TITLE: What is meant when an author refers to a "diagonal embedding"?
QUESTION [1 upvotes]: What is meant when an author refers to a "diagonal embedding"? Specifically, it is referring to a map
$$A\hookrightarrow PGL_2(\mathbb{Q}_5)\times(SU(2)\times\mathbb{R}^*)$$
w.r.t. a chosen real place, where the domain $A$ is the set of Hamilton quaternions over $\mathbb{Q}$ whose entries and inverse's entries all lie in $\mathbb{Z}[1/5]$. My question is: what exactly does this embedding do to a given domain element $w+xi+yj+zk$? Thanks!

REPLY [2 votes]: If you have $f:A\to B$ and $g:A\to C$, then the diagonal imbedding $\delta:A\to B\times C$ is $\delta(a)=(f(a),g(a))$. The same applies with more than just two targets.
EDIT: more details... Here, we can actually look at a somewhat simpler scenario of maps from all rational Hamiltonian quaternions $Q$ (not restricting denominators) to the indicated spaces. First, we note that the full algebra $\mathbb H$ of Hamiltonian quaternions has multiplicative group isomorphic to $SU(2)\times \mathbb R^\times$. (A small exercise.)
A map to $GL_2(\mathbb Q_5$ comes from the fact that $Q\otimes_{\mathbb Q} \mathbb Q_5\approx M_2(\mathbb Q_5)$. That is, the rational Hamiltonian quaternions "split" over $\mathbb Q_5$. A choice of coordinates, together with the natural map $Q\to Q\otimes_{\mathbb Q} \mathbb Q_5$, give an imbedding of $Q^\times\to GL_2(\mathbb Q_5)$.
So we have a diagonal map $Q^\times \to GL_2(\mathbb Q_5)\times (SU(2)\times \mathbb R^\times)$. Ok. It is not clear to me, and is in fact doubtful, that the map factors through the projectivized $GL_2$.
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