Vector-valued Differential Form
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Vector-valued Differential Form

In mathematics, a vector-valued differential form on a manifold M is a differential form on M with values in a vector space V. More generally, it is a differential form with values in some vector bundle E over M. Ordinary differential forms can be viewed as R-valued differential forms.

An important case of vector-valued differential forms are Lie algebra-valued forms. (A connection form is an example of such a form.)

## Definition

Let M be a smooth manifold and E -> M be a smooth vector bundle over M. We denote the space of smooth sections of a bundle E by ?(E). An E-valued differential form of degree p is a smooth section of the tensor product bundle of E with ?p(T *M), the p-th exterior power of the cotangent bundle of M. The space of such forms is denoted by

${\displaystyle \Omega ^{p}(M,E)=\Gamma (E\otimes \Lambda ^{p}T^{*}M).}$

Because ? is a strong monoidal functor,[1] this can also be interpreted as

${\displaystyle \Gamma (E\otimes \Lambda ^{p}T^{*}M)=\Gamma (E)\otimes _{\Omega ^{0}(M)}\Gamma (\Lambda ^{p}T^{*}M)=\Gamma (E)\otimes _{\Omega ^{0}(M)}\Omega ^{p}(M),}$

where the latter two tensor products are the tensor product of modules over the ring ?0(M) of smooth R-valued functions on M (see the seventh example here). By convention, an E-valued 0-form is just a section of the bundle E. That is,

${\displaystyle \Omega ^{0}(M,E)=\Gamma (E).\,}$

Equivalently, an E-valued differential form can be defined as a bundle morphism

${\displaystyle TM\otimes \cdots \otimes TM\to E}$

which is totally skew-symmetric.

Let V be a fixed vector space. A V-valued differential form of degree p is a differential form of degree p with values in the trivial bundle M × V. The space of such forms is denoted ?p(M, V). When V = R one recovers the definition of an ordinary differential form. If V is finite-dimensional, then one can show that the natural homomorphism

${\displaystyle \Omega ^{p}(M)\otimes _{\mathbb {R} }V\to \Omega ^{p}(M,V),}$

where the first tensor product is of vector spaces over R, is an isomorphism.[2]

## Operations on vector-valued forms

### Pullback

One can define the pullback of vector-valued forms by smooth maps just as for ordinary forms. The pullback of an E-valued form on N by a smooth map ? : M -> N is an (?*E)-valued form on M, where ?*E is the pullback bundle of E by ?.

The formula is given just as in the ordinary case. For any E-valued p-form ? on N the pullback ?*? is given by

${\displaystyle (\varphi ^{*}\omega )_{x}(v_{1},\cdots ,v_{p})=\omega _{\varphi (x)}(\mathrm {d} \varphi _{x}(v_{1}),\cdots ,\mathrm {d} \varphi _{x}(v_{p})).}$

### Wedge product

Just as for ordinary differential forms, one can define a wedge product of vector-valued forms. The wedge product of an E1-valued p-form with an E2-valued q-form is naturally an (E1E2)-valued (p+q)-form:

${\displaystyle \wedge :\Omega ^{p}(M,E_{1})\times \Omega ^{q}(M,E_{2})\to \Omega ^{p+q}(M,E_{1}\otimes E_{2}).}$

The definition is just as for ordinary forms with the exception that real multiplication is replaced with the tensor product:

${\displaystyle (\omega \wedge \eta )(v_{1},\cdots ,v_{p+q})={\frac {1}{p!q!}}\sum _{\sigma \in S_{p+q}}\operatorname {sgn}(\sigma )\omega (v_{\sigma (1)},\cdots ,v_{\sigma (p)})\otimes \eta (v_{\sigma (p+1)},\cdots ,v_{\sigma (p+q)}).}$

In particular, the wedge product of an ordinary (R-valued) p-form with an E-valued q-form is naturally an E-valued (p+q)-form (since the tensor product of E with the trivial bundle M × R is naturally isomorphic to E). For ? ? ?p(M) and ? ? ?q(M, E) one has the usual commutativity relation:

