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If we consider the group of upper triangular matrices $B=\bigl(\begin{smallmatrix} a&b\\ 0&a^{-1} \end{smallmatrix} \bigr)$ where $a$ and $b$ are either real or complex and $a\neq1$, then the left Haar measure is given by $a^{-2}\,da \,db$.

While I understand that this measure is invariant with respect to left translation, I am a little confused as to why the factor of $a^{-2}$ is necessary. Any clarifications would be appreciated, Thank you!

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If $a \neq 1$ what's the identity of this group? – Qiaochu Yuan Jun 29 '12 at 15:41
$a\ne0$ I presume. – anon Jun 29 '12 at 15:48

Let's look at left translation: $\bigl(\begin{smallmatrix} A&B\\ 0&A^{-1} \end{smallmatrix} \bigr)\bigl(\begin{smallmatrix} a&b\\ 0&a^{-1} \end{smallmatrix} \bigr)=\bigl(\begin{smallmatrix} Aa&Ab+Ba^{-1}\\ 0&(Aa)^{-1} \end{smallmatrix} \bigr)$, i.e. $a_{new}=Aa$, $b_{new}=Ab+Ba^{-1}$, hence the Jacobian of $(a,b)\mapsto(a_{new},b_{new})$ is $A^2$. The measure $da\,db$ is thus not invariant (Jacobian is not identically $1$); the factor $a^{-2}$ compensates for this.

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