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barycentric subdivision
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Recall that an abstract $n$simplex is an abstract simplicial complex $K$ such that $\bigcup K\in K$ and the cardinality of $\bigcup K$ is $n+1$. It can be identified as the powerset (minus the empty set element) of a set $V_{K}$ of elements $v_{1},\ldots,v_{n}$, called the vertices of $K$.
The barycentric subdivision of an abstract simplex $K$ is the construction of a certain abstract simplicial complex $K^{{\prime}}$ from $K$. $K^{{\prime}}$ itself is called the barycentric subdivision of $K$. Before giving the general construction, let us describe some simple cases, specifically, when $n=1,2,$ and $3$ and when $V_{K}$ is embedded in some ambient Euclidean space $\mathbb{R}^{m}$ where $m\geq n$:

When $n=1$, $V_{K}$ is just a onepoint space. We define the barycentric subdivision of $K$ to be $K$ itself: $K^{{\prime}}=K$.

When $n=2$, $V_{K}$ consists of two distinct points $v_{1}$ and $v_{2}$. Take the midpoint $w$ of $v_{1}$ and $v_{2}$. Then the barycentric subdivision $K^{{\prime}}$ is defined to be the union of powersets of $\{v_{1},w\}$ and $\{v_{2},w\}$. Abstracting this process, $K^{{\prime}}$ can be thought of as the union of all the $2$simplices having $w$ as a vertex.

When $n=3$, $V_{K}$ consists of three distinct noncollinear points, $v_{1},v_{2},v_{3}$. We may picture them as the vertices of a triangle. From each of these vertices, draw a line connecting the vertex to the midpoint of the opposite side. This construction creates six smaller triangles, such that each pair either intersect at a common vertex or a common side. The barycentric subdivision $K^{{\prime}}$ is defined as the union of each of these smaller triangles considered as a simplex. In other words, $K^{{\prime}}$ is the union of the powerset of the threepoint sets that correspond to these smaller triangles.
Abstracting this process, we may take $w$ as the barycenter of $v_{1},v_{2},$ and $v_{3}$. Label $w=w(123)$, so
$w(123)=\frac{1}{3}(v_{1}+v_{2}+v_{3}).$ Next, take $w(12),w(23),$ and $w(13)$ as the midpoints of $(v_{1},v_{2})$, $(v_{2},v_{3})$ and $(v_{1},v_{3})$. Finally, relabel $v_{1},v_{2},v_{3}$ as $w(1),w(2),w(3)$. We form six threepoint sets:
$W(32):=\{w(123),w(12),w(1)\},\quad W(31):=\{w(123),w(12),w(2)\},$ $W(23):=\{w(123),w(13),w(1)\},\quad W(21):=\{w(123),w(13),w(3)\},$ $W(13):=\{w(123),w(23),w(2)\},\quad W(12):=\{w(123),w(23),w(3)\}.$ The formation of any one of these sets goes as follows:
(a) it must contain the barycenter $w(123)$;
(b) from $w(123)$, pick one of the midpoints $w(12),w(23),w(13)$ to be the next point in the set;
(c)
The logic behind the labeling $ab$ of $W(ab)$ comes from the fact that the labeling of the midpoint in $W(ab)$ does not contain $a$ and the labeling of the initial vertex in $W(ab)$ does not contain $b$.
Finally, we form $K^{{\prime}}$ as the union of the powersets of $W(ab)$’s.
In the last example, one can abstract the construction one step further. Since each labelled point is the barycenter of at least one of the initial vertices $v_{i}$, we can uniquely identify any nonempty subset $V$ of $V_{K}$ with the labelled point that is the barycenter of the point(s) in $V$. Then each $W(ab)$ above can be identified as a maximal chain (ordered by inclusion) in $K$ with $\varnothing$ deleted.
This suggests the general construction of the barycentric subdivision of an abstract $n$simplex.
Definition. Let $K$ be an abstract $n$simplex. Order $K$ by inclusion $\subseteq$. Let
$\mathcal{C}_{K}:=\big\{P(C)\mid C\mbox{ is a maximal chain in }K\big\}.$ 
The barycentric subdivision $K^{{\prime}}$ of $K$ is:
$K^{{\prime}}=\bigcup\mathcal{C}_{K}\{\varnothing\}.$ 
It is easy to see that every maximal chain in $K$ is an $(n1)$simplex whose powerset is an $n$simplex (so isomorphic to $K$). In addition, the barycentric subdivision $K^{{\prime}}$ of $K$ is a simplicial complex with $n!$ maximal simplices, each of which is isomorphic to $K$.
Remark. This definition can be generalized to include the barycentric subdivision of an abstract simplicial complex. If $K$ is an abstract simplicial complex, then the barycentric subdivision $K^{{\prime}}$ of $K$ is the union of the barycentric subdivisions of the individual maximal simplicies in $K$. Below are two examples:

