Barycentric subdivision of an affine $n$-simplex $\Sigma^n$ definition? From Rotman's Algebraic Topology:

The barycentric subdivision of an affince $n$-simplex $\Sigma^n$, denoted by $\text{Sd} \space \Sigma^n$, is a family of affine $n$-simplexes defined inductively for $n \ge 0$:
$(i) \space \text{Sd } \Sigma^0 = \Sigma^0$
$(ii)$ If $\phi_0, \phi_1, \dots, \phi_{n+1}$ are the $n$-faces of $\Sigma^{n+1}$ and if $b$ is the barycenter of $\Sigma^{n+1}$, then $\text{Sd } \Sigma^{n+1}$ consists of all the $(n+1)$- simplexes spanned by $b$ and $n$-simplexes in $\text{Sd } \phi_i, i = 0, \dots, n+1.$

My question is:
What does " $(n+1)$- simplexes spanned by $b$ and $n$-simplexes in $\text{Sd } \phi_i, i = 0, \dots, n+1.$" mean?

I can see that if $\Sigma' = \Sigma^1 = [e_0, e_1]$, then $\phi'_0 = e_1, \phi'_1 = e_0$, $b = \frac{e_0 + e_1} {2}$ and $\text{Sd } \Sigma'$ consists of all $1$-simplexes spanned by $b$ and $0$-simplexes in $\phi'_i$.  This seems to imply that $\text{Sd }\Sigma' = \{[e_0, b],[b,e_1]\}$
Now suppose I have $\Sigma = \Sigma^2 = [e_0, e_1, e_2]$ a $2$-simplex.  Then the $\phi_0 = [e_1, e_2], \phi_1 = [e_0, e_2], \phi_2 = [e_0, e_1]$, and $b$ is just the barycenter of the triangle.  Then the simplexes in $\Sigma$ should be the $2$-simplexes spanned by $b$ and the $1$-simplexes of $\text{Sd }\phi_i$
What is a $2$-simplex spanned by a $0$-simplex and a $1$-simplex?  Is there a definition of this?  The only time I've seen span used is for affine sets spanned by a finite set of points.
 A: An $n$-simplex $\Sigma^n = [p_0,\ldots,p_n]$ is the convex hull (= span) of a set of $n+1$ points $p_0,\ldots,p_n$ in general position. This means that the affine subspace A($p_0,\ldots,p_n) = A(\Sigma^n)$ spanned by $p_0,\ldots,p_n$ is $n$-dimensional (which is equivalent to $p_1-p_0,\ldots,p_n-p_0$ being linearly independent). For $i = 0,\ldots,n$, the $i$-th face of $\Sigma^n$ is $\phi_i = \partial_i \Sigma^n = [p_0,\ldots,\hat{p}_i,\ldots,p_{n+1}]$, where the vertex $p_i$ has been omitted. Note that a $0$-simplex $\Sigma^0 = [p_0]$  has only one face $\partial_0 \Sigma^0$ which is the span of a set of zero points, i.e. we have $\partial_0 \Sigma^0 = \emptyset$. Moreover, the barycenter of $\Sigma^0 = [p_0]$ is the point $p_0$.
If $b \notin A(\Sigma)$, then let $b * \Sigma$ denote the convex hull of $b,p_0,\ldots,p_n$. This is an $(n+1)$-simplex which we call the span of $b$ and $\Sigma$.
Now let $\Sigma^{n+1} = [p_0,\ldots,p_{n+1}]$. Each face $\phi_i$ determines a unique $n$-dimensional affine subspaces $A_i =A(\phi_i)$. If $b_i$ is the barycenter of $\phi_i$, then $\text{Sd} \phi_i$ has $n$-simplices $\beta_{ij} = b_i * \phi_{ij}$, where $\phi_{ij}$ is the $j$-th face of $\phi_i$. Note that $A(\beta_{ij}) = A_i$. The cases $n = 0, 1$ are treated explicitly in the edit below.
If $b$ is the barycenter of $\Sigma^{n+1}$, then the $(n+1)$-simplexes spanned by $b$ and the $n$-simplexes in $\text{Sd } \phi_i, i = 0, \dots, n+1$, are the simplices $\sigma_{ij} = b * \beta_{ij}$. Note that $b \notin A(\beta_{ij}) = A_i$.
Edited:
For $n = 0$ we have $\Sigma^1 = [p_0,p_1]$. Its faces are the $0$-simplices $\phi_0 = [p_1], \phi_1 = [p_0]$. The $0$-simplex $\phi_i$ has only one face $\phi_{i0}$ which is the empty set. If $b_i$ denotes the barycenter of $\phi_i$, we have $b_0 = p_1, b_1 =p_0$ and $\text{Sd} \phi_i$ has only one $0$-simplex $b_i * \emptyset = [b_i]$. That is, $\text{Sd} \phi_0 = [p_1] = \phi_0, \text{Sd} \phi_1 = [p_0] = \phi_1$.  Note that this is nothing else than $(i)$ in your question. Therefore, as the barycentric subdivision of  $\Sigma^1$ we get the simplices $b * [p_1] = [b, p_1]$ and $b * [p_0] = [p_0,b]$.
For $n = 1$ we have $\Sigma^2 = [p_0,p_1,p_2]$. Its faces are the $1$-simplices $\phi_0 = [p_1,p_2]$, $\phi_1 = [p_0,p_2]$, $\phi_2 = [p_0,p_1]$. The $1$-simplex $\phi_i$ has two $0$-simplices as faces. Explicitly, $\phi_{00} = [p_2]$, $\phi_{01} = [p_1]$, $\phi_{10} = [p_2]$, $\phi_{11} = [p_0]$, $\phi_{20} = [p_1]$, $\phi_{21} = [p_0]$. With barycenters $b_i$ of $\phi_i$ we get six $1$-simplices $\beta_{00} = [b_0,p_2]$, $\beta_{01} = [p_1,b_0]$, $\beta_{10} = [p_0,b_1]$, $\beta_{11} = [b_1,p_2]$, $\beta_{20} = [p_0,b_2]$, $\beta_{21} = [b_2,p_1]$. This gives six $2$-simplices $\sigma_{00} = [b,b_0,p_2]$, $\sigma_{01} = [b,p_1,b_0]$, $\sigma_{10} = [b,p_0,b_1]$, $\sigma_{11} = [b,b_1,p_2]$, $\sigma_{20} = [b,p_0,b_2]$, $\sigma_{21} = [b,b_2,p_1]$. Drawing a picture is helpful.
