Math 2.0

K-theory Definition

Fredholm module

Builds on C*-algebraCompact operator

Definition

Let AA be a C*-algebra. Let HH be a separable Hilbert space and ρ ⁣:A→B(H)\rho \colon A \to B(H) a ∗*-representation. An odd Fredholm module over AA is a triple (H,ρ,F)(H, \rho, F) with F∈B(H)F \in B(H) such that for all a∈Aa \in A the operators (F2−1)ρ(a),(F−F∗)ρ(a),[F,ρ(a)]=Fρ(a)−ρ(a)F(F^2 - 1)\rho(a), \qquad (F - F^*)\rho(a), \qquad [F, \rho(a)] = F \rho(a) - \rho(a) F are compact. An even Fredholm module is such a triple together with a grading, a self-adjoint unitary γ∈B(H)\gamma \in B(H), such that γρ(a)=ρ(a)γ\gamma \rho(a) = \rho(a) \gamma for all a∈Aa \in A and γF=−Fγ\gamma F = -F \gamma.

Remarks

Example: let A=C(𝕋)A = C(\mathbb{T}) act on H=L2(𝕋)H = L^2(\mathbb{T}) by multiplication, and let F=2P−1F = 2P - 1, where PP is the orthogonal projection onto the Hardy space H2H^2 spanned by znz^n, n≥0n \ge 0. Then FF is a self-adjoint unitary, and [P,Mf][P, M_f] is compact for continuous ff (it is of finite rank for trigonometric polynomials), so this is an odd Fredholm module. Non-example: with A=C(𝕋)A = C(\mathbb{T}) on L2(𝕋)L^2(\mathbb{T}), let F=JF = J be the flip (Jf)(z)=f(zˉ)(J f)(z) = f(\bar{z}). Then FF is a self-adjoint unitary, but [F,Mz]=(Mzˉ−Mz)J[F, M_z] = (M_{\bar{z}} - M_z) J, and this is not compact, because multiplication by a nonzero continuous function on L2(𝕋)L^2(\mathbb{T}) is never compact. The module is degenerate if the three operators above are zero for all aa; degenerate modules represent zero in K-homology.

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Further reading

  1. Web https://en.wikipedia.org/wiki/Fredholm_module
About this page
TagXNDDW5R577
URIm2://XNDDW5R577
Kinddefinition
WritingW0 stub
MSC19K33, 58B34, 46L87
DefinesFredholm module
UsesC*-algebra, Compact operator
Changed3 November 2026

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