Is it possible to classify finite dimensional vector bundles in terms of Fredholm operators?

Question

Let ${\mathcal K}$ be the space of Fredholm operators on a Hilbert space. It is well known that ${\mathcal K}$ represents $K$-theory. Let ${\mathcal K}_0$ be the path component of ${\mathcal K}$ of operators of index zero. Then ${\mathcal K}_0$ is a model for the space that algebraic topologists usually call $BU$ - the classifying space of the infinite unitary group.

My question is: is it possible to realise the filtration of $BU$ by the subspaces $BU(n)$ in terms of Fredholm operators?

An obvious idea is to consider the subspace of ${\mathcal K}_0$ consisting of operators whose kernel/cokernel has rank at most $n$. Is this space weakly homotopy equivalent to $BU(n)$?

Hopefully this question is interesting enough on its own (I am prepared to find out that it is stupid). My motivation is to understand better Jesse McKeown's answer to my previous question. This is my attempt to understand the statement that the space of subspaces of a Hilbert space of corank at most $n$ is a model for $BU(n)$. If there are other ways to make it precise, I would be very interested in learning about it.

EDIT: I think that Tyler Lawson's negative answer to the previous question makes it very likely that the answer to this question is negative as well.

Answer 1

The problem with my remark from a couple of years ago is that the cokernel can also "move around". Let us therefore fix this.

Let $W\subset \mathbb{H}$ be a finite, say $l$, dimensional subspace of a separable infinite dimensional Hilbert space. I assume it is known that the space of $l$ dimensional subspaces of the Hilbert space is a model for $BU(l)$.

Let $\mathcal{O}^W=\{A\in \mathcal{K}_0\,|\, A(\mathbb{H})+W=\mathbb H\}$ be the space of Fredholm operators whose image is transverse to $W$. The space $\mathcal{O}^W$ is an open subspace of the space of Fredholm operators. There is a map $\mathcal{O}^W\rightarrow BU(l)$, that maps an operator $A$ to the subspace $A^{-1}(W)$. Note that the operator $A$ sends $A^{-1}(W)^\perp$ isomorphically to $W^\perp$. There are $GL(A^{-1}(W)^\perp,W^\perp)$ many such choices to do this. This suggest that $$ GL(\mathbb H)\rightarrow \mathcal O^W\rightarrow BU(l) $$ is a fiber bundle. But by Kuiper's theorem $GL(\mathbb H)$ is contractible, so we have the required (weak) homotopy equivalence.

Now how does this fit into a grander story? You can take a sequence of finite dimensional $W_i$ that include in each other and $\mathbb{H}=\overline{\bigcup W_i}$. For example one can take $W_i$ to be the span of the first $i$ vectors in a basis of $\mathbb H$. Let us assume this. Then $\mathcal O^{W_i}\subset \mathcal O^{W_{j}}$ if $i\leq j$. This is the analogue of $BU(i)\subset BU(j)$. And $\bigcup _i \mathcal O^{W_i}=\mathcal K_0$.

I should credit Dan Freed's lecture notes https://web.ma.utexas.edu/users/dafr/M392C-2015/index.html.