Prove that $\operatorname{SL}(n,\Bbb R)$ is connected.

Solution 1:

Exercise 2.M.8(a) (Artin's Algebra, 2nd edition). The group $SL_n(\mathbb{R})$ is generated by elementary matrices of the first type (see Exercise 2.4.8). Use this fact to prove that $SL_n(\mathbb{R})$ is path-connected.

Let $\sim$ be the binary operation corresponding to path-connectivity in $SL_n(\mathbb{R})$; by my answer here, $\sim$ is an equivalence relation.

In order to show $SL_n(\mathbb{R})$ is path-connected, it suffices to show $A\sim I_n$ for all $A\in SL_n(\mathbb{R})$. But by my answer here, $A$ can be written as a (possibly empty) product of elementary matrices of the first type, so it in fact suffices to prove that$$E_{uv}(a)M\sim M$$for all $M\in SL_n(\mathbb{R})$ and Type 1 elementary matrices $E_{uv}(a)$ ($1\le u,\,v\le n$) of the form$$I_n + [a[(i,\,j) = (u,\,v)]]_{i,\,j\,=\,1}^n.$$Yet$$M\to E_{uv}(b)M$$simply adds $b$ times row $j$ to row $i$, i.e. takes $r_i$ to $r_i+br_j$. For fixed $u$, $v$, $M$, this map is continuous in $b$ (and preserves the determinant), so the continuous function$$X(t) = E_{uv}(ta)M$$over $[0,1]$ takes$$X(0) = M \to X(1) = E_{uv}(a)M$$while remaining inside $SL_n(\mathbb{R})$, as desired.

Solution 2:

Hint: prove that if two matrices can be transformed one into another using row-echelon transformation, then they are connected.

as we focus on elements of $SL_n$, we only need to prove that transvections $L_i \to l_i + aL_j$ connect elements.

let $A\in SL_n$, $B$ is the image of $A$ under the transvection $L_i \to L_i + aL_j$.

Then $$ \gamma: [0,1]\to SL_n $$defined by "$\gamma(t)$ is the image of $A$ under the transvection $L_i \to L_i + taL_j$ " is continuous, and such as $\gamma(0) = A$, $\gamma(1) = B$ (also, check that for every $t$, $\gamma (t)\in SL_n$). Hence $A,B$ are path connected.

Solution 3:

$\textbf{Lemma: }$Let $Y$ be connected and $f\colon X\to Y$ be surjective continuous map having connected fibers. If $f$ is open then $X$ also connected.

Proof. We can prove this by contradiction. If possible, write $X=U\bigsqcup V$ where $U,V$ are non-empty open subsets of $X$. Then, $f(U),f(V)$ are open subsets of $Y$ such that $f(U)\cup f(V)=Y$. Now, $Y$ is connected implies $f(U)\cap f(V)\not=\emptyset.$ Take, $y\in f(U)\cap f(V)$, then $f^{-1}(y)\cap U\not =\emptyset$ and $f^{-1}(y)\cap V\not=\emptyset$, contradicts to the fact that fibers are connected sets. $\square$

$\textbf{Theorem.}$ $\text{GL}^+(n,\Bbb R):=\big\{A\in \text{M}(n,\Bbb R): \det(A)>0\big\}$ is connected.

Proof. We will prove $\text{GL}^+(n,\Bbb R)$ is connected by induction on $n$.

Consider the map $p\colon \text{M}(n,\Bbb R)=\Bbb R^n\times \text{M}\big(n\times (n-1),\Bbb R\big)\longrightarrow \Bbb R^n$ given by $p(A)=Ae_1$. That is $p$ sends $A\in \text{M}(n,\Bbb R)$ to its first column. Note that $p$ is a projection map, hence open and continuous.

Note that, $\text{GL}^+(1,\Bbb R)=(0,\infty)$. Now, let $f\colon \text{GL}^+(n,\Bbb R)\longrightarrow \Bbb R^n-\{0\}$ be the restriction of $p$ to the $\text{GL}^+(n,\Bbb R)\subseteq_{\text{open}}\text{M}(n,\Bbb R)$. So, $f$ is also open continuous.

Next, $f^{-1}(e_1)=\Bbb R^{n-1}\times \text{GL}^+(n-1,\Bbb R)$, which is a connected set by induction. For $y\in \Bbb R^n-\{0\}$ choose $B\in \text{GL}^+(n,\Bbb R)$ with $f(B)=y$. Then, $f^{-1}(y)=\big\{B\cdot C:C\in f^{-1}(e_1)\big\}$. Hence each fibre of $f$ is connected. So, $\text{GL}^+(n,\Bbb R)$ is connected by the above Lemma. $\square$

$\textbf{Theorem.}$ $\text{SL}(n,\Bbb R)$ is connected.

Proof. To prove $\text{SL}(n,\Bbb R)$ is connected, consider the continuous surjective map $$\Psi\colon \text{GL}^+(n,\Bbb R)\ni A\longmapsto \frac{A}{\det A}\in \text{SL}(n,\Bbb R).$$ Recall that continuous image of a connected set is connected, so we are done by the previous theorem. $\square$

Solution 4:

If you know the connectedness of $SO(n)$, there is an another approach.

Let $X \in SL(n) $ be given. By polar decomposition $X$ can be written as a product $X=UP$, where $U$ is an orthogonal matrix and $P$ is a positive-definite symmetric matrix. Here $\text{det}(U)=\text{det}(P)=1$ since $\text{det}(X)=1$. By the spectral theorem, $P=V^{-1}\mbox{diag}(\lambda_1, ... ,\lambda_n)V$ for some $V \in SO(n)$ and $\lambda_j \in (0, \infty)$.

This shows that the continuous map $$SO(n) \times SO(n) \times M \rightarrow SL(n)$$ given by $$\left( U, V, (x_1, ..., x_n) \right) \mapsto UV^{-1}\mbox{diag}(x_1, ..., x_n)V $$ is surjective, where $M$ denotes the embedded submanifold $\{(x_1, ..., x_n) \in \mathbb{R}^n_{>0} : x_1x_2...x_n=1\}$ of $\mathbb{R}^n_{>0}= \{(x_1, ..., x_n): x_1, ..., x_n >0 \}$.

To see M is (path) connected, let $(p_1, ..., p_n) \in M$ be given. Then $(p_1, ..., p_n)$ and $(1, p_1p_2, ...,p_n)$ is joined by the path

$$ t \mapsto (\frac{p_1}{\text{exp}(t \log p_1)}, \exp(t \log p_1)p_2, ... , p_n) $$

Inductively, we can construct a path from $(p_1, ..., p_n)$ to $(1, ..., 1)$.

As a proudct of connected spaces, $SO(n) \times SO(n) \times M$ is connected. Hence $SL(n)$ connected because it is a continuous image of a connected space.