I am working on a model for topological KO-theory which is represented by explicit spaces of orthogonal Clifford module extensions. That is, assuming $M$ compact, $KO^{-n+1}(M) := [M,X\_n]$ where the spaces $X\_n$ are defined as follows:


Fix the background inner product space $\mathbb{R}^{\infty}$ with the standard inner product, and a background Clifford module structure, i.e., orthogonal operators $e\_1,\ldots,e\_n$ such that $e\_i^2 = -I$ and $e\_i e\_j = -e\_j e\_i$ for $i \neq j$. By Bott periodicity it suffices to do this only up to $n=8$. Then let $X\_n = \{f \in O(\mathbb{R}^\infty) | f^2 = -I, fe\_k = -e\_kf \text{ for }k < n, \dim\ker(f-e\_n)^\perp < \infty\}$.
This is the space of operators that play the role of $e\_n$, i.e. define extensions of the module structure given by $e\_1, \ldots, e\_{n-1}$, additionally required to be skew-symmetric and differ from $e\_n$ in only finite dimensions. This turns out to give the correct spaces in the $KO$-spectrum.


Given two maps $f,g:M \to X\_n$ we define their sum as follows: let
$\rho: \mathbb{R}^\infty \oplus \mathbb{R}^\infty \to \mathbb{R}^\infty$
be a module isomorphism. We can choose a particular $\rho$ so that it "shuffles" together irreducibles. Then $f \boxplus g = \rho(f \oplus g)\rho^{-1}$
is the map from $M$ to $X\_n$ that represents the sum in k-theory. Morally it is just the pointwise direct sum.


My question is this: what operation should play the role of the inverse? That is, given a map
$f:M \to X\_n$
what is $\operatorname{Inv}(f)$ such that $f \boxplus \operatorname{Inv}(f)$ is homotopic to the constant map to $e\_n$? I suspect that something like $\operatorname{Inv}(f) := -f$ ought to do it, but I can't nail down a proof (and $-f$ on its own violates the finite-dimension condition). I know this is a fairly specific setup but any hints at all on how to define the inverse for k-theory represented by spaces, especially in the context of Clifford modules, would be a great help. Please feel free to make any simplifying assumptions or be imprecise.


Note that this definition is similar to Karoubi's definition using what he calls "gradations", but he works with triples $(E,\eta\_1,\eta\_2)$ where $E$ is the background module and $\eta\_1,\eta\_2$ are gradations (analogous to extensions). I would like to find out what swapping $\eta\_1$ and $\eta\_2$ corresponds to in terms of a single map into a colimit of spaces of gradations.



What is the inverse in K-theory represented by Clifford module extensions?
For what it's worth, I did eventually get the answer. The idea is to notice that if $\overline{W}$ denotes the background Clifford module $W$ (above $\mathbb{R}^n$) endowed with the opposite module structure, i.e., where the operators $e\_i$ act instead by $-e\_i$, then for any map $f:M \to X\_n(W)$, the map $f \oplus -f: X(n,W \oplus \overline{W})$ is contractible by the homotopy
$$ \begin{bmatrix}f & 0 \\ 0 & -f\end{bmatrix} \cos\left(\frac{\pi}{2}t\right) + \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix} \sin\left(\frac{\pi}{2}t\right). $$
Then one notices that, for instance, the operator $e\_{n+1}$ gives an isomorphism $W \to \overline{W}$ since $e\_{n+1}e\_k = (-e\_k)e\_{n+1}$ for $k=1,\ldots,n$. So the inverse of $f$ is given by conjugating $-f$ by $e\_{n+1}$:
$$\mathrm{Inv}(f) = e\_{n+1} f e\_{n+1}.$$
Since the shuffle sum induces a homeomorphism $X(n,W) \simeq X(n,W \oplus W)$, $f \boxplus e\_{n+1}fe\_{n+1}$ is contractible.


