Barycentric subdivision preserves geometric realization












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$begingroup$


I have the following definitions:



Definition 1: A simplicial complex $K$ is a family of finite nonempty subsets of a set $V_k$ (the elements of $V_k$ are called vertices) such that:



1) if $vin V_k$, then ${v}in K$.



2) if $Sin K$ and $emptyset neq Tsubseteq S$, then $Tin K$.



The elements of $K$ are called simplexes.



Definition 2: The geometric realization of a simplicial complex $K$ is defined as $|K|= {alpha: V_K to [0,1] ; | ; alpha^{-1}((0,1]) in K , ;sum_{vin V_k} alpha(v)=1 }$.



Definition 3: The barycentric subdivision of a simplicial complex $K$ is defined as the simplicial complex $sd(K)$ such that $V_{sd(K)}=K$ and a simplex in $sd(K)$ has the form ${s_0, s_1, ...,s_q}$, where $s_i in K $ and $s_{i-1} subset s_i $, ($s_{i-1}$ is a proper subset of $s_i$ ), for all $i$.



I'm asked to show that $|sd(K)|$ is homeomorphic to $|K|$.



I already know that for $S={v_0,v_1,...,v_q} in K$, $|S|={alpha in |K| ; | ; alpha^{-1} ((0,1]) subseteq S}$ is homeomorphic to $Delta^q = {(t_0,t_1,...,t_q)in mathbb{R}^{n+1} ; | ; t_i geq 0, sum_{i=0}^{n} t_i =1}$ (actually, the homeomorphism is an isometry).



Here, the topology we are using is the one induced by the metric $d:|K|$x$|K|to mathbb{R}$ such that $d(alpha, beta) = sqrt{sum_{vin V_K} (alpha (v) - beta (v))^2}$.



Now, I'm trying to construct an homeomorphism from $|K|$ to $|sd(K)|$, but everything I can think of, just doesn't work. I would be very thankful if you could give me a hint on how to construct this homeomorphism, or if you could tell me another way to proceed on this. Thank you so much!










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  • $begingroup$
    You are not going to solve this by just thinking formalistically about the definitions like this. Do you have a geometric picture of what these definitions mean?
    $endgroup$
    – Eric Wofsey
    Dec 8 '18 at 21:54










  • $begingroup$
    I have intuition when $V_k$ is a finite set of points in $R^n$, because $|K|$ is going to be the convex hull of this points. I'm thinking about $V_k= {v_0, v_1, v_2}$, and $K={{v_0},{v_1},{v_2},{v_0,v_1},{v_0,v_2},{v_1,v_2} }$, then $|K|$ is a filled triangle and, for example, ${v_0,v_1}$ represents the midpoint of the line between $v_0$ and $v_1$. That's why the geometric realization of barycentric subdivision is obtained by "adding" this midpoints as vertices, and now, simplexes here are going to be the little triangles formed when taking the baricenter of this triangle.
    $endgroup$
    – user392559
    Dec 8 '18 at 22:09










  • $begingroup$
    Now, I don't really have intuition when $V_k$ is an infinite set, and I don't even know how to the finite case formally.
    $endgroup$
    – user392559
    Dec 8 '18 at 22:11










  • $begingroup$
    Oh, hmm, your definition of the topology on $|K|$ is nonstandard when $V_K$ is infinite. That makes it fairly annoying to prove you have a homeomorphism: it's not bad to construct a continuous bijection $|sd(K)|to |K|$, but proving that the inverse is continuous will be a pain if $V_K$ is infinite.
    $endgroup$
    – Eric Wofsey
    Dec 8 '18 at 22:38










  • $begingroup$
    (The standard topology to put on $|K|$ is the topology that makes it the colimit of all its simplices. This means that to check that the inverse map $|K|to |sd(K)|$ is continuous, you just have to check that it is continuous when restricted to each simplex of $|K|$. But this is then automatic since a continuous bijection between compact Hausdorff spaces always has continuous inverse.)
    $endgroup$
    – Eric Wofsey
    Dec 8 '18 at 22:44
















1












$begingroup$


I have the following definitions:



Definition 1: A simplicial complex $K$ is a family of finite nonempty subsets of a set $V_k$ (the elements of $V_k$ are called vertices) such that:



1) if $vin V_k$, then ${v}in K$.



2) if $Sin K$ and $emptyset neq Tsubseteq S$, then $Tin K$.



The elements of $K$ are called simplexes.



Definition 2: The geometric realization of a simplicial complex $K$ is defined as $|K|= {alpha: V_K to [0,1] ; | ; alpha^{-1}((0,1]) in K , ;sum_{vin V_k} alpha(v)=1 }$.



Definition 3: The barycentric subdivision of a simplicial complex $K$ is defined as the simplicial complex $sd(K)$ such that $V_{sd(K)}=K$ and a simplex in $sd(K)$ has the form ${s_0, s_1, ...,s_q}$, where $s_i in K $ and $s_{i-1} subset s_i $, ($s_{i-1}$ is a proper subset of $s_i$ ), for all $i$.



I'm asked to show that $|sd(K)|$ is homeomorphic to $|K|$.



I already know that for $S={v_0,v_1,...,v_q} in K$, $|S|={alpha in |K| ; | ; alpha^{-1} ((0,1]) subseteq S}$ is homeomorphic to $Delta^q = {(t_0,t_1,...,t_q)in mathbb{R}^{n+1} ; | ; t_i geq 0, sum_{i=0}^{n} t_i =1}$ (actually, the homeomorphism is an isometry).



Here, the topology we are using is the one induced by the metric $d:|K|$x$|K|to mathbb{R}$ such that $d(alpha, beta) = sqrt{sum_{vin V_K} (alpha (v) - beta (v))^2}$.



Now, I'm trying to construct an homeomorphism from $|K|$ to $|sd(K)|$, but everything I can think of, just doesn't work. I would be very thankful if you could give me a hint on how to construct this homeomorphism, or if you could tell me another way to proceed on this. Thank you so much!










share|cite|improve this question









$endgroup$












  • $begingroup$
    You are not going to solve this by just thinking formalistically about the definitions like this. Do you have a geometric picture of what these definitions mean?
    $endgroup$
    – Eric Wofsey
    Dec 8 '18 at 21:54










  • $begingroup$
    I have intuition when $V_k$ is a finite set of points in $R^n$, because $|K|$ is going to be the convex hull of this points. I'm thinking about $V_k= {v_0, v_1, v_2}$, and $K={{v_0},{v_1},{v_2},{v_0,v_1},{v_0,v_2},{v_1,v_2} }$, then $|K|$ is a filled triangle and, for example, ${v_0,v_1}$ represents the midpoint of the line between $v_0$ and $v_1$. That's why the geometric realization of barycentric subdivision is obtained by "adding" this midpoints as vertices, and now, simplexes here are going to be the little triangles formed when taking the baricenter of this triangle.
    $endgroup$
    – user392559
    Dec 8 '18 at 22:09










  • $begingroup$
    Now, I don't really have intuition when $V_k$ is an infinite set, and I don't even know how to the finite case formally.
    $endgroup$
    – user392559
    Dec 8 '18 at 22:11










  • $begingroup$
    Oh, hmm, your definition of the topology on $|K|$ is nonstandard when $V_K$ is infinite. That makes it fairly annoying to prove you have a homeomorphism: it's not bad to construct a continuous bijection $|sd(K)|to |K|$, but proving that the inverse is continuous will be a pain if $V_K$ is infinite.
    $endgroup$
    – Eric Wofsey
    Dec 8 '18 at 22:38










  • $begingroup$
    (The standard topology to put on $|K|$ is the topology that makes it the colimit of all its simplices. This means that to check that the inverse map $|K|to |sd(K)|$ is continuous, you just have to check that it is continuous when restricted to each simplex of $|K|$. But this is then automatic since a continuous bijection between compact Hausdorff spaces always has continuous inverse.)
    $endgroup$
    – Eric Wofsey
    Dec 8 '18 at 22:44














1












1








1





$begingroup$


I have the following definitions:



Definition 1: A simplicial complex $K$ is a family of finite nonempty subsets of a set $V_k$ (the elements of $V_k$ are called vertices) such that:



1) if $vin V_k$, then ${v}in K$.



2) if $Sin K$ and $emptyset neq Tsubseteq S$, then $Tin K$.



The elements of $K$ are called simplexes.



Definition 2: The geometric realization of a simplicial complex $K$ is defined as $|K|= {alpha: V_K to [0,1] ; | ; alpha^{-1}((0,1]) in K , ;sum_{vin V_k} alpha(v)=1 }$.



Definition 3: The barycentric subdivision of a simplicial complex $K$ is defined as the simplicial complex $sd(K)$ such that $V_{sd(K)}=K$ and a simplex in $sd(K)$ has the form ${s_0, s_1, ...,s_q}$, where $s_i in K $ and $s_{i-1} subset s_i $, ($s_{i-1}$ is a proper subset of $s_i$ ), for all $i$.



I'm asked to show that $|sd(K)|$ is homeomorphic to $|K|$.



I already know that for $S={v_0,v_1,...,v_q} in K$, $|S|={alpha in |K| ; | ; alpha^{-1} ((0,1]) subseteq S}$ is homeomorphic to $Delta^q = {(t_0,t_1,...,t_q)in mathbb{R}^{n+1} ; | ; t_i geq 0, sum_{i=0}^{n} t_i =1}$ (actually, the homeomorphism is an isometry).



Here, the topology we are using is the one induced by the metric $d:|K|$x$|K|to mathbb{R}$ such that $d(alpha, beta) = sqrt{sum_{vin V_K} (alpha (v) - beta (v))^2}$.



Now, I'm trying to construct an homeomorphism from $|K|$ to $|sd(K)|$, but everything I can think of, just doesn't work. I would be very thankful if you could give me a hint on how to construct this homeomorphism, or if you could tell me another way to proceed on this. Thank you so much!










share|cite|improve this question









$endgroup$




I have the following definitions:



Definition 1: A simplicial complex $K$ is a family of finite nonempty subsets of a set $V_k$ (the elements of $V_k$ are called vertices) such that:



1) if $vin V_k$, then ${v}in K$.



2) if $Sin K$ and $emptyset neq Tsubseteq S$, then $Tin K$.



The elements of $K$ are called simplexes.



Definition 2: The geometric realization of a simplicial complex $K$ is defined as $|K|= {alpha: V_K to [0,1] ; | ; alpha^{-1}((0,1]) in K , ;sum_{vin V_k} alpha(v)=1 }$.



Definition 3: The barycentric subdivision of a simplicial complex $K$ is defined as the simplicial complex $sd(K)$ such that $V_{sd(K)}=K$ and a simplex in $sd(K)$ has the form ${s_0, s_1, ...,s_q}$, where $s_i in K $ and $s_{i-1} subset s_i $, ($s_{i-1}$ is a proper subset of $s_i$ ), for all $i$.



I'm asked to show that $|sd(K)|$ is homeomorphic to $|K|$.



I already know that for $S={v_0,v_1,...,v_q} in K$, $|S|={alpha in |K| ; | ; alpha^{-1} ((0,1]) subseteq S}$ is homeomorphic to $Delta^q = {(t_0,t_1,...,t_q)in mathbb{R}^{n+1} ; | ; t_i geq 0, sum_{i=0}^{n} t_i =1}$ (actually, the homeomorphism is an isometry).



