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I am trying to solve the following problem:

Let $M$ and $N$ be smooth manifolds, and suppose $P:Mto N$ is a surjective smooth submersion with connected fibers. Show that given $alpha in Omega^k(M)$ there exists $betain Omega^k(N)$ such that $alpha=P^*beta$ if and only if $iota_X alpha=0$ and $L_X alpha =0$ for every $X in ker(dP)$.

Solution

Let $alpha=P^*beta$, so by definition of pullback we have that:
begin{align}
alpha_x(X)=beta_{P(x)}(dP(X))
end{align} for $x in M$ and $Xin T_xM$
If we take $X$ such that $Xin ker(dP)$, then $iota_X alpha = 0$.

To prove that $L_X alpha =0$, we will use Cartan's magic formula:
begin{align}
L_Xalpha= iota_X dalpha +d(iota_X alpha)
end{align}
We already have that the second term is zero for $Xin ker(dP)$.
Using the fact that $d(P^*beta)=P^*dbeta$, and the definition of pullback we get that the first term in Cartan's formula becomes $dbeta_{P(x)}(dP(X))=0$ if $Xin ker(dP)$.

Thus we have that $L_X alpha =0$ for all $Xin ker(dP)$.

Conversely, let us define the pullback in terms of local coordinates with $P:Mto N$ as $P(x_1,...,x_n,...,x_m)=(x_1,...,x_n)$.

Given $alpha in Omega^k(M)$, we can write it as
begin{align}
alpha = sum_{1}^{m}f_I(x_1,...,x_m) dx_I
end{align}
for $I=(x_1,...,x_m)$. By hypothesis, $iota_X alpha =0$ for all $Xin ker(dP)$. By the way we defined the pullback, the expression $Xin ker(dP)$ means that $X=c_iX_i$, where $X_i=0$ for $n<ileq m$.

The second condition $L_X alpha =0$, with cartan's formula and the first condition means that $iota_X dalpha=0$. Replacing we get:
begin{align}
dalpha = sum_{I,j}^{m}dfrac{f_I(x_1,...,x_n)}{partial(x_i)}dx_jwedge dx_I
end{align}
Evaluating on $Xin ker(dP)$, we get that the change of $f_I$ in the direction of $x_i$ is zero for $n<ileq m$. Then our conditions on alpha means that $f_I(x_1, ...,x_m)= f_I(x_1,...,x_n,0,...,0)$ for all $Xin ker(dP)$.

Let $g_J= f_Icirc P$, so we have that:
begin{align}
P^*alpha = sum_{I}^{m} f_Icirc P(x_1,...,x_m) dx_I=sum_{J}^{n} g_J(x_1,...,x_n) dx_J=beta
end{align}

Thus completing the proof.

I want to check this proof as I am not confident on the second part. I think I am missing something on how the $dx_i$ with $n<ileq m$ dissappear. Also I think I am messing something up with the indices but I can't say what exactly.
My professor explained me the main idea of the second part of the proof. I think I understand the general idea, but not enough to put it in rigorous terms.

It doesn't look so good to me. We don't necessarily have $k=1$. Let's organize ourselves:

• If $alpha = P^*beta$, then $X in ker {rm d}P require{cancel}$ says that $$iota_Xalpha(X_1,ldots, X_{k-1}) = alpha(X,X_1,ldots,X_k) = beta(cancelto{0}{{rm d}P(X)},{rm d}P(X_1),ldots, {rm d}P(X_{k-1})) = 0,$$so $iota_Xalpha=0$. Also, we have that $$begin{align} mathcal{L}_Xalpha(X_1,ldots,X_k) &= X(alpha(X_1,ldots,X_k)) - sum_{i=1}^n alpha(X_1,ldots,[X,X_i],ldots, X_k) \ &= X(beta({rm d}P(X_1),ldots, {rm d}P(X_k))) - sum_{i=1}^n beta({rm d}P(X_1),ldots,{rm d}P([X,X_i]),ldots, {rm d}P(X_k)) \ &= cancelto{0}{X(beta({rm d}P(X_1),ldots, {rm d}P(X_k)))} - sum_{i=1}^n beta({rm d}P(X_1),ldots,[cancelto{0}{{rm d}P(X)},{rm d}P(X_i)]),ldots, {rm d}P(X_k))\ &= 0.end{align}$$
  


• Assume $iota_Xalpha=0$ and $mathcal{L}_Xalpha=0$ for all $X in ker {rm d}P$. I don't have much time to go over all details now, but here's the idea: define $beta$ locally using the charts given by the constant rank theorem -- $mathcal{L}_Xalpha=0$ will allow us to define $beta$ along the whole chart around $P(x)$ in $N$ via the flows of the vector fields $partial_i$, with $n+1 leq i leq m$, since $(mathcal{L}_{partial_i}alpha)_{i_1cdots i_k} = partial alpha_{i_1 cdots i_k}/partial x^i$ (which you seem to have done, when writing $g_{i_1 cdots i_k}(x_1,ldots, x_n) = f_{i_1cdots i_k}(x_1,ldots,x_n,0ldots,0)$), and the condition $iota_Xalpha=0$ will say that the resulting $beta$ is well-defined (it is not clear to me what you mean by $c_iX_i$, are you missing a sum?).