928: Intersection of Simplicial Complexes Is Simplicial Complex, and Underlying Space of Intersection Is Contained in but Not Necessarily Equal to Intersection of Underlying Spaces of Constituents

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description/proof of that intersection of simplicial complexes is simplicial complex, and underlying space of intersection is contained in but not necessarily equal to intersection of underlying spaces of constituents

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Topics

About: vectors space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Structured Description
• 2: Natural Language Description
• 3: Proof
• 4: Note

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Starting Context

• The reader knows a definition of simplicial complex.
• The reader admits the proposition that for any 2 vectors spaces that share the operations on the intersection, the intersection is a vectors space with the restriction of the shared operations.

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Target Context

• The reader will have a description and a proof of the proposition that the intersection of any 2 simplicial complexes is a simplicial complex, and the underlying space of the intersection is contained in but not necessarily equal to the intersection of the underlying spaces of the constituent simplicial complexes.

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Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$V_1$: $\in \{\text{ the real vectors spaces }\}$
$V_2$: $\in \{\text{ the real vectors spaces }\}$
$C_1$: $\in \{\text{ the simplicial complexes on }{V}_1\}$
$C_2$: $\in \{\text{ the simplicial complexes on }{V}_2\}$
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Statements:
$V_1$ and $V_2$ share the operations on $V_1\cap V_2$
$⟹$
(
$V_1\cap V_2\in \{\text{ the real vectors spaces }\}$ with the restriction of the shared operations
$\wedge$
$C_1\cap C_2\in \{\text{ the simplicial complexes on }{V}_1\cap V_2\}$
$\wedge$
$|C_1\cap C_2|\subseteq |C_1|\cap |C_2|$
$\wedge$
Not necessarily $|C_1\cap C_2|=|C_1|\cap |C_2|$
)
//

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2: Natural Language Description

For any real vectors spaces, $V_1,V_2$, such that $V_1$ and $V_2$ share the operations on $V_1\cap V_2$, any simplicial complexes, $C_1$, on $V_1$ and $C_2$, on $V_2$, $V_1\cap V_2$ is a vectors space with the restriction of the shared operations, $C_1\cap C_2$ is a simplicial complex on $V_1\cap V_2$, and $|C_1\cap C_2|\subseteq |C_1|\cap |C_2|$, but not necessarily $|C_1\cap C_2|=|C_1|\cap |C_2|$.

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3: Proof

Whole Strategy: Step 1: see that $V_1\cap V_2$ is a real vectors space with the restriction of the shared operations; Step 2: see that each $S\in C_1\cap C_2$ is an affine simplex on $V_1\cap V_2$; Step 3: see that each face of $S$ is contained in $C_1\cap C_2$; Step 4: see that for each $S_1,S_2\in C_1\cap C_2$, $S_1\cap S_2$ is a face of $S_1$ and is a face of $S_2$; Step 5: see that $|C_1\cap C_2|\subseteq |C_1|\cap |C_2|$; Step 6: see an example that $|C_1\cap C_2|\ne |C_1|\cap |C_2|$.

Step 1:

$V_1\cap V_2$ is a real vectors space with the restriction of the shared operations, by the proposition that for any 2 vectors spaces that share the operations on the intersection, the intersection is a vectors space with the restriction of the shared operations.

Step 2:

For each $S\in C_1\cap C_2$, $S\in C_1$ and $S\in C_2$. $S\subseteq V_1$ and $S\subseteq V_2$, which means that $S\subseteq V_1\cap V_2$. $S$ is an affine simplex in $V_1\cap V_2$, because $V_1\cap V_2$ is a vectors subspace of $V_1$, and the set of the vertexes of $S$ is affine-independent on $V_1\cap V_2$.

Step 3:

Each face of $S$ is contained in $C_1$, because $S$ is an element of $C_1$, and likewise is contained in $C_2$. So, each face of $S$ is contained in $C_1\cap C_2$.

Step 4:

For each $S_1,S_2\in C_1\cap C_2$, $S_1\in C_1$ and $S_2\in C_1$. So, $S_1\cap S_2$ is a face of $S_1$. Likewise, $S_1\cap S_2$ is a face of $S_2$.

So, $C_1\cap C_2$ is a simplicial complex on $V_1\cap V_2$.

Step 5:

Let us prove that $|C_1\cap C_2|\subseteq |C_1|\cap |C_2|$.

For each point, $p\in |C_1\cap C_2|$, there is an $S\in C_1\cap C_2$ such that $p\in S$; as $S\in C_1$, $p\in |C_1|$, and likewise, $p\in |C_2|$; so, $p\in |C_1|\cap |C_2|$.

Step 6:

As an example of not $|C_1\cap C_2|=|C_1|\cap |C_2|$, let $V_1=V_2={\mathbb{R}}^2$ and $C_1$ consist of a 2-simplex and $C_2$ consist of a 2-simplex that shares a vertex and only a part of an edge (which contains the vertex) of the former simplex. Then, $C_1\cap C_2$ consists of the 0-simplex as the shared vertex, and $|C_1\cap C_2|$ consists of only the vertex, while $|C_1|\cap |C_2|$ consists of the shared part of the edge. The issue is that as the latter 2-simplex is not any element of $C_1$, the intersection of the 2 2-simplexes does not need to be a face of the former 2-simplex, so, does not need to be contained in $C_1$.

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4: Note

The union, $C_1\cup C_2$, is not necessarily a simplicial complex, as is shown in another article.

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References

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