Here, the topology we are using is the one induced by the metric $d:|K|$x$|K|to mathbb{R}$ such that $d(alpha, beta) = sqrt{sum_{vin V_K} (alpha (v) - beta (v))^2}$.



Now, I'm trying to construct an homeomorphism from $|K|$ to $|sd(K)|$, but everything I can think of, just doesn't work. I would be very thankful if you could give me a hint on how to construct this homeomorphism, or if you could tell me another way to proceed on this. Thank you so much!







algebraic-topology homology-cohomology simplex simplicial-complex






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asked Dec 8 '18 at 21:44









user392559user392559

37618




37618












  • $begingroup$
    You are not going to solve this by just thinking formalistically about the definitions like this. Do you have a geometric picture of what these definitions mean?
    $endgroup$
    – Eric Wofsey
    Dec 8 '18 at 21:54










  • $begingroup$
    I have intuition when $V_k$ is a finite set of points in $R^n$, because $|K|$ is going to be the convex hull of this points. I'm thinking about $V_k= {v_0, v_1, v_2}$, and $K={{v_0},{v_1},{v_2},{v_0,v_1},{v_0,v_2},{v_1,v_2} }$, then $|K|$ is a filled triangle and, for example, ${v_0,v_1}$ represents the midpoint of the line between $v_0$ and $v_1$. That's why the geometric realization of barycentric subdivision is obtained by "adding" this midpoints as vertices, and now, simplexes here are going to be the little triangles formed when taking the baricenter of this triangle.
    $endgroup$
    – user392559
    Dec 8 '18 at 22:09










  • $begingroup$
    Now, I don't really have intuition when $V_k$ is an infinite set, and I don't even know how to the finite case formally.
    $endgroup$
    – user392559
    Dec 8 '18 at 22:11










  • $begingroup$
    Oh, hmm, your definition of the topology on $|K|$ is nonstandard when $V_K$ is infinite. That makes it fairly annoying to prove you have a homeomorphism: it's not bad to construct a continuous bijection $|sd(K)|to |K|$, but proving that the inverse is continuous will be a pain if $V_K$ is infinite.
    $endgroup$
    – Eric Wofsey
    Dec 8 '18 at 22:38










  • $begingroup$
    (The standard topology to put on $|K|$ is the topology that makes it the colimit of all its simplices. This means that to check that the inverse map $|K|to |sd(K)|$ is continuous, you just have to check that it is continuous when restricted to each simplex of $|K|$. But this is then automatic since a continuous bijection between compact Hausdorff spaces always has continuous inverse.)
    $endgroup$
    – Eric Wofsey
    Dec 8 '18 at 22:44


















  • $begingroup$
    You are not going to solve this by just thinking formalistically about the definitions like this. Do you have a geometric picture of what these definitions mean?
    $endgroup$
    – Eric Wofsey
    Dec 8 '18 at 21:54










  • $begingroup$
    I have intuition when $V_k$ is a finite set of points in $R^n$, because $|K|$ is going to be the convex hull of this points. I'm thinking about $V_k= {v_0, v_1, v_2}$, and $K={{v_0},{v_1},{v_2},{v_0,v_1},{v_0,v_2},{v_1,v_2} }$, then $|K|$ is a filled triangle and, for example, ${v_0,v_1}$ represents the midpoint of the line between $v_0$ and $v_1$. That's why the geometric realization of barycentric subdivision is obtained by "adding" this midpoints as vertices, and now, simplexes here are going to be the little triangles formed when taking the baricenter of this triangle.
    $endgroup$
    – user392559
    Dec 8 '18 at 22:09










  • $begingroup$
    Now, I don't really have intuition when $V_k$ is an infinite set, and I don't even know how to the finite case formally.
    $endgroup$
    – user392559
    Dec 8 '18 at 22:11










