Free, open-source online mathematics for students, teachers and workers

The exterior product $\wedge$ is inherited in cohomology:

$$\wedge:H^k(M)\times H^l(M)\longrightarrow H^{k+l}(M)$$
$$[\alpha]\wedge[\beta]=[\alpha\wedge\beta]$$

For $\wedge$ to be well-defined, we have to check two properties:

  1. If $\alpha$ and $\beta$ are closed, $\alpha\wedge\beta$ is also closed:

    $$\mathrm{d} \alpha=0,\,\mathrm{d} \beta=0\Longrightarrow\mathrm{d}(\alpha\wedge\beta)=\mathrm{d} \alpha\wedge\beta+(-1)^k\alpha\wedge\mathrm{d}\beta=0$$

    Therefore,

    $$\wedge:\Omega^k(M)\times \Omega^l(M)\longrightarrow \Omega^{k+l}(M)$$

    induces

    $$\wedge:Z^k(M)\times Z^l(M)\longrightarrow Z^{k+l}(M)$$

    and also

    $$\wedge:Z^k(M)\times Z^l(M)\longrightarrow \left[Z^{k+l}(M)/B^{k+l}(M)=H^{k+l}(M)\right]$$

  2. If moreover $\alpha$ or $\beta$ is exact, $\alpha\wedge\beta$ is also exact:

    $$\alpha=\mathrm{d} \gamma\Longrightarrow\alpha\wedge\beta=\mathrm{d}\gamma\wedge\beta=\mathrm{d}(\gamma\wedge\beta)-(-1)^{k-1} \gamma\wedge\mathrm{d}\beta=\mathrm{d}(\gamma\wedge\beta)$$

    Therefore,

    $$\wedge:Z^k(M)\times Z^l(M)\longrightarrow \left[Z^{k+l}(M)/B^{k+l}(M)=H^{k+l}(M)\right]$$

    induces

    $$\wedge:\left[Z^k(M)/B^k(M)=H^k(M)\right]\times \left[Z^l(M)/B^l(M)=H^l(M)\right]\longrightarrow \left[Z^{k+l}(M)/B^{k+l}(M)=H^{k+l}(M)\right]$$