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Well, the question is in the title.

Is it true, that given a smooth manifold $M$, the following isomorphism holds:

$$ \Gamma(\Lambda(T^*M)) \cong \Lambda(\Gamma(T^*M)) $$ $\Gamma$ - smooth sections functor, $\Lambda$ - exterior algebra functor. Base ring for both - $C^\infty(M)$.

Motivation: I'm kind of confused, because some sources define differential forms as sections of exterior bundle, others define forms as elements of exerior algebra of sections:

For example:

Are those equivalent?

Sergey Dylda
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    They are equivalent but the proof is not trivial. See the answer to https://math.stackexchange.com/questions/492166/global-sections-of-a-tensor-product-of-vector-bundles-on-a-smooth-manifold for a sketch of the proof in the case of tensor product (instead of exterior algebra). You can find a full proof in Conlon's "Differentiable Manifolds". – levap Apr 27 '20 at 00:35

2 Answers2

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Working with smooth manifolds, these are equivalent notions because you can use partitions of unity to extend local $k$-forms to be global ones, thus writing a global $k$-form as the sum of the wedge products of global $1$-forms.

The equivalence is false in the holomorphic category, not surprisingly. The simplest example I can think of is a smooth quartic ($K3$) surface $X\subset\Bbb CP^3$. It has trivial canonical bundle, so $\Lambda^2 T^*X$ is the trivial bundle, hence has global sections. On the other hand, $T^*X$ has no global holomorphic sections other than $0$ because $h^0(X,\Omega^1_X) = h^{1,0}(X) = h^{0,1}(X) = h^1(X,\mathscr O_X) = 0$.

Ted Shifrin
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By definition, a differential form of order $k$ is nothing but a section of $\Lambda^k \left(T^{*} M\right).$ It is natural unlike $\Gamma\left(\Lambda\left(T^{*} M\right)\right)$, because you want differential forms to form a bundle, so any differential form should be a section of some bundle. In particular, a differential forms of order 1 are a space $\Gamma(T^{*} M).$ Now let us describe $\Lambda^k \Gamma(T^{*} M)$, namely: since any 1-differential form has a local basis $dx_{i},$ then any element of $\Lambda^k \Gamma(T^{*} M)$ has a local basis $dx_{i_1} \wedge .. \wedge dx_{i_k}$, therefore this is a space of $k$-differential forms. Thus, you have an identification between these two spaces.

iou
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  • Sorry, I do not understand: your first sentence says that differential form is a section of exterior bundle, and second says that module of sections is unnatural to consider. And what about non-homogeneous differential forms? (mixed order) – Sergey Dylda Apr 26 '20 at 23:01
  • I could not conclude from your response if $\Gamma(\Lambda(T^M)) \cong \Lambda(\Gamma(T^M))$ or not. If you can expand your response and finish it by clear conclusion, I can mark it as an answer. – Sergey Dylda Apr 26 '20 at 23:06