  • $begingroup$
    Oh, hmm, your definition of the topology on $|K|$ is nonstandard when $V_K$ is infinite. That makes it fairly annoying to prove you have a homeomorphism: it's not bad to construct a continuous bijection $|sd(K)|to |K|$, but proving that the inverse is continuous will be a pain if $V_K$ is infinite.
    $endgroup$
    – Eric Wofsey
    Dec 8 '18 at 22:38










  • $begingroup$
    (The standard topology to put on $|K|$ is the topology that makes it the colimit of all its simplices. This means that to check that the inverse map $|K|to |sd(K)|$ is continuous, you just have to check that it is continuous when restricted to each simplex of $|K|$. But this is then automatic since a continuous bijection between compact Hausdorff spaces always has continuous inverse.)
    $endgroup$
    – Eric Wofsey
    Dec 8 '18 at 22:44
















$begingroup$
You are not going to solve this by just thinking formalistically about the definitions like this. Do you have a geometric picture of what these definitions mean?
$endgroup$
– Eric Wofsey
Dec 8 '18 at 21:54




$begingroup$
You are not going to solve this by just thinking formalistically about the definitions like this. Do you have a geometric picture of what these definitions mean?
$endgroup$
– Eric Wofsey
Dec 8 '18 at 21:54












$begingroup$
I have intuition when $V_k$ is a finite set of points in $R^n$, because $|K|$ is going to be the convex hull of this points. I'm thinking about $V_k= {v_0, v_1, v_2}$, and $K={{v_0},{v_1},{v_2},{v_0,v_1},{v_0,v_2},{v_1,v_2} }$, then $|K|$ is a filled triangle and, for example, ${v_0,v_1}$ represents the midpoint of the line between $v_0$ and $v_1$. That's why the geometric realization of barycentric subdivision is obtained by "adding" this midpoints as vertices, and now, simplexes here are going to be the little triangles formed when taking the baricenter of this triangle.
$endgroup$
– user392559
Dec 8 '18 at 22:09




$begingroup$
I have intuition when $V_k$ is a finite set of points in $R^n$, because $|K|$ is going to be the convex hull of this points. I'm thinking about $V_k= {v_0, v_1, v_2}$, and $K={{v_0},{v_1},{v_2},{v_0,v_1},{v_0,v_2},{v_1,v_2} }$, then $|K|$ is a filled triangle and, for example, ${v_0,v_1}$ represents the midpoint of the line between $v_0$ and $v_1$. That's why the geometric realization of barycentric subdivision is obtained by "adding" this midpoints as vertices, and now, simplexes here are going to be the little triangles formed when taking the baricenter of this triangle.
$endgroup$
– user392559
Dec 8 '18 at 22:09












$begingroup$
Now, I don't really have intuition when $V_k$ is an infinite set, and I don't even know how to the finite case formally.
$endgroup$
– user392559
Dec 8 '18 at 22:11




$begingroup$
Now, I don't really have intuition when $V_k$ is an infinite set, and I don't even know how to the finite case formally.
$endgroup$
– user392559
Dec 8 '18 at 22:11












$begingroup$
Oh, hmm, your definition of the topology on $|K|$ is nonstandard when $V_K$ is infinite. That makes it fairly annoying to prove you have a homeomorphism: it's not bad to construct a continuous bijection $|sd(K)|to |K|$, but proving that the inverse is continuous will be a pain if $V_K$ is infinite.
$endgroup$
– Eric Wofsey
Dec 8 '18 at 22:38




$begingroup$
Oh, hmm, your definition of the topology on $|K|$ is nonstandard when $V_K$ is infinite. That makes it fairly annoying to prove you have a homeomorphism: it's not bad to construct a continuous bijection $|sd(K)|to |K|$, but proving that the inverse is continuous will be a pain if $V_K$ is infinite.
$endgroup$
– Eric Wofsey
Dec 8 '18 at 22:38












$begingroup$
(The standard topology to put on $|K|$ is the topology that makes it the colimit of all its simplices. This means that to check that the inverse map $|K|to |sd(K)|$ is continuous, you just have to check that it is continuous when restricted to each simplex of $|K|$. But this is then automatic since a continuous bijection between compact Hausdorff spaces always has continuous inverse.)
$endgroup$
– Eric Wofsey
Dec 8 '18 at 22:44




$begingroup$
(The standard topology to put on $|K|$ is the topology that makes it the colimit of all its simplices. This means that to check that the inverse map $|K|to |sd(K)|$ is continuous, you just have to check that it is continuous when restricted to each simplex of $|K|$. But this is then automatic since a continuous bijection between compact Hausdorff spaces always has continuous inverse.)
$endgroup$
– Eric Wofsey
Dec 8 '18 at 22:44










1 Answer
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oldest

votes


















1












$begingroup$

Let us think of a function $alpha:V_Kto [0,1]$ as a formal sum $sum_{vin V_K}alpha(v)cdot v$. The homeomorphism $|sd(K)|to |K|$ is then given by mapping $Sin V_{sd(K)}=K$ to the formal sum $sum_{vin S}frac{1}{|S|}cdot v$, and "extending linearly". In other words, a point $alphain|sd(K)|$ can be written as a formal sum $sum_{Sin K} alpha(S)cdot S$ and then we map this formal sum to $|K|$ by replacing each $S$ with $sum_{vin S}frac{1}{|S|}cdot v$ to get the formal sum $sum_{Sin K}sum_{vin S}frac{1}{|S|}cdot v$ which we can think of as a formal sum of elements of $V_K$. I'll leave it to you to show that this is a homeomorphism. (As a hint for showing it is a bijection, given a point of $|K|$, you can figure out what simplex of $|sd(K)|$ it is in the image of by ordering its nonzero coefficients from smallest to biggest. For instance, a sum like $0.2a+0.3b+0.5c$ for three vertices $a,b,c$ can be split up as $0.6(frac{1}{3}a+frac{1}{3}b+frac{1}{3}c)+0.2(frac{1}{2}b+frac{1}{2}c)+0.2c$, to write it as a convex combination of expressions of the form $sum_{vin S}frac{1}{|S|}cdot v$.)



The intuition here is that for each geometric simplex $|S|$ in $|K|$, we add a new vertex at the barycenter of $S$, the point $sum_{vin S}frac{1}{|S|}cdot v$ that is the average of all its vertices. We can then take convex combinations of the barycenters of $|S|$ and its faces to subdivide $|S|$ into many smaller simplices. For a very simple example, suppose $K$ is just a line segment, with two vertices and an edge connecting them. Then $sd(K)$ has three vertices: two are just the vertices that $K$ had, but one of them comes from the edge. We think of this new vertex as being the midpoint of the edge of $K$, which we can connect to the two vertices of $K$ to split the edge into two smaller edges. So, $|sd(K)|$ is just two line segments joined together, which is homeomorphic to a single line segment $|K|$.






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    $begingroup$

    Let us think of a function $alpha:V_Kto [0,1]$ as a formal sum $sum_{vin V_K}alpha(v)cdot v$. The homeomorphism $|sd(K)|to |K|$ is then given by mapping $Sin V_{sd(K)}=K$ to the formal sum $sum_{vin S}frac{1}{|S|}cdot v$, and "extending linearly". In other words, a point $alphain|sd(K)|$ can be written as a formal sum $sum_{Sin K} alpha(S)cdot S$ and then we map this formal sum to $|K|$ by replacing each $S$ with $sum_{vin S}frac{1}{|S|}cdot v$ to get the formal sum $sum_{Sin K}sum_{vin S}frac{1}{|S|}cdot v$ which we can think of as a formal sum of elements of $V_K$. I'll leave it to you to show that this is a homeomorphism. (As a hint for showing it is a bijection, given a point of $|K|$, you can figure out what simplex of $|sd(K)|$ it is in the image of by ordering its nonzero coefficients from smallest to biggest. For instance, a sum like $0.2a+0.3b+0.5c$ for three vertices $a,b,c$ can be split up as $0.6(frac{1}{3}a+frac{1}{3}b+frac{1}{3}c)+0.2(frac{1}{2}b+frac{1}{2}c)+0.2c$, to write it as a convex combination of expressions of the form $sum_{vin S}frac{1}{|S|}cdot v$.)



