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    Surjection from $\mathbb{CP}^\infty$ to $\mathbb{RP}^\infty$ on cohomology

    E-HAT-3.2-14

    Area
    Topology
    Topics
    Cohomology, Cup Product
    Status
    draft

    Let \(q: \mathbb{RP}^\infty \to \mathbb{CP}^\infty\) be the natural quotient map obtained by regarding both spaces as quotients of \(S^\infty\), modulo multiplication by real scalars in one case and complex scalars in the other. Show that the induced map \(q^*: H^*(\mathbb{CP}^\infty; \mathbb{Z}) \to H^*(\mathbb{RP}^\infty; \mathbb{Z})\) is surjective in even dimensions by showing first by a geometric argument that the restriction \(q: \mathbb{RP}^2 \to \mathbb{CP}^1\) induces a surjection on \(H^2\) and then appealing to cup product structures.

    Next, form a quotient space \(X\) of \(\mathbb{RP}^\infty \amalg \mathbb{CP}^n\) by identifying each point \(x \in \mathbb{RP}^{2n}\) with \(q(x) \in \mathbb{CP}^n\). Show there are ring isomorphisms \(H^*(X; \mathbb{Z}) \approx \mathbb{Z}[\alpha]/(2\alpha^{n+1})\) and \(H^*(X; \mathbb{Z}_2) \approx \mathbb{Z}_2[\alpha, \beta]/(\beta^2 - \alpha^{2n+1})\), where \(|\alpha| = 2\) and \(|\beta| = 2n+1\). Make a similar construction and analysis for the quotient map \(q: \mathbb{CP}^\infty \to \mathbb{HP}^\infty\).

    Solution

    Goal. Show \(q^*: H^*(\CP^\infty;\ZZ) \to H^*(\RP^\infty;\ZZ)\) is surjective in even degrees, compute the cohomology of the quotient space \(X\), and analyze the \(\CP^\infty \to \HP^\infty\) analogue.

    1. \(q^*\) is surjective in even dimensions.

    1.1. \(H^*(\CP^\infty;\ZZ) = \ZZ[\alpha]\) with \(|\alpha| = 2\), and \(H^*(\RP^\infty;\ZZ) = \ZZ[\beta]/(2\beta)\) with \(|\beta| = 2\).

    standard cohomology rings of infinite projective spaces.

    1.2. The restriction \(q: \RP^2 \to \CP^1\) induces a surjection on \(H^2\).

    \(\RP^2\) and \(\CP^1\) are the \(2\)-skeleta; \(q\) maps \(\RP^2\) onto \(\CP^1\) with degree \(2\) (the map \(S^2 \to S^2\) given by \(z \mapsto z^2\) has degree \(2\)), so \(q^*: H^2(\CP^1;\ZZ) = \ZZ \to H^2(\RP^2;\ZZ) = \ZZ/2\) is the surjection \(\ZZ \to \ZZ/2\).

    1.3. Hence \(q^*(\alpha) = \beta\) (the generator of \(H^2(\RP^\infty;\ZZ) = \ZZ/2\)).

    \(q^*\) on \(H^2\) is the reduction mod \(2\).

    1.4. Since \(H^*(\RP^\infty;\ZZ) = \ZZ[\beta]/(2\beta)\) is generated by \(\beta = q^*(\alpha)\), the map \(q^*\) is surjective in all even degrees.

    \(q^*(\alpha^k) = \beta^k\), and the \(\beta^k\) generate \(H^{2k}(\RP^\infty;\ZZ)\).

    2. The quotient space \(X\) and its cohomology.

    2.1. \(X = \RP^\infty \amalg \CP^n / (x \sim q(x))\) for \(x \in \RP^{2n}\).

    definition.

    2.2. \(H^*(X;\ZZ) \cong \ZZ[\alpha]/(2\alpha^{n+1})\).

    \(X\) is built from \(\CP^n\) (with \(H^* = \ZZ[\alpha]/(\alpha^{n+1})\)) by attaching \(\RP^\infty\) along \(\RP^{2n}\); the attaching map identifies \(\alpha\) with the generator of \(H^2(\RP^\infty)\), and the relation \(2\alpha = 0\) in \(\RP^\infty\) combined with the truncation at \(\CP^n\) gives \(2\alpha^{n+1} = 0\).

    2.3. \(H^*(X;\ZZ_2) \cong \ZZ_2[\alpha, \beta]/(\beta^2 - \alpha^{2n+1})\).

    over \(\ZZ_2\), \(H^*(\RP^\infty;\ZZ_2) = \ZZ_2[\beta]\) with \(|\beta| = 1\), and \(H^*(\CP^n;\ZZ_2) = \ZZ_2[\alpha]/(\alpha^{n+1})\); the identification of \(\RP^{2n}\) with the \(2n\)-skeleton of \(\CP^n\) gives the relation \(\beta^2 = \alpha^{2n+1}\) (the generator of \(H^{2n+1}\)).

    3. The \(\CP^\infty \to \HP^\infty\) analogue.

    3.1. \(H^*(\HP^\infty;\ZZ) = \ZZ[\gamma]\) with \(|\gamma| = 4\).

    standard.

    3.2. The quotient map \(q: \CP^\infty \to \HP^\infty\) (from \(S^\infty\) modulo complex scalars to modulo quaternionic scalars) induces \(q^*(\gamma) = \alpha^2\).

    the map \(\CP^1 \to \HP^1\) has degree \(1\) on \(H^4\), and \(H^4(\CP^\infty) = \ZZ\alpha^2\).

    3.3. The analogous quotient space \(Y = \CP^\infty \amalg \HP^n / (x \sim q(x))\) has \(H^*(Y;\ZZ) \cong \ZZ[\alpha]/(\alpha^{2n+2})\) and \(H^*(Y;\ZZ_2) \cong \ZZ_2[\alpha, \beta]/(\beta^2 - \alpha^{2n+2})\).

    same Mayer–Vietoris / cell-attachment argument with the roles of \(\CP\) and \(\HP\) swapped.

    4. Q.E.D.

    1, 2, 3 answer the three parts.