TITLE: $\operatorname{STop}_{n,n-2}\simeq S^1$?
QUESTION [5 upvotes]: $\DeclareMathOperator\STop{STop}$I am interested in any information about the homotopy type of the groups $\STop_{n,j}$ of homeomorphisms of $R^n$ preserving orientation and pointwise $R^j\subset R^n$. It is easy to see that $\STop_{n,n-1}$ is contractible, being homeomorphic to the square of the group of relative-to-the-boundary homeomorphisms of a disc. But what about $\STop_{n,n-2}$? How does it compare to $S^1$? At least for $n=3$ the answer is probably known.

REPLY [7 votes]: The general result in this direction is due to Kirby-Siebenmann, Theorem B of Normal bundles for codimension 2 locally flat imbeddings: the map
$$\mathrm{SO}(2) \longrightarrow \mathrm{STop}_{n,n-2}$$
is $(n-2)$-connected. (They state this for $n \neq 4$ but it is true in this case as well by the corresponding result for codimension 2 normal bundles due to Freedman-Quinn.)