Problem Sheet 3.

You can find Problem Sheet #3 here.

Please, submit some of your solutions by Thursday 29th at 9:00.
You can do that by email or by leaving them under my door (sala 327).

Reference: Chapter 2 from Gualtieri's PhD thesis.

You can find the second chapter of Marco Gualtieri's thesis here.

Although these are just the preliminares, his thesis is one of the foundational texts of this theory and a perfect reference for us at this point. You can review the things we saw at class and we discussed on the blog. Pay special attention to the concepts we discussed at class but not here, like type (in Section 2.2) and the statement of Proposition 2.25, which will be very important for us.

Note that we have not dealt with Sections 2.7 and 2.8, as well as with other parts of the chapter.

L6a: Going complex for generalized complex structures!

We have been discussing maximal isotropic subspaces in Problem Sheet #2 because we proved that they are a way of looking at linear gcs. However, linear gcs are equivalent to maximal isotropic subspaces of \( (V \oplus V^*)_\mathbb{C} \), whereas we have been dealing with maximal isotropic subspaces of \(V\oplus V^*\).

Well, the things we did do not rely on the fact that \( V \) was a real vector space, so they apply for both cases. It is good we did for real vector spaces in case you want to work on Dirac geometry. From now on, we will deal with complex differential forms \( \wedge^\bullet V^*_\mathbb{C} \), where the pairing is defined exactly in the same way.

You can read more about this in the reference I am posting next, where you will find several very important facts that were mentioned in class but not on this blog.

L5b: Differential forms as spinors.

We are going to rebrand the differential forms as spinors. In L4b we saw how \( \textrm{Spin}(V\oplus V^*) \) sits inside \( Cl(V\oplus V^*) \). In other words, the Clifford product is giving a representation \( \textrm{Spin}(V\oplus V^*) \to \textrm{End}(Cl(V\oplus V^*)) \). One can prove that this representation decomposes into \( 2^n \) subrepresentations which are all isomorphic to $$ Cl(V^*)\det(V) = \wedge^\bullet V^* \det(V)\subset Cl(V\oplus V^*).$$

Each of these is the spin representation of the spin group, a representation that does not descend to \(\textrm{SO}(V\oplus V^*) \) (recall that \(\textrm{SO}(V\oplus V^*) \) is not simply connected!). The elements of the spin representation are called spinors. Moreover, the spin representation splits into two irreducible half-spin representations: \( \wedge^{ev} V^* \det(V) \), \( \wedge^{od} V^* \det(V) \), of even and odd spinors.

 Also in L4b, we related the action of \( V \oplus V^* \) on \( \wedge^\bullet V^* \) with the Clifford product of \( Cl(V\oplus V^*) \) on \( Cl(V^*)\det(V) \subset Cl(V\oplus V^*) \). Having this in mind, we will call the differential forms (without the factor \( \det(V) \) spinors.

One last but very important thing about spinors is that they come equipped with a pairing. For differential forms this map $$ (,) : \wedge^\bullet V^* \otimes \wedge^\bullet V^* \to det V $$ is given by $$(\phi, \psi) = (\phi^T\wedge \psi)_{top},$$ where \( \phantom{.}^T \) is the anti-automorphism of the Clifford algebra extending the map \( (v_1 \ldots v_n)^T=v_n\ldots v_1\) and \( top \) denotes the top-degree component of the differential form. Moreover, this pairing is \( \textrm{Spin}_0(V\oplus V^*) \)-invariant.

L5a. The B-field acting on differential forms.

Last time we stopped soon after we got the map.
$$d\kappa: \mathfrak{so}(V\oplus V^*)\subset Cl(V\oplus V^*) \to \mathfrak{so}(V\oplus V^*)$$
Let us see how \( B \) in the left-hand side \(\mathfrak{so}(V\oplus V^*)\) sits inside the right-hand side \( \mathfrak{so}(V\oplus V^*)\), that is, in the Clifford algebra.

