I think I need some hints about a proof I am currently reading in order to understand it. This question is similar to the construction used in Lemma 17.9 in the book "Introduction to smooth manifolds" by John Lee. Unfortunately, I am currently not at the level to understand his notation, although it would probably answer my question. Let $\omega: M \times [0,1]$ be a differential 2-form ( $M$ is a smooth manifold), then I want to transform it into a 1-form on $M$ by a linear map $I : \Omega^2(M \times [0,1]) \rightarrow \Omega^1(M).$

So the idea is somehow to integrate with an operator $I$ over the $[0,1]$ part in order to get the right differential form.

First, I notice that each $2-$ form on $M \times[0,1]$ is necessarily of the form $\omega = \omega_1 \wedge dt + \omega_2,$ where $\omega_1$ is a 1-form depending only on coordinates of the manifold, $t$ is the coordinate of the interval here and $\omega_2$ is a 2-form depending only on coordinates of $M$. Clearly, by skew symmetry of the wedge product, there cannot be a 2-form with $dt \wedge dt.$

Now, the big question is: How do I define $I(\omega)$ to get the following results; cause I am currently reading some lecture notes (not online unfortunately) where exactly such a construction is used, and it is said that its integration with respect to the $[0,1]$ part, but it is not explicitely mentioned how it is defined, so what this operator explicitly does to a form on $M \times [0,1]$. Altogether, I am looking for an explanation of what this operator $I$ exactly does. Let me provide you with the calculation given in the notes: So $\omega_1 = \sum_{i=1}^{n} a_i(x,t) dx_i,$ where $x_1,...,x_n$ are coordinates of the manifold $M$ and $\omega_2 = \sum_{i<j} b_{i,j}(x,t)dx_i \wedge dx_j$

Now, it is said that $I(\omega ) = \int_0^1 \omega_{p-1} dt.$ So somehow this operator does not affect $\omega_p$ if I get this correctly(maybe because there is no $dt$). Furthermore, for $d \omega = d \omega_{p-1} \wedge dt + d \omega_{p}$ which can be written in coordinates as $$ d\omega_{p-1} \wedge dt= \sum_{i=1}^{n} \sum_{j=1}^{n} \partial_{x_j} a_i(x,t) dx_j\wedge dx_i \wedge dt$$

and $$d \omega_p = \sum_{i<j} \left(\sum_{k=1}^{n} \partial_{x_k}b_{i,j}(x,t) dx_k \wedge dx_i \wedge dx_j + \partial_t b_{i,j}(x,t)dt\wedge dx_i \wedge dx_j \right),$$ we get by linearity $I(d\omega) = I( d \omega_p) + I (d\omega_{p-1}\wedge dt)$ which shall altogether give us: $d(I(\omega))-I(d\omega) = (-1) (\sum_{i<j} b_{i,j}(x,1)-b_{i,j}(x,0)dx_i\wedge dx_j).$

This is what the lecture notes contain, can we tell from this how $I$ exactly acts on forms on $M \times[0,1]$?

Especially, why this operator apparently gives zero, if the form does not depend on $t$ via $dt$, although $a,b$ depend on $t$ in general?

If anything is unclear, please let me know.


You have things sort of confused. For a $2$-form $\omega$, represented as $\omega = dt\wedge\omega_1 + \omega_2$ (I prefer to put the $dt$ first to spare sign problems), we define $$I(\omega) = I(dt\wedge\omega_1) = \sum_{i=1}^n\big(\int_0^1 a_i(x,t)dt\big)dx_i.$$ However, to prove your formula, we need to define the operator $I$, in general, on $p$-forms, and we do this analogously. In local coordinates $(x_1,\dots,x_n)$, write a general $p$-form $\omega$, similarly to the above, as $\omega=dt\wedge \omega_1 + \omega_2$, where $\omega_1$ is a $(p-1)$-form involving only $dx_i$'s, and $\omega_2$ is a $p$-form involving only $dx_i$'s. Write $\omega_1 = \sum\limits_I a_I(x,t)dx_I$, where these are multi-indices of length $p-1$. Then we set, once again, $$I(\omega)= I(dt\wedge\omega_1) = \sum_I\big(\int_0^1 a_I(x,t)dt\big)dx_I.$$

Now it will follow that for any $p$-form $\omega$ on $M\times [0,1]$, $d(I(\omega))+I(d\omega) = \iota_1^*\omega - \iota_0^*\omega$, where $\iota_t(x)=(x,t) \in M\times [0,1]$.

  • $\begingroup$ thank you. that was really illuminating $\endgroup$ – RealAnalysis May 18 '15 at 21:32
  • $\begingroup$ You're most welcome. $\endgroup$ – Ted Shifrin May 18 '15 at 23:48

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