Knowlpedia

Definition

de Rham complex

The cochain complex of smooth differential forms with the exterior derivative.

Definition

For a MM, the de Rham complex is the

0⟶Ω0(M)→dΩ1(M)→dΩ2(M)→d⋯ ,0\longrightarrow\Omega^0(M)\xrightarrow{d}\Omega^1(M)\xrightarrow{d}\Omega^2(M) \xrightarrow{d}\cdots,

where Ωk(M)\Omega^k(M) is the of smooth and each differential is the . The identity d∘d=0d\circ d=0 makes this a cochain complex. Its degree-kk cocycles are closed forms, its coboundaries are exact forms, and its cohomology is the de Rham cohomology of MM.

Differential graded algebra structure

The makes Ω∙(M)\Omega^\bullet(M) a graded-commutative algebra, while the exterior derivative obeys

d(α∧β)=dα∧β+(−1)deg⁡αα∧dβ.d(\alpha\wedge\beta)=d\alpha\wedge\beta+(-1)^{\deg\alpha}\alpha\wedge d\beta.

Thus the de Rham complex is more specifically a commutative differential graded algebra, not merely a sequence of vector spaces and .

Cohomology and topology

The quotient

HdRk(M)=ker⁡(d:Ωk→Ωk+1)/im⁡(d:Ωk−1→Ωk)H_{\mathrm{dR}}^k(M)=\ker(d:\Omega^k\to\Omega^{k+1})/ \operatorname{im}(d:\Omega^{k-1}\to\Omega^k)

is the . De Rham's theorem identifies it with singular cohomology with real coefficients, so a complex built from smooth forms recovers a topological invariant Bott and Tu, Chapter I.

Functoriality

A f:M→Nf:M\to N induces f∗:Ωk(N)→Ωk(M)f^*:\Omega^k(N)\to\Omega^k(M). Since f∗d=df∗f^*d=df^* and pullback preserves wedge products, f∗f^* is a morphism of differential graded algebras and induces the contravariant map on de Rham cohomology.

References
  1. R. Bott and L. W. Tu, Differential Forms in Algebraic Topology, Springer, 1982. Springer DOI record. Relevant: Chapter I.
  2. L. W. Tu, An Introduction to Manifolds, 2nd ed., Springer, 2011. Springer DOI record. Relevant: chapters on differential forms and de Rham cohomology.