Self-adjoint Fredholm operators #
This file proves that a self-adjoint Fredholm operator on a Hilbert space has index zero. More generally, the same conclusion holds whenever an operator and its adjoint have the same kernel. The closed range of a Fredholm operator has orthogonal complement equal to the kernel of its adjoint. Orthogonal decomposition therefore identifies the cokernel with that kernel; under the kernel-equality hypothesis, it identifies the cokernel with the original kernel.
Main declarations #
ContinuousLinearMap.IsFredholm.cokerEquivKerOfKerAdjointEq: identify the cokernel with the kernel when the operator and its adjoint have equal kernels.ContinuousLinearMap.index_eq_zero_of_ker_adjoint_eq: the corresponding index-zero criterion.ContinuousLinearMap.index_eq_zero_of_isSelfAdjoint: a self-adjoint Fredholm operator has index zero.ContinuousLinearMap.index_eq_zero_of_isSymmetric: the same result in terms of symmetry of the underlying linear map.
This is the elementary self-adjoint index computation in the Fredholm package needed by the nonlinear-analysis substrate of the analytic Heegaard Floer roadmap. The convention and argument follow McDuff--Salamon, J-holomorphic Curves and Symplectic Topology, Appendix A.1.
The cokernel of a Fredholm operator is linearly equivalent to its kernel if the operator and its adjoint have the same kernel.
Equations
- hT.cokerEquivKerOfKerAdjointEq hker = (↑(T.cokerEquivKerAdjoint ⋯)).trans (LinearEquiv.ofEq (↑(ContinuousLinearMap.adjoint T)).ker (↑T).ker hker)
Instances For
The cokernel of a self-adjoint Fredholm operator is linearly equivalent to its kernel.
Equations
- hT.cokerEquivKer hself = hT.cokerEquivKerOfKerAdjointEq ⋯
Instances For
A Fredholm operator has index zero if it and its adjoint have the same kernel.
A self-adjoint Fredholm operator on a Hilbert space has Fredholm index zero.
A symmetric Fredholm operator on a Hilbert space has Fredholm index zero.