Problem Sheet 7.

You can find Problem Sheet #7 here.

Please, submit some of your solutions by Thursday 26th at 9:00.
You can do that by email or by leaving them under my door (sala 327).

Problem Sheet 6.

You can find Problem Sheet #6 here.

Please, submit some of your solutions by Monday 23rd at 9:00.
You can do that by email or by leaving them under my door (sala 327).

Problem Sheet 5.

You can find Problem Sheet #5 here.

Please, submit some of your solutions by Thursday 12th at 9:00.
You can do that by email or by leaving them under my door (sala 327).

Problem Sheet 4.

You can find Problem Sheet #4 here.

Please, submit some of your solutions by Thursday 5th at 9:00.
You can do that by email or by leaving them under my door (sala 327).

L9a. The definition of a Courant algebroid.

We defined the Dorfman bracket in L7a. We call it bracket although it is not skew-symmetric. One may want to have a proper skew-symmetric bracket. To do so, we just have to skew-symmetrize: $$ [[ v, w ]] = \frac{1}{2}( [v,w] - [w,v] ), $$  in order to get the Courant bracket. They are indeed closely related: $$ [v,w] = [[v,w]] + D\langle v,w\rangle $$

In Problem Sheet 3 we have checked many properties that the Dorfman and the Courant bracket satisfy. They all fit in an abstract object called Courant algebroid. Let us give two definitions of this object, one using a non-skew-symmetric but well behaved Dorfman bracket, and the other one using a skew-symmetric Courant bracket.

A Courant algebroid \( (E,\langle\cdot,\cdot\rangle,[\cdot,\cdot],\pi) \) over a manifold \(M\) consists of a vector bundle \(E\to M\) together with a non-degenerate symmetric bilinear form \(\langle\cdot,\cdot\rangle\) on \(E\), a DORFMAN bracket \([\cdot,\cdot]\) on the sections \(\mathcal{C}^\infty(E)\) and a bundle map \(\pi:E\to TM\) such that the following properties are satisfied:
 (D1): \([v,[w,w']]=[[v,w],w'] + [w,[v,w']],\)
(D2): \(\pi([v,w])=[\pi(v),\pi(w)]\),
(D3): \([v,fw]=f[v,w]+(\pi(v)f)w\),
(D4): \( \pi(v)\langle w,w'\rangle = \langle [v,w], w' \rangle + \langle w, [v,w'] ,w'\rangle \),
(D5): \( [v,v]=\pi^* d\langle v, v\rangle \),
  for any \(v,w,w' \in \Gamma(E)\), \(f,g\in \mathcal{C}^\infty(M)\), where \(D: \mathcal{C}^\infty(M)\to \mathcal{C}^\infty(E)\) is defined by $$\langle D f, v \rangle=\frac{1}{2}\pi(v)(f).$$

A Courant algebroid \( (E,\langle\cdot,\cdot\rangle,[[\cdot,\cdot]],\pi) \) over a manifold \(M\) consists of a vector bundle \(E\to M\) together with a non-degenerate symmetric bilinear form \(\langle\cdot,\cdot\rangle\) on \(E\), a skew-symmetric COURANT bracket \([[\cdot,\cdot]]\) on the sections \(\mathcal{C}^\infty(E)\) and a bundle map \(\pi:E\to TM\) such that the following properties are satisfied:
 (C1): (sorry about how this looks with the double bracket) \([[v,[[w,w']]]]=[[[[v,w]],w']] + [[w,[[v,w']]]] - \frac{1}{3} D( \langle [[v,w]],w'\rangle + \langle [[w,w']], v \rangle + \langle [[w',v]],w \rangle),\)
(C2): \(\pi([[v,w]])=[[\pi(v),\pi(w)]]\),
(C3): \([[v,fw]]=f[[v,w]]+(\pi(v)f)w - \langle v,w \rangle D f\),
(C4): \( \pi(v)\langle w,w'\rangle = \langle [[v,w]]+D\langle v,w\rangle, w' \rangle + \langle w, [[v,w']] + D\langle v,w'\rangle \rangle \),
(C5): \(\pi\circ D =0\), and consequently, \(\langle D f, D g \rangle = 0\),
  for any \(v,w,w' \in \Gamma(E)\), \(f,g\in \mathcal{C}^\infty(M)\), where \(D: \mathcal{C}^\infty(M)\to \mathcal{C}^\infty(E)\) is defined by $$\langle D f, v \rangle=\frac{1}{2}\pi(v)(f).$$

L8a. The group of generalized diffeomorphisms.