    The intuition here is that for each geometric simplex $|S|$ in $|K|$, we add a new vertex at the barycenter of $S$, the point $sum_{vin S}frac{1}{|S|}cdot v$ that is the average of all its vertices. We can then take convex combinations of the barycenters of $|S|$ and its faces to subdivide $|S|$ into many smaller simplices. For a very simple example, suppose $K$ is just a line segment, with two vertices and an edge connecting them. Then $sd(K)$ has three vertices: two are just the vertices that $K$ had, but one of them comes from the edge. We think of this new vertex as being the midpoint of the edge of $K$, which we can connect to the two vertices of $K$ to split the edge into two smaller edges. So, $|sd(K)|$ is just two line segments joined together, which is homeomorphic to a single line segment $|K|$.






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      1












      $begingroup$

      Let us think of a function $alpha:V_Kto [0,1]$ as a formal sum $sum_{vin V_K}alpha(v)cdot v$. The homeomorphism $|sd(K)|to |K|$ is then given by mapping $Sin V_{sd(K)}=K$ to the formal sum $sum_{vin S}frac{1}{|S|}cdot v$, and "extending linearly". In other words, a point $alphain|sd(K)|$ can be written as a formal sum $sum_{Sin K} alpha(S)cdot S$ and then we map this formal sum to $|K|$ by replacing each $S$ with $sum_{vin S}frac{1}{|S|}cdot v$ to get the formal sum $sum_{Sin K}sum_{vin S}frac{1}{|S|}cdot v$ which we can think of as a formal sum of elements of $V_K$. I'll leave it to you to show that this is a homeomorphism. (As a hint for showing it is a bijection, given a point of $|K|$, you can figure out what simplex of $|sd(K)|$ it is in the image of by ordering its nonzero coefficients from smallest to biggest. For instance, a sum like $0.2a+0.3b+0.5c$ for three vertices $a,b,c$ can be split up as $0.6(frac{1}{3}a+frac{1}{3}b+frac{1}{3}c)+0.2(frac{1}{2}b+frac{1}{2}c)+0.2c$, to write it as a convex combination of expressions of the form $sum_{vin S}frac{1}{|S|}cdot v$.)



      The intuition here is that for each geometric simplex $|S|$ in $|K|$, we add a new vertex at the barycenter of $S$, the point $sum_{vin S}frac{1}{|S|}cdot v$ that is the average of all its vertices. We can then take convex combinations of the barycenters of $|S|$ and its faces to subdivide $|S|$ into many smaller simplices. For a very simple example, suppose $K$ is just a line segment, with two vertices and an edge connecting them. Then $sd(K)$ has three vertices: two are just the vertices that $K$ had, but one of them comes from the edge. We think of this new vertex as being the midpoint of the edge of $K$, which we can connect to the two vertices of $K$ to split the edge into two smaller edges. So, $|sd(K)|$ is just two line segments joined together, which is homeomorphic to a single line segment $|K|$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Let us think of a function $alpha:V_Kto [0,1]$ as a formal sum $sum_{vin V_K}alpha(v)cdot v$. The homeomorphism $|sd(K)|to |K|$ is then given by mapping $Sin V_{sd(K)}=K$ to the formal sum $sum_{vin S}frac{1}{|S|}cdot v$, and "extending linearly". In other words, a point $alphain|sd(K)|$ can be written as a formal sum $sum_{Sin K} alpha(S)cdot S$ and then we map this formal sum to $|K|$ by replacing each $S$ with $sum_{vin S}frac{1}{|S|}cdot v$ to get the formal sum $sum_{Sin K}sum_{vin S}frac{1}{|S|}cdot v$ which we can think of as a formal sum of elements of $V_K$. I'll leave it to you to show that this is a homeomorphism. (As a hint for showing it is a bijection, given a point of $|K|$, you can figure out what simplex of $|sd(K)|$ it is in the image of by ordering its nonzero coefficients from smallest to biggest. For instance, a sum like $0.2a+0.3b+0.5c$ for three vertices $a,b,c$ can be split up as $0.6(frac{1}{3}a+frac{1}{3}b+frac{1}{3}c)+0.2(frac{1}{2}b+frac{1}{2}c)+0.2c$, to write it as a convex combination of expressions of the form $sum_{vin S}frac{1}{|S|}cdot v$.)



