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Let $v=(x,y{)}^T$ and $B(\rho )=\left(\begin{array}{cc}\cosh \rho & \sinh \rho \\ \sinh \rho & \cosh \rho \end{array}\right)$. Matrix multiplication givesThus $G$ is a semidirect product of the translation plane by the one-parameter Lorentz boost group. Differentiating left multiplication at the identity gives the left-invariant vector fieldsThey obey $[X_0,X_1]=X_2$, $[X_0,X_2]=X_1$ and $[X_1,X_2]=0$.

Their dual coframe is $d\rho$, $\cosh \rho dx-\sinh \rho dy$, and $-\sinh \rho dx+\cosh \rho dy$. The Haar measure from a left-invariant coframe is their absolute wedge product. Since $\det B(\rho )=1$,One can also check this directly: left multiplication has Jacobian determinant one. Right multiplication has a block triangular Jacobian with determinant one too, so the same measure is both left and right Haar measure.