Generalized diffeomorphisms are orthogonal automorphisms \( F \) of \(T+T^*\) covering a diffeomorphism \( f \) and preserving the Courant bracket.

We have proved that the group of generalized diffeomorphisms is the semidirect product
$$ \textrm{Diff}(M) \subset \Omega^2_{cl}(M), $$ where the diffeomorphisms \( f \) act by pushforward on \( T + T^* \), and a closed \(2\)-form \(B\) acts as a \( B \)-field: \( X+\xi\mapsto X+\xi+i_X B \). The \( B \)-fields are a new symmetry, not present in classical geometry, and they will play a very important role.

Note that a diffeomorphism and a \(B\)-field do not commute: $$  \exp(B) \circ g_*= g_* \exp(g^*B) .$$

If you look at automorphisms  of  \(T\) covering a diffeomorphism \( f \) and preserving the Lie bracket, you only get the differential of the diffeomorphisms, so this approach is consistent with the symmetries of classical differential geometry.

L7a. A bracket for T+T*.

The Lie bracket of two vector fields \( X, Y \in \mathcal{C}^\infty(T) \) is usually defined as the only vector field \( [X,Y] \) such that $$[X,Y](f)=X(Y(f)) - Y(X(f)).$$ This works very nicely for vector fields, since they are derivations of functions. One checks that \(f\mapsto X(Y(f)) - Y(X(f)) \) is a derivation and we call that derivation \( [X,Y] \).

The Lie bracket is also the only one such that, for any differential form $\phi$, $$ i_{[X,Y]} \phi = [ \mathcal{L}_X, i_Y] \phi = [ [d,i_X],i_Y] \phi,$$ where the bracket on the leftmost term is the Lie bracket that we are defining, while the other brackets are supercommutators ( \( [x,y]=xy-(-1)^{|x||y|}yx \) ).

We would like now to generalize this. We do want a bracket for the sections of \( T+T^* \). Our starting poing should be how they act on a differential form: $$(X+\xi)\cdot \phi = i_X \phi + \xi\wedge \phi .$$ If we only look at functions, we will not be succesful. Whereas $X(f)=i_X df$ is again a function, $(X+\xi)\cdot df $ is not. We must use the second approach. Define the bracket $[X+\xi,Y+\eta]$ as the only generalized vector field satisfying $$ i_{[X+\xi,Y+\eta]} \phi = [ [d,(X+\xi)\cdot],(Y+\eta)\cdot] \phi$$ for any differential form $\phi$. This is called a derived bracket and there is a general theory behind it.

Since the expression we are using to derive the bracket is bilinear on the fields, we can first compute the terms \( [X,Y], [X,\eta], [\xi,Y], [\xi,\eta] \) as we did in the class. The result is the Dorfman bracket. $$[X+\xi,Y+\eta] = [X,Y] + \mathcal{L}_X \eta - i_Y d\xi .$$

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 \( \textrm{O} \).

Problem Sheet 2

You can find Problem Sheet #2 here.

Please, submit some of your solutions by Wednesday 21st at 9:00.
You can do that by email or by leaving them under my door (sala 327).

L3b. Generalized linear algebra: the symmetries (III).

We have just seen the decomposition
$$ \mathfrak{o}(V\oplus V^*) = \textrm{End}(V)\oplus \wedge^2 V^* \oplus \wedge^2 V.$$
By looking at these elements as matrices, we have
$$  \left( \begin{array}{cc} A & \beta \\ B & -A^T \end{array} \right),$$
with \( A \in \textrm{End}(V) \), \( B\in \wedge^2 V^* \), \( \beta \in \wedge^2 V \).

