I am learning differential geometry, and I have difficulty understanding the construction of the exterior algebra of an $n$-dimensional vector space $V$.

We have the wedge product $$\wedge:\Lambda^k(V^\ast)\times\Lambda^l(V^\ast)\to\Lambda^{k+l}(V^\ast)$$ defined as $$\omega\wedge\eta=\frac{(k+l)!}{k!l!}\operatorname{Alt}(\omega\otimes\eta)$$ and that's all right. Then one just define the exterior algebra to be the direct sum $$\Lambda(V^\ast)=\bigoplus_{k=0}^n\Lambda^k(V^\ast),$$ and that is supposed to be an algebra. But $\wedge$ is defined only on $\Lambda^k(V^\ast)\times\Lambda^l(V^\ast)$, so how does it act on a general element? Component wise?

Question: If $(\omega_0,\ldots,\omega_n)\in\Lambda(V^*)$ and $(\eta_0,\ldots,\eta_n)\in\Lambda(V^*)$, what is $$(\omega_0,\ldots,\omega_n)\wedge(\eta_0,\ldots,\eta_n)?$$

  • $\begingroup$ What do you mean that $(\omega_1, \dots, \omega_n) \in \Lambda (V^*)$? If you mean to write these as members of the direct sum, with $\omega_i \in \Lambda^i(V^*)$, then that wedge product would just be $\sum_{i,j} \omega_i \wedge \eta_j$, right? $\endgroup$ – shalop Mar 19 '15 at 0:33

Each of $\omega_i$ and $\eta_j$ are elements of a particular $\Lambda^k$. It is perhaps more enlightening to write $$ \omega_0+\omega_1+\cdots +\omega_n. $$ Do the similar thing for the $\eta_j$. Then, you multiply using the distributive property and the fact that you know the wedge product for each pairing of $\omega_i$ with $\eta_j$.

  • $\begingroup$ Ok, thank you. So it make sense to write, for example, $3+dx-5dx\wedge dy$? My book was not so clear about this... $\endgroup$ – user224761 Mar 19 '15 at 0:43
  • $\begingroup$ @user224761 It does make sense to write that. $\endgroup$ – Joe Johnson 126 Mar 19 '15 at 1:23

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