Given a map \(f: S^{2n} \to S^{2n}\), show that there is some point \(x \in S^{2n}\) with either \(f(x) = x\) or \(f(x) = -x\). Deduce that every map \(\mathbb{RP}^{2n} \to \mathbb{RP}^{2n}\) has a fixed point. Construct maps \(\mathbb{RP}^{2n-1} \to \mathbb{RP}^{2n-1}\) without fixed points from linear transformations \(\mathbb{R}^{2n} \to \mathbb{R}^{2n}\) without eigenvectors.
Solution
Goal. Show every map \(f: S^{2n} \to S^{2n}\) has a fixed point or an antipodal point; deduce every map \(\RP^{2n} \to \RP^{2n}\) has a fixed point; and construct fixed-point-free maps of \(\RP^{2n-1}\).
1. Every map \(f: S^{2n} \to S^{2n}\) has \(f(x) = x\) or \(f(x) = -x\) for some \(x\).
1.1. If \(f(x) \neq x\) for all \(x\), then \(f\) has no fixed point, so \(\deg f = (-1)^{2n+1} = -1\).
a fixed-point-free map of \(S^m\) has degree \((-1)^{m+1}\) (the antipodal map \(x \mapsto -x\) has degree \((-1)^{m+1}\), and any fixed-point-free map is homotopic to it).
1.2. If also \(f(x) \neq -x\) for all \(x\), then \(-f\) has no fixed point, so \(\deg(-f) = -1\).
\(-f\) is fixed-point-free, so by 2.1 its degree is \((-1)^{2n+1} = -1\).
1.3. But \(\deg(-f) = (-1)^{2n+1}\deg f = -\deg f = -(-1) = 1\), contradiction.
the antipodal map has degree \((-1)^{2n+1} = -1\), and \(\deg(-f) = \deg(\text{antipodal} \circ f) = (-1)^{2n+1}\deg f = -\deg f\).
1.4. Hence \(f(x) = x\) or \(f(x) = -x\) for some \(x\).
2.3 contradicts the assumption that neither holds.
2. Every map \(\RP^{2n} \to \RP^{2n}\) has a fixed point.
2.1. A map \(g: \RP^{2n} \to \RP^{2n}\) lifts to a map \(\tilde g: S^{2n} \to S^{2n}\).
\(\RP^{2n}\) has universal cover \(S^{2n}\); since \(\pi_1(\RP^{2n}) = \ZZ/2\) and \(g_*\) maps \(\ZZ/2\) into itself, the lift exists (the induced map on \(\pi_1\) is either identity or zero, both compatible with the covering).
2.2. By 1, \(\tilde g(x) = x\) or \(\tilde g(x) = -x\) for some \(x\).
apply 1 to \(\tilde g\).
2.3. In either case \(g([x]) = [x]\).
\([x] = [-x]\) in \(\RP^{2n}\), and \(g([x]) = [\tilde g(x)] = [x]\) or \([-x] = [x]\).
3. Fixed-point-free maps of \(\RP^{2n-1}\).
3.1. A linear map \(T: \RR^{2n} \to \RR^{2n}\) without real eigenvectors induces a map \(\RP^{2n-1} \to \RP^{2n-1}\).
\(T\) sends lines to lines (it is linear), and no line is fixed because a fixed line would be an eigenspace, i.e. an eigenvector.
3.2. Example: \(T(x_1, \dots, x_{2n}) = (-x_2, x_1, -x_4, x_3, \dots, -x_{2n}, x_{2n-1})\), a block rotation by \(90^\circ\) in each coordinate pair.
\(T\) has no real eigenvectors (its eigenvalues are \(\pm i\)), so the induced map on \(\RP^{2n-1}\) has no fixed point.
4. Q.E.D.
1, 2, and 3 are the three requested statements.