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Let $(E,M,p)$ be a smooth n dimensional vector bundle. Then $(TE,TM,Dp)$ is a 2n dimensional vector bundle. We restrict this bundle to $M\subset TM$. We denote this restricted bundle by $F$, as a 2n bundle over $M$. In this situation we say that $E$ is the integral of $F$ and write $E=\int F$. In this question we search for a converse statement:

Question:

Assume that $F$ is a 2n dimensional smooth bundle over a manifold $M$. Is there a smooth n-bundle $E$ over $M$ such that $E=\int F$? What obstructions would appear? Assume that $E_{1}$ and $E_{2}$ are two solutions of $E=\int F$. What can be said about the difference $E_{1}-E_{2}$ as an element of K-theory?

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As in my answer to your previous question, $F=E\oplus E$. This should answer the rest (as much as it is possible to answer). A simple obstruction is that the total Stiefel--Whitney class should be a square. The same about the total rational Pontrjagin class. The class should be divisible by two in the $K$-group. I'm not sure that there are simple sufficient condition. As to the last question, it's unique up to $2$-torsion: $2(E_2-E_2)=0$.

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