In order to ask the question in the title more precisely, let me recall some standard stuff introduced in [1; Atiyah, Bott, Shapiro].
Suppose $X$ is a compact CW complex and $V \to X$ is an oriented real vector bundle, carrying a positive-definite quadratic form on each fibre. There is a natural K-group associated to $V$, namely the group completion of equivalence classes of bundles of $Z/(2)$-graded Clifford modules of $V$. I'll follow the notation of [1] and denote this group by $M(V)$. In fact this group comes in two flavours: real and complex, according to whether we have bundles of real or complex modules (even though $V$ is always real). I'll denote both by $M(V)$ since it won't matter.
By giving $V\oplus 1$ the natural positive-definite form on each fibre, the inclusion $V \to V \oplus 1$, $v \mapsto (v, 0)$ induces a natural map:
$M(V\oplus 1) \to M(V)$A key ingredient of [1] is the construction of a natural map:
$\partial : M(V) \to \tilde K(Th(V))$where $Th(V)$ is the Thom space of $V$ and $K$ denotes topological $K$-theory (real or complex according to the flavour of $M(V)$). There are various simple (though very ingenious) geometric constructions of $\partial$ and they rely on the fact that:
$q(v) = 0 \iff v = 0 \qquad (*)$for a positive-definite real quadratic form $q$. Furthermore, the following sequence combining the above maps is exact at $M(V)$:
$M(V\oplus 1) \to M(V) \to \tilde K(Th(V))$I am interested in the map corresponding to $\partial$ when $V$ is a complex vector bundle carrying a non-degenerate quadratic form on each fibre. Note that a complex vector bundle carries a non-degenerate quadratic form iff $V \simeq V^*$, or equivalently, iff it is the complexification of a real vector bundle. Also since property $(*)$ fails for any complex quadratic form, it cannot be that the complex situation is directly analogous to the real situation.
Perhaps the right thing to do is simply to forget complex structures and fall back to the real case. Indeed if we choose a real structure $\bar V \simeq V$ that is compatible with the in-place quadratic form in the sense that the induced map $\bar V \simeq V^*$ defines a Hermitian metric then, forgetting the complex structure on $V$, we are left with an oriented real vector bundle carrying positive definite quadratic form and so we can use the real $\partial$ map to obtain elements of $\tilde K(Th(V))$. However thanks to the data that $V$ is carrying, there is a natural Thom isomorphism
$\Phi : \tilde K(Th(V)) \simeq K(X)$(This follows from Theorem 12.3 of [1] together with the embedding $U(m) \hookrightarrow Spin^c(2m)$ or alternatively [2, page 387].) Thus we have a map:
$M(V) \to K(X)$(We had to choose a real structure to define this map but I could believe this washes out.) Note that there is an obvious map $M(V) \to K(X)$ since geometric representatives for elements of $M(V)$ are differences of bundles of Clifford modules so if we just forget the module structure we obtain elements of $K(X)$. I thus come at last to direct questions:
- Is $\Phi \circ \partial$ the same as the "obvious" map I just mentioned?
- Is the map $M(V) \to K(X)$ just described the "right" one to use when $V$ is a complex vector bundle?
- Assuming affirmative answers, what is the image of the map $M(V) \to K(X)$? (In my intended application, this would provide a convenient topological restriction on which bundles can support a Clifford module structure.)
Frankly, I think I'm making a meal of this and the complex case should be far simpler than the real case so I'm hoping somebody out there can illuminate me before I spend more time taking the long route to what I expect is a very simple answer. Nevertheless I'll add a few final words which, for me at least, put the situation in perspective.
Bear with me, for sketchy though my understanding of topological K-theory may be, I have an even sketchier understanding of algebraic K-theory. Now it seems to me that the ABS construction (i.e., the map $\partial$) can be understood from the point of view of algebraic K-theory as a connecting homomorphism in a long exact sequence associated to a homomorphism between two rings. According to [3, III exercise 4.5], if $\phi : A \to B$ is a homomorphism of rings then there are groups $K_i(\phi)$ fitting into a doubly-infinite long exact sequence:
$\cdots \to K_0(A) \to K_0(B) \to K_{-1}(\phi) \to K_{-1}(A) \to \cdots$For appropriate choices of $A$ and $B$ (roughly, sections of the bundle of Clifford algebras associated to $V$ and $V\oplus 1$ with some twisting to take care of the fact that $M(V)$ is defined in terms of graded modules) we'll have:
$K_0(A) \simeq M(V\oplus 1)$ and $K_0(B) \simeq M(V)$and so there should be a natural map $\phi$ and a natural map: $M(V) \to K_{-1}(\phi)$. I thus expect that when $V$ is real we have $K_{-1}(\phi) \simeq \tilde K(Th(V))$ and that this is the ABS map. Since this algebraic construction appears to go through in case when $V$ is complex, I thus expect that I could use this to work answer all my questions (and ideally to get a geometric construction for the map as ABS provide in the real case) but as I have said I think there must be simpler routes to the truth.
Incidentally, I expect working out the next term in the long exact sequence should make it easier to identify the image of $\partial$ since it will be the kernel of the next map in the exact sequence so I will add the question:
Assuming the above (or something like it) is correct, what is $K_{-1}(A)$ and what is the map $K_{-1}(\phi) \to K_{-1}(A)$?
After quite a few hours mulling / digging, I feel like I'm taking a very long route so I am hoping proper experts can point the way. Thanks!
[1] M.F. Atiyah, R. Bott, A. Shapiro. Clifford modules. Topology, 3(suppl. 1):3-38, 1964.
[2] H.B. Lawson, M. Michelson. Spin geometry, Princeton University Press, 1989.
[3] C. Weibel, The K-book: an introduction to algebraic K-theory, Graduate Studies in Math. vol. 145, AMS, 2013.
