# On Proposition 2.6 Gualtieri Thesis Generalized complex geometry

I'm working with Gualtieri's thesis about Generalized complex Geometry and I don't understand the proof of the Proposition 2.6 (p. 7). It says Every maximal isotropic subspace (maximal totally null subspaces) of $V\oplus V^*$ can be express as $L(E,\alpha)$ for some appropriate $E\subseteq V$ and 2-form $\alpha\in\Lambda^2(E)$. (Recall

$$L(E,\alpha)=\{X+\xi\in E\oplus V^* : \xi|_E = i_\alpha X \}.$$

In the proof, he defines $E=\pi_V (L)$ and

\begin{array}{rcl} \alpha: E & \longrightarrow & E^* \\ X & \longmapsto & \Psi(\pi_{V^*}(\pi_V^{-1}(X)\cap L)) , \end{array}

where $\pi$ are the canonical projections onto $V$ and $V^*$ and $\Psi:E^*\rightarrow V^*/\operatorname{Ann} E$ is the isomorphism he mentions (he doesn't use the isomorphism explicitly even they says it is necessary).

I don't understand what the map does so I can't prove it is skew. Can you help me? I wish to understand what the map does, but my actual goal is to prove it is skew

$$\alpha(X)(Y)+\alpha(Y)(X)=0 \qquad \forall X,Y\in E$$

(I've picked some tags but the real ones don't exist. We have neither Generalize geometry and Dirac structures)

EDIT:

I haven't done any progress but I can show some example I have thougt about to show you I'm working in it.

Let $V$ be a 3-dimensional vector space spanned by $\{E_i\}$ and let $\{\epsilon_j\}$ be its dual basis ($\epsilon_j(E_i)=\delta_{ij}$).

The set $\{E_i,\epsilon_j\}$ is a basis for $V\oplus V^*$. Here we consider the indefinite product

$$\langle X+\xi, Y+\eta \rangle = \eta(X)+ \xi (Y) .$$

The subspaces

$$L_1 = \operatorname{span} \{E_1+\epsilon_2, E_1+\epsilon_3, \epsilon_3 \}$$

and

$$L_2 = \operatorname{span} \{E_1+\epsilon_3, E_2+\epsilon_3, \epsilon_3 \}$$

are both maximal isotropic subspaces.

The problem in both cases is $L$ can be descomposed into $W\oplus \operatorname{Ann}W$, $W=\pi_V(L)$. So it is easy to see $\alpha=0$. This kind of examples are very easy and I can't think of anythinc more avdanced. Some ideas?

SOLUTION?

I think I can give an alternative proof which constructs explicitly such a form. Ii would be like this:

Let $L$ be a maximal isotropic subspaces spanned by the vectors $\{E_1+\xi_1,E_2+\xi_2,\dots ,E_n+\xi_n\}$. As above, define $W=\pi_V(L)$. This subspaces will be spanned by some of the $E_i$'s. Call $B$ sucha set. For the shake of simplicity also call

$$I=\{i: E_i\in B\}$$

Let $\{\theta_i\}$ be a dual basis for $B$:

$$\theta_i(E_j)= \delta_{ij} \qquad \forall i,j\in I$$

Now, define $\alpha\in\Lambda^2(W)$ as follows:

For $E_1$: if $E_1\in B$, let $\alpha_1= \iota^*(\epsilon_1)\wedge\theta^1$, where $\iota^*:V^*\rightarrow W^*$ is the dual of the inclusion. Otherwise $\alpha_1=0$.

Next, if $E_2\in B$ let $\alpha_2=\alpha_1+ \iota^*(\epsilon_2)\wedge \theta_2$. If moreover $E_2=E_1$, set $\alpha_2 = \frac{1}{2}\alpha_1 + \frac{1}{2}\iota^*(\epsilon_2)\wedge \theta_2.$

At the end we should have a $2$-form

$$\alpha= \sum_{i\in I} w_i \iota^*(\epsilon_i)\wedge \theta_i,$$

where $w_i$ is 1 over the times each $E_i$ appears in the span of $L$. I claim this $\alpha$ is what I'm looking for, but I don't know how to prove it. My problem is I need to show that, if $E_1+\epsilon_1$ and $E_1+\epsilon_2$ span $L$, then $\iota^*(\epsilon_1)=\iota^*(\epsilon_2)$.

Clearly, for each $i\in B$, $\epsilon_i=i_{E_i}\alpha$.

• Working on a similar question, I have appreciated your "for the shake of simplicity" :) Sep 12 '20 at 7:31

Ok. I think I have an answer for my question. My question was about (firstly) about the definition of $\alpha$. So I start explaining it a little:

$$\alpha(X)=\pi_{V^*}(\pi^{-1}_V(X)\cap L)$$

is the set of all 1-forms $\xi$ such that $X+\xi\in L$. Let $\xi,\xi'\in\alpha(X)$ two such forms. Then

$$\xi-\xi'=X+\xi-X-\xi' \in L$$

and because of that

$$\langle\xi-\xi',Y+\eta\rangle = \frac{1}{2}(\xi-\xi')(Y) = 0 \qquad \forall Y\in W.$$

Thus $\xi-\xi'\in\operatorname{Ann} W$. This result means the set $\alpha(X)$ is an equivalence class of the set $E/\operatorname{Ann}W$ and using $\Psi$ can be identified with a 1-form in $W^*$. Now, since $\xi|_W$ is an element of $W^*$, it can be expressed as an equivalence class $[\eta]$, for some $\eta\in V^*$. On the other hand, the isomorphism $\Psi$ is simply $[\eta] \mapsto \eta|_W$. Hence $\xi|_W$ is related with $[\xi]$ and it must be equal to $\alpha(X)$.

Now I'm going to prove the skew-symmetry: let $X+\xi,Y+\eta\in L$:

$$\langle X-Y+\xi-\eta \rangle =(\xi-\eta)(X-Y) = 0 \Longrightarrow \xi(X)+\eta(Y) = \xi(Y)+\eta(X).$$

Thus

$$\alpha(X)(Y)+\alpha(Y)(X)=\xi(Y)+\eta(X)=\xi(X)+\eta(Y)=0.$$

And this is how the proof goes.

Now, can I give a different proof? Well, it is possible to state it quite different, although the background is the same. For example, you can define $\alpha$ as follows: for each $X\in W$, $\alpha(X)=\xi|_W$, where $\xi\in V^*$ is such that $X+\xi\in L$. This $\alpha$ is well-defined since if $\xi'$ is another form such that $X+\xi'\in L$, then, as before, $\xi-\xi'\in\operatorname{Ann} W$, so $\xi|_W=\xi'|_W$. However, this is the same as before but using different words. The proof of the skew-symmetry would be the same.

And finally my alternative'' proof. That proof constructs $\alpha$ explicitly. I had the problem that I didn't know how to prove that if $E_1+\varepsilon_1,E_1+\varepsilon_2\in L$, then $i^*(\varepsilon_1)=i^*(\varepsilon_2)$. Now I know and in fact, I have already proved it.

This is my answer. I think my ideas are right. Maybe I haven't express myself well, but I wanted to explain all the possibilities. Nevertheless, the bounty is still open, and I will accept extra answers.