Simplicial set represented by an (unordered) setSimplicial space and its simplicial replacement?Hodge star...
Simplicial set represented by an (unordered) set
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$begingroup$
Let $X$ be a (finite if you want) set and form the simplicial set $F^{bullet}(X)$ with
$$
F^{n}(X) = mathrm{Hom}_{mathrm{set}} ([n], X)
$$
where the right hand side denotes arbitrary maps of sets (of course
it wouldn't make sense to say order preserving as $X$ doesn't come
with an order).
I'm wondering about a description of $F^{bullet}(X)$. For example if $X = {0,1}$ then there are 2 0-simplices, may as well call them $[0] and [1]$ and 2 1-simplices $[0, 1]$ and $[1,0]$ glued together to form a copy of $S^1$.
Edit: as pointed out in Goodwillie's answer, this is not the end of the story, there are way more higher dimensional non-degenerate simplices.
Is there an analogous description when $X = {0, 1, 2}$?
A closely related question is whether there's a right adjoint to the
forgetful functor from the simplex category $Delta$ (finite ordered
sets) to, say, finite (unordered) sets -- and if so what is it.
Example where such simplicial sets arise: given a map of topological spaces $f: X
to Y$ we can always form a
simplicial object $mathcal{S}^{bullet}(f)$ with
$$
mathcal{S}^{n} = prodnolimits_{X}^{n} = underbrace{X times_{Y}
cdots times_{Y} X}_{ntext{ times }}
$$
with face and degeneracy maps given by projections and diagonals
respectively. Taking connected components gives a simplicial set.
When $Y$ is the union $bigcup_{i=1}^{N} H_{i}$ of the coordinate
hyperplanes in $mathbb{C}^{N}$ and $f: X=coprod_{i=1}^{N} H_{i} to
bigcup_{i=1}^{N} H_{i}=Y$ is the obvious map, I believe the simplicial
set we get is $F^{bullet}({1, dots, n})$.
ag.algebraic-geometry at.algebraic-topology ct.category-theory simplicial-stuff hyperplane-arrangements
asked 5 hours ago
cgodfreycgodfrey
35819
$endgroup$
add a comment |
$begingroup$
Let $X$ be a (finite if you want) set and form the simplicial set $F^{bullet}(X)$ with
$$
F^{n}(X) = mathrm{Hom}_{mathrm{set}} ([n], X)
$$
where the right hand side denotes arbitrary maps of sets (of course
it wouldn't make sense to say order preserving as $X$ doesn't come
with an order).
I'm wondering about a description of $F^{bullet}(X)$. For example if $X = {0,1}$ then there are 2 0-simplices, may as well call them $[0] and [1]$ and 2 1-simplices $[0, 1]$ and $[1,0]$ glued together to form a copy of $S^1$.
Edit: as pointed out in Goodwillie's answer, this is not the end of the story, there are way more higher dimensional non-degenerate simplices.
Is there an analogous description when $X = {0, 1, 2}$?
A closely related question is whether there's a right adjoint to the
forgetful functor from the simplex category $Delta$ (finite ordered
sets) to, say, finite (unordered) sets -- and if so what is it.
Example where such simplicial sets arise: given a map of topological spaces $f: X
to Y$ we can always form a
simplicial object $mathcal{S}^{bullet}(f)$ with
$$
mathcal{S}^{n} = prodnolimits_{X}^{n} = underbrace{X times_{Y}
cdots times_{Y} X}_{ntext{ times }}
$$
with face and degeneracy maps given by projections and diagonals
respectively. Taking connected components gives a simplicial set.
When $Y$ is the union $bigcup_{i=1}^{N} H_{i}$ of the coordinate
hyperplanes in $mathbb{C}^{N}$ and $f: X=coprod_{i=1}^{N} H_{i} to
bigcup_{i=1}^{N} H_{i}=Y$ is the obvious map, I believe the simplicial
set we get is $F^{bullet}({1, dots, n})$.
ag.algebraic-geometry at.algebraic-topology ct.category-theory simplicial-stuff hyperplane-arrangements
asked 5 hours ago
cgodfreycgodfrey
35819
$endgroup$
add a comment |
$begingroup$
Let $X$ be a (finite if you want) set and form the simplicial set $F^{bullet}(X)$ with
$$
F^{n}(X) = mathrm{Hom}_{mathrm{set}} ([n], X)
$$
where the right hand side denotes arbitrary maