## Branch cuts

Let $GL_n^+(\mathbb{R})$ denote the group of $n\times n$ real matrices with positive determinant. Topologically, $GL_n^+(\mathbb{R})$ is connected, and $$ \pi_1(GL_2^+(\mathbb{R})) = \mathbb{Z}$$ $$ \pi_1(GL_n^+(\mathbb{R})) = \mathbb{Z}/2,\;\;n\geq 3$$ These identities are perhaps more well-known for the homotopy-equivalent $SO_n(\mathbb{R})$; in fact, for the rest of my question, $GL_n^+(\mathbb{R})$ may be replaced by $SO_n(\mathbb{R})$.

I want to find a closed submanifold $C\subset GL_n^+(\mathbb{R})$, such that the complement $C^c$ is connected and simply-connected. The idea is that $C$ cuts $GL_n^+(\mathbb{R})$ in such a way that it kills the fundamental group without disconnecting the space.

This is essentially the same problem as choosing a fundamental domain for the action of $\pi_1(GL_n^+(\mathbb{R}))$ on the universal cover.

## Dependence on a subspace

Certainly, many such cuts exist, but I would like a construction which depends continuously on one extra piece of data. That data is a choice of an oriented, codimension-1 subspace $V$ of $\mathbb{R}^n$, which gives an embedding $GL_{n-1}^+(\mathbb{R})\subset GL_{n}^+(\mathbb{R})$ (up to $GL_{n-1}^+(\mathbb{R})$-conjugation).

This is easy for $n=2$. In this case, let $C$ be the subset of $GL_2^+(\mathbb{R})$ which sends $V$ to $V$ and preserves orientation. If $V$ is the first coordinate subspace of $\mathbb{R}^2$, then $C$ is the subspace of upper triangular matrices with positive diagonal entries.

However, the same trick doesn't work for $n\geq 3$. If we let $C$ be the subset of $GL_n^+(\mathbb{R})$ which sends $V$ to $V$ and preserves orientation, then $C^c$ is homotopy-equivalent to $GL_{n-1}^+(\mathbb{R})$, and so it is not simply-connected.

## The Question

Is there a smart way to cut $GL_n^+(\mathbb{R})$ when $n\geq3$?