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Assume that $E$ is a Riemannian vector bundle, then its structure group is reduced to $O(n)$. Then the structure group of $E \oplus E$ is reduced to $D(O(n) \oplus O(n)) \subset Sp(2n)$ where $$D(O(n) \oplus O(n))=\{ A\oplus A\mid A \in O(n)\}$$

So there is a natural symplectic structure $\omega $ on each fiber of $E \oplus E$.(We use a maximal atlas of trivialization charts for $E \oplus E$ whose structure group is contained in $D(O(n) \oplus O(n))$.

let $T:E \to E \oplus E$ be an arbitrary morphism. We pull back $\omega $ to $T^*(\omega)$. So $T^*(\omega)$ is a $2$- form on each fiber of $E$.

The above construction can be applied to $E=TM$ where $M$ is a compact Riemannian $2$ dimensional manifold.

Associated with a compact Riemannian surface $(M,g)$, we define a a quantity $q(M,g)$ as follows: $$q(M,g) =\sup_T |\int_M T^*(\omega)|$$ where $T$ varies on all isometric embedding of $TM$ into $TM \oplus TM$ where the latter is equipped with the product metric.

Is this quantity finite? Is it independent of choosing a maximal trivialization atlas for $TM \oplus TM$ whose transition maps are contained in $D(O(n) \oplus O(n))$?

If the answer of the above question is "yes", what would be an interpretation of this quantity $q(M,g)$?

Note that the same construction can be repeated for an arbitrary compact $2k$ dimensional manifold by integration of ${(T^*(\omega))}^k$ on $M$.

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    $\begingroup$ Reminds me of Courant algebroid constructions. Is that what you're going for? $\endgroup$
    – AHusain
    May 29, 2017 at 17:15
  • $\begingroup$ @AHusain thanks for your comment. I am not aware of such construction. Is this construction related to my question? I search in wikipedia for Courant algebroid. Thank you for informing me of that. $\endgroup$ May 29, 2017 at 18:56
  • $\begingroup$ Actually, it is almost the opposite of this question. There you typically want the map $E \to E \bigoplus E$ (turn one into $E^\vee$) to give pullback $0$ rather than as big as possible. $\endgroup$
    – AHusain
    May 29, 2017 at 19:18
  • $\begingroup$ @AHusain Are you referring to the folowing concept?en.wikipedia.org/wiki/Courant_algebroid $\endgroup$ May 29, 2017 at 20:01
  • $\begingroup$ @AHusain I really do not understand the relation to my question. $\endgroup$ May 29, 2017 at 20:02

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