological degree is just an integer whereas a homology group has more algebraic structure. In general, given $f:X \to R$ you can associate to each critical point $c$ the relative homology of $(X^c,X^c\setminus \{c\})$ where $X^c$ is given by $x$ in $X$ with $f(x) \leq f(c)$. $\endgroup$
    – Vidit Nanda
    Commented Sep 11, 2014 at 17:57
  • $\begingroup$ I do not know many things about the topological degree, i know that the critical groups of a functionnal at an isolated critical point is given by $C_k(\varphi,p)=H_k(\varphi^c\cap U,\varphi^c\cap U\setminus \lbrace p\rbrace), ~k\in\mathbb{N},$ and i know the relation between the leray-schauder index and the critical groups for a compact perturbation of identity $i(\varphi',p)=\sum_{k=0}^{\infty}(-1)^k \dim C_k(\varphi,p)$ $\endgroup$
    – Vrouvrou
    Commented Sep 11, 2014 at 18:04
  • $\begingroup$ Iknow that the critical groups distiguishe between the different critical point $\endgroup$
    – Vrouvrou
    Commented Sep 11, 2014 at 18:06
  • $\begingroup$ Right, so the Leray Schauder index is the Euler characteristic of the critical groups. It is easy to come up with examples of spaces which have the same Euler characteristic but completely different homology: in that sense, the critical groups are more refined invariants. $\endgroup$
    – Vidit Nanda
    Commented Sep 11, 2014 at 18:46
  • $\begingroup$ I don't realy understand can you give me an exaple please? and what about :"better estimate of the number of the solutions " thank you $\endgroup$
    – Vrouvrou
    Commented Sep 11, 2014 at 18:53

2 Answers 2

Reset to default
5
$\begingroup$

To see why Morse theory gives more info in the variational case consider system of equations on a torus $T^n$:

$$f_1(\theta_1,\dotsc, \theta_n)=\cdots=f_n(\theta_1,\dotsc, \theta_n)=0.$$

where $f_,\dotsc, f_n$ are smooth functions. Degree theory cannot say much about this system. However if

$$ (f_1,\dotsc, f_n)= \nabla g (\theta_1,\dotsc, \theta_n) $$

for some smooth function $g$, then Morse theory predicts that system of equations

$$ \nabla g=0 $$

has at least $n+1$ solutions. If we a priori know that all the solutions are nondegenerate, then we can conclude that that are at least $2^n$ solutions.

Look at the simplest case when $f(\theta)$ is a smooth $2\pi$-periodic function. The equation $f(\theta)=0$ may have no solutions, but the equation $f'(\theta)=0$ has at least two solutions.

The variational nature of the equation $f'(\theta)=0$ adds some subtle features which Morse theory speculates to produce more refined results.

Improve this answer
$\endgroup$
4
  • $\begingroup$ and from the relation between critical groups and Leray-schauder index, the fact that the Leray-schauder index is the Euler characteristic of the critical groups how it give us that Relative homology groups and critical groups provide both a finer structure and better estimate of the number of solutions? thank you $\endgroup$
    – Vrouvrou
    Commented Sep 12, 2014 at 10:10
  • $\begingroup$ i dont clearly understand your unswer because i don't know exactly how to find solutions using degree theory $\endgroup$
    – Vrouvrou
    Commented Sep 12, 2014 at 10:12
  • 2
    $\begingroup$ That's the point. Degree theory is ineffective in this example, while Morse theory has plenty to say. The lower bounds on the number of solutions of $\nabla g=0$ are obtained using homology theory. $\endgroup$
    – Liviu Nicolaescu
    Commented Sep 12, 2014 at 11:58
  • $\begingroup$ Small question please, i don't know degree but why we can't use it in this case ? thank you $\endgroup$
    – Vrouvrou
    Commented Sep 13, 2014 at 6:13
3
$\begingroup$

Here is an example to illustrate the difference between the two approaches to find critical points. Consider the two-dimensional sphere $S^2$ and a function $f \in C^2(S^2 \to \mathbb{R})$ and assume that $f$ has two nondegenerate minimum points. The question is whether you can find other critical points.

You can apply degree theory to the map $F = \nabla f \in C^1 (S^2, TS^2)$. The nondegenerate critical points give you two nondegenerates zeroes of the vector field $F$, and that is consistent with the degree theory for maps in $C^1 (S^2, TS^2)$ for which the sum of the degrees should be $2$.

On the other hand, if you apply Morse theory, the Morse inequalities give you that $f$ should have two additional critical points corresponding to a maximum and a saddle point.

An explanationis that, in the two-dimensional case, degree theory does not distinguish between a maximum and a minimum point, whereas Morse theory does. In general, in dimension higher than $1$, the degree of a zero of the gradient around a critical point is a weaker information then the Morse index of the function at a critical point.

Improve this answer
$\endgroup$
11
  • $\begingroup$ What about the number of solutions please ? "better estimate of the number of solutions " $\endgroup$
    – Vrouvrou
    Commented Sep 12, 2014 at 10:55
  • $\begingroup$ With degree theory, you cannot prove the existence of any additional solution, whereas with Morse theory you prove the existence of two additional solutions. $\endgroup$
    – Jean Van Schaftingen
    Commented Sep 12, 2014 at 12:30
  • $\begingroup$ @ Jean I think that in your example you need to replace $S^2$ with a surface $\Sigma$ of positive genus so that $H_1(\Sigma)\neq 0$. On $S^2$ there exist Morse functions with exactly two critical points, a minimum and a maximum and no saddle points. (Take for example the height function.) $\endgroup$
    – Liviu Nicolaescu
    Commented Sep 12, 2014 at 13:01
  • 2
    $\begingroup$ @LiviuNicolaescu I am assuming in my example that I have already two minimum points, and I am searching for additional critical points. $\endgroup$
    – Jean Van Schaftingen
    Commented Sep 12, 2014 at 13:20
  • $\begingroup$ @JeanVanSchaftingen please can you tell me how you defin $f:S^2\rightarrow \mathbb{R}$ and why it has two critical point ? thank you $\endgroup$
    – Vrouvrou
    Commented Sep 12, 2014 at 13:59

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .