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? – Omnomnomnom Sep 2 '15 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. – Cheerful Parsnip Sep 2 '15 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 $I = \left(\matrix{1 & 0 \\0 & 1}\right)$.