# 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 $$n\geq3$$?

• 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.