# Branch cuts of $GL_n^+(\mathbb{R})$

## 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 $n\geq3$?

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You can cut $SO_3$ along the matrices that have trace $-1$, i.e. the matrices that have $-1$ as an eigenvalue, i.e. the matrices that have $-1$ as a double eigenvalue. Topologically this is the same as cutting $\mathbb RP^3$ along $\mathbb RP^2$. – Tom Goodwillie Jul 6 '11 at 2:20

There is a Poincare duality isomorphism $H_k(SO_n\setminus C)=H^{d-k}(SO_n,C)$, where $d=\dim(SO_n)=(n^2-n)/2$. This works with any coefficients because $SO_n$ is oriented, but let us use $\mathbb{Z}/2$; then it is known that $H_*(SO_n)$ is an exterior algebra with generators $x_1,\dotsc,x_{n-1}$ where $|x_i|=i$, and $H^*(SO_n)$ is dual to this. It follows that both $H^d(SO_n)$ and $H^{d-1}(SO_n)$ have rank one.
You want $H_0(SO_n\setminus C)=\mathbb{Z}/2$ and $H_1(SO_n\setminus C)=0$, so you need $H^{d-1}(C)=H^{d-1}(SO_n)=\mathbb{Z}/2$ and $H^d(C)=0$. More specifically, you need the restriction map $H^{d-1}(SO_n)\to H^{d-1}(C)$, or the dual map $H_{d-1}(C)\to H_{d-1}(SO_n)$, to be an isomorphism. Note that all monomials in $x_1,\dotsc,x_{n-2}$ are carried on $SO_{n-1}$, and the individual generators $x_1,\dotsc,x_{n-1}$ are all carried on a standard copy of $\mathbb{R}P^{n-1}$ in $SO_n$; we need to arrange for $C$ to carry the product $x_2x_3\dotsb x_{n-1}$. That's as far as I see for the moment.