Verify that the formula \(f(z_1, \cdots, z_{2k}) = (\bar{z}_2, -\bar{z}_1, \bar{z}_4, -\bar{z}_3, \cdots, \bar{z}_{2k}, -\bar{z}_{2k-1})\) defines a map \(f: \mathbb{C}^{2k} \to \mathbb{C}^{2k}\) inducing a quotient map \(\mathbb{CP}^{2k-1} \to \mathbb{CP}^{2k-1}\) without fixed points.
Solution
Define \[ F(z_1,\ldots,z_{2k})=(\bar z_2,-\bar z_1,\ldots,\bar z_{2k},-\bar z_{2k-1}). \]
1. The map \(F\) is conjugate-linear and satisfies\[ F(\lambda z)=\bar\lambda F(z), \qquad F^2=-\operatorname{id}. \]
Both identities follow by direct calculation on each coordinate pair \((z_{2j-1},z_{2j})\): \[ (z_{2j-1},z_{2j})\mapsto(\bar z_{2j},-\bar z_{2j-1}) \mapsto(-z_{2j-1},-z_{2j}). \]
2. Hence \(F\) induces a well-defined map\[ f:\mathbb{CP}^{2k-1}\to\mathbb{CP}^{2k-1}, \qquad f([z])=[F(z)]. \]
If \([z]=[\lambda z]\) with \(\lambda\ne0\), then \[ F(\lambda z)=\bar\lambda F(z), \] so \(F(\lambda z)\) and \(F(z)\) determine the same complex line. Also \(F(z)\ne0\) for \(z\ne0\) because \(F^2=-\operatorname{id}\).
3. The map \(f\) has no fixed point.
If \([z]\) were fixed, then for some \(\lambda\in\mathbb C^*\) we would have \[ F(z)=\lambda z. \] Applying \(F\) and using conjugate-linearity gives \[ -z=F^2(z)=F(\lambda z)=\bar\lambda F(z)=|\lambda|^2z. \] Since \(z\ne0\), this would imply \(|\lambda|^2=-1\), impossible.
Thus the displayed conjugate-linear map induces a fixed-point-free self-map of \(\mathbb{CP}^{2k-1}\).