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Fix a complex separable infinite-dimensional Hilbert space $H$. It is well known that the space of (bounded) Fredholm operators $Fred(H)$ with the norm topology is a classifying space for the topological $K$-theory functor, i.e. by the Atiyah-Jänich theorem, for any finite CW-complex $X$, we have $[X, Fred(H)] \cong K(X)$ and the map is given by the index homomorphism. The addition on the left hand side is coming directly from structure on $Fred(H)$, and I am interested in a specific map $Fred(H)\times Fred(H)\to Fred(H)$ which implements it.

In order to define the "blocksum" of two operators $A$, $B$, chose a unitary isomorphism $\rho\colon H \to H\oplus H$ and define $A\oplus B := \rho^* \begin{pmatrix} A & 0 \\ 0 & B\end{pmatrix}\rho$. Using the contractibility of the infinite-dimensional unitary group, one sees that this does not depend on the specific choice of $\rho$ up to homotopy and furthermore is associative and commutative up to homotopy. It therefore equips $Fred(H)$ with an associative and commutative $H$-space structure.

From the definition of the index homomorphism, it is clear that this induces addition in $K$-theory on finite CW-complexes - kernels and cokernels behave additively under $\oplus$. Furthermore, let $X$ be a finite CW-complex and denote by $const_1\colon X\to Fred(H)$ the constant map to the identity operator. Then it is also clear that any map $f\colon X \to Fred(H)$ must be homotopic to $f\oplus const_1$, since these maps correspond to the same $K$-theory class under the index homomorphism. In particular, the map $A \mapsto A\oplus 1$ induces the identity on all homotopy groups of $Fred(H)$.

Now here is my question: Can this be done globally, i.e. is the identity operator a unit for the $H$-space structure given by the blocksum? In other words, is there a homotopy $H\colon Fred(H)\times I \to Fred(H)$ starting at $H_0=Id_{Fred(H)}$ and ending at $H_1\colon A\mapsto A\oplus 1$, which somehow gets rid of the $\oplus 1$? It seems that all attempts to write down such a homotopy need at some point that we only consider a compact subset of operators, and I also do not see an abstract reason why it needs to exist.

Improve this question
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    $\begingroup$ Well an abstract reason why it does exist is that $K^0(X)$ is an abelian group functorially in $X$, and so Yoneda forces the unit (and sum, and inverse) maps to exist. I suspect an explicit construction can be done along the line of the contraction of the identity of $S^∞$ but I'll confess not having thought it through $\endgroup$
    – Denis Nardin
    Commented Feb 28, 2019 at 17:53
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    $\begingroup$ I do not think that Yoneda works here, since it is not clear that block sum induces addition over non-compact spaces, in particular over $Fred(H)$ itself. $\endgroup$
    – Benedikt Hunger
    Commented Mar 1, 2019 at 5:25
  • $\begingroup$ @BenediktHunger I guess the question is whether the linear isometries operad (whose n-th space is given by the space of isometric embeddings of $H^{\oplus n}$ into $H$) is an $E_\infty$-operad, because it seems to me that the blocksum produces an action of this operad on $Fred(H)$. But, again, I haven't thought it through $\endgroup$
    – Denis Nardin
    Commented Mar 1, 2019 at 9:22

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