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 Γ(''E''). An ''E''-valued differential form of degree ''p'' is a smooth section of the tensor product bundle of ''E'' with Λ^''p''(''T'' ^∗''M''), the ''p''-th exterior power of the cotangent bundle of ''M''. The space of such forms is denoted by
:$\Omega^p(M,E) = \Gamma(E\otimes\Lambda^pT^*M).$
Because Γ is a strong monoidal functor, this can also be interpreted as
:$\Gamma(E\otimes\Lambda^pT^*M) = \Gamma(E) \otimes_ \Gamma(\Lambda^pT^*M) = \Gamma(E) \otimes_ \Omega^p(M),$
where the latter two tensor products are the tensor product of modules over the ring Ω⁰(''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,
:$\Omega^0(M,E) = \Gamma(E).\,$
Equivalently, an ''E''-valued differential form can be defined as a bundle morphism
:$TM\otimes\cdots\otimes TM \to E$
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 Ω^''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
:$\Omega^p(M) \otimes_\mathbb V \to \Omega^p(M,V),$
where the first tensor product is of vector spaces over R, is an isomorphism.

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 φ : ''M'' → ''N'' is an (φ*''E'')-valued form on ''M'', where φ*''E'' is the pullback bundle of ''E'' by φ.

The formula is given just as in the ordinary case. For any ''E''-valued ''p''-form ω on ''N'' the pullback φ*ω is given by
:$(\varphi^*\omega)_x(v_1,\cdots, v_p) = \omega_(\mathrm d\varphi_x(v_1),\cdots,\mathrm d\varphi_x(v_p)).$

Wedge product

Just as for ordinary differential forms, one can define a wedge product of vector-valued forms. The wedge product of an ''E''₁-valued ''p''-form with an ''E''₂-valued ''q''-form is naturally an (''E''₁⊗''E''₂)-valued (''p''+''q'')-form:
:$\wedge : \Omega^p(M,E_1) \times \Omega^q(M,E_2) \to \Omega^(M,E_1\otimes E_2).$
The definition is just as for ordinary forms with the exception that real multiplication is replaced with the tensor product:
:$(\omega\wedge\eta)(v_1,\cdots,v_) = \frac\sum_\sgn(\sigma)\omega(v_,\cdots,v_)\otimes \eta(v_,\cdots,v_).$
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 ω ∈ Ω^''p''(''M'') and η ∈ Ω^''q''(''M'', ''E'') one has the usual commutativity relation:
:$\omega\wedge\eta = (-1)^\eta\wedge\omega.$

In general, the wedge product of two ''E''-valued forms is ''not'' another ''E''-valued form, but rather an (''E''⊗''E'')-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
:$\Omega(M,E) = \bigoplus_^\Omega^p(M,E)$
becomes a graded-commutative associative algebra. If the fibers of ''E'' are not commutative then Ω(''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 is a basis for ''V'' then the differential of a ''V''-valued ''p''-form ω = ω^α''e''_α is given by
:$d\omega = (d\omega^\alpha)e_\alpha.\,$
The exterior derivative on ''V''-valued forms is completely characterized by the usual relations:
:$\begin &d(\omega+\eta) = d\omega + d\eta\\ &d(\omega\wedge\eta) = d\omega\wedge\eta + (-1)^p\,\omega\wedge d\eta\qquad(p=\deg\omega)\\ &d(d\omega) = 0. \end$
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:
:$\nabla : \Omega^0(M,E) \to \Omega^1(M,E).$
If ''E'' is equipped with a connection ∇ then there is a unique covariant exterior derivative
:$d_\nabla: \Omega^p(M,E) \to \Omega^(M,E)$
extending ∇. The covariant exterior derivative is characterized by linearity and the equation
:$d_\nabla(\omega\wedge\eta) = d_\nabla\omega\wedge\eta + (-1)^p\,\omega\wedge d\eta$
where ω is a ''E''-valued ''p''-form and η 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 ''π'' : F(''E'') → ''M'' be the (associated) frame bundle of ''E'', which is a principal GL_''k''(R) bundle over ''M''. The pullback of ''E'' by ''π'' is canonically isomorphic to F(''E'') ×_ρ R^''k'' via the inverse of 'u'', ''v''→''u''(''v''), where ρ is the standard representation. Therefore, the pullback by ''π'' of an ''E''-valued form on ''M'' determines an R^''k''-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 GL_''k''(R) on F(''E'') × R^''k'' and vanishes on vertical vectors (tangent vectors to F(''E'') which lie in the kernel of d''π''). 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 ''π'' : ''P'' → ''M'' be a (smooth) principal ''G''-bundle and let ''V'' be a fixed vector space together with a representation ''ρ'' : ''G'' → GL(''V''). A basic or tensorial form on ''P'' of type ρ is a ''V''-valued form ω on ''P'' which is equivariant and horizontal in the sense that
#$(R_g)^*\omega = \rho(g^)\omega\,$ for all ''g'' ∈ ''G'', and
#$\omega(v_1, \ldots, v_p) = 0$ whenever at least one of the ''v''_''i'' are vertical (i.e., d''π''(''v''_''i'') = 0).
Here ''R''_''g'' 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 ρ is the adjoint representation of ''G'' on the Lie algebra, then the connection form ω satisfies the first condition (but not the second). The associated curvature form Ω satisfies both; hence Ω is a tensorial form of adjoint type. The "difference" of two connection forms is a tensorial form.

Given ''P'' and ''ρ'' as above one can construct the associated vector bundle ''E'' = ''P'' ×_''ρ'' ''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 $\overline$ on ''M'' with values in ''E'', define φ on ''P'' fiberwise by, say at ''u'',
:$\phi = u^\pi^*\overline$
where ''u'' is viewed as a linear isomorphism $V \overset\to E_ = (\pi^*E)_u, v \mapsto$, v/math>. φ is then a tensorial form of type ρ. Conversely, given a tensorial form φ of type ρ, the same formula defines an ''E''-valued form $\overline$ on ''M'' (cf. the Chern–Weil homomorphism.) In particular, there is a natural isomorphism of vector spaces
:$\Gamma(M, E) \simeq \, \, \overline \leftrightarrow f$.

Example: Let ''E'' be the tangent bundle of ''M''. Then identity bundle map id_''E'': ''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 id_''E''. Denoted by θ, 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
:$\nabla \overline = \overline.$

In particular for zero-forms,
:$\nabla: \Gamma(M, E) \to \Gamma(M, T^*M \otimes E)$.

This is exactly the covariant derivative for the connection on the vector bundle ''E''.

Examples

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

Notes

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References

* Shoshichi Kobayashi and Katsumi Nomizu (1963) Foundations of Differential Geometry, Vol. 1, Wiley Interscience.

Differential forms
Vector bundles