${\displaystyle \omega \wedge \eta =(-1)^{pq}\eta \wedge \omega .}$

In general, the wedge product of two E-valued forms is not another E-valued form, but rather an (EE)-valued form. However, if E is an algebra bundle (i.e. a bundle of algebras rather than just vector spaces) one can compose with multiplication in E to obtain an E-valued form. If E is a bundle of commutative, associative algebras then, with this modified wedge product, the set of all E-valued differential forms

${\displaystyle \Omega (M,E)=\bigoplus _{p=0}^{\dim M}\Omega ^{p}(M,E)}$

becomes a graded-commutative associative algebra. If the fibers of E are not commutative then ?(M,E) will not be graded-commutative.

### Exterior derivative

For any vector space V there is a natural exterior derivative on the space of V-valued forms. This is just the ordinary exterior derivative acting component-wise relative to any basis of V. Explicitly, if {e?} is a basis for V then the differential of a V-valued p-form ? = ??e? is given by

${\displaystyle d\omega =(d\omega ^{\alpha })e_{\alpha }.\,}$

The exterior derivative on V-valued forms is completely characterized by the usual relations:

{\displaystyle {\begin{aligned}&d(\omega +\eta )=d\omega +d\eta \\&d(\omega \wedge \eta )=d\omega \wedge \eta +(-1)^{p}\,\omega \wedge d\eta \qquad (p=\deg \omega )\\&d(d\omega )=0.\end{aligned}}}

More generally, the above remarks apply to E-valued forms where E is any flat vector bundle over M (i.e. a vector bundle whose transition functions are constant). The exterior derivative is defined as above on any local trivialization of E.

If E is not flat then there is no natural notion of an exterior derivative acting on E-valued forms. What is needed is a choice of connection on E. A connection on E is a linear differential operator taking sections of E to E-valued one forms:

${\displaystyle \nabla :\Omega ^{0}(M,E)\to \Omega ^{1}(M,E).}$

If E is equipped with a connection ? then there is a unique covariant exterior derivative

${\displaystyle d_{\nabla }:\Omega ^{p}(M,E)\to \Omega ^{p+1}(M,E)}$

extending ?. The covariant exterior derivative is characterized by linearity and the equation

${\displaystyle d_{\nabla }(\omega \wedge \eta )=d_{\nabla }\omega \wedge \eta +(-1)^{p}\,\omega \wedge d\eta }$

where ? is a E-valued p-form and ? is an ordinary q-form. In general, one need not have d?2 = 0. In fact, this happens if and only if the connection ? is flat (i.e. has vanishing curvature).

## Basic or tensorial forms on principal bundles

Let E -> M be a smooth vector bundle of rank k over M and let ? : F(E) -> M be the (associated) frame bundle of E, which is a principal GLk(R) bundle over M. The pullback of E by ? is canonically isomorphic to F(E) ×?Rk via the inverse of [u, v] ->u(v), where ? is the standard representation. Therefore, the pullback by ? of an E-valued form on M determines an Rk-valued form on F(E). It is not hard to check that this pulled back form is right-equivariant with respect to the natural action of GLk(R) on F(E) × Rk and vanishes on vertical vectors (tangent vectors to F(E) which lie in the kernel of d?). Such vector-valued forms on F(E) are important enough to warrant special terminology: they are called basic or tensorial forms on F(E).

Let ? : P -> M be a (smooth) principal G-bundle and let V be a fixed vector space together with a representation ? : G -> GL(V). A basic or tensorial form on P of type ? is a V-valued form ? on P which is equivariant and horizontal in the sense that

1. ${\displaystyle (R_{g})^{*}\omega =\rho (g^{-1})\omega \,}$ for all g ? G, and
2. ${\displaystyle \omega (v_{1},\ldots ,v_{p})=0}$ whenever at least one of the vi are vertical (i.e., d?(vi) = 0).

