Search the corpus

    Kronecker pairing

    D-VP4LC

    Let \(X\) be a topological space, \(R\) a commutative ring, and \(n\geq 0\). For a singular cochain \(\varphi\in C^n(X;R) = \Hom_R(C_n(X;R), R)\) and a singular chain \(\alpha\in C_n(X;R)\), put \(\inner{\varphi}{\alpha} \coloneqq \varphi(\alpha)\). The Kronecker pairing, also called the Kronecker product, is the \(R\)-bilinear map \[ \inner{\wait}{\wait}\colon H^n(X; R) \cross H_n(X; R) \to R, \qquad \inner{[\varphi]}{[\alpha]} \coloneqq \varphi(\alpha), \] for a cocycle \(\varphi\) and a cycle \(\alpha\). Its adjoint is the evaluation homomorphism \[ h\colon H^n(X; R) \to \Hom_R(H_n(X;R), R), \qquad h([\varphi])([\alpha]) \coloneqq \varphi(\alpha). \]

    Let \(X\) be a topological space, \(R\) a commutative ring, and \(n\geq 0\).

    1. For \(\psi\in C^{n-1}(X;R)\) and \(\alpha\in C_n(X;R)\), \(\inner{\delta\psi}{\alpha} = \inner{\psi}{\del\alpha}\); hence the Kronecker pairing is well defined on classes.

    2. For a continuous map \(f\colon X\to Y\), \(\beta\in H^n(Y;R)\), and \(x \in H_n(X;R)\), \(\inner{f^*\beta}{x} = \inner{\beta}{f_*x}\).

    3. If \(R\) is a principal ideal domain, there is a split short exact sequence \[ 0 \to \Ext_R(H_{n-1}(X;R), R) \to H^n(X;R) \xrightarrow{h} \Hom_R(H_n(X;R), R) \to 0, \] so \(h\) is surjective, and \(h\) is an isomorphism if and only if \(\Ext_R(H_{n-1}(X;R), R) = 0\).

    The evaluation homomorphism \(h\) need not be injective. For \(X = \RR P^2\), \(R = \ZZ\), and \(n = 2\), one has \(H_1(\RR P^2;\ZZ)\cong\ZZ/2\) and \(H_2(\RR P^2;\ZZ) = 0\), so \(H^2(\RR P^2;\ZZ)\cong\Ext_\ZZ(\ZZ/2,\ZZ)\cong\ZZ/2\) while \(\Hom_\ZZ(H_2(\RR P^2;\ZZ),\ZZ) = 0\).

    [@Hat02].

    Guide Appearances

    What links to this