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Let $X$ be an oriented compact manifold of dimension $2k$. Suppose that a compact Lie group $G$ acts differentiably on $X$ in an orientation-preserving way. Let $B$ be the ${\mathbb R}$-bilinear form on $H^k(X; {\mathbb R})$ be given by cup product evaluated on the fundamental class of $X$, i.e. $B(x,y)=(x\cup y)[X]$. Why does the induced action of $G$ on $H^k(X; {\mathbb R})$ preserve $B$?

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    $\begingroup$ Because all products are natural? $\endgroup$
    – Alex Degtyarev
    Commented Oct 17, 2016 at 22:16
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    $\begingroup$ Besides naturality, we also need that $g^*[X]=[X]$ for $g \in G$, which is because $G$ acts by orientation-preserving automorphisms. $\endgroup$
    – HeinrichD
    Commented Oct 17, 2016 at 22:17
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    $\begingroup$ You're both right (it's natural in the setting of oriented manifolds, rather than besides naturality...) $\endgroup$
    – YCor
    Commented Oct 18, 2016 at 2:09

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