# $GL_n(\mathbb{R})$ is the union of two path connected subsets (Michael Artin's Algebra).

I have that $SL_n(\mathbb{R})$ is path connected and generated by elementary matrices of the form $I + a e_{i,j} (i \neq j)$, where $e_{i,j}$ is the matrix with $1$ at position $(i,j)$ and zero elsewhere. I also have that $GL_n(\mathbb{R})$ is path connected and generated by elementary matrices of the above type together with ones of the form $I + c e_{i,i}$. So clearly I already have a union of two path connected subsets, but I was wondering if there is a smaller second set, for instance, matrices that have at least one factor of the second type. Hints are welcome.

• $I+ce_{1,1}$ with $c\ne-1$ should suffice. $A\mapsto (I+ce_{1,1})A$ is a homeomorphims of $SL_n$ with the set of matrices of determinant $c+1$, and $(A,c)\mapsto (I+ce_{1,1})A$ is a homeomorphism of $SL_n\times(\mathbb R\setminus\{-1\})$ with $G_n$. – Hagen von Eitzen Apr 5 '13 at 21:32
• @EnjoysMath What I meant is, your second set is wrong in the first place. The two components have to be matrices of respectively positive and negative determinants, but your matrices $I+ce_{i,i}$ mess up the signs. You should fix your second set first, by separating into the two subsets, one with $c>-1$ and the other with $c<-1$. – user1551 Apr 5 '13 at 21:43
• Since $G = GL_n(\mathbb{R})$ is path connected, can't I just choose any subset such as $(G - SL_n(\mathbb{R}))$ ? – Shine On You Crazy Diamond Apr 5 '13 at 21:52
• @enjoysMath $GL_n(\mathbb{R})$ is not connected as $\det(GL_n(\mathbb{R}))$ is not connected but the determinant is continuous – Dominic Michaelis Apr 5 '13 at 21:53
• It is connected: $X(t) = (I + ct e_{i,i})$ is a path from $I$ to an elementary matrix of the second type. $Y(t) = (I + at e_{i,j})$ is a path from $I$ to an elementary matrix of the first type. If $X_1, \ldots, X_k$ are paths from $I$ to $A_1, \dots , A_k$, resp, then $X_1\cdots X_k$ is a path from $I$ to $A_1 \cdots A_k$. Since $G = GL_n(\mathbb{R})$ is generated by elementary matrices of the first and second type, we have that for all $A \in G$, there is a path from $I$ to $A$ and thus $G$ is path connected. QED – Shine On You Crazy Diamond Apr 5 '13 at 22:00

At first i wanted to mention that $GL_n(\mathbb{R})$ is not connected, the determinant is continuous and the image of a connected space under a continuous function is a connected space . But the image is $\mathbb{R}\setminus \{0\}$ which is surely not connected..

On the other hand $GL_n(\mathbb{R})$ is surely $$GL_n(\mathbb{R})=GL_n^- (\mathbb{R}) \cup GL_n^+ (\mathbb{R})$$ Now think of Row reduced Echolon form and you will find continuous generators, for each of them.

For the generating system of $GL_n^+ (\mathbb{R})$ you make the following: \begin{align*} B_{b}^\lambda=(b_{k,l})_{1\leq k,l\leq n}= \left\{ \begin{array}{ll} \lambda & k=i \text{ and } l=j\\ 0 & \text{ else }\\ \end{array} \right. \end{align*} \begin{align*} \gamma_{b}(t)=E_n+t\cdot B_{b}^\lambda \end{align*}

\begin{align*} C(t)=(c_{k,l}(t))_{1\leq k,l \leq n}&=\left\{ \begin{array}{cl} 1 & k=l \text{ with } k,l\neq i,j,\\ -\sin{(\pi \cdot t)} & k=i,\ l=j\\ \sin{(\pi \cdot t)} & k=j, \ l=i\\ \cos{(\pi \cdot t)} & k=l=i \\ \cos{(\pi \cdot t)} & k=l=j \\ 0 & \text{else}\\ \end{array} \right.\\ \gamma_{c)}(t)&=C(t) \end{align*}

\begin{align*} A(t)=(a_{kl})_{1\leq k,l\leq n}= \left\{ \begin{array}{ll} 1 & k=l\neq i\\ 1+t\cdot (\lambda-1) & k=l=i\\ 0 & \text{else}\\ \end{array} \right. \end{align*} with $\lambda > -1$

$t$ is always in $[0,1]$.

• Dear Dominic, what do you mean by <<continuous generators>> of $\text{GL}_n(\Bbb{R})$? Regards, – user38268 Apr 5 '13 at 23:32

Exercise 2.M.8(b) (Artin's Algebra, 2nd edition). Show that $GL_n(\mathbb{R})$ is a union of two path-connected subsets, and describe them.

Let $\sim_1$, $\sim_2$ be the binary operations corresponding to path-connectivity in $SL_n(\mathbb{R})$, $GL_n(\mathbb{R})$, respectively; by my answer here, $\sim_1$, $\sim_2$ are equivalence relations.

In the same spirit as my answer here, we consider the map$$M\to f(c,M)$$that multiplies the first row of $M$ by $c$, and note that $f(c,M)$ is continuous in $c$ for fixed $M$ (but multiplies the determinant by $c$). If $c>0$ and $M\in GL_n(\mathbb{R})$, the continuous function$$X(t) = f(1 + (c-1)t\,,M)$$on $[0,1]$ takes$$X(0)=M \to X(1) = f(c,\,M),$$ while keeping$$1+(c-1)t\ne 0$$at all times (we only hit the interval $[1,c]$) and thus remaining inside $GL_n(\mathbb{R})$. Hence,$$f(c,M)\sim_2 M.$$Yet any $B\in GL_n(\mathbb{R})$ can be written as $f(\epsilon\det{B},\,\epsilon A)$ for some $\epsilon\in\{-1,1\}$ and $A\in \epsilon SL_n$ such that$$\epsilon\det{B}>0,$$so$$B\sim_2 \epsilon A\sim_2 \epsilon I_n$$by here and the trivial observation$$P\sim_1 Q\iff dP\sim_2 dQ$$ for nonzero scalars $d$. Thus,$$S_+ = \{B\in GL_n(\mathbb{R}): \det(B)>0\},\text{ }S_- = \{B\in GL_n(\mathbb{R}): \det(B)<0\}$$are path-connected subsets covering $GL_n(\mathbb{R})$.

On the other hand, $B_+\in S_+$ and $B_-\in S_-$ can not be connected in $GL_n(\mathbb{R})$. Suppose otherwise, so some continuous function$$X:[0,1]\to\mathbb{R}^{n\times n}$$takes$$X(0)=B_+,\text{ }X(1)=B_-$$while staying inside $GL_n(\mathbb{R})$. Then $\det{X(t)}$, a polynomial in the $n^2$ continuous component functions of $X$, is continuous itself. However, $$\det{X(0)}>0>\det{X(1)},$$so $\det{X(t)}$ must vanish somewhere by the Intermediate Value Theorem. This is clearly absurd, so we indeed have $$B_+\not\sim_2 B_-,$$and that $\sim_2$ partitions $GL_n(\mathbb{R})$ into $S_+$ and $S_-$.

$GL_n(\mathbb R)$ is not path connected. Think about determinants.