Do not forget that we did this because we wanted to talk about \( \textrm{O}(V\oplus V^*) \). The way we have to pass information from the Lie algebra to the Lie group is the exponential. Since we are dealing with groups of matrices, the exponential is just the usual exponential \( e^A = \sum_{i=0}^{+\infty} \frac{A^i}{i!} \).

If we take \( A\in  \textrm{End}(V) \), we get the element
$$ \left( \begin{array}{cc} e^A & 0 \\ 0 & ((e^A)^{*})^{-1} \end{array}   \right) = \left( \begin{array}{cc} P & 0 \\ 0 & ((P)^{*})^{-1} \end{array}   \right) \in \textrm{O}(V\oplus V^*), $$
where \(P \in \textrm{GL}^+(V) \) ( \(\textrm{det}P>0 \) ), although if the determinant of \( P \) is negative, the matrix you can see is clearly still in  \( \textrm{O}(V\oplus V^*) \). This is showing that, as we could expect, the generalized symmetries contain the classical symmetries.

Let us look for something new. Starting with \( B\in \wedge^2 V^* \) or \( \beta \in \wedge^2 V \) we get the \( B \)-fields and the \( \beta \)-fields:
$$ \left( \begin{array}{cc} 1 & 0 \\ B & 1 \end{array}   \right), \qquad \qquad \qquad \left( \begin{array}{cc} 1 & \beta \\ 0 & 1 \end{array}   \right).$$
Notice that the \( B \)-fields, which act on \( X+\xi \in V\oplus V^* \) by $$X+\xi \mapsto X+\xi+i_X B,$$ preserve the projection to \( V \). We will soon forget about \( \beta \) -fields and focus on \(B \)-fields.

L3a. Generalized linear algebra: the symmetries (II).

Here you have the file that Hudson wrote about the symmetries of \( V\oplus V^* \). He uses the regular value theorem to prove that \( \textrm{O}(V\oplus V^*) \) is a Lie group, computes the Lie algebra (by using the differential of the map he was using) and then looks at the Lie algebra structure. (If anyone wants to add r ask anything, remember that you can comment on the posts or send me files.)



Just for clarity, let me put here the way I was justifying what is the equation describing the Lie algebra \( \mathfrak{o}(V\oplus V^*) \). Since \( \textrm{O}(V\oplus V^*) \) is a linear Lie group, we can describe its Lie algebra \( \mathfrak{o}(V\oplus V^*) \)  by taking derivatives of curves inside \( \textrm{O}(V\oplus V^*) \) passing by \( \textrm{Id} \) at time \( 0 \).
$$  \mathfrak{o}(V\oplus V^*) = \{ c'(0) \; : \; c:(-\varepsilon,\varepsilon)\to \textrm{O}(V\oplus V^*), c(0)= \textrm{Id} \}.$$
This will tell us what the condition for the elements in \( \mathfrak{o}(V\oplus V^*) \) is.

The condition for the elements \( A\in \textrm{O}(V\oplus V^*) \) is \( \langle Av,Aw \rangle = \langle v,w \rangle \) for any \( v,w \in V\oplus V^* \). Recall that the pairing is \( \langle X+\xi, Y+\eta \rangle = \frac{1}{2} (i_X \eta + i_Y \xi ) \), so we can write
$$ \langle v,w \rangle = v^T  C w,$$
for \( N = \left( \begin{array}{cc} 0 & \frac{1}{2} \\ \frac{1}{2} & 0 \end{array} \right)  \). In other words, \( A\in \textrm{O}(V\oplus V^*) \) if f \( A^T N A = N \).

Now, given a curve \( c(t) \) in \( \textrm{O}(V\oplus V^*) \), it satisfies \( c(t)^T N c(t) = N \). Since we are interested in \( c'(0) \), we differentiate:
$$ c'(t)^T N c(t) + c(t)^T N c'(t) = 0,$$
which at \( t=0 \), gives \( A^T N + NA = 0 \).

