Let $k$ be a field with $char(k) \neq 2$ and let $V$ be a vector space over $k$. Let $B(\cdot ,\cdot )$ be a symmetric bilinear form on $V$ (not necessarily non-degenerate). Assume that $V$ has a basis $\{e(i)|i \in I\}$ where $I$ is totally ordered. Let $\wedge V$ be the exterior algebra of $V$ and let $Cliff(V,B)$ be the Clifford algebra of the pair $(V,B)$.

Question 1.

I know that $\wedge V$ has a basis $\{e(i_1) \wedge e(i_2) \wedge \cdot \cdot \cdot \wedge e(i_r)|r \ge 0, i_1 \gt i_2 \gt \cdot \cdot \cdot \gt i_r\}$ if $dim V \lt \infty $. I heard that this is ture even if $dim V = \infty$. How can we prove this statement?

Question 2.

I know that $Cliff(V,B) \simeq \wedge V$ as vector spaces (not necessarily as $k$-algebras) if $dim V \lt \infty$. Now, is this statement true when $dim V = \infty$?

Question 3.

Is the set $\{e(i_1)e(i_2)\cdot \cdot \cdot e(i_r)|r \ge 0, i_1 \gt i_2 \gt \cdot \cdot \cdot \gt i_r\}$ one of the bases of $Cliff(V,B)$?

If you know good references about this issue, please tell me.


Since the exterior algebra is just a special case of a Clifford algebra with bilinear form $0$, we'll answer both of those at the same time.

Suppose you think you have a nontrivial linear combination of the elements. Look at the finitely many $e_i$'s in the combination, and consider the finite dimensional subspace of $V$ spanned by them. Restricting the bilinear form, you can generate a finite dimensional Clifford algebra, and the elements are linearly independent, a contradiction.

As for question 2, you now know that given a basis for $V$, the derived "wedge" basis has the same number of elements as the Clifford basis, so the generated algebras have the same dimension.

  • $\begingroup$ Thank you for the explanation. I understood that the infinite-dimensional case is not so different from the finite-dimensional case. $\endgroup$ – user356126 Aug 31 '16 at 9:00

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