# Lefschetz number

For the unit sphere $S^n \subset \mathbb{R}^{n+1}$ let $f : S^n \to S^n$ be the map reversing the signs of all but one coordinate, $$f(x_0, x_1, \dots, x_n) = (x_0, -x_1, \dots, -x_n):$$ Compute the Lefschetz number $L(f)$.

So I am actually wondering, that if I should use the sterographic projection, or I can just use the local parametrization $\psi: \mathbb{R}^m \mapsto S^m \subseteq \mathbb{R}^{m+1}$ that $$\psi(x_1, \dots, x_n) = (\sqrt{1 - x_1^2 - \dots - x_n^2}, x_1, x_2, \dots, x_n)?$$

• What definition are you using for Lefchetz number? I only know of it as the trace of the induced map on homology groups with rational coefficients. In this case you can see that this definition coincides with the Brouwer degree of $f.$
– user17794
Commented Jul 24, 2013 at 5:25

Because $(x_0, x_1, \dots, x_n)$ sits on the unit sphere, $x_0 = \sqrt{1 - x_1^2 - \dots - x_n^2}$. Since $x_0$ is dependent of $x_1, \dots, x_n$, I can simply drop it as rewrite $f$ as: $$\tilde f(\tilde x_1, \dots, \tilde x_n) = (-\tilde x_1, \dots, - \tilde x_n).$$
Then $$df_x - I = \begin{pmatrix} -2 & 0 & \cdots & 0 \\ 0 & -2 & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & -2 \end{pmatrix}.$$ Hence, $$\det(df_x - I) = (-2)^n.$$ Since the sign of $L_x(f)$ equals the sign of $\det(df_x - I)$, we have $L_x(f) = 1$ if $n$ is even, and $L_x(f) = -1$ is $n$ is odd.
But since $df_x$ is independent of $x$, we have $L(f) = 2L_x(f)$, since clearly, there are two fixed points $(1,0,\dots, 0)$ and $(-1,0,\dots, 0)$. Hence $L(f) = 2$ if $n$ is even, and $L(f) = -2$ is $n$ is odd.