        The intuition here is that for each geometric simplex $|S|$ in $|K|$, we add a new vertex at the barycenter of $S$, the point $sum_{vin S}frac{1}{|S|}cdot v$ that is the average of all its vertices. We can then take convex combinations of the barycenters of $|S|$ and its faces to subdivide $|S|$ into many smaller simplices. For a very simple example, suppose $K$ is just a line segment, with two vertices and an edge connecting them. Then $sd(K)$ has three vertices: two are just the vertices that $K$ had, but one of them comes from the edge. We think of this new vertex as being the midpoint of the edge of $K$, which we can connect to the two vertices of $K$ to split the edge into two smaller edges. So, $|sd(K)|$ is just two line segments joined together, which is homeomorphic to a single line segment $|K|$.






        share|cite|improve this answer









        $endgroup$



        Let us think of a function $alpha:V_Kto [0,1]$ as a formal sum $sum_{vin V_K}alpha(v)cdot v$. The homeomorphism $|sd(K)|to |K|$ is then given by mapping $Sin V_{sd(K)}=K$ to the formal sum $sum_{vin S}frac{1}{|S|}cdot v$, and "extending linearly". In other words, a point $alphain|sd(K)|$ can be written as a formal sum $sum_{Sin K} alpha(S)cdot S$ and then we map this formal sum to $|K|$ by replacing each $S$ with $sum_{vin S}frac{1}{|S|}cdot v$ to get the formal sum $sum_{Sin K}sum_{vin S}frac{1}{|S|}cdot v$ which we can think of as a formal sum of elements of $V_K$. I'll leave it to you to show that this is a homeomorphism. (As a hint for showing it is a bijection, given a point of $|K|$, you can figure out what simplex of $|sd(K)|$ it is in the image of by ordering its nonzero coefficients from smallest to biggest. For instance, a sum like $0.2a+0.3b+0.5c$ for three vertices $a,b,c$ can be split up as $0.6(frac{1}{3}a+frac{1}{3}b+frac{1}{3}c)+0.2(frac{1}{2}b+frac{1}{2}c)+0.2c$, to write it as a convex combination of expressions of the form $sum_{vin S}frac{1}{|S|}cdot v$.)



        The intuition here is that for each geometric simplex $|S|$ in $|K|$, we add a new vertex at the barycenter of $S$, the point $sum_{vin S}frac{1}{|S|}cdot v$ that is the average of all its vertices. We can then take convex combinations of the barycenters of $|S|$ and its faces to subdivide $|S|$ into many smaller simplices. For a very simple example, suppose $K$ is just a line segment, with two vertices and an edge connecting them. Then $sd(K)$ has three vertices: two are just the vertices that $K$ had, but one of them comes from the edge. We think of this new vertex as being the midpoint of the edge of $K$, which we can connect to the two vertices of $K$ to split the edge into two smaller edges. So, $|sd(K)|$ is just two line segments joined together, which is homeomorphic to a single line segment $|K|$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 '18 at 22:17









        Eric WofseyEric Wofsey

        185k14214341




        185k14214341






























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