We study the K-homology of the rotation algebras $A_{\theta}$ using the six-term cyclic sequence for the K-homology of a crossed product by ${\bf Z}$. In the case that $\theta$ is irrational, we use Pimsner and Voiculescu's work on AF-embeddings of the $A_{\theta}$ to search for the missing generator of the even K-homology.

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 self-adjoint 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 self-adjoint operator. The path $$ D_t^u:=(1-t) D+tuDu^\ast=D+tB $$ is a ``continuous'' path of unbounded self-adjoint ``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 norm-continuous path of (bounded) self-adjoint 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 half-integer: $$ \alpha (X)={1\over {\tilde {C}_m}}\Tr \Bigl(X\bigl(1+(D+B)^2\bigr)^{-m}\Bigr) $$ 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 right-hand side. We also prove a bounded finitely summable version of the form: $$ \sf(\{F_t\})={1\over C_n}\int_0^1\Tr\Bigl({d\over dt}(F_t)(1-F_t^2)^n\Bigr)\,dt $$ for $n\geq{{p-1}\over 2}$ 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.