I am faced with the following issue that I do not understand but seems contradictory, coming from the book of Roberts and Schmidt about $GSp(4)$. Consider a local non-archimedean field $F$, let $p$ be its maximal ideal, $\mathcal{O}$ its ring of integers and $G=GSp(4, F)$. We are interested in the following Klingen congruence subgroup $$K = \left( \begin{array}{cccc} \mathcal{O} & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ p & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ p & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ p & p & p & \mathcal{O} \end{array} \right) $$
(from now on all the subgroups written in this matrix-entries form is meant to be their intersection with $GSp(4, F)$. I am interested in computing the index of this subgroup in the maximal compact subgroup $K_0$ (where all the entries are integers).
Iwahori decomposition We have $$ K = \left( \begin{array}{cccc} 1 & & & \\ p & 1 & & \\ p & & 1 & \\ p & p & p & 1 \end{array} \right) \left( \begin{array}{cccc} \mathcal{O}^\times & & & \\ & \mathcal{O} & \mathcal{O} & \\ & \mathcal{O} & \mathcal{O} & \\ & & & \mathcal{O}^\times \end{array} \right) \left( \begin{array}{cccc} 1 & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ & 1 & & \mathcal{O} \\ & & 1 & \mathcal{O} \\ & & & 1 \end{array} \right) $$
so that in particular by decomposing the left subgroup we should obtain $$ \left( \begin{array}{cccc} \mathcal{O} & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ \mathcal{O} & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ \mathcal{O} & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ \mathcal{O} & \mathcal{O} & \mathcal{O} & \mathcal{O} \end{array} \right) = \bigsqcup_{a, b, c \in \mathcal{O}/p} \left( \begin{array}{cccc} 1 & & & \\ a & 1 & & \\ b & & 1 & \\ c & b & -a & 1 \end{array} \right) K $$
(where the fact that the entries on the right are this way comes from the conditions of belonging to $GSp(4)$). So that in particular the index should be, writting $N(p)$ for the norm of $p$,
$$[K_0:K] =N(p)^3$$
Bruhat decomposition On the other hand, introducing the subgroup $$ Q = \left( \begin{array}{cccc} \mathcal{O} & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ & \mathcal{O} & \mathcal{O} & \mathcal{O} \\ & & & \mathcal{O} \end{array} \right) $$
the Bruhat decomposition yields that for any field $k$, $$ GSp(4, k) = Q \sqcup Qx \left( \begin{array}{cccc} 1 & k & & \\ & 1 & & \\ & & 1 & k \\ & & & 1 \end{array} \right) \sqcup Qxy \left( \begin{array}{cccc} 1 & & k & \\ & 1 & k & k \\ & & 1 & \\ & & & 1 \end{array} \right) \sqcup Qxyx \left( \begin{array}{cccc} 1 & k & k & k\\ & 1 & & k\\ & & 1 & k \\ & & & 1 \end{array} \right) $$
where the transformations $x$ and $y$ are defined by $$x= \left( \begin{array}{cccc} & 1& & \\ 1 & & & \\ & & & 1 \\ & &1 & \end{array} \right) $$ $$y = \left( \begin{array}{cccc} 1 & & &\\ & & 1 & \\ & -1 & & \\ & & & 1 \end{array} \right) $$
In particular if $k$ is the finite field with $N(p)$ elements, the index we search for is exactly the cardinality of $GSp(4,k) / Q$, and this one is $(1+N(p))(1+N(p)^2)$, so that we should say
$$[K_0:K] =(1+N(p))(1+N(p)^2)$$
Here is the question following from this discussion:
Both results are different, what is happening?
Your Iwahori-decomposition computation is a bit too free with quotient computations. It would work fine for vector spaces, which is, in some sense, why you get the correct leading term in your count; but there are additional subtleties on the group level that it does not take into account—among other things, that some of the entries denoted by $\mathcal O$ can be $0$, but some cannot!
The same computation would suggest that the quotient of $\operatorname{GL}_2(k)$ by its Borel subgroup of upper-triangular matrices should have as coset representatives the lower uni-triangular matrices $u_-(c) = \begin{pmatrix} 1 & 0 \\ c & 1 \end{pmatrix}$, whereas actually there is another coset, the one containing $w_0 = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$. Actually a better way to think about it is that most coset representatives are of the form $u_+(c)w_0$, where $u_+(c)$ is the transpose of $u_-(c)$, and only the trivial coset is not of this form; it just happens that $u_+(c)w_0$ and $u_-(c^{-1})$ lie in the same coset when $c \ne 0$.
Your count via the Bruhat decomposition shows what’s going on here: you get $(1 + N(p))(1 + N(p)^2) = 1 + N(p) + N(p)^2 + N(p)^3$, which suggests, correctly, that we have stratified $\mathbb P^3k$ by affine spaces of the obvious dimensions, corresponding to the Weyl-group elements $1$, $x$, $x y$, and $x y x$ (which are minimal-length representatives in the quotient by an appropriate parabolic subgroup, which is doubtless what led you to choose them!).