By looking at \(A\) as a block matrix,
$$ A =  \left( \begin{array}{cc} E & \beta \\ B & F \end{array} \right),$$
the condition \( A^T N + NA = 0 \) gives \( F = - E^T \), \( B+B^T =0\) and \( \beta + \beta^T = 0 \). This means that an element in \( \textrm{O}(V\oplus V^*) \) consists of  \(E\in \textrm{End}(V)\) (acting as \(-E^T \) on \(V^* \) ), \( B\in \wedge^2 V^* \) and  \(\beta\in\wedge^2 V \).

We thus have
$$ \mathfrak{o}(V\oplus V^*) = \textrm{End}(V)\oplus \wedge^2 V^* \oplus \wedge^2 V, $$
where the direct sum refers to vector spaces, not to Lie algebras! (as Hudson showed).

L2a. Linear generalized complex structures.

We want to define linear generalized complex structures (linear gcs) as an analogue to linear complex structures, but \( V \oplus V^* \) has something that \( V \) does not have: a canonical pairing. This pairing must be preserved by the the linear gcs we are defining right now.

Definition. A linear gcs on \( V \) is a linear endomorphism
$$ \mathcal{J}: V\oplus V^* \to V\oplus V^*$$ such that \( \mathcal{J}^2= - \textrm{Id} \) which preserves the pairing:
$$ \langle \mathcal{J}v, \mathcal{J}w \rangle = \langle v,w \rangle. $$
We saw that a complex structure on a vector space \( V \) of dimension \(n=2m\) can alternatively be seen as a complex subspace \( L \subset V_{\mathbb{C}} \) such that \( \textrm{dim}_\mathbb{C} V = m \) and \( L\cap \overline{L} = 0 \). Since a linear gcs on \( V \) is, in particular, a usual complex structure on \( V\oplus V^* \), we get a complex subspace \( L \) of \( (V\oplus V^*)_\mathbb{C} \) such that \( \textrm{dim}_\mathbb{C} V = n \) and \( L\cap \overline{L} = 0 \). But the linear gcs satisfies an extra condition, \( \mathcal{J} \) preserves the pairing. How does this affect \( L \subset (V\oplus V^*)_{\mathbb{C}} \)?

First, given two elements \( l,l' \in L \), we have
$$ \langle \mathcal{J}l, \mathcal{J}l' \rangle = \langle il,il' \rangle = - \langle l, l' \rangle,$$ so \( \langle l,l' \rangle = 0\) for any \( l,l' \in L\). This means that \( L \) must be isotropic. Is this the only condition? Conversely, if \( L \) is isotropic, we check that the corresponding \( \mathcal{J} \) preserves the metric. Notice that $$ V = \{ z + \overline{z} \; :\; z\in V_{\mathbb{C}} \}=\{ z + \overline{z} \; :\; z\in L\oplus \overline{L} \}= \{ z + \overline{z} \; :\; z\in L \}.$$ We thus have, for \( v=a+\overline{a}, w=b+\overline{b} \in V\),
$$ \langle \mathcal{J}(a+\overline{a}), \mathcal{J}(b+\overline{b}) \rangle = \langle ia - i\overline{a}, ib-i\overline{b} \rangle=\langle a,\overline{b}\rangle + \langle \overline{a},b \rangle=\langle a+\overline{a},  b+\overline{b}\rangle.$$

We have just proved a characterization of a linear gcs:

• A linear gcs on \( V \) is given by a complex isotropic subspace \( L \subset (V\oplus V^*)_\mathbb{C} \) such that \( \dim_\mathbb{C} L = n \) and \( L \cap \overline{L} = 0 \).

