# Blocksum induces a unital H-space structure on the space of Fredholm operators

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.

• 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 – Denis Nardin Feb 28 at 17:53
• 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. – Benedikt Hunger Mar 1 at 5:25
• @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 – Denis Nardin Mar 1 at 9:22