Let $G$ be an $SL(n, F)$ for a non archimedean local field $F$ with Iwahori subgroup $I$. Let $\mu$ be the Haar measure of $G$.

What are the properties of the function $w\mapsto \mu(IwI)/\mu(I)$ for $w$ being an element of the normalizer $N$ of the maximal Levi subgroup of $G$?

Note that for $d i$ being the Haar measure of $I$, that $$ \int\limits_G f(g) d \mu(g) = \sum\limits_{x \in N / N \cap I} ( \mu(IwI)/mu(I) )^{-1} \int\limits_I \int\limits_I f(i_1 w i_2) d i_1 d i_2 .$$

I would like to have a reference for the above fact and perhaps also the value of $mu(IwI)/mu(I) $ at least for $n=2$ and also for similar situation of $PGL(n)$.

I suspect that the fact that $(I,N)$ is a $BN$ pair and that the Iwahori decomposition is available for $I$ should give everything and also give some nice properties for the function $w\mapsto \mu(IwI)/mu(I)$.

Naively, one could argue that the function is multiplicative by the following computation $$ \mu(Iw_1w_2I)/\mu(I) = \mu(Iw_1Iw_2I)/\mu(Iw_1I) \cdot \mu(I w_1 I) / \mu( I )$$ $$ = \mu(Iw_2I)/\mu(I) \cdot \mu(I w_1 I) / \mu( I ),$$ which is wrong. What is the correct relation? $\geq$, if $w_1$ satisfies $w_1^2 = 1$?

Let $G$ be $SL(n, F)$ for a non-archimedean local field $F$ with Iwahori subgroup $I$. Let $\mu$ be the Haar measure of $G$.

What are the properties of the function $w\mapsto \mu(IwI)/\mu(I)$ for $w$ being an element of the normalizer $N$ of the maximal Levi subgroup of $G$?

Let $d i$ denote the Haar measure on $I$. We have that $$ \int\limits_G f(g) d \mu(g) = \sum\limits_{x \in N / N \cap I} ( \mu(IwI)/\mu(I) )^{-1} \int\limits_I \int\limits_I f(i_1 w i_2) d i_1 d i_2 .$$

I would like to have a reference for the above fact and perhaps also the value of $\mu(IwI)/\mu(I) $ at least for $n=2$ and also for similar situation of $PGL(n)$.

I suspect that the fact that $(I,N)$ is a $BN$ pair and that the Iwahori decomposition is available for $I$ should give everything and also give some nice properties for the function $w\mapsto \mu(IwI)/mu(I)$.

Perhaps one suggestion: Naively, one could argue that the function is multiplicative by the following computation $$ \mu(Iw_1w_2I)/\mu(I) = \mu(Iw_1Iw_2I)/\mu(Iw_1I) \cdot \mu(I w_1 I) / \mu( I )$$ $$ = \mu(Iw_2I)/\mu(I) \cdot \mu(I w_1 I) / \mu( I ),$$ which is wrong. What is the correct relation? $\geq$, if $w_1$ satisfies $w_1^2 = 1$?

Let $G$ be an $SL(n, F)$ for a non archimedean local field $F$ with Iwahori subgroup $I$. Let $\mu$ be the Haar measure of $G$.

What are the properties of the function $w\mapsto \mu(IwI)/mu(I)$ for $w$ being an element of the normalizer $N$ of the maximal Levi subgroup of $G$?

Note that for $d i$ being the Haar measure of $I$, that $$ \int\limits_G f(g) d \mu(g) = \sum\limits_{x \in N / N \cap I} ( \mu(IwI)/mu(I) )^{-1} \int\limits_I \int\limits_I f(i_1 w i_2) d i_1 d i_2 .$$

I would like to have a reference for the above fact and perhaps also the value of $mu(IwI)/mu(I) $ at least for $n=2$ and also for similar situation of $PGL(n)$.

I suspect that the fact that $(I,N)$ is a $BN$ pair and that the Iwahori decomposition is available for $I$ should give everything and also give some nice properties for the function $w\mapsto \mu(IwI)/mu(I)$.

Let $G$ be an $SL(n, F)$ for a non archimedean local field $F$ with Iwahori subgroup $I$. Let $\mu$ be the Haar measure of $G$.

What are the properties of the function $w\mapsto \mu(IwI)/\mu(I)$ for $w$ being an element of the normalizer $N$ of the maximal Levi subgroup of $G$?

Note that for $d i$ being the Haar measure of $I$, that $$ \int\limits_G f(g) d \mu(g) = \sum\limits_{x \in N / N \cap I} ( \mu(IwI)/mu(I) )^{-1} \int\limits_I \int\limits_I f(i_1 w i_2) d i_1 d i_2 .$$

I would like to have a reference for the above fact and perhaps also the value of $mu(IwI)/mu(I) $ at least for $n=2$ and also for similar situation of $PGL(n)$.

I suspect that the fact that $(I,N)$ is a $BN$ pair and that the Iwahori decomposition is available for $I$ should give everything and also give some nice properties for the function $w\mapsto \mu(IwI)/mu(I)$.

Naively, one could argue that the function is multiplicative by the following computation $$ \mu(Iw_1w_2I)/\mu(I) = \mu(Iw_1Iw_2I)/\mu(Iw_1I) \cdot \mu(I w_1 I) / \mu( I )$$ $$ = \mu(Iw_2I)/\mu(I) \cdot \mu(I w_1 I) / \mu( I ),$$ which is wrong. What is the correct relation? $\geq$, if $w_1$ satisfies $w_1^2 = 1$?