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 Λp(T ∗M), the p-th exterior power of the cotangent bundle of M. The space of such forms is denoted by

Ω p ( M , E ) = Γ ( E ⊗ Λ p T ∗ M ) . {\displaystyle \Omega ^{p}(M,E)=\Gamma (E\otimes \Lambda ^{p}T^{*}M).}

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

Γ ( E ⊗ Λ p T ∗ M ) = Γ ( E ) ⊗ Ω 0 ( M ) Γ ( Λ p T ∗ M ) = Γ ( E ) ⊗ Ω 0 ( M ) Ω p ( M ) , {\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 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,

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

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

T M ⊗ ⋯ ⊗ T M → E {\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 Ωp(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

Ω p ( M ) ⊗ R V → Ω p ( M , V ) , {\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 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

( φ ∗ ω ) x ( v 1 , ⋯ , v p ) = ω φ ( x ) ( d φ x ( v 1 ) , ⋯ , d φ x ( v p ) ) . {\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 (E1⊗E2)-valued (p+q)-form:

∧ : Ω p ( M , E 1 ) × Ω q ( M , E 2 ) → Ω p + q ( M , E 1 ⊗ E 2 ) . {\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:

( ω ∧ η ) ( v 1 , ⋯ , v p + q ) = 1 p ! q ! ∑ σ ∈ S p + q sgn ⁡ ( σ ) ω ( v σ ( 1 ) , ⋯ , v σ ( p ) ) ⊗ η ( v σ ( p + 1 ) , ⋯ , v σ ( p + q ) ) . {\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 ω ∈ Ωp(M) and η ∈ Ωq(M, E) one has the usual commutativity relation:

ω ∧ η = ( − 1 ) p q η ∧ ω . {\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 (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

Ω ( M , E ) = ⨁ p = 0 dim ⁡ M Ω p ( M , E ) {\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 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 {eα} is a basis for V then the differential of a V-valued p-form ω = ωαeα is given by

d ω = ( d ω α ) e α . {\displaystyle d\omega =(d\omega ^{\alpha })e_{\alpha }.\,}

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

d ( ω + η ) = d ω + d η d ( ω ∧ η ) = d ω ∧ η + ( − 1 ) p ω ∧ d η ( p = deg ⁡ ω ) d ( d ω ) = 0. {\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:

∇ : Ω 0 ( M , E ) → Ω 1 ( M , E ) . {\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

d ∇ : Ω p ( M , E ) → Ω p + 1 ( M , E ) {\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

d ∇ ( ω ∧ η ) = d ∇ ω ∧ η + ( − 1 ) p ω ∧ d η {\displaystyle 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 GLk(R) bundle over M. The pullback of E by π is canonically isomorphic to F(E) ×ρ Rk 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 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 dπ). 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

  1. ( R g ) ∗ ω = ρ ( g − 1 ) ω {\displaystyle (R_{g})^{*}\omega =\rho (g^{-1})\omega \,} for all g ∈ G, and
  2. ω ( v 1 , … , v p ) = 0 {\displaystyle \omega (v_{1},\ldots ,v_{p})=0} whenever at least one of the vi are vertical (i.e., dπ(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 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 ϕ ¯ {\displaystyle {\overline {\phi }}} on M with values in E, define φ on P fiberwise by, say at u,

ϕ = u − 1 π ∗ ϕ ¯ {\displaystyle \phi =u^{-1}\pi ^{*}{\overline {\phi }}}

where u is viewed as a linear isomorphism V → ≃ E π ( u ) = ( π ∗ E ) u , v ↦ [ u , v ] {\displaystyle V{\overset {\simeq }{\to }}E_{\pi (u)}=(\pi ^{*}E)_{u},v\mapsto [u,v]} . 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 ϕ ¯ {\displaystyle {\overline {\phi }}} on M (cf. the Chern–Weil homomorphism.) In particular, there is a natural isomorphism of vector spaces

Γ ( M , E ) ≃ { f : P → V | f ( u g ) = ρ ( g ) − 1 f ( u ) } , f ¯ ↔ f {\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 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

∇ ϕ ¯ = D ϕ ¯ . {\displaystyle \nabla {\overline {\phi }}={\overline {D\phi }}.}

In particular for zero-forms,

∇ : Γ ( M , E ) → Γ ( M , T ∗ M ⊗ E ) {\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 27 October 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
    Ω p ( M , V ) = Ω 0 ( M , V ) ⊗ Ω 0 ( M ) Ω p ( M ) , {\displaystyle \Omega ^{p}(M,V)=\Omega ^{0}(M,V)\otimes _{\Omega ^{0}(M)}\Omega ^{p}(M),}
    and that because R {\displaystyle \mathbb {R} } is a sub-ring of Ω0(M) via the constant functions,
    Ω 0 ( M , V ) ⊗ Ω 0 ( M ) Ω p ( M ) = ( V ⊗ R Ω 0 ( M ) ) ⊗ Ω 0 ( M ) Ω p ( M ) = V ⊗ R ( Ω 0 ( M ) ⊗ Ω 0 ( M ) Ω p ( M ) ) = V ⊗ R Ω p ( M ) . {\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: D ( f ϕ ) = D f ⊗ ϕ + f D ϕ {\displaystyle D(f\phi )=Df\otimes \phi +fD\phi } for any scalar-valued tensorial zero-form f and any tensorial zero-form φ of type ρ, 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.

References