Let $E \to M$ be a smooth fiber bundle. Instead of differential forms defined on the whole tangent bundle $TE$ one could also consider forms on the vertical tangent bundle $VE$, i.e. forms defined on every fiber $E_m$ varying smoothly with $m \in M$. Is this notation of 'differential form along the fiber' discussed somewhere in the literature?

Clearly the basic notations of inner product and wedge product still make sense (with some modification). However, differentiation is now possible in two directions (differentiation along the fiber and with respect to the base) and thus should lead to two exterior differentials.

Remark: One approach would be to choose a connection on the bundle and then extend the fiber differential form by $0$ to the horizontal space and thereby get a bona-fide form on $E$. Besides the dependence on the connection this construction has another disadvantage. A fiber differential form, which is closed in fiber direction, does not need to have a closed extension.

  • $\begingroup$ I think people just deal with this ad-hoc when it comes up, which I've mostly seen in variations of complex structures on compact manifolds. For example, the differential-geometric approach to Weil-Petersson metrics goes through this sort of stuff (see Nannicini, Schumacher and Siu's work), and I think Griffith's original proof of transversality did too (Voisin does something like that in her Hodge theory book). Basically, you've got the right idea and can work out whatever you need. $\endgroup$ Mar 25, 2015 at 15:26
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    $\begingroup$ Yes. For any submersion of (smooth, real-analytic, complex-analytic, rigid-analytic, etc.) manifolds $f:X \rightarrow Y$ there is the notion of "relative differential forms" $\Omega^n_{X/Y}$ that is given by exterior powers of the vector bundle $\Omega^1_{X/Y} = {\rm{coker}}(f^{\ast}(\Omega^1_Y) \hookrightarrow \Omega^1_X)$ whose fiber-rank at a point $x \in X$ is the dimension of the fiber of $f$ through $x$. This notion is used very often in relative algebraic and complex-analytic geometry. $\endgroup$
    – user74230
    Mar 25, 2015 at 16:54
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    $\begingroup$ @user74230, could you turn your comment into an answer? This question is worth an actual answer and you have one to give. $\endgroup$ Mar 26, 2015 at 10:30
  • $\begingroup$ Another way to say what user74230 said is that given a subspace $W\subset V$ let $W^\perp \subset V$ be its annihilator. Then $W^* = V^*/W^\perp$. Therefore, at each point $p$ in the bundle $E$, $k$-forms on the fiber are given by $\Lambda^kT^*_pF = \Lambda^k(T^*_pE/(T_pF)^\perp).$ In particular, no connection is needed. $\endgroup$
    – Deane Yang
    Mar 26, 2015 at 11:57

1 Answer 1


The differential forms on the total space that vanish when restricted to the fibers give you a differential ideal of $\Omega^*(E)$, the powers of this ideal give a filtration which give rise to a spectral sequence. Some people call this the differential Leray-Serre spectral sequence, but I'm not shure if this is the standard name. Unfortunately I also don't know any written references, I learned this from Luca Vitagliano and Alexandre Vinogradov.

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    $\begingroup$ Luca Vitagliano pointed me to the following reference: A. HATTORI, Spectral sequence in the de Rham cohomology of fibre bundles, J. Fac. Sci. Univ. Tokyo, Sect. 1, 8 (1960), 289-331. $\endgroup$ Mar 26, 2015 at 14:10

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