Note that when we talk about the pairing or metric on \( (V\oplus V^*)_\mathbb{C} \) -we need it to talk about isotropy-, we are extending the pairing linearly: $$\langle a+ib, c+id \rangle = \langle a, b \rangle  - \langle c,d \rangle + i ( \langle a,d\rangle + \langle b, c \rangle).$$

Actually, \( n \) is the maximum possible dimension of an isotropic space in a complex metric space of dimension \( 2n \), so a linear gcs is given by a maximal isotropic subspace \( L \subset (V\oplus V^*)_\mathbb{C} \) such that \( L \cap \overline{L} = 0 \). We will make some comments about this fact later.

Reference about the Lie algebra of a linear Lie group.

Here you have a few pages from the book "Lie groups beyond an introduction" by Knapp. I hope they help.

L1d. Generalized linear algebra: the symmetries.

Once we have a vector space \( V \oplus V^* \) that comes with a pairing of signature \( (n,n) \), we would like to know what its symmetries are.

By symmetries we mean bijections preserving the structure. For a set without any structure we have that the symmetries are all the possible permutations of that set. For a vector space, the symmetries are the (linear) automorphisms, \( \textrm{Aut}(V) \) or \( \textrm{GL}(V) \), permutations compatible with the sum and the product by scalars. For \( V\oplus V^* \) we have
$$ \textrm{O}(V\oplus V^*) = \{ A\in GL(V\oplus V^*) : \langle Av,Aw \rangle = \langle v,w \rangle \textrm{ for } v,w \in V\oplus V^* \}, $$
i.e., bijections preserving the vector space structure that also preserve the metric. The groups \( \textrm{GL}(V) \) and \( \textrm{O}(V\oplus V^*)\) are Lie groups: they are both group and manifold in a nice compatible way. Neither a group or a manifold are objects easy to handle, so we will do something similar to what we do in geometry: linearize, look at the tangent space (in this case of the identity of the group)!

Let's look then at the Lie algebra \( \mathfrak{o}(V\oplus V^* ) \). What does it satisfy? How can we describe its elements?

L1c. Generalized linear algebra: the pairing.

Instead of doing geometry, we were doing linear algebra. Let us now generalize this linear algebra before doing any geometry. This means looking at \( V \oplus V^* \).

Notice that we have not added an arbitrary vector space to \( V \), we have added its dual space \( V^* \) and we cannot dismiss this. How do we keep track of this fact in a nice way? By defining a (canonical) pairing.

This is a good moment to introduce some conventions. We will normally use the capital letters \( X,Y \) for elements of \( V \), and the greek letters \( \xi, \eta \) (xi and eta) for elements of \( V^* \). For the elements of \( V\oplus V^* \) we will use \( v, w \). Thus, \( v=X+\xi, w=Y+\eta \)  are the usual elements of  \( V\oplus V^* \).

And finally the pairing \( \langle \; , \; \rangle \):
$$ \langle X+\xi, Y+\eta \rangle = \frac{1}{2} (i_X \eta + i_Y \xi ).$$
Note that \( \langle X+\xi, X+\xi \rangle = i_X \xi \).

This pairing has signature \( (n,n) \). Given bases \( \{ e_1,\ldots,e_n\} \) and \( \{ e^1, \ldots, e^n \} \) of \( V \)  and \( V^* \), the matrix of the pairing is written as
$$ \left( \begin{array}{cc} 0 & \frac{1}{2}\textrm{Id} \\ \frac{1}{2}\textrm{Id} & 0 \end{array} \right) $$
for the basis \( \{ e_1,\ldots,e_n, e^1, \ldots, e^n \} \) of \( V \oplus V^* \), or as
$$ \left( \begin{array}{cc} \textrm{Id} & 0 \\ 0 & -\textrm{Id} \end{array} \right) $$ for the basis  \( \{ e_1+e^1,\ldots,e_n+e^n, e^1-e^1,\ldots,e_n-e^n \} \), from where we clearly see the signature.

Note that the pairing between any two elements of V, or any two elements of V^* is zero. This is what we call being isotropic.

Definition:  A vector subspace \( L \) of a metric vector space \( (W,\langle\;,\;\rangle ) \) is said to be isotropic if \( \langle x, y \rangle = 0 \) for all \( x,y \in L \).