Here Rg denotes the right action of G on P for some g ? G. Note that for 0-forms the second condition is vacuously true.

• Example: If ? is the adjoint representation of G on the Lie algebra, then the connection form ? satisfies the first condition (but not the second). The associated curvature form ? satisfies both; hence ? is a tensorial form of adjoint type. The "difference" of two connection forms is a tensorial form.

Given P and ? as above one can construct the associated vector bundle E = P ×?V. Tensorial q-forms on P are in a natural one-to-one correspondence with E-valued q-forms on M. As in the case of the principal bundle F(E) above, given a q-form ${\displaystyle {\overline {\phi }}}$ on M with values in E, define ? on P fiberwise by, say at u,

${\displaystyle \phi =u^{-1}\pi ^{*}{\overline {\phi }}}$

where u is viewed as a linear isomorphism ${\displaystyle V{\overset {\simeq }{\to }}E_{\pi (u)}=(\pi ^{*}E)_{u},v\mapsto [u,v]}$. ? is then a tensorial form of type ?. Conversely, given a tensorial form ? of type ?, the same formula defines an E-valued form ${\displaystyle {\overline {\phi }}}$ on M (cf. the Chern-Weil homomorphism.) In particular, there is a natural isomorphism of vector spaces

${\displaystyle \Gamma (M,E)\simeq \{f:P\to V|f(ug)=\rho (g)^{-1}f(u)\},\,{\overline {f}}\leftrightarrow f}$.
• Example: Let E be the tangent bundle of M. Then identity bundle map idE: E ->E is an E-valued one form on M. The tautological one-form is a unique one-form on the frame bundle of E that corresponds to idE. Denoted by ?, it is a tensorial form of standard type.

Now, suppose there is a connection on P so that there is an exterior covariant differentiation D on (various) vector-valued forms on P. Through the above correspondence, D also acts on E-valued forms: define ? by

${\displaystyle \nabla {\overline {\phi }}={\overline {D\phi }}.}$

In particular for zero-forms,

${\displaystyle \nabla :\Gamma (M,E)\to \Gamma (M,T^{*}M\otimes E)}$.

This is exactly the covariant derivative for the connection on the vector bundle E.[3]

## Examples

Siegel modular forms arise as vector-valued differential forms on Siegel modular varieties.[4]

## Notes

1. ^ "Global sections of a tensor product of vector bundles on a smooth manifold". math.stackexchange.com. Retrieved 2014.
2. ^ Proof: One can verify this for p=0 by turning a basis for V into a set of constant functions to V, which allows the construction of an inverse to the above homomorphism. The general case can be proved by noting that
${\displaystyle \Omega ^{p}(M,V)=\Omega ^{0}(M,V)\otimes _{\Omega ^{0}(M)}\Omega ^{p}(M),}$
and that because ${\displaystyle \mathbb {R} }$ is a sub-ring of ?0(M) via the constant functions,
${\displaystyle \Omega ^{0}(M,V)\otimes _{\Omega ^{0}(M)}\Omega ^{p}(M)=(V\otimes _{\mathbb {R} }\Omega ^{0}(M))\otimes _{\Omega ^{0}(M)}\Omega ^{p}(M)=V\otimes _{\mathbb {R} }(\Omega ^{0}(M)\otimes _{\Omega ^{0}(M)}\Omega ^{p}(M))=V\otimes _{\mathbb {R} }\Omega ^{p}(M).}$
3. ^ Proof: ${\displaystyle D(f\phi )=Df\otimes \phi +fD\phi }$ for any scalar-valued tensorial zero-form f and any tensorial zero-form ? of type ?, and Df = df since f descends to a function on M; cf. this Lemma 2.
4. ^ Hulek, Klaus; Sankaran, G. K. (2002). "The Geometry of Siegel Modular Varieties". Advanced Studies in Pure Mathematics. 35: 89-156.