I am asking this question from my possibly defected memory, so the things below may not be accurate.

I want to know how many different definitions of the odd Chern character form using differential geometry, and the proofs of their equivalence.

The first definition of the odd Chern character form as

$$\textrm{ch}^{\textrm{odd}}(g)=\textrm{CS}(d, d+g^{-1}dg),$$

where $\textrm{CS}$ is the Chern-Simons form. Here we regard $K^{-1}(X)$ as given by maps $g$ from $X$ to the stable unitary group.

The second definition of the odd Chern character form, by regarding $K^{-1}(X)$ is generated by complex vector bundles with automorphisms, should be given by some sort of integrating the (even) Chern character form. I cannot make this precise because I don't know a reference for this.

The third definition of the odd Chern character is defined in terms of the (even) Chern character form so that it is compatible with suspension and desuspension between $K^0$ and $K^{-1}$. Again I think it's more or less the same as the second definition (I could be wrong).

Since the Chern-Simons form can be given by integrating the (even) Chern character form, I think the these definitions are (of course) essentially the same. Anyway I would like to see proofs of the equivalence of these three (or more) definitions or or any reference contains such proofs.


  • $\begingroup$ I think you are right, they are all equivalent. Maybe I write a full answer later. So far just one comment: A vector bundle on the (reduced) suspension $SX$ is a vector bundle $V\to X\times S^1$ that is trivial on $X\times\{*\}$ (and on $\{*\}\times S^1$). It is therefore given by gluing a trivial vector bundle on $X\times[0,1]$ using a clutching function $g\colon X\to G$ (with $g(*)=e$) to identify the fibres over $X\times\{0\}$ and $X\times\{1\}$. Pick the trivial connection $d$ on $X\times\{0\}$, then the induced connection on $X\times\{1\}$ is $g^{-1}\circ d\circ g=d+g^{-1}\,dg$. $\endgroup$ – Sebastian Goette Jun 19 '16 at 14:56
  • $\begingroup$ It seems that your second possibility describes $K^0(X\times S^1)$ and not $K^0(X_+\wedge S^1)=K^1(X)$. But maybe you want a specific equivalence relation, so that you get $K^1(X)$ in the end? $\endgroup$ – Sebastian Goette Jun 24 '16 at 8:42
  • $\begingroup$ You are right that the second definition of $K^{-1}(X)$ includes an equivalence relation. It is generated by complex vector bundles with automorphisms, and two generators $(E, U^E)$, $(F, U^F)$ are equivalent if there are "stably isomorphic", besides the bundle part has to be isomorphic, the autormorphism part has to respect pullback by the isomorphism. For the odd Chern character form in this case, put a connection $\nabla^E$ on $E\to X$. Then $\textrm{CS}(\nabla^E, U\circ\nabla^E\circ U^{-1})$ is the odd Chern character form. Thus it is the same as what you said in above. $\endgroup$ – GRR Jun 24 '16 at 10:20

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