
Fredholm modules and spectral flow
An {\it odd unbounded\/} (respectively, $p${\it summable})
{\it Fredholm module\/} for a unital Banach $\ast$algebra, $A$, is a pair $(H,D)$
where $A$ is represented on the Hilbert space, $H$, and $D$ is an unbounded
selfadjoint operator on $H$ satisfying:
\item{(1)} $(1+D^2)^{1}$ is compact (respectively, $\Trace\bigl((1+D^2)^{(p/2)}\bigr)
<\infty$), and
\item{(2)} $\{a\in A\mid [D,a]$ is bounded$\}$ is a dense
$\ast$subalgebra of $A$.
If $u$ is a unitary in the dense $\ast$subalgebra mentioned in (2) then
$$
uDu^\ast=D+u[D,u^{\ast}]=D+B
$$
where $B$ is a bounded selfadjoint operator. The path
$$
D_t^u:=(1t) D+tuDu^\ast=D+tB
$$
is a ``continuous'' path of unbounded selfadjoint ``Fredholm'' operators.
More precisely, we show that
$$
F_t^u:=D_t^u \bigl(1+(D_t^u)^2\bigr)^{{1\over 2}}
$$
is a normcontinuous path of (bounded) selfadjoint Fredholm
operators. The {\it spectral flow\/} of this path $\{F_t^u\}$ (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,
$$
\sf(\{D_t^u\}):=\sf(\{F_t^u\}),
$$
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,
$B_{\sa}(H)$, 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 $X\in T_B(B_{\sa}(H))$ is given by an $X$ in $B_{\sa}(H)$ as the
derivative at $B$ along the curve $t\mapsto B+tX$ in the manifold.
Then we show that for $m$ a sufficiently large halfinteger:
$$
\alpha (X)={1\over {\tilde {C}_m}}\Tr \Bigl(X\bigl(1+(D+B)^2\bigr)^{m}\Bigr)
$$
is a closed $1$form. For any piecewise smooth path $\{D_t=D+B_t\}$ with
$D_0$ and $D_1$ unitarily equivalent we show that
$$
\sf(\{D_t\})={1\over {\tilde {C}_m}} \int_0^1\Tr \Bigl({d\over {dt}}
(D_t)(1+D_t^2)^{m}\Bigr)\,dt
$$
the integral of the $1$form $\alpha$. If $D_0$ and $D_1$ are not unitarily
equivalent, we must add a pair of correction terms to the righthand
side. We also prove a bounded finitely summable version of the form:
$$
\sf(\{F_t\})={1\over C_n}\int_0^1\Tr\Bigl({d\over dt}(F_t)(1F_t^2)^n\Bigr)\,dt
$$
for $n\geq{{p1}\over 2}$ an integer. The unbounded case is proved by
reducing to the bounded case via the map $D\mapsto F=D(1+D^2
)^{{1\over 2}}$. We prove simultaneously a type II version of our
results.
Categories:46L80, 19K33, 47A30, 47A55 