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 \( f\circ i = j \).
$$(X+\xi)\cdot \phi = i_X \phi + \xi\wedge \phi .$$
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 \( f\circ i = j \).
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