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Given a fiber bundle $f: E\rightarrow M$ with connected fibers we call the image $f^*(\Omega^k(M))\subset \Omega^k(E)$ the subspace of basic forms. Clearly, for any vertical vector field $X$ on $E$ we have that the interior product $i_X(f^*\omega)$ and the Lie derivative $L_X(f^*\omega)$ vanish for all $\omega \in \Omega^k(M)$. Is the converse true? That is, if $\alpha \in \Omega^k(E)$ is a form such that $i_X(\alpha)=0$ and $L_X(\alpha)=0$ for all vertical vector fields $X$ on $E$, is it true that $\alpha$ is a basic form? I believe so, but I am not sure how to prove it. Thanks for your help.

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  • $\begingroup$ As stated, no. You need at least that the fibres $f^{-1}(m)$ are connected. Otherwise consider a double cover of $S^1$ by itself, it is a fibre bundle with fibre consisting of two points. The set of vertical vector fields is the empty set. But the lift of an $\Omega^0(S^1)$ form (function) to the double cover must be $\pi$ periodic, which is not the whole space of $2\pi$ periodic functions. $\endgroup$ Commented Oct 4, 2011 at 9:26

2 Answers 2

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First, this can be checked locally, so we may as well assume $E = F\times M$.

Use coordinates $x_i$ on $F$ and $y_j$ on $M$. Then a $k-$form is given by $\alpha =\sum f_{IJ} dx^I\wedge dy^J$. Here, $I = \{i_1,...,i_s\}$ and $dx^I$ means $dx^{i_1}\wedge...\wedge dx^{i_s}$ and we have $|I|+|J|=k$.

The goal is to show that if $i_X(\alpha) = 0$ for all vertical $X$, then $f_{IJ} = 0$ whenever $I\neq \emptyset$.

So fix any $I\neq \emptyset$. Suppose $i_1\in I$. Let $X = \frac{\partial}{\partial x^{i_1}}$, the dual vector do $dx^{i_1}$. Then $0 = i_X(\alpha)$ so $ 0 = i_X(\alpha)(\frac{\partial}{\partial x^{I-i_1}}, \frac{\partial}{\partial x^{J}}) = \pm f_{IJ}$.

Thus the condition guarantees that $f_{IJ} = 0$ whenever $I\neq \emptyset$, so the only terms which appear in $\alpha$ are of the form $f_J dy^J$.

The only problem now is that $f_J$ could depend on the $F$ factor. This is where the Lie bracket term will come in.

The Lie bracket of a form is given by $L_Y(\alpha) = i_Y d\alpha + d i_Y(\alpha)$.

Taking $Y$ vertical, this reduces to $L_Y(\alpha) = i_Y(d\alpha)$ since we've just shown that $i_Y( \alpha) = 0$.

Now, $d\alpha = \sum \frac{\partial{f_J}}{{\partial x^i}} dx^i\wedge dy^J + \sum \frac{\partial{f_J}}{{\partial y^j}} dy^j\wedge dy^J$.

Just as before, by utilizing a dual basis, we can show that since $0 = L_X(\alpha) = di_X(\alpha)$, that $\frac{\partial{f_J}}{\partial x^i} = 0$.

But this implies that each $f_J$ only depends on the $y^j$ coordinates. It's now clear that $\alpha$ is the pull back of something, mainly of $\sum f_J dy^J$, which makes sense because $f$ is really only a function on $M$.

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  • $\begingroup$ I'd love to see a coordinate free proof if anyone comes up with one. This one is messier than I'd like. $\endgroup$ Commented Oct 4, 2011 at 2:26
  • $\begingroup$ Thanks. But just one detail, why is it sufficient to check locally? If we construct such form locally, why do we get a well defined global form in the base? The local constructions might not agree in the intersections of the trivializations. $\endgroup$
    – Manuel
    Commented Oct 4, 2011 at 14:09
  • $\begingroup$ You're right that a priori we couldn't check locally, so that was a mistake on my part. However, in my last paragraph, I should have said "$\alpha$ is the pull back of someting unique", and the uniqueness justifies patching all these together. $\endgroup$ Commented Oct 4, 2011 at 14:14
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    $\begingroup$ @AndréMuchon: The OP has identified a subset $P\subseteq \Omega(E)$ with the property that $f^\ast:\Omega(M)\rightarrow P\subseteq \Omega(E)$, and he/she asks if $f^\ast$ surjects onto $P$. Now, given a trivializing cover $U_\alpha$ of$ M$, we may decompose $E$ as a union of open sets of the form $f^{-1}(U_\alpha)\cong F\times U_\alpha$. Now, given $\omega \in P$, $\omega$ is determined by its restrictions to each $f^{-1}(U_\alpha)$. My answer shows that for any $\omega \in P$, $\omega|_{f^{-1}(U_\alpha)}$ is in the image of $f^\ast|_{U_\alpha}$...... $\endgroup$ Commented Sep 14, 2020 at 2:07
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    $\begingroup$ Say $\omega|_{f^{-1}(U_\alpha)} = f^\ast(\eta_\alpha)$ for some $\eta_\alpha$ defined on $U_\alpha$. Further, because $f^\ast$ is injective, $\eta_\alpha$ is uniquely defined. Suppose $U_\alpha\cap U_\beta \neq \emptyset$, then on $U_\alpha\cap U_\beta$, $\eta_\alpha = \eta_\beta$ by uniqueness. Hence, the $\eta_\alpha$ patch together to a form $\eta\in\Omega(M)$ and $f^\ast(\eta) = \omega$. $\endgroup$ Commented Sep 14, 2020 at 2:10
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If you added a connectedness assumption (which is essential, as I mentioned in my comment to your main question), what you desire follows from

