Let $G$ be a reductive algebraic group over a complete non-archimedean field $k$. We know that maximal compacts are exactly the same as maximal parahorics when the Iwahori is open compact subgroup of $G$.
My question is exactly when is the Iwahori open compact subgroup? Does it hold for any reductive group or do we need simply connected or some simplicity assumption?
Another question is, could it be that maximal compacts are same as maximal parahorics even when the Iwahori is not open compact?
First I guess that the right setting is the case where $G$ is locally compact for its natural topology coming from the topology of $k$. So you need to assume that $k$ is locally compact, i.e. that its residue field is finite.
In Bruhat-Tits theory, by definition, an Iwahori subgroup is a particular case of a parahoric subgroup and such a group is open and compact.
If as an algebraic group over the algebraic closure of $k$, $G$ is simply connected, then the maximal compact subgroups of $G$ are parahoric subgroups. Conversely, a parahoric subgroup is a maximal subgroup of $G$ is and only if it is maximal among parahoric subgroup. In fact, the maximal compact subgroups of $G$ are the parahoric subgroups whose fixators in the Bruhat-Tits buildings are $0$-dimensional facet.
If $G$ is not simply connected, e.g. ${\rm PGL}(n,k)$, then the maximal compact subgroups are not parahoric subgroups in general : they may be bigger. For instance an Iwahori subgroup $I$ of ${\rm PGL}(n,k)$ is formed of those matrices whose representatives in ${\rm GL}(n,k)$ have coefficients in the ring of integers ${\mathfrak o}_k$ and are upper triangular modulo ${\mathfrak p}_k$, the maximal ideal of ${\mathfrak o}_k$. Now the normalizer $\tilde I$ of $I$ in ${\rm GL}(n,k)$ is a maximal compact subgroup which is not a parahoric subgroup. It has the form ${\tilde I} =\langle \Pi\rangle \ltimes I$, where $\Pi$ is the following element : one representative is the endomorphism $f$ of $k^n$, defined on the canonical basis $(e_1 ,...,e_n )$ by $f(e_i )=e_{i-1}$, $i=2,...,n$, $f(e_1 )=\varpi_k e_n $ ($\varpi_k$ is a uniformizer of $k$).
