@ThomasRot I think the space of Fredholm operators with (dim ker A, dim coker A) = (k,c) fixed has the homotopy type of BGL(k) x BGL(c): first, there is a fibration F_{k,c} -> Gr(k,H) x Gr(c,H') given by sending an operator to the pair of kernel and cokernel. The fiber fixes these, and we may identify the space of such Fredholm operators by rewriting them as maps ker A + H_0 -> coker A + H'_0...
where the only non-zero component is H_0 -> H'_0, which is an isomorphism; now we apply contractibility of GL(H_0, H'_0).
Fix the index to be zero for convenience of conversation. Then what we are really doing is choosing two subspaces of whatever dimension. The top stratum is the space of invertible operators, which you know to be contractible; the next stratum is BGL(1) x BGL(1)... The interesting thing is to see how the strata relate to one another. My feeling is that because the process of picking up kernel "trades kernel for cokernel", this should be intertwining the topology of the factors BGL(k) and BGL(c).
So the answer should be something like: infinite dimensions give you too much wiggle room and you can pull a swindle to show GL(H) is contractible. Choosing Fredholm operators of fixed index and kernel-dimension reduces you to an almost finite-dimensional situation (you're really just choosing what subspace is the kernel and what subspace is the cokernel). And then the process of picking up and dropping kernel should intertwine the topology of the kernel and cokernel sides.
Does the regular slice filtration have a universal property (among functorial filtrations on genuine $G$-spectra)? I feel like I've heard "the regular slice filtration is the universal $G$-commutative filtration" somewhere, but I don't know 1) if that was in a dream or if someone actually said that, or 2) exactly what that means