Prop Let $f:M\to N$ be a submersion such that for every $y\in N$, $f^{-1}(y)$ is a connected submanifold of $M$. Then if $\mu\in\Omega^k(M)$ is a differential form satisfying $i_Y\mu = 0$ and $i_Y d\mu = 0$ for any $Y\in \ker df$, then there is a unique $\nu\in\Omega^k(N)$ such that $\mu = f^*\nu$.

Sketch of proof It suffices to show that the differential form defined by $\nu(f_*X_1,\ldots ,f_*X_k) = \mu(X_1,\ldots, X_k)$ is well-defined. By the assumption that $i_Y\mu = 0$ if $y\in \ker df$, at a fixed base point $x\in M$ you have that if $f_*X_1 = f_*X'_1$, $\mu(X_1,\ldots, X_k) = \mu(X'_1, X_2,\ldots,X_k)$. Now, it is clear that if $X$ is a vector field on $M$ such that its pushforward is well-defined (in the sense that there exists a vectorfield $Z$ on $N$ such that if $x_1,x_2\in f^{-1}(y)$, $df(X)|_{x_1} = df(X)|_{x_2} = Z|_y$) and if $Y$ is a vector field in $\ker df$, we have that $[X,Y] \in \ker df$. This can be easily checked by acting $[X,Y]$ on the pullback of functions defined on $N$. From this you can conclude that using $i_Yd\mu = 0$, for any vector fields $X_1, \ldots, X_k$ such that the pushforward is well-defined, $\mu(X_1,\ldots, X_k)$ is constant along connected components of $f^{-1}(y)$. q.e.d.


Addendum to address comments: Consider the scalar $\mu(X_1,\ldots, X_k)$ on $M$. Let $Y$ be a vertical vector field, then we have that $$ Y(\mu(X_1,\ldots,X_k)) = (\mathcal{L}_Y\mu)(X_1,\ldots,X_k) + \mu([Y,X_1],X_2,\ldots,X_k) + \cdots + \mu(X_1,\ldots, X_{k-1}, [Y,X_k]) $$ using Leibniz rule for Lie derivatives. So local constancy of the scalar field needs that $\mathcal{L}_Y\mu \equiv 0$ and that $[Y,X_l]$ is vertical.

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  • $\begingroup$ @Wille, I don't understand some details of your proof. For example, where did you use the fact that $[X,Y]$ is vertical if $X$ has a well defined pushforward and $Y$ is vertical? Also, what is $\alpha$? Is it a form in $N$ or $M$? Are $X_i$ vector fields on $N$ or $M$? I am a bit confused. Thanks for the clarification! $\endgroup$
    – Manuel
    Commented Oct 4, 2011 at 13:52
  • $\begingroup$ Also, how is $\nu$ defined for arbitrary vector fields $Y_1,...,Y_k$ on $N$? I feel that what you are trying to say is that these $Y_i$ can be lifted to $M$ and that the definition on $\nu$ is independent of such lift. But I can't follow your argument. Thanks again. $\endgroup$
    – Manuel
    Commented Oct 4, 2011 at 14:19
  • $\begingroup$ Oops, $\alpha$ is supposed to be $\mu$. That was a typo. I'll go fix that now. By the assumption that $f$ is a submersion every $Y\in T_yN$ has at least one pre-image in $T_xM$ for $x\in f^{-1}(y)$. So it suffices to show that the definition using preimages give the same value regardless of which representative you choose. $\endgroup$ Commented Oct 4, 2011 at 15:09
  • $\begingroup$ Sorry to resurrect this, but I'd like to check if I have understood this correctly. To initially define $\nu$ you need to arbitrarily fix for each $n\in N$ a point $m\in\pi^{-1}(n)$ (that presumably requires the axiom of choice) and define \begin{equation*} \nu_{n}( u_1,\ldots,u_k)=\mu_{m} (\hat u_1,\ldots,\hat u_k) \end{equation*} where $\hat u_i$ is any lift of $u_i$, right? In other words, the lift is arbitrary, but the point on the fibre needs to be fixed. The rest of the proof basically shows that the value of $\nu$ is independent of the point chosen on the fibre, right? $\endgroup$
    – GFR
    Commented Jan 26, 2018 at 10:40
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    $\begingroup$ @GFR: another way to think about this is to think about the tangent bundle map $\nabla f: TM \to TN$ induced by $f$. The end results are the same. $\endgroup$ Commented Jan 26, 2018 at 15:47

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