What is the exterior algebra? 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 $$(\omega_0,\ldots,\omega_n)\wedge(\eta_0,\ldots,\eta_n)?$$

 A: 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 $\eta_j$.
A: *

*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 $n<i+j$:
$$\omega\wedge\eta=\sum_{i+j\leq n}\omega_i\wedge\eta_j$$
