# How to find a path in $SL(2)$?

$SL(2,\mathbb{R})$ is path-connected. Therefore, for all $A,B \in SL(2,\mathbb{R})$ there is a path $\varphi:[0,1]\rightarrow SL(2,\mathbb{R})$, connecting both matrices. I would like to know, how I can actually construct such a path for two given matrices?

• Are you comfortable with Jordan canonical form? Sep 2, 2015 at 18:43
• @Omnomnomnom: you probably are already aware, but if you use Jordan form, you need to make sure your path preserves the fact that the matrix has real entries. Sep 2, 2015 at 21:50

Hint.

The result is true in dimension $n \ge 1$: the special linear group $SL(n,\mathbb R)$ is path connected.

One proof uses the fact that $SL(n,\mathbb R)$ is generated by the transvections $T_{i,j}(\lambda)$ where $i,j \in \{1,\dots,n\}$, $i \neq j$ and $\lambda \in \mathbb R$.

Based on that, you can write any matrix $A \in SL(n,\mathbb R)$ as $$A = \prod_{C \in X} T_C(\lambda_C)$$ where $X$ is a family of elements of $\{1,\dots,n\}^2$ and $\lambda_C \in \mathbb R$ for $C \in X$.

Then you can define a path $$\varphi : t \in [0,1] \mapsto A(t)=\prod_{C \in X} T_C(t\lambda_C)$$ between the identity matrix $I_n$ and $A$. The path is lying in $SL(n,\mathbb R)$ as a product of transvections is an element of $SL(n,\mathbb R)$, which ends the proof.

For the record $T_{i,j}(\lambda)=I_n +\lambda E_{i,j}$ where $E_{i,j}$ is the matrix with all coefficients equal to zero except the one at row $i$ and column $j$ for which the coefficient is equal to one.

Here's a fairly specific geometric construction for the 2-dimensional case. Without loss of generality we can assume $B$ is the identity matrix $I$ (since given a path from $A$ to $I$ and a path from $B$ to $I$, we can easily construct a path from $A$ to $B$). Now $A = \left(\matrix{a& b \\ c & d}\right)$ represents the linear transformation $v \mapsto vA$ that sends $(1, 0)$ to $(a, b)$ and $(0, 1)$ to $(c, d)$. So the rotation $R_{\alpha}$ through some angle $\alpha$ with $0 \le \alpha \le \pi$ will give us that $A' = AR_{\alpha} = \left(\matrix{a' & b'\\0 & d'}\right)$ is upper triangular with $a', d' > 0$ and $a'd' = 1$ ($\alpha$ is the angle between the non-zero vector $(c, d)$ and the positive half of the $y$-axis). So we have a path $t \mapsto AR_{t\alpha}$ from $A$ to $A'$. Now $A'$ represent the composite of dilations by $a'$ and $d'$ along the axes composed with a shear. So we have a path $t \mapsto \left(\matrix{(1-t)a' + t & (1-t)b'\\0 & \frac{1}{(1-t)a' + t}}\right)$ that continuously eliminates the dilation factors and the shear taking $A'$ to the identity matrix $I = \left(\matrix{1 & 0 \\0 & 1}\right)$.