L4b. Meeting the Clifford algebra.

We saw the action of $V\oplus V^*$ on $\wedge^\bullet V^*$:
$$(X+\xi)\cdot \phi = i_X \phi + \xi\wedge \phi .$$
Notice what happens when we act twice by the same element:
$$(X+\xi)^2\cdot \phi = (X+\xi)\cdot (i_X \phi + \xi\wedge \phi) = i_X \xi \phi - \xi\wedge i_X\phi+ \xi\wedge i_X\phi = i_X \xi \phi ,$$ $$(X+\xi)^2\cdot \phi = \langle X+\xi,X+\xi \rangle \phi,$$
Interesting! So, actually, if you extend this (linear!) action to tuples of vectors, i.e., to the tensor algebra $\bigotimes^\bullet (V\oplus V^*)$, we see that the ideal generated by the elements $v\otimes v - \langle v, v \rangle 1$, for $v\in V\oplus V^*$, acts trivially. This means that we have an action of the algebra
$$\frac{ \bigotimes^\bullet (V\oplus V^*)}{gen(v\otimes v - \langle v, v \rangle 1)}.$$
This algebra has indeed a name, the Clifford algebra $Cl(V\oplus V^*)$ of the metric vector space $(V\oplus V^*, \langle,\rangle)$. The product is just juxtaposition, like the tensor product, and then you can reduce the way you write an element by using the rule $v\otimes v - \langle v, v \rangle 1$.

Remark: If we had done this with the action of $V$ on $\wedge^\bullet V^*$ given by $i_X \phi$, we would have obtained an action of the algebra $\frac{ \bigotimes^\bullet (V\oplus V^*)}{gen(v\otimes v)},$ that is, $\wedge^\bullet V$.
So we would have got the map $i: \wedge^\bullet V \to \textrm{End}(\wedge^\bullet V^*)$.

There are many nice things about the Clifford algebra. Let us talk about $Cl(W)$ for a metric vector space $W$. Notice that the field, whatever field is, and the vector space $W$ itself  are contained in $Cl(W)$.
Take now an orthogonal basis $\{w_i \}$ of  $W$, we have that $w_i^2=\langle w_i,w_i \rangle$. Look at $(w_i+w_j)^2$. On the one hand, this is $w_i^2 + w_i w_j + w_j w_i + w_j$. On the other hand, $\langle w_i + w_j, w_i + w_j \rangle = w_i^2 + w_j^2$. This means $w_i w_j = - w_j w_i$.
This is telling us that, as a vector space!, the Clifford algebra is isomorphic to the exterior algebra $\wedge^\bullet W^*$ , so its dimension is $2^{\textrm{dim W} }$. Actually, $Cl(W)$ recovers the exterior algebra when the metric on $W$ is identically zero.

Wait a moment, this means $\wedge^\bullet V^* = Cl(V^*)$, since $V^*$ is isotropic. So, we actually have an action
$$Cl(V\oplus V^*) \otimes Cl(V^*) \to Cl(V^*).$$
You probably believe me if I tell you that $Cl(V^*)$ is a subalgebra of $Cl(V\oplus V^*)$. How is this action related to the Clifford product? First, let us derive some identities for the union of dual bases $\{e_i\}\cup \{e^i \}$, which is a basis  of  $V\oplus V^*$:
$$e_i^2 = 0, \qquad (e^i)^2 = 0 \qquad e_ie^i = 1-e^ie_i, \qquad e_i e^j = - e^j e_i.$$ Now, let us see if there is any relation to the Clifford product. For any vector $e_1$ and the differential form $1 \in \wedge^\bullet V^*$, our initial action is $i_{e_1} 1 = 0$, while the Clifford product is $e_11=e_1$. They are not the same, but they are actually related. How? The answer is that $\wedge^\bullet V^*$ is isomorphic to $Cl(V^*)\cdot \textrm{det} V \subset Cl(V\oplus V^*)$, where $\textrm{det} V \subset Cl(V) \subset Cl(V\oplus V^*)$ is a one-dimensional vector space generated by $e_1 \ldots e_n$. Let us check naively that we do not have the same issue as before. The element $1 \in \wedge^\bullet V^*$ corresponds to $e_1 \ldots e_n \in Cl(V\oplus V^*)$. The action of $e_1$ by the Clifford product is $e_1 e_1 \ldots e_n = 0 = 0 ( e_1 \ldots e_n)\in Cl(V\oplus V^*)$, so the corresponding form is $0$ and everything fits. Another example. The element $e^1 \in \wedge^\bullet V^*$ corresponds to $e^1 e_1 \ldots e_n  \in Cl(V^*)\cdot \textrm{det}$. The action of $e_1$ by Clifford product is
$$e_1 \cdot (e^1 e_1 \ldots e_n) = (1-e^1 e_1)e_1\ldots e_n = 1 e_1\ldots e_n,$$
which corresponds to $1 \in \wedge^\bullet V^*$. One can formally check that this works.

Finding the Clifford algebra is good news. To start with, we can invoke the following result.

Proposition. The subset of  $Cl(V\oplus V^*)$ given by $$\textrm{Spin}(V\oplus V^* )  :=  \{ v_1\ldots v_r \; : \; v_i\in V\oplus V^*, \langle v_i,v_i \rangle = \pm 1, r \textrm{ even } \}$$ together with the Clifford product has the structure of a Lie group. This Lie group is a double cover of $\textrm{SO}(V\oplus V^*)$ via the homomorphism
$$\kappa: \textrm{Spin}(V\oplus V^* ) \to \textrm{SO}(V\oplus V^*)$$ $$\kappa(x)(v)=xvx^{-1},\; x\in \textrm{Spin}(V\oplus V^*), v\in V\oplus V^*.$$

By taking the differential of the covering map $\kappa$,
$$d\kappa: \mathfrak{so}(V\oplus V^*)\subset Cl(V\oplus V^*) \to \mathfrak{so}(V\oplus V^*)$$ $$d\kappa_x(v)=xv-vx=[x,v],\;  x\in \mathfrak{so}(V\oplus V^* ), v\in V\oplus V^* ,$$ we are getting an isomorphism of two different realizations of $\mathfrak{so}(V\oplus V^*)$. On the right-hand side we have the usual $\mathfrak{so}(V\oplus V^* )$. On the left-hand side we have $\mathfrak{so}(V\oplus V^*)$ inside the Clifford algebra. Why should $\mathfrak{so}(V\oplus V^*)$ be inside the Clifford algebra? Well, we have claimed that the spin group sits inside the, let us say vector space, $Cl(V\oplus V^*)$. The tangent space of the spin group at the identity is sitting inside the tangent space of $Cl(V\oplus V^*)$ at the identity, which is $Cl(V\oplus V^*)$.


Anyway, the map $d\kappa$ will hopefully convince us how the spin group fits there.



Remark: universal property of the Clifford algebra. Let $i:W\to Cl(W)$ be the inclusion, given any associative algebra $A$ and any linear map $W \to A$ such that $j(v)^2 = \langle v, v \rangle 1_A$ for all $v \in V$, there is a unique algebra homomorphism $f: Cl(W) \to A$ such that $f\circ i = j$.