1
$\begingroup$

Let $M$ be a closed manifold with holomorphic cell decomposition (if it is complex), or at least with only even cohomology. In particular, its Euler characteristic is equal to the sum of its Betti number.

Let $\phi$ be a diffeomorphism which is isotopic to the identity. As far a I understand, Lefschetz theorem implies that $\phi$ must have at least one fixed point. Moreover, its Lefschetz number is equal to the sum of Betti numbers of $M$.

Is it straightforward then that the number of fixed points of $\phi$, counted with multiplicity, is at least the sum of Betti numbers ? I guess my problem lies in the fact that I don't really know how to relate the Lefschetz number and the multiplicity of the fixed points.

$\endgroup$
  • 1
    $\begingroup$ The Lefschetz number is not the sum of Betti numbers. It is the Euler characteristic, i.e., the alternating sum of Betti numbers. $\endgroup$ – ThiKu Oct 17 '18 at 14:23
  • 1
    $\begingroup$ @ThiKu: yes, but the assumption is that $M$ has a decomposition by even-dimensional cells, so there are no odd degree Betti numbers. $\endgroup$ – R. van Dobben de Bruyn Oct 17 '18 at 14:24
  • $\begingroup$ Thanks for your comments. I think that my problem actually lies in the fact that I don't really understand the relation between the Lefschetz number of $\phi$ and the total multiplicity of its fixed points. $\endgroup$ – BrianT Oct 17 '18 at 14:27
  • 1
    $\begingroup$ What is the definition of a holomorphic cell decomposition? $\endgroup$ – Nick L Oct 17 '18 at 14:35
  • 7
    $\begingroup$ The only notion of multiplicity I know for fixed points is the fixed point index. The sum of the fixed points counted with their fixed point index EQUALS the Lefschetz number, whenever the set of fixed points is finite. That's the Lefschetz trace formula. $\endgroup$ – Dan Petersen Oct 25 '18 at 7:14
-1
$\begingroup$

I think this is not true in general. It is classical (converse of Poincare-Hopf) that every closed smooth manifold with nonzero Euler characteristic has a vector field which vanishes only at one point. If we deform the identity along this vector field we get a diffeomorphism $f$ isotopic to the identity with a single fixed point.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.