Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Good evening,

In differential forms (in the proof of the naturality of the exterior derivative), I don't get why if $h\in \Lambda^0(U)$ and $f^*$ is the pullback then,

$f^*dh=d(f^*h)$.

I wrote $dh=\frac{\partial h}{\partial x^i}dx^i$ so (?) $f^*dh(v)=\frac{\partial h}{\partial f^i}df^i(df(v))$ but I don't know if this is correct and I can't continue...

Thanks a lot for your help

share|improve this question
add comment

2 Answers

up vote 3 down vote accepted

Good evening!

If we define the pullback map as $$f^*(\sum g_I dx_{i_1} \ldots dx_{i_p}):= \sum (g_I \circ f) df_{i_1} \ldots df_{i_p}$$

Then on one hand we have:

$$d(f^*h)=d(h \circ f) $$

And on the other hand (using the $\textit{Chain Rule}$):

$$f^*(dh)= f^*(\sum \frac{\partial h}{\partial x_i} dx_i)= \sum ((\frac{\partial h}{ \partial x_i} \circ f) df_i) = d(h \circ f)$$

As we wanted. This result can be generalised to n-forms and interpreted as ''the pullback commutes with the exterior derivative''.

If you need more help, consult Bott and Lu's $\textit{Differential Forms}$ book.

share|improve this answer
add comment

Let $f: M \longrightarrow N$ be a smooth map between the manifolds $M$ and $N$, and let $h: N \longrightarrow \Bbb R$ be a smooth map. Then by definition, for any point $x \in M$ and any vector $V \in T_x M$, we have $$(f^\ast dh)_x(V) = dh_{f(x)}(df_x(V)).$$ On the other hand, since $f^\ast h = h \circ f$, by the chain rule we have that $$d(f^\ast h)_x(V) = d(h \circ f)_x(V) = (dh_{f(x)} \circ df_x)(V) = dh_{f(x)}(df_x(V)).$$ Hence pullbacks commute with the exterior derivative in the case asked about in your question (of course this holds for forms of arbitrary degree as well).

share|improve this answer
    
Thank you very much! –  Nre Jan 10 '13 at 21:50
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.