1
$\begingroup$

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 $$\psi(x_1, \dots, x_n) = (\sqrt{1 - x_1^2 - \dots - x_n^2}, x_1, x_2, \dots, x_n)?$$

$\endgroup$
1
  • $\begingroup$ 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 $f.$ $\endgroup$
    – user17794
    Commented Jul 24, 2013 at 5:25

1 Answer 1

3
$\begingroup$

I am not too convinced with myself that the intersection can be treated locally, hence sterographic projection is not necessary, so I can just drop the first coordinate.

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.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .