What is the exterior algebra?

Question

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 $$(\omega_0,\ldots,\omega_n)\wedge(\eta_0,\ldots,\eta_n)?$$

Answer 1

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 $\eta_j$.

Answer 2

• The equation $$\Lambda=\bigoplus_{m=0}^n\Lambda^m$$ means that $\Lambda^m\subset\Lambda$ and the function \begin{align} \Lambda^0\times\cdots\times \Lambda^n&\to\Lambda\\ (\omega_0,\ldots,\omega_n)&\mapsto \omega_0+\ldots+\omega_n \end{align} is bijective, not necessarely that $\Lambda=\Lambda^0\times\cdots\times \Lambda^n$.
• If $\omega=\omega_0+\ldots+\omega_n\in\Lambda$ and $\eta=\eta_0+\ldots+\eta_n\in\Lambda$, then $$\omega\wedge\eta=\sum_{i=0}^n\omega_i\wedge\sum_{j=0}^n\eta_j=\sum_{i=0}^n\sum_{j=0}^n\omega_i\wedge\eta_j$$ since the wedge product is bilinear. In addition, you can use the fact that $\omega_i\wedge\eta_j=0$ if $n<i+j$: $$\omega\wedge\eta=\sum_{i+j\leq n}\omega_i\wedge\eta_j$$