I have $F\colon \Bbb R^2\to \Bbb R^2$ given by $(x,y)\mapsto (xy,x^2-3y^2)$. It's easy to see that such map induces a map $f\colon \Bbb R P^1\to \Bbb R P^1$. I'm asked to compute its degree and its Lefschetz number.

So the fixed point I found of $f$ are $[0;1]$, $[2;1]$ and $[-2;1]$. Notice that they are in the domain of the local chart $[x;1]\mapsto x$ so we can work with this chart only.

A quick computation of the differential in this chart gives us that $$d_{[x;1]}f=-\dfrac{x^2+3}{(x^2-3)^2}$$ If we plug in our fixed point we see that everyone of them is Lefschetz, with local lefschetz number $-1$ and therefore the Lefschetz number is $-3$.

From this we see that the degree should be $4$, thanks to the characterisation of the Lefschetz number as alternate sum of trace of maps into homology.

But a direct computation of the degree gives us a different answer. In fact the value $[0,1]$ is regular, since its preimages $[0;1]$ and $[1;0]$ are regular points. A standard computations gives that the degree is $-2$. This would imply that the Lefschetz number is $3$ and not $-3$.

Where am I missing something?

EDIT: the answer below suggests that I should compute the local Lefschtez number with $I-d_pf$ and not $d_pf-I$. Why is that the case? According to Guillemin-Pollack the second formula is the correct one.

  • 3
    $\begingroup$ Your question prompted me to look through books and to think about this more carefully. Guillemin & Pollack, Hirsch, Bott & Tu — all use $\det(df_x - I)$ to define the local Lefschetz number. The Poincaré-Hopf theorem, which computes the sum of the indices of a vector field $\vec v$ as the intersection number $I(Z,\vec v) = I(Z,Z) = \chi(M)$ ($Z$ being the zero section of the tangent bundle). We all expect the Lefschetz number to be $I(\Delta,\text{gr}(f))$ by analogy. Well, that does not agree with the homological definition of the Lefschetz number when $\dim M$ is odd. $\endgroup$ Aug 24, 2017 at 22:48
  • 2
    $\begingroup$ One needs either to consider $I(\text{gr(f)},\Delta)$ or to reverse the graph (as Godbillon does), considering the mapping $x\rightsquigarrow (f(x),x)$. This will then be consistent with the homological definition of Lefschetz number, but this necessitates the local Lefschetz number's being given by $\det(I-df_x)$, as Math536 says in his answer. Sigh! $\endgroup$ Aug 24, 2017 at 22:51

1 Answer 1


The local Lefschetz number of $f$ at $[x;1]$ (where $x \in \{0,-2,2\}$) is given by $$ L(f,[x;1]) = \mathrm{sgn} ( \det (1 - d_{[x;1]}f) ) = \mathrm{sgn} \left( 1 + \frac{x^2+3}{(x^2 - 3)^2} \right) = 1. $$ Thus the Lefschetz number of $f$ is $3$.

Some books define the local Lefschetz number by $$(1) \quad L(f,p) = \mathrm{sgn} (\det( d_pf - 1)) $$ instead of $$(2) \quad L(f,p) = \mathrm{sgn} (\det( 1 - d_pf)). $$ These two definitions are not equivalent: they agree for even-dimensional manifolds, but differ by a sign for odd-dimensional manifolds. If you want the Lefschetz number of $f$ (defined as an alternating sum of traces of induced maps on homology) to be equal to the sum of its local Lefschetz numbers, you need to use definition (2).

Lefschetz fixed-point formula: Let $M^n$ be a closed smooth manifold and $f:M \to M$ be a smooth map with nondegenerate fixed points. Then $$ \sum_{i=0}^n (-1)^i \mathrm{Tr}(H_i(f)) = \sum_{p \in \mathrm{Fix}(f)} \mathrm{sgn}(\det(1 - d_pf))$$

  • $\begingroup$ Dear Math536, I thought that the local lefschetz number was given by $\det(d_pf-1)$. Why did you put a $-$ everywhere? $\endgroup$
    – Luigi M
    Aug 23, 2017 at 12:48

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