Is exterior algebra an example of an algebra over a field?

Question

An algebra over a field is a pair $(V,\cdot)$ where $V$ is a vector space over a field $F$ with a bilinear product $\cdot:V\times V\to V$. In the Szekeres' book on Mathematical Physics, he explains exterior algebra $\bigwedge(V)$ over a vector space $V$ as an example of that definition. But I'm feeling that the wedge product is not closed, meaning that if $u,v\in V$, then $u\wedge v$ does not belong to $V$. Am I feeling right? If so, Is there some structure which the exterior algebra is an example of (something like "algebra over a vector space", I googled it but didn't find any references)?

Answer 1

The exterior algebra (generated by a vector space $V$) is an algebra. It is the quotient of the tensor algebra $TV$ by the relation $xy+yx=0$ for all $x,y\in V$. I think you are confusing what is the algebra and what is the vector space. $V$, in this context, does not form an algebra under the wedge product. However, the vector space $\Lambda(V)$ does. We have a bilinear map $$ \wedge:\Lambda(V)\times \Lambda(V)\to \Lambda(V) $$ The elements of $\Lambda(V)$ look very different than those of $V$. In fact, if $\{e_{i}\}$ is a basis of $V$, an homogeneous element of degree $n$ of $\Lambda(V)$ will look like a linear combination $$ \sum_{i_1\leq i_2\leq ...\leq i_n\leq k} \alpha_{i_1i_2... i_n}e_{i_1}\wedge e_{i_2}\wedge ... \wedge e_{i_n}, $$ where $V$ has dimension $k$.

Answer 2

First of all, $V$ in $\wedge V$ is a vector space, not vector field (just to fix the terminology). As egreg pointed out, if $u,v\in V$, then $u\wedge v$ is not an element of $V$, but it is an element of $\wedge^2 V\subset \wedge V$. The algebra $\wedge V$ is really constructed in such a way that you will have a correctly defined operation $\wedge$ on it. In a certain sense, it is the smallest algebra which contains vector space $V$ and has a skew-commutative multiplication.

You can read very detailed explanation of the object $\wedge V$ on Wikipedia.