# Prove that $p-$form $\omega$ on $M\times N$ is $\delta \pi(\alpha)$ iff $i(X)\omega=0$ and $L_X\omega=0$

Let $M$ be connected and let $\pi:M\times N \rightarrow N$ be the natural projection. Prove that $p-$form $\omega$ on $M\times N$ is $\delta \pi(\alpha)$ for some $p-$form $\alpha$ on $N$ if and only if $i(X)\omega=0$ and $L_X\omega=0$ for every vector field $X$ on $M\times N$ for which $d\pi(X(m,n))=0$ at each point $(m,n)\in M\times N$

Here, $i(X)\omega$ is interior multiplication, $L_x\omega$ denotes the lie derivative of $\omega$ and $\delta \pi$ is the pullback of $\pi$.

I am not at all comfortable with the notion of interior multiplication. In my book, interior multiplication is defined as the transpose of the exterior product. That is,

Let $u\in \Lambda(V)$ and defined the endomorphism $\epsilon(u)$ by $\epsilon(u)(v)=u\wedge v$. Denote the transpose of $\epsilon(u)$ by $i(u)$. Then $i(u)$ is called interior multiplication by $u$ and it satisfies $(i(u)v^*,w)=(v^*,u\wedge w)$, where $v^*\in \Lambda(V^*)$ and $w\in \Lambda(V)$.

We define interior multiplcation by vector fields pointwise. That is,

$(i(X)\omega)_m=i(X_m)\omega_m$

The lie derivative of a differential form with respect to a vector field is defined as follows:

$(L_X\omega))_m=\frac{d}{dt}|_{t=0}(\delta X_t(\omega_{X_t(m)})),$ where $\delta X_t$ is the pullback.

Inner multiplication and lie derivatives are related by the following identity:

$L_X=i(X)\circ d+d\circ i(X)$

I think I don't understand how interior multiplication and lie derivatives work. Here is my attempt at the problem.

Suppose $\omega=\delta \pi(\alpha)$ and let $X$ be a vector field on $M\times N$ for which $d\pi(X)=0$ for all $(m,n) \in M\times N$. If we can show that $i(X)\omega=0$, then the identity will show that $L_x(\omega)=0$. Fix a point $p=(m,n)\in M\times N$ and $w \in \Lambda(M\times N)$. Then $(i(X_p)\omega_p,w)=(\omega_p,X_p\wedge w)=(\delta \pi(\alpha)|_p,X_p\wedge w)=(\alpha(d\pi),X_p \wedge w).$ I need to show that this is zero.

I have basically no idea about the other direction. Any help would be much appreciated.

(1) $M,\ N$ has charts $x,\ y$ Then if $\omega$ is a pull back then $\omega= (f\circ \pi) (x,y) dy_{1}\cdots dy_p$ where $f: N\rightarrow {\bf R}$ (For convenience, we can write it) So $$\partial_{x_i} (f\circ \pi )= df\ d\pi \partial_{x_i} =0$$

If $d\pi X=0$, then $X=g_i(x,y)\partial_{x_i}$ So $i_X\omega =0$ In further since $L_X=i_Xd +di_X$ so that $$L_X\omega = i_Xd\omega = i_X \bigg( \partial_{y_i}(f\circ \pi) dy_i dy_{1}\cdots dy_p \bigg) =0$$

(2) $\omega=gdx_1\cdots dx_k dy_{k+1}\cdots dy_p+f dy_{1}\cdots dy_p$ where $1\leq k\leq p$ Since $i_X\omega =0$, then for $d\pi\ X=0$, $g=0$

Now we must prove that $f$ is independent of $x$ : If $X=\partial_{x_i}$, then $0=L_X\omega =i_Xdf dy_{1}\cdots dy_p$ That is $\partial_{x_i} f=0$

• This doesn't seem correct. In (2) you need to consider terms with at least one $dx_i$. – Ted Shifrin Apr 19 '16 at 7:00
• You are right I see I will correct it – HK Lee Apr 19 '16 at 7:11
• @HKLee how is $\omega=(f\circ \pi)(x,y)dy_1\ldots dy_p$? – user210552 Apr 19 '16 at 7:47
• Note that $(x,y)$ is chart for $M\times N$ so that $\omega$ is a combination of wedge products of $dx_i,\ dy_j$ But $\pi^\ast dy_i=dy_i$ so that $dx_i$ can not appear – HK Lee Apr 19 '16 at 7:56
• @HKLee ah alright...also could you explain why $i_X\omega=0$? – user210552 Apr 19 '16 at 7:58