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)


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

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


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

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?


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

  • $\begingroup$ Working on a similar question, I have appreciated your "for the shake of simplicity" :) $\endgroup$ – Jean Marie Sep 12 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 $X+\xi,Y+\eta\in L$:

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


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

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.