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


I have this paragraph from K.C. Chang [Infinite dimensional Morse theory](http://books.google.sk/books?id=wPo_AQAAIAAJ)



> 
> 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.



Morse theory Vs degree theory
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.


