Subalgebra of Clifford algebra

Question

Let $Cl(V\oplus V^{*})$ the Clifford algebra of $V\oplus V^{*}$ with natural pairing. How can check a subset $ Λ^{•} (V^{*}) $ is a subalgebra of Clifford algebra $Cl(V\oplus V^{*})$?

Answer 1

The natural quadratic form on $V\oplus V^*$ is

$$Q(v, \phi) = \phi(v)$$

where $v\in V$ and $\phi \in V^*$. It is an example of hyperbolic/split/metabolic quadratic space.

Elements of $V^*$ are naturally embedded into $V\oplus V^*$ via

$$\phi \mapsto (0, \phi)$$

and the value of the quadratic form on those is

$$Q(0,\phi) = \phi(0) = 0$$

Thus, the embedded copy of $V^*$ is a subspace with zero quadratic form. The Clifford algebra of such a space is, by definition, the exterior algebra. Since Clifford algebra is functorial, embedding of quadratic spaces gives algebra embeddings; thus, the Clifford algebra of $V^*$ with a zero quadratic form, which is $\Lambda(V^*)$, is embedded into $\operatorname{Cl}(V \oplus V^*)$. This is exactly the subalgebra generated by the embedded copy of $V^*$ via the diagram of embeddings

$$ \begin{matrix} V^* & \rightarrow & V \oplus V^* \\ \downarrow & & \downarrow \\ \Lambda(V^*) & \rightarrow & \operatorname{Cl}(V \oplus V^*) \end{matrix} $$

The same reasoning applies to $V$ embedded into $V \oplus V^*$ and $\Lambda(V)$ embedded into $\operatorname{Cl}(V \oplus V^*)$.