L1b. Complex and symplectic str. on a vector space.

We do geometry by looking together at all the tangent spaces of a manifold: the tangent bundle \( T\). We will do generalized geometry by looking at \( T + T^* \). But before generalizing and before doing any geometry, we'd better understand what happens at a single point. Looking at the tangent space at a point is just looking at a vector space. Let us call this vector space \( V \), and let \( n \) be its dimension.

We used almost complex structures as an example when motivating generalized geometry, so let us talk now about (linear) complex structures on a vector space and let us look at them in many possible ways. This will help us to have different points of view when we generalize them.

• A complex structure on \( V \) is given by an endomorphism \( J: V \to V \) such that \( J^2 = - \textrm{Id} \). This implies that the dimension of \( V \) must be even, \(n=2m \), why?
  Well, by looking at the determinant of the endomorphism \( J \), we have \( \textrm{det}(J)^2=\textrm{det}(J^2)=\textrm{det}(-\textrm{Id})=(-1)^n \), so \( n \) must be even, say, \( n=2m \).

The endormophism \( J \) is completely determined by the \( +i \)-eigenspace as an endormophism of \( V_\mathbb{C} \) and the fact that it is a real endormorphism. This suggests another way of describing a complex structure.

• A complex structure on \( V \) is given by a complex subspace \( L \subset V_\mathbb{C} \) such that \( \dim_\mathbb{C} L = m \) and \( L \cap \overline{L} = 0 \).

There is yet another way of describing a complex structure that will be helpful for us. It is based on the definition using the complex subspace \( L \). Take a basis \( \{ \partial_{z_1}, \ldots, \partial_{z_m} \} \) of \( L \). Then \( \{ {\partial}_{\overline{z}_1}, \ldots, {\partial}_{\overline{z}_m} \} \) is a basis of \( \overline{L} \), and we have the dual basis \( \{ d\overline{z}_1, \ldots, d\overline{z}_m \} \) of \( \overline{L^*}  \). The element \( \phi = d\overline{z}_1\wedge \ldots\wedge d\overline{z}_m \) determines \( L \) as follows:
$$ L = \textrm{Ann}(\phi) = \{ X \in V_\mathbb{C} : i_X \phi = 0 \},$$
where \( i_X \) denotes the interior derivative. This inspires the third way. 

• A complex structure on \( V \) is given by an element \( \phi \in \wedge^m V^*_\mathbb{C} \) satisfying... what does it satisfy?

Good! These three will be enough, but also necessary. Make sure you also understand how to pass from one to the others.


Time for symplectic structures. You are probably working on this now. At first glance, we have that \( \omega \), seen as a skew map \( V \to V^* \) and as an element of \( \wedge^2 V^* \), give analogues to \( J\) and \( \phi \). What about an analogue for \( L \)? Check how you recover \( L \) from \( J \) in a complex structure!

L1a. Motivation: what is generalized geometry?

Remember the first time you did geometry. I bet it was two-dimensional geometry. You worked on the real plane and you measured distances, angles, areas... The real plane was the simplest place to do that. A one-dimensional line was rather abstract and the three-dimensional space felt way too big at that moment.

Later in life you came across surfaces. If you wanted to do something similar to what you did on the real plane, say, measure an angle between two curves on the surface, how did you do it? Well, at the intersection of these two curves you put a plane, the tangent plane, and each curve defined a vector on that plane (the tangent vector of the curve at that point). Then you could measure the angle between those two vectors as you did on the real plane!

Actually, what I have described is the way most of differential geometry has been created. The tangent plane of a surface or the tangent space of a manifold \(M\) have been used to define metrics, complex structures, symplectic structures... How many times have you seen \( TM \) around?

Generalized geometry proposes replacing \( TM \) by \( TM \oplus T^*M \) and redoing geometry from the beginning. Let us show what this means with an example. In order to make it shorter, let me use \( T \) for \( TM \),  \( T^* \) for \( T^*M \) and use \( + \) instead of \( \oplus \) from now on.

