Transpose in $ {SL}(2,\mathbb{R})$

Question

Let $SL(2, \Bbb{R})$ denote the group of special invertible $2\times 2$ matrices over $\mathbb{R}$. As a locally compact group, the Haar measure of $SL(2, \Bbb{R})$ is computable through Iwasawa decomposition by $ \int_{\mathbb{R}} \int_{\mathbb{R}_{>0}} \int_{\Bbb{T}} d\theta \; da/a^2\; dn$ where $$ A(n,a,\theta) = \left[ \begin{array}{l l} 1 & n \\ 0 & 1 \end{array} \right] \left[ \begin{array}{l l} \sqrt{a} & 0 \\ 0 & \frac{1}{\sqrt{a}} \end{array} \right] R_\theta. $$ Also it is known that $SL(2, \mathbb{R})$ is a unimodular group. The question that I have is about the transpose action and its relation to the Haar measure, in other words how can one write $$ \int_{SL(2,\mathbb{R})} f(A^t) dA $$ using $ \int_{SL(2, \mathbb{R})} dA $?

Answer 1

Let us write $$\int f(A)\, d\mu(A) := \int f(A^t) \, dA.$$

Since $dA$ is left-invariant, we have

$$ \int f(AB) \, d\mu(A) = \int f(A^t B) \, dA = \int f((B^t A)^t) \, dA = \int f(A^t) \, dA = \int f(A) \, d\mu(A). $$

In short, we have shown that $\mu$ is right-invariant. It is easy to see that $\int |f| \, d\mu < \infty$ for $f \in C_c(SL(\Bbb{R}^2))$, so that $\mu$ is a right Haar integral.

But the same is true for the usual Haar measure, since (as you say) $SL(\Bbb{R}^2)$ is unimodular.

Hence, $d\mu = c\cdot dA$ for some $c \in (0,\infty)$. But if we take any $f \in C_c$ with $f \geq 0$ and $f \not \equiv 0$ and set $g(A) := f(A) + f(A^t)$, then $g \in C_c$ with $g \geq 0$ and $g \not \equiv 0$ and $g(A) = g(A^t)$. Hence,

$$ c \cdot \int g(A) \, dA=\int g \, d\mu = \int g(A^t) \, dA = \int g(A) \, dA, $$

so that $c = 1$, since $\int g(A) \, dA > 0$.

All in all, we have shown $d\mu = dA$ and hence

$$ \int f(A^t) \, dA = \int f \, d\mu = \int f(A) \, dA. $$