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As I learn from the comments to this question a non elliptic operator on a compact manifold can not be a fredholm operator.

However, I learn also from the conversations in the same post that the hypoelliptic operators can be Fredholm but non elliptic operators

On the other hand it seems that the following operator is fredholm(At least, algebraically):

For a Morse function $F:S^{2}\to \mathbb{R}$ which has only two singular points(one max at N and one min at S), define $D:C^{\infty}(S^{2})\to C^{\infty}(S^{2})$ with $D(g)= \nabla F. \nabla g$.

The simple dynamical structure of the gradient vector field $\nabla F$ easily shows that the kernel of $D$ is one dimensional and codimension of the range is 2. So apparently we have a fredholm operator.

Is not this a contradictory situation?

It is written in the comments to the above question that this is a fredholm opeartor, algebrically but not analytically. Can one help me to understand this statement?

Ali Taghavi
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    I can see that the kernel of $D$ contains only constant functions, but how do you argue that the codimension of the range is 2? – Lukas Geyer May 15 '17 at 16:36
  • @LukasGeyer Assume that $h$ is an smooth function on the spher with $h(P)=h(N)=0$. We deined $D(g)=X.g$ the derivation of $g$ along the trajectories of $X$ where $X$ has jyst two singularity at $N$ and $P$. We denote by $\phi_{t}$ the flow of the vector field $X$. Them $g(x)= \int_{-\infty } ^{+\infty} h(\phi_{t}(x))dt$ satisfies $X.g=h$. this obviousely shows that the codimension of the range of $D$ is two. Note that the integral is convergent since we assumed that the singularities are hyperbolic, so they attract the orbits exponentially. – Ali Taghavi May 15 '17 at 16:50
  • @LukasGeyer The singularities are hyperbolic since we choos $X$ as a gradient vector field associated with a morse function with a Maximum and a minimum. – Ali Taghavi May 15 '17 at 16:57
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    Just to elaborate, if you assume that the max and min are 1 and 0, then it seems you will only get maps in your image whose integral along the gradient flow lines of $F$ are 1. However, a generic smooth function will not have the same integral along each of the flow lines, indicating that the codimension of the range is infinite. – Lukas Geyer May 15 '17 at 17:16
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    Ali, your function $g$ will in general not have a continuous extension to $P$ and $N$, unless I am missing something. – Lukas Geyer May 15 '17 at 17:19
  • @LukasGeyer Thank you very much for your answer. I was mistaken.I did the same mistake as in the incorrect proposition of the following note: The proposition says the codimension of the range IS EQUALE to the number of limit cycles but the true version is that the codimension of the range is an upper bound for the number of limit cycles: https://arxiv.org/pdf/math/0408037.pdf – Ali Taghavi May 15 '17 at 17:52
  • The true version is indicated in the second part of the following note:https://arxiv.org/abs/1302.0001 – Ali Taghavi May 15 '17 at 17:55
  • The situation is also discussed in a talk by Loic Teyssier https://cms.math.ca/Events/Toulouse2004/abs/ss7.html#lt – Ali Taghavi May 15 '17 at 17:55
  • @LukasGeyer For a related post please see the following: https://mathoverflow.net/questions/164059/codimension-of-the-range-of-certain-linear-operators – Ali Taghavi May 15 '17 at 17:57
  • @LukasGeyer What function algebra is appropriate for action of the operator $D$? That is what algebra $A$ of functions is invariant under $D$ with the property that $A$ seperates compact subsets of the plane and the codimension of $D$ is finite? – Ali Taghavi May 15 '17 at 18:00
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    Ali: Sorry, I have no idea what the appropriate function algebra would be. – Lukas Geyer May 16 '17 at 00:25

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