Take the simple \(B\)-field \( e^i \wedge e^j  \in \wedge^2 V^*\subset  \mathfrak{so}(V\oplus
V^*)\). This element is completely determined by its action on \( V \): \(e_i\mapsto e^j\), \(e_j\mapsto -e^i\), \(e_k\mapsto 0\) for \(k\neq i,j\). After some observation, we find \( d\kappa^{-1}(e^i\wedge e^j)=e^je^i \), since $$(e^je^i)e_i - e_i(e^je^i)=(e^ie_i+e_ie^i)e^j=e^j,$$ and \( e^j
e^i \) vanishes when acting on \( e_k \) for \( k\neq i,j\).  We then have that a general \( B=\frac{1}{2} \sum B_{ij} e^i\wedge e^j\in \mathfrak{so}(V\oplus V^*) \) corresponds to \( \frac{1}{2}\sum B_{ij} e^je^i \) via \( d\kappa^{-1}\).

So... what? Why do we want to look at \( \mathfrak{so}(V\oplus V^*)\) inside the Clifford algebra? Well, we had an action of the Clifford algebra on \( \wedge^\bullet V^*\). Keeping it simple, without giving details about the \( Cl(V^*)\cdot \textrm{det}(V) \) business, a \( B \in \mathfrak{so}(V\oplus V^*) \)-field  acts on the exterior algebra by $$B\cdot \phi = \frac{1}{2} \sum B_{ij} e^j\wedge (e_i\wedge \phi)= -B\wedge \phi.$$
If we look at \( \exp(B)\in \textrm{Spin}(V\oplus V^*) \) acting on \( \wedge^\bullet V^* \), we have
$$ \exp(B)\cdot \phi=e^{-B}\wedge \phi = (1-B+\frac{1}{2}B^2-\ldots)\wedge \phi.$$

Note that this action matches what we were doing with the maximal isotropic subspaces \( L(E,\epsilon) \). We saw in Problem Sheet #2 that if \( L(E,\epsilon) = \textrm{Ann}(\phi) \), \( e^B L(E,\epsilon) = \textrm{Ann}(e^{-B}\wedge \phi ) \). By applying \(\exp(B)\in \textrm{Spin}(V\oplus V^*)\) to the differential form \( v\cdot \phi \), we have
$$\exp(B)(v\cdot \phi) = \exp(B)(v) \cdot \exp(B)\phi = e^B v \cdot (e^{-B}\wedge \phi ),$$
so if \( v \in \textrm{Ann}(\phi) \), \(e^Bv\in \Ann(e^{-B}\wedge \phi).

Remark: notice the difference between \( \exp(B)\in \textrm{Spin}(V\oplus V^*) \) and \( e^B=\sum_{n=0}^{+\infty} \frac{B^n}{n!} \). The former is the element of the group, the second is the way we write its action explicitely, either on vectors by \( \exp(B)(X+\xi) =  e^B (X+\xi) = X+\xi+i_XB \) or on differential forms by \( \exp(B)(\phi) = e^B \wedge \phi \).

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 \( f\circ i = j \).

L4a: A word about orientation.

 When you have a vector space \( W \) with a metric and you want to talk about \( \textrm{SO}(W) \) instead of \( \textrm{O}(W) \). You do need an orientation, i.e., an element, up to positive multiples, of \( \wedge^{\dim W} W \). Of course, this orientation is invisible when you look at them as matrices, since you are choosing a basis, say \( \{w_i\} \), and the basis is giving you an orientation \(  w_1\wedge  \ldots \wedge w_n  \). If you choose a different basis, say \( \{-w_1\}\cup \{ w_2,\ldots,w_n \} \), you may get the opposite orientation, as it happens in this example: \( -w_1\wedge \ldots \wedge w_n \).
 
What about \( V\oplus V^* \)? Apart from a canonical pairing, there is a canonical orientation on \( V\oplus V^* \)! We have to give an element of
$$ \wedge^{(2\dim V)} (V\oplus V^*) = \wedge^{\dim V} V \otimes \wedge^{\dim V^*} V^*.$$ The canonical pairing between \( \wedge^{\dim V} V \) and \( \wedge^{\dim V^*} V^*\) gives that element. If you want, it is \( 1 \in \mathbb{R} \equiv \wedge^{\dim V} V \otimes \wedge^{\dim V^*} V^*,\)
where the isomorphism is given by the canonical pairing.

We will talk about \( \textrm{SO}(V\oplus V^*) \). Notice that the Lie algebra \( \mathfrak{so}(V\oplus V^*)\) is exactly \(\mathfrak{o}(V\oplus V^*) \), as it only depends on a neighbourhood of the identity element, which is the same for \( \textrm{SO} \) and \( \textrm{O} \).