# Haar mesure on $\mathrm{GL}_{d}(F)$

$$\DeclareMathOperator\GL{GL}$$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 $$\mu_{M_{d}(F)}(A)$$ ?

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