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 $n\geq3$?

  • 4
    $\begingroup$ 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 $\mathbb RP^2$. $\endgroup$ – 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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.