In this example (pictured above), the maximal simplices of $K$ consist of a triangle, and two line segments.
\pspicture(0,1)(8,3) \pspolygon[fillstyle=solid,fillcolor=lightblue](0,0)(0,3)(3,0) \psline(0,3)(3,0) \psline(3,0)(3,3) \psline(0,3)(3,3) \psdots(0,0) \psdots(0,3) \psdots(3,0) \psdots(3,3) \rput[b](1.5,1)$K$ \pspolygon[fillstyle=solid,fillcolor=lightblue](5,0)(5,3)(8,0) \psline(5,3)(8,0) \psline(5,0)(6.5,1.5) \psline(5,3)(6.5,0) \psline(8,0)(5,1.5) \psline(5,3)(8,3) \psline(8,0)(8,3) \psdots(5,0) \psdots(5,3) \psdots(8,0) \psdots(8,3) \psdots(5,1.5) \psdots(6.5,0) \psdots(6.5,1.5) \psdots(8,1.5) \psdots(6.5,3) \psdots(6,1) \rput[b](6.5,1)$K^{{\prime}}$

Here (pictured below), the maximal simplices are two triangles meeting at a common edge.
\pspicture(0,1)(8,3) \pspolygon[fillstyle=solid,fillcolor=lightblue](0,0)(0,3)(3,0) \pspolygon[fillstyle=solid,fillcolor=purple](0,3)(3,3)(3,0) \psdots(0,0) \psdots(0,3) \psdots(3,0) \psdots(3,3) \psline(0,3)(3,0) \rput[b](1.5,1)$K$ \pspolygon[fillstyle=solid,fillcolor=lightblue](5,0)(5,3)(8,0) \pspolygon[fillstyle=solid,fillcolor=purple](5,3)(8,3)(8,0) \psdots(5,0) \psdots(5,3) \psdots(8,0) \psdots(8,3) \psdots(5,1.5) \psdots(6.5,0) \psdots(6.5,1.5) \psdots(8,1.5) \psdots(6.5,3) \psdots(6,1) \psdots(7,2) \psline(5,3)(8,0) \psline(5,0)(8,3) \psline(5,3)(6.5,0) \psline(8,0)(5,1.5) \psline(5,3)(8,1.5) \psline(8,0)(6.5,3) \rput[b](6.5,1)$K^{{\prime}}$
In both examples, the vertex sets of the original simplicial complexes are the same.
It can be shown that the barycentric subdivision $K^{{\prime}}$ of an abstract simplicial complex $K$ can be constructed as follows: $V_{{K^{{\prime}}}}:=\{S\mid S\in K\}$ is the set of vertices of $K^{{\prime}}$, and $T\in K^{{\prime}}$ iff $T$ is a chain of simplexes in $K$: $T=\{S_{1},\ldots,S_{{n(T)}}\}$, $S_{i}\subset S_{j}$ for $i<j$.
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