Ok, first generalization. An almost complex structure is defined as a bundle map \( J: T \to T \) such that \( J^2=-\textrm{Id} \), so we will define a generalized almost complex structure as a map $$  \mathcal{J}: T +  T^* \to T + T^* $$ such that \( \mathcal{J}^2=-\textrm{Id} \), and which satisfies another property that we omit for now.

One can see that almost complex structures fit in our new generalized setting. Given \( J \), we produce $$ \mathcal{J} = \left( \begin{array}{cc} -J & 0 \\ 0 & J^* \end{array} \right) $$ by using the diagonal entries.


What if we use the off-diagonal entries? We need a map \( T\to T^* \) and a map \( T^* \to T \). What about a pre-symplectic structure \( \omega \) and its inverse?
$$ \mathcal{J}_\omega = \left( \begin{array}{cc} 0 & -\omega^{-1} \\ \omega & 0 \end{array} \right).$$ Thus, at least intuitively for now, both complex and symplectic structures are particular cases of generalized complex structures. Generalized geometry provides, in first place, an unifying way of looking at previous structures.

But not only that. Generalized geometry also introduces genuinely new structures: there are compact manifolds which are neither complex nor symplectic, but still admit a generalized complex structure, for instance \( 3\mathbb{C} P^2 \# 19 \overline{\mathbb{C} P^2} \).

A third remarkable phenomenon in generalized geometry is the revival of previously known but somehow forgotten structures. This is the case of generalized Kahler structures. Generalized Kahler geometry was defined and shown to be equivalent to a bihermitian geometry defined in 1984. The statement of this equivalence was indeed followed by many publications about this subject.



Let us put some names to this. Generalized Geometry, as we are presenting it, was introduced by Nigel Hitchin around 2002. He dealt with Generalized Calabi-Yau manifolds and his PhD student Marco Gualtieri developed Generalized Complex and Kahler Geometry. Then, many other people and more of Hitchin's students, like Gil Cavalcanti, kept working on this. Actually, Cavalcanti found, together with Gualtieri, the example \( 3\mathbb{C} P^2 \# 19 \overline{\mathbb{C} P^2} \). However, as we will see, many of the objects we will be dealing with were already there. Generalized complex structures are, in particular, Dirac structures, a concept introduced by Ted Courant and Alan Weinstein around 1986, which unified Poisson structures and closed \(2\)-forms. We will also talk about the Dorfmann product, introduced by Irene Dorfman herself in 1987, and about Courant algebroids, a structure introduced by Weinstein together with Zhang-Ju Liu and Ping Xu in 1997. As you can see, all of these were ready before the 21st century!

Problem Sheet 1

You can find Problem Sheet #1 here.

Please, submit some of your solutions by Tuesday 13th by 18:00.
You can do that by email or by leaving them under my door (sala 327).
Of course, do not come on purpose to IMPA in order to submit.
If you keep on working, submit more solutions on Wednesday by 9:00.

About the blog

Just a bit of information about the blog:

• Each post will start with a code like L3b, which means Lecture number 3, second post (b). 
• The contents of the posts are not meant to be a full record of the lectures, but they will focus on the most important items. Many details and proof are left as problems.
• The posts are not in their final version. I will keep editing them, especially after we deal with the problems of the Problem Sheet and we understand a bit better what was going on.
• Please, use the comments to ask questions, suggest alternatives, share the way you understood something... make the blog yours!
• Please, use anonymous comments to give some feedback about the course. These will not be published, but I really appreciate them. In order to make the course better, I need to know how you are feeling about it!

Let's start with the course!

Basic information

When? Mondays, Wednesdays and Thursdays from 12th January to 26th February 2015 (apart from Carnival break).

What time? 10:00 - 12:00.

Where? Room 347, IMPA.

Syllabus: here.

Course notes: there will be no official notes. Instead, we will work with several references. These references, together with some information about each lecture, will be posted in this blog.

Grading: the grade will be based on the participation during the class and some assignments. The students who do not want to be graded in this way can choose to take an exam.