Does a Bratelli diagram have to label the dimension of its nodes? I'm reading about Bratelli diagrams, and am seeing what look to be two different definitions. When converting a direct sequence of finite-dimensional $C^*$-algebras $A_n$ into a Bratelli diagram, some authors will label each vertex with the "dimension" $d$ of the factor $M_d(\mathbb{C}) \subseteq A_n$ it corresponds to, while others won't. So, for example, if I had the sequences $M_1(\mathbb{C}) \hookrightarrow M_2(\mathbb{C}) \hookrightarrow M_3(\mathbb{C}) \hookrightarrow \cdots$ and $M_2(\mathbb{C}) \hookrightarrow M_4(\mathbb{C}) \hookrightarrow M_6(\mathbb{C}) \hookrightarrow \cdots$, and I wanted to make their Bratelli diagrams, then for the first convention, I would have the diagrams
$\require{AMScd}$
\begin{CD}
\stackrel{1}{\cdot} @>>> \stackrel{2}{\cdot} @>>> \stackrel{3}{\cdot} @>>> \cdots ,
\end{CD}
\begin{CD}
\stackrel{2}{\cdot} @>>> \stackrel{4}{\cdot} @>>> \stackrel{6}{\cdot} @>>> \cdots ,
\end{CD}
respectively, but for the second convention, I'd just have the diagram
\begin{CD}
\cdot @>>> \cdot @>>> \cdot @>>> \cdots
\end{CD}
for both. But it's not clear to me how I could look at the latter and recover the direct sequence, because it just tells me how many factors I have at each step and how many copies of them I can fit into the factors of the next step. Is the idea that given this "dimensionless" Bratelli diagram, I could just choose the sizes of the factors of $A_n$ to be whatever I wanted as long as they were large enough to fit however many copies of the previous step's factors, and I would still get the same limit $C^*$-algebra? If so, is there any intuition for why we have this degree of freedom?
Thanks for your help!

I cannot comment much because I have not seen what you mention. In general, you need to write the sizes. Otherwise
$\require{AMScd}$
\begin{CD}
\cdot @>>> \cdot @>>> \cdot @>>> \cdots
\end{CD}
could mean
\begin{CD}
\cdot @>>> \stackrel1\cdot @>>> \stackrel1\cdot @>>> \stackrel1\cdot @>>>\cdots
\end{CD}
so isomorphic to $\mathbb C$, while
\begin{CD}
\cdot @>>> \stackrel1\cdot @>>> \stackrel2\cdot @>>> \stackrel3\cdot @>>>\cdots
\end{CD}
would give you $K(H)$. I suppose it could be that in some cases the context makes the sizes obvious?
There is one case where the sizes are deduced, which is the case where the embeddings are prescribed to be unital and the initial algebra $\mathbb C$. In such case the dimension is precisely the sum of the origins of the arrows.  For instance, this graph

woud automatically represent this:

