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    Maps $\mathbb{RP}^n\to\mathbb{RP}^m$ for $n>m$ and the Borsuk--Ulam theorem

    E-HAT-3.2-3

    Area
    Topology
    Topics
    Cohomology, Cup Product
    Status
    draft

    1. Using the cup product structure, show there is no map \(\mathbb{RP}^n \to \mathbb{RP}^m\) inducing a nontrivial map \(H^1(\mathbb{RP}^m; \mathbb{Z}_2) \to H^1(\mathbb{RP}^n; \mathbb{Z}_2)\) if \(n > m\). What is the corresponding result for maps \(\mathbb{CP}^n \to \mathbb{CP}^m\)?

    2. Prove the Borsuk–Ulam theorem by the following argument. Suppose on the contrary that \(f: S^n \to \mathbb{R}^n\) satisfies \(f(x) \neq f(-x)\) for all \(x\). Then define \(g: S^n \to S^{n-1}\) by \(g(x) = \bigl(f(x) - f(-x)\bigr) / \bigl|f(x) - f(-x)\bigr|\), so \(g(-x) = -g(x)\) and \(g\) induces a map \(\mathbb{RP}^n \to \mathbb{RP}^{n-1}\). Show that part (a) applies to this map.

    Solution

    With \(\mathbb Z_2\) coefficients, \[ H^*(\mathbb{RP}^r;\mathbb Z_2)\cong \mathbb Z_2[\alpha]/(\alpha^{r+1}),\qquad |\alpha|=1. \] Suppose \(f:\mathbb{RP}^n\to\mathbb{RP}^m\) with \(n>m\) induces a nonzero map on \(H^1\). Since both \(H^1\) groups are one-dimensional, if \(\alpha_m,\alpha_n\) denote the generators then \[ f^*(\alpha_m)=\alpha_n. \] Hence \[ 0=f^*(\alpha_m^{m+1})=\alpha_n^{m+1}, \] contradicting \(m+1\le n\), since \(\alpha_n^{m+1}\ne0\). Thus every such map is zero on \(H^1\).

    Similarly \[ H^*(\mathbb{CP}^r;\mathbb Z)\cong \mathbb Z[\beta]/(\beta^{r+1}),\qquad |\beta|=2. \] If \(n>m\) and \(f:\mathbb{CP}^n\to\mathbb{CP}^m\) has \(f^*(\beta_m)=d\beta_n\), then \[ 0=f^*(\beta_m^{m+1})=d^{m+1}\beta_n^{m+1}. \] Since \(H^{2m+2}(\mathbb{CP}^n;\mathbb Z)\) is torsion-free and \(\beta_n^{m+1}\ne0\), this forces \(d=0\). Hence \(f^*:H^2\to H^2\) is zero.

    For Borsuk–Ulam, suppose \(f:S^n\to\mathbb R^n\) satisfies \(f(x)\ne f(-x)\) for all \(x\). Then \[ g(x)=\frac{f(x)-f(-x)}{|f(x)-f(-x)|} \] defines an antipodal-equivariant map \(g:S^n\to S^{n-1}\), so it descends to \[ \bar g:\mathbb{RP}^n\to\mathbb{RP}^{n-1}. \] The map \(\bar g\) is nontrivial on \(H^1(-;\mathbb Z_2)\): on fundamental groups it sends the nontrivial loop in \(\mathbb{RP}^n\) to the nontrivial loop in \(\mathbb{RP}^{n-1}\), because an equivariant lift joins antipodal endpoints upstairs. This contradicts the first part with \(m=n-1\). Therefore some \(x\) satisfies \(f(x)=f(-x)\).

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