Let M be a smooth manifold. Can every k-form $\omega$ on M be written as a sum of k-forms, that are wedge products of 1-forms, i.e. $\omega = \sum_{i=0}^n \alpha_1^{(i)} \wedge \ldots \wedge \alpha_k^{(i)} $, where $\alpha_l^{(i)} \in \Omega^1(M) $ ? If M is compact, one can cover M be finitely many charts and use a partition of unity to see that this holds...but how about the general case?
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5$\begingroup$ Just a comment: being coverable by finitely many charts is a much weaker condition on a manifold than compactness. This, or conditions closely related to it, is often called being "of finite type". In my experience even the noncompact manifolds one studies are typically of finite type. (Nevertheless your question is a perfectly reasonable one, and I don't know the answer off the top of my head.) $\endgroup$– Pete L. ClarkCommented Mar 1, 2011 at 21:34
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$\begingroup$ I vaguely recall something like countable partitions of unity, and most authors pose some countability conditions on manifolds. Do these fit together? $\endgroup$– darij grinbergCommented Mar 1, 2011 at 21:37
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$\begingroup$ OK, more precisely: A second-countable manifold has a countable base $\mathcal T$. Every chart, being an open set, is the union of some subfamily of $\mathcal T$. Taking the union of all of these subfamilies (over all charts) and throwing away equal sets, we obtain a subfamily $\mathcal S$ of $\mathcal T$ such that every set in $\mathcal S$ is included in a chart and the sets in $\mathcal S$ cover all of the manifold. This family $\mathcal S$, being a subfamily of the countable $\mathcal T$, is countable. Now, use a countable partition of unity. Am I right? $\endgroup$– darij grinbergCommented Mar 1, 2011 at 21:41
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$\begingroup$ The answer is yes for silly reasons -- given any form $\omega$, multiply it by a partition of unity where your bump functions have compact support. You can do what you ask in every compact set, so you can do it on your manifold. Your "sum" will be locally finite, so it's defined. But perhaps you want a sum that's globally finite? $\endgroup$– Ryan BudneyCommented Mar 1, 2011 at 21:56
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$\begingroup$ I think you can tweak the argument above to be a globally finite sum by choosing your partition of unity carefully, and then re-arranging the terms in the sum carefully to make the global sum finite. $\endgroup$– Ryan BudneyCommented Mar 1, 2011 at 21:57
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3 Answers
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The answer is yes and you need transversality in some form. Here we use Whitneys embedding theorem, but a weaker statement would suffice. Embed $M \subset \mathbb{R}^m$. Thus there is a monomorphism $TM \to M \times \mathbb{R}^m$ of vector bundles, dualizing to an epimorphism $M \times \mathbb{R}^m \to T^{\ast} M$. Let $a_1,\ldots ,a_m$ be the images of the basis vectors. You can write any $k$-form on $M$ as a linear combination with $C^{\infty} (M)$-coefficients of the forms $a_{i_1} \wedge \ldots a_{i_k}$. Done.
answered Mar 1, 2011 at 22:03
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Johannes' answer can be upgraded to the following statement:
Let $M$ be a second countable and Hausdorff manifold (who cares about others?) and $\pi_i\colon E_i \longrightarrow M$ vector bundles for $i = 1, \ldots, N$. Then the canonical map \begin{equation} \Gamma^\infty(E_1) \otimes_{C^\infty(M)} \cdots \otimes_{C^\infty(M)} \Gamma^\infty(E_N) \longrightarrow \Gamma^\infty(E_1 \otimes \cdots \otimes E_N) \end{equation} is an isomorphism of $C^\infty(M)$-modules. The same holds if you do some symmetrization/antisymmetrization in tensor powers of a single vector bundle in addition.
My favorite argument uses the smooth version of Serre-Swan's theorem: the sections $\Gamma^\infty(E)$ are a finitely generated and projective module over $C^\infty(M)$ and any finitely generated projective module is (up to iso) of that form. This can e.g. be proved along the same lines as Johannes' argument by embedding the tangent bundle of $E$ into some big $\mathbb{R}^n$ and observe that $E$ is naturally identifiable with the vertical subbundle of $TE$ viewed as bundle over $M$.
Then the above statement is shown by noting that the map is clearly injective (and well-defined). The surjectivity is the harder part. But the tensor product on the left hand side is again finitely generated and projective, so it has to be the space of smooth sections of some vector bundle. Then the injectivity gives that this vector bundle can be included as a subbundle of $E_1 \otimes \cdots \otimes E_N$. Counting fibre dimensions shows that they coincide...
The main point is that Serre Swan also holds in the non-compact case. An alternative proof of this can be found e.g. in Well's book in complex differential geometry.
answered Mar 2, 2011 at 8:32
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$\begingroup$ Does your argument gives information on the number of simple forms which sum up to our given form? For example is it obvious that every $2$-form on a $3$ manifold is a sum of 3 simple two forms?(a simple form is a product of 1 forms)? $\endgroup$ Commented May 28, 2018 at 7:45
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1$\begingroup$ One can get upper bounds from the dimension of $M$ and the ranks of the vector bundles in question: I think that $\Gamma^\infty(E)$ has a dual basis (in the sense of projective modules) of at most $(n+1)k$ elements, where $n$ is the dimension of $M$ and $k$ the rank of $E$. For tensor- and wedge-products one can then do the combinatorics to get upper bounds on the number of elements in the dual bases. $\endgroup$ Commented May 28, 2018 at 8:26
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$\begingroup$ Yes. Thank you.regarding orientable 3 manifolds every 2-form is a sum of three 2-firms since the manifold is parallelizable. But it would be interesting to find a counter example of non orientable case. Or a 2- form on $S^4$ which is not a combiation of six simple 2-forms. $\endgroup$ Commented May 28, 2018 at 9:31
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The same argument as in comapct case works if you use in addition Ostrand's theorem on colored dimension
answered Mar 2, 2011 at 0:43