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

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

The problem is I know only topological groups from Munkres only. Again Just started fundamental groups. So if anyone can explain to me how it is true in a lucid language and in an easy way such that it remains in my boundary of knowledge then it would be a great help. I have mentioned what I know. Again if tag this in the wrong field. Please forgive me.

• Since you asked, my opinion is that you don't need the algebraic topology tag. Commented Nov 7, 2014 at 16:40
• On the other hand a linear-algebra tag would be very appropriate since, as the answer of @mookid shows, this is a consequence of basic facts in basic linear algebra. Commented Nov 7, 2014 at 20:56
• Could one show a stronger statement, that $SL_{n}(\mathbb{R})$ is path connected? Commented Sep 9, 2023 at 16:22

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.

$$\textbf{Lemma: }$$Let $$Y$$ be connected and $$f\colon X\to Y$$ be surjective continuous map having connected fibers. If $$f$$ is open, then $$X$$ is 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 $$f(U)\cap f(V)\not=\varnothing.$$ Take $$y\in f(U)\cap f(V)$$, then $$f^{-1}(y)\cap U\not =\varnothing$$ and $$f^{-1}(y)\cap V\not=\varnothing$$, 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 $$f\colon \text{GL}^+(n,\Bbb R)\longrightarrow \Bbb R^n\setminus\{0\}$$ 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 fiber 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. $$\square$$

• $\mathrm{det}(A/\mathrm{det}(A))=1/\mathrm{det}(A)^{n-1}$. Commented Jul 6, 2022 at 20:33

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.

• How do I do that? Using topological group definition? Commented Nov 7, 2014 at 16:48
• the transformations have explicit forms. Use the 'path-connected implies connected' implication. Commented Nov 7, 2014 at 16:49
• can you please write this answer ilaborately then atleast I get an idea for this kind of questions. Commented Nov 7, 2014 at 16:49
• Because I am new to deal with these questions. Thats why I am requesting you. Please it will be a great help. Commented Nov 7, 2014 at 16:52
• @user152715 I wrote the hint part. Now it is your turn to finish the proof. Tell me if you have other issues. Commented Nov 7, 2014 at 16:55

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 product 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.