I've seen in literature a statement like 'there exists an open and dense Bruhat cell'. In $GL(2,F)$ for example, where $F$ is a p-adic field, let $\omega=\begin{pmatrix} & 1 \\ 1 & \end{pmatrix}$, then $A_2N_2\omega N_2$ is the big cell, here $A_2$ is the diagonal matrices, $N_2$ is the upper triangular unipotents.
Question: Is $A_2N_2\omega N_2$ an open and dense subset of $GL(2)$ in p-adic topology, or in Zariski topology?
I asked this because, if we want to find a sequence of elements in this cell approaching the identity, compute
$$\begin{pmatrix} a & \\ & b\end{pmatrix}\begin{pmatrix} 1 & x \\ & 1 \end{pmatrix}\begin{pmatrix} & 1 \\ 1 & \end{pmatrix}\begin{pmatrix} 1 & y \\ & 1 \end{pmatrix}=\begin{pmatrix} ax & a+axy \\ b & by \end{pmatrix}$$
If this product goes to the identity matrix in the p-adic topology, we will get $b\to 0$, $ax\to 1$, $by\to 1$, and $a+axy\to 0$, from which we will have a contradiction.
Does this mean the big cell is not dense in the p-adic topology? Or I'm misunderstanding things.
Thank you very much for any comment.
The big Bruhat cell is exactly the complement of the Borel subgroup, so it is defined by the non-vanishing of an algebraic function on $G$ (the bottom left entry); thus it is certainly open and dense in both the Zariski and $p$-adic topologies.
PS. You might find the identity $$ \begin{pmatrix} r & s \\ t & u \end{pmatrix} = \begin{pmatrix} s - ru/t & \\ & t \end{pmatrix}\begin{pmatrix} 1& -rt/(ru-st) \\ & 1 \end{pmatrix}\begin{pmatrix} & 1 \\ 1 & \end{pmatrix}\begin{pmatrix} 1 & u/t \\ & 1 \end{pmatrix} $$ helpful. Note that as $t \to 0$ then lots of things diverge.