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I am confused about the definition of a differential form on a manifold.

The definition I have comes from Bott and Tu and is as follows:

A differential form, $\omega$, on a manifold $M$ is a collection of forms $\omega_U$ for $U$ in the atlas defining $M$, which are compatible in the following sense:

$i^*\omega_U=j^*\omega_V$ where $i,j$ are the inclusion maps.

I am confused as to what exactly $\omega_u$ is. Is it the pull back by a chart of a form on Euclidean space?

Moreover, how can I fail the compatibility criterion? It seems to me like it should always be true.

I think I simple example would really help me understand but I can't find one.

Thanks for your help,

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see if you can construct everything for the unit sphere in $\mathbb R^3.$ everything can be written using the ambient $x,y,z$ coordinates –  Will Jagy Nov 19 '13 at 5:51
    
You may just think of an $n$ form on $M$ as a collection of $\omega_U$ ($n$-form on open coordinate $U$) such that $f^*\omega_V = \omega_U$, where $f :U' \to V'$ is the change of coordinate. –  John Nov 19 '13 at 6:02

2 Answers 2

up vote 1 down vote accepted

A differential form $\omega$ of degree $k$ on a real manifold $M$ of dimension $n$ is, formally, a section of the $p$-th exterior power of the cotangent bundle $T^{*}M$ over $M$, in symbols:

$$\omega\in \Gamma(\wedge^k T^{*}M).$$

This means that $\omega$ is a smooth map $M\rightarrow \wedge^k T^{*}M$ which, at any point $p\in M$ is given by

$$\omega_U(p):=\omega_{U,i_1,\dots,i_k}(p)dx^{i_1}\wedge\dots\wedge dx^{i_k}, $$

where $(x^{1},\dots, x^{n})$ denote local coordinates in the open set $U$ centered at $p\in M$. In other words, the differential form $\omega$ can be seen as a collection of "local" forms like above; any coordinate transformation $f: U\rightarrow V$ induces a transformation $\omega_V=f^{*}\circ\omega_U$ as pointed out by @John. As often happens in differential geometry, one starts with local data and gives a rule to glue them into a global structure (when possible).

$f^{*}\omega$ is the pull-back of $\omega$ along $f$: you can find its definition on every textbook. Essentially, the coordinate transformation formula reduces to an identity involving the Jacobian matrix of $f$.

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I am still a bit confused. Why is the existence of an n-form different to that of an k-form? My current understanding of a differential form is a smooth function, tensor a product of 'dx', where the 'dx' are formal symbols. So if I have a (k-1)-form, why can I not turn this into a k-form by adding another dx to the product? –  Bates Nov 19 '13 at 13:07
    
If $n$ is the real dimension of your manifold $M$, then differential forms can be of degree $0,1,...,n$. If you choose any $k\in \{0,...,n\}$ you can apply the considerations above. The proper way to turn a $k-1$ form into a $k$ form is expressed by a specific operation called exterior differentiation. –  Avitus Nov 19 '13 at 13:25
    
I think I am asking a silly question but, if $f(X)dx_1...dx_{k-1}$ is a (k-1)-form then is $f(X)dx_1...dx_{k-1}d_k$ a k-form? –  Bates Nov 19 '13 at 15:49
    
yes, it is true: $k$ is the length of the string $dx_1\wedge...\wedge dx_k$ in the local coordinates expression of the diff. form $\omega$. In my answer the diff. form is of degree $k$. –  Avitus Nov 19 '13 at 16:08
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In that case, doesn't the existence of a nowhere vanishing function (0-form) imply the existence of a nowhere vanishing k-form for all k<n+1? I can just multiply by the desired number if dx's. –  Bates Nov 19 '13 at 16:26

This is just an intuitive opinion. I believe this works the same way as the differentiable structure definition works for a manifold. We want a maximal atlas that is the collection of all compatible atlases given any representative atlas. This probably tries to do something similar for a form. In as many words, as Avitus notes, this is an attempt to relate the local and global existences.

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