Algebra over a field generated by a vector space?

Question

I have just learned about Clifford algebras and am a bit confused by the construction. My understanding is that there is a universal construction, but here is the one I was given for now (Szekeres's A Course in Modern Mathematical Physics):

Let $V$ be a real vector space with (an in general non-Euclidean) inner product $u\cdot v$, and $e_1,..., e_n$ an orthonormal basis such that the matrix of inner product components in this basis is given by $$g_{ij} = \begin{cases} \pm 1 & i=j \\\ 0 & i\neq j \end{cases}.$$ The Clifford algebra associated with this inner product space, denoted $\mathcal{C}_g$, is defined as the associative algebra generated by $1, e_1, e_2, . . . , e_n$ with the product rules $$e_i e_j +e_j e_i =2g_{ij}1, \, 1e_i =e_i1=e_i.$$

In particular, I am confused about this notion of an algebra "generated by" $V$. What does this mean? I am familiar with the definition of an algebra over a field as a vector space (say, $V$) with a product defined thereon, but $\mathcal{C}_g$ seems to be "bigger" than $V$ (as a set) given that it's later shown that $\dim \mathcal{C}_g = 2^n$ (whereas $\dim V = n$). That is, I am familiar with turning $V$ into an algebra over a field, so I'm not sure what's going on here (for example, I assume $1$ here is the multiplicative identity in $V$, but when turning $V$ into an algebra over a field I am in the past only familiar with defining structure constants with respect to a basis, and $1$ is not in our given basis here!).

In the end, I'm wondering what is the definition of algebra generated by a vector space being used? I couldn't find anything on Wikipedia which was intelligible to me.

Answer 1

Given a set $X$, you can form the free algebra $\mathbf{R}\langle X \rangle$. This algebra satisfies the usual universal property ascribed to "free" algebraic structures. In this case $\mathcal{C}_g$ is constructed by taking $\mathbf{R}\langle e_1,\ldots,e_n \rangle$ and quotienting out the ideal generated by the elements $e_i e_j + e_je_i - 2g_{ij}$ for all $i,j$. If we abuse notation and write $e_i$ to talk about its image in the quotient, then we can say that $\mathcal{C}_g$ is generated by $e_1,\ldots,e_n$ and $e_i e_j + e_j e_i = 2g_{ij}$ for all $i,j$. Informally speaking, $\mathcal{C}_g$ is the "largest" algebra containing the $e_i$ subject to the given relations.

Answer 2

If no other constraints are imposed, then the free (associative) algebra generated by $V$ over $\mathbb{C}$ is the ring usually written $\mathbb{C}[[e_1, e_2, ..., e_n]]$. The elements are "non-commutative polynomials" with coefficients in $\mathbb{C}$ (so $e_1 e_2$ is different from $e_2 e_1$), with the operations defined in the obvious way. If we impose additional relations such as $e_i e_j + e_j e_i = 2 g_{ij} 1 $ then we take the quotient $\mathbb{C}[[e_1, e_2, ..., e_n]]$ by the two-sided ideal generated by all the elements $e_i e_j + e_j e_i - 2 g_{ij} 1$.

Informally, we would just say that we introduce multiplication on elements of $V$ with the relations $e_i e_j + e_j e_i = 2 g_{ij}$, under the minimum constraints possible (but still forming an algebra over $\mathbb{C}$, i.e. a ring and a vector space over $\mathbb{C}$ which are compatible). Thus a product like $e_1 e_2$ is just some new element (not in $V$), because there is nothing forcing it to be equal to any existing element of $V$. We could call it $e_{12}$. Similarly $e_1 e_2 e_3$ is a new element (which we could call $e_{123}$), and similarly for any product of $e$'s, where the indices are in order, with no duplicates.

But $e_1 e_1$ equals the constant $g_{ii}$, according to the relations given, and a product in the "wrong" order like $e_2 e_1$ can be rewritten as $-e_1 e_2 = -e_{12}$ using the relations.. So the Clifford algebra has dimension $2^n$ because it is generated by all products of $e$'s where the indices are in order and there are no duplicates. These products correspond to subsets of $\{1, 2, \ldots, n\}$ and there are $2^n$ such subsets. Any product of $e$'s can be put into a "standard form", with $e$'s in order, by using the given relations.