Given a Dirac operator $D$ on a closed odd-dimensional manifold $M$, I've sometimes heard it said that the Fredholm index of $D$ vanishes because it is an ungraded self-adjoint operator, so that $\dim\ker D=\dim\ker D^*$.
However, it seems to me that this is not the correct way to think about things. For example, when one works with more general versions of the index, the odd-dimensional index does not necessarily vanish. Say $M$ is not necessarily compact but there is an action of a group $G$ on $M$ with compact quotient $M/G$. Then the index of $D$ lies in $K_1(C^*(G))$, where $C^*(G)$ is the group $C^*$-algebra of $G$. If one thinks of the index of $D$ as a difference between the kernel and the cokernel (in the sense of finitely generated projective $C^*(G)$-modules), then this would vanish also. But the index in this case should not always vanish.
The technical definition of the index in the odd-dimensional case is given in terms of the exponential map in $K$-theory. I would like to understand this more intuitively, much like how the boundary map in the even-dimensional case can be understood as giving the difference $\dim\ker D-\dim\ker D^*$.
It seems to me that the correct way to understand the odd-dimensional index should somehow involve suspensions and Toeplitz operators; but I cannot piece together exactly how the story should go. So, along these lines, I would like to ask a slightly vague question.
Question: How should one understand the index of Dirac operators on odd-dimensional manifolds?
