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$\DeclareMathOperator\GL{GL}$Let $F$ be a $\mathfrak{p}$-adic field and $\mathscr{O}_{F}$ its valuation ring. For any measurable subset of $M_{d}(F)$ such as $$ A= \left( \begin{array}{ccc} a_{11}+t^{\alpha_{11}}\mathscr{O}_{F} & \cdots & a_{1d}+t^{\alpha_{1d}}\mathscr{O}_{F} \\ \vdots & \ddots & \vdots \\ a_{d1}+t^{\alpha_{d1}}\mathscr{O}_{F} & \cdots & a_{dd}+t^{\alpha_{dd}}\mathscr{O}_{F} \end{array} \right), $$ we define $\mu_{M_{d}(F)}(A)=(\frac{1}{q})^{\sum_{i, j}\alpha_{ij}}$, where $t$ is an uniformizer of $F$ and $q$ is the order of residue field of $F$. It is well-known that $\mu_{M_{d}(F)}$ is invariant for additive, that is, for any $a\in M_{d}(F)$, $\mu_{M_{d}(F)}(a+A)=\mu_{M_{d}(F)}(A)$.

I want to know a relation between $\mu_{M_{d}(F)}$ and a left (or right) Haar measure on $\GL_{d}(F)$. Let $\mu$ be the left Haar measure on $\GL_{d}(F)$ such that $\mu(\GL_{d}(\mathscr{O}))=1$. The measurable subset $A$ on $M_{d}(F)$ as above, when $A\subset \GL_{d}(F),$ what is the relationship between the value of $\mu(A)$ and $\mu_{M_{d}(F)}(A)$ ?

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  • $\begingroup$ Isn't it just the standard for writing the multiplicative Haar measure from the additive one? See my next comment. I vote to close! $\endgroup$
    – Bugs Bunny
    Mar 17, 2021 at 12:10
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    $\begingroup$ $d\mu (x) = \alpha |\det (x)|^{-d} d \mu_{M_d} (x)$. Find the constant $\alpha$, then integrate. $\endgroup$
    – Bugs Bunny
    Mar 17, 2021 at 12:12
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    $\begingroup$ "Standard" often means "the thing that everyone in my area of math knows". Instead of closing, just leave a nice answer! $\endgroup$ Mar 17, 2021 at 12:34

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