An odd unbounded (respectively,
$p$
-summable) Fredholm module for a unital Banach
$*$
-algebra,
$A$
, is a pair
$(H,D)$
where
$A$
is represented on the Hilbert space,
$H$
, and
$D$
is an unbounded self-adjoint operator on
$H$
satisfying:
(1)
${{(1+{{D}^{2}})}^{-1}}$
is compact (respectively, Trace
$\left( {{\left( 1+{{D}^{2}} \right)}^{-(p/2)}} \right)<\infty $
, and
(2)
$\{a\in A|\,\,[D,a]\,\,\text{is}\,\,\text{bounded}\}\text{ }$
is a dense
$*$
- subalgebra of
$A$
.
If
$u$
is a unitary in the dense
$*$
-subalgebra mentioned in (2) then
$$uD{{u}^{*}}=D+u[D,{{u}^{*}}]=D+B$$
where
$B$
is a bounded self-adjoint operator. The path
$$D_{t}^{u}:=(1-t)D+tuD{{u}^{*}}=D+tB$$
is a “continuous” path of unbounded self-adjoint “Fredholm” operators. More precisely, we show that
$$F_{t}^{u}:=D_{t}^{u}{{\left( 1+{{\left( D_{t}^{u} \right)}^{2}} \right)}^{-\frac{1}{2}}}$$
is a norm-continuous path of (bounded) self-adjoint Fredholm operators. The spectral flow of this path
$\left\{ F_{t}^{u} \right\}$
(or
$\{D_{t}^{u}\}$
) is roughly speaking the net number of eigenvalues that pass through 0 in the positive direction as
$t$
runs from 0 to 1. This integer,
$$\text{sf}\left( \left\{ D_{t}^{u} \right\} \right):=\text{sf}\left( \left\{ F_{t}^{u} \right\} \right),$$
recovers the pairing of the
$K$
-homology class
$[D]$
with the
$K$
-theory class
$[u]$
.
We use I.M. Singer's idea (as did E. Getzler in the
$\theta $
-summable case) to consider the operator
$B$
as a parameter in the Banach manifold,
${{B}_{\text{sa}}}\left( H \right)$
, so that spectral flow can be exhibited as the integral of a closed 1-form on this manifold. Now, for
$B$
in our manifold, any
$X\,\in \,{{T}_{B}}({{B}_{\text{sa}}}(H))$
is given by an
$X$
in
${{B}_{\text{sa}}}(H)$
as the derivative at
$B$
along the curve
$t\,\mapsto \,B\,+\,tX$
in the manifold. Then we show that for
$m$
a sufficiently large half-integer:
$$\alpha (X)=\frac{1}{{{{\tilde{C}}}_{m}}}\text{Tr}\left( X{{\left( 1+{{\left( D+B \right)}^{2}} \right)}^{-m}} \right)$$
is a closed 1-form. For any piecewise smooth path
$\left\{ {{D}_{t}}=D+{{B}_{t}} \right\}$
with
${{D}_{0}}$
and
${{D}_{1}}$
unitarily equivalent we show that
$$\text{sf}\left( \left\{ {{D}_{t}} \right\} \right)=\frac{1}{{{{\tilde{C}}}_{m}}}\int_{\text{0}}^{\text{1}}{\text{T}}\text{r}\left( \frac{d}{dt}\left( {{D}_{t}} \right){{\left( 1+D_{1}^{2} \right)}^{-m}} \right)dt$$
the integral of the 1-form
$\alpha $
. If
${{D}_{0}}$
and
${{D}_{1}}$
are not unitarily equivalent, wemust add a pair of correction terms to the right-hand side. We also prove a bounded finitely summable version of the form:
$$\text{sf}\left( \left\{ {{F}_{t}} \right\} \right)=\frac{1}{{{C}_{n}}}\int_{\text{0}}^{\text{1}}{\text{T}}\text{r}\left( \frac{d}{dt}\left( {{F}_{t}} \right){{\left( 1-{{F}_{t}}^{2} \right)}^{n}} \right)dt$$
for
$n\ge \frac{p-1}{2}$
an integer. The unbounded case is proved by reducing to the bounded case via the map
$D\mapsto F=D{{(1+{{D}^{2}})}^{-\frac{1}{2}}}$
We prove simultaneously a type II version of our results.