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There are different models of odd $K$-theory. In one case, one takes the group $U=\lim\limits_{\longrightarrow}U(n)$ as classifying space. Similarly, if $\mathcal U$ denotes the unitary group of a separable Hilbert space, one can consider the subspace $$\mathcal U_0=\bigl\{\,g\in\mathcal U\bigm|g+\mathrm{id}\text{ is Fredholm}\,\bigr\}\;.$$

In another model, one considers instead the space $\mathcal F$ of selfadjoint Fredholm operators on a Hilbert space that have an infinite number of positive as well as of negative eigenvalues. Both spaces are related by a Cayley transformation $$\mathcal F\owns D\mapsto \frac{D-i}{D+i}\in\mathcal U_0\subset\mathcal U_0\;.$$

Characteristic classes are cohomology classes on these classifying spaces. It is well-known that the cohomology ring is the exterior algebra over $\mathbb Z$ with one generator $\gamma_k$ in each odd degree $2k-1$. There is a de Rham type description for maps $u\colon M\to U$ of these generators of the form $$\gamma_k(u)=c_k\,\mathrm{tr}\bigl((u^{-1}\,du)^{2k-1}\bigr)$$ with appropriate constants $c_k\in\mathbb C$.

On the other hand, for a map $D\colon X\to\mathcal F$, the characteristic classes of $[D]\in K^1(X)$ should be determined entirely by the way that $\mathrm{spec}(D)\cap(-\varepsilon,\varepsilon)$ (and maybe the corresponding eigenspaces) behave for some small $\varepsilon>0$.

My question is, how are these descriptions related? In there some book/article that makes characteristic classes of selfadjoint Fredholm operators more explicit and relates them to those of maps to $U$?

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  • $\begingroup$ Is it obvious that $\mathcal U _{0}$ is a subgroup? $\endgroup$ Mar 9, 2016 at 18:41
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    $\begingroup$ @AliTaghavi I am not entirely sure. But maybe it does not have to be a subgroup to be a valid classifying space. I think you get a subgroup if you demand that $1$ is the only essential eigenvalue. $\endgroup$ Mar 9, 2016 at 19:00

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I know something about it when $X$ is compact. Since $U$ is the classifying space of $K^1$, for $x\in K^1(X)$, there exists $u\in C(X, U)$, such that $[u]=x$. In fact, we could choose $N$ large enough, such that $u\in C(X, U(N))$.

For $(b,t,v)\in X\times [0,1]\times \mathbb{C}^N$, the relation $(b,0,v)\sim (b,1, u(b)v)$ makes a vector bundle $W$ over $X\times S^1$. Let $U=X\times S^1\times \mathbb{C}^N$ be the trivial bundle. Then $[W]-[U]\in K^0(X\times S^1)$ corresponds to $[u]\in K^1(X)$.

Let $\nabla^u=d+t\cdot u^{-1}du$. Then $\nabla^u$ is a connection on $W$ over $X\times S^1$ under the paste above. $$\mathrm{ch}([u]):=\int_{S^1}\mathrm{ch}(W)=\left[\int_{S^1}\mathrm{Tr}\left[\exp\left(\frac{(\nabla^{u})^2}{2\pi i}\right)\right]\right].$$ (There is a sign notation here for the integral on odd dimensional manifold.) Calculate it carefully, $$\mathrm{ch}([u])= \mathrm{Tr}\left[\sum_{n≥0}\frac{(−1)^n} {(2πi)^{n+1}}\frac{n!}{ (2n + 1)!}(u^{−1}du)^{2n+1}\right].$$ This is the familiar form in the paper of Getzler: The odd chern character in cyclic homology and spectral flow

For Fredholm type interpretation, we could only consider the Dirac type case. Let $\pi:Z\rightarrow X$ be a submersion with odd-dimensional compact Spin fiber $Y$. Let $E$ be a complex vector bundle over $Z$. We have the fiberwise Dirac operator $D_Y^E$ and $\mathrm{ind}(D_Y^E)\in K^1(X)$.

For any $x\in K^1(X)$, there exists such submersion with appropriate geometric structure such that $\mathrm{ind}(D_Y^E)=x$. See Section 4.2.3 in the book Index theory, eta forms and Deligne cohomology. In fact, choose submersion $X\times S^1\rightarrow X$ above, we have $\mathrm{ind}(D_{S^1}^W)=[u]\in K^1(X)$.

For the Caylay transform, maybe you could find something in Section 5 of the paper of Melrose-Piazza: An index theorem for families of Dirac operators on odd-dimensional manifolds with boundary.

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  • $\begingroup$ Thanks for the nice list of references. The equation $\operatorname{ind}(D^W_{S^1})=[u]\in K^1(X)$ is already pretty close to what I am looking for. But is there a nice relation to the Cayley transform? The transformation in Melrose-Piazza's article seems to be something different. $\endgroup$ May 19, 2016 at 20:43
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    $\begingroup$ @sebastian-goette Sorry that I made a mistake. The transformation in Melrose-Piazza's paper is the relation to the definition of $K^1$ group in Atiyah's paper: Index theory for skew-adjoint Fredholm operators, not the Cayley transform. I don't know the relation to the Cayley transform. Maybe there is something in Quillen's article: Superconnection character forms and the Cayley transform. But I haven't read this. $\endgroup$
    – Bo Liu
    May 20, 2016 at 9:53

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