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Dear community,

there are so-called orientation maps $a:MSpin\to ko$ and $b:MSpin^c \to k$, "defined" in ABS's paper "Clifford modules". Unfortunately I am not familiar with representation theory.

Let $c:MSpin \to MSpin^c$ resp. $d:ko \to k$ denote the obvious maps given by considering a spin- as a spin$^c$-manifold resp. complexification.

Is it true that $b\circ c=d\circ a$?

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1 Answer 1

up vote 8 down vote accepted

A point in the $n$th space of $MSpin$ is an $n$-dimensional manifold equipped with a spin structure. In other words, it is a manifold equipped with a bundle of bimodules between the Clifford algebra of $\mathbb R^n$ and the Clifford algebra of $TM$. That bundle of bimodules is called the spinor bundle of $M$.

Similarly, a point in the $n$th space of $MSpin^c$ is an $n$-dimensional manifold equipped with a bundle of bimodules between the Clifford algebra of $\mathbb C^n$ and the Clifford algebra of $TM\otimes_{\mathbb R}\mathbb C$.

The map $MSpin\to MSpin^c$ is given by complexifying the bimodule.


A point in $n$th space of $KO$ (allow me to take non-connective $K$-theory - the result for connective $K$-theory then follows readily) is given by a real Hilbert space equipped with an action of $Cliff(\mathbb R^n)$, and an odd skew-adjoint clifford-linear Fredholm operator.

Similarly, a point in the $n$th space of $K$ is given by a complex Hilbert space with an action of $Cliff(\mathbb C^n)$, and an odd skew-adjoint clifford-linear Fredholm operator.

The map $KO\to K$ is again complexification.


The ABS orientation (which is not constructed in ABS) sends a spin manifold $M$, now also equipped with a metric, to the Hilbert space of $L^2$ sections of the spinor bundle, equipped the the obvious $Cliff(\mathbb R^n)$-action. The Fredholm operator is the Dirac operator constructed from (the connection associated to) the metric and the $Cliff(TM)$-action.

The spin-c version is identical. It is then obvious by construction that the diagram you asked about is commutative.


I should say that the above argument is completely hand-wavy...
I actually don't know in which paper/textbook the ABS orientation is defined as a map of spectra (and I would like to know -- so if someone knows, please tell me). My guess is that, regardless of the approach taken, once you see the definition, it is completely obvious that the diagram is commutative.

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In A symmetric ring spectrum representing KO-theory, Topology 40 (2001), 299-308, Michael Joachim says that he gives an explicit model of $MSpin\to KO$ in his thesis. I have not seen his thesis; but you might want to ask him about it. –  Charles Rezk Jun 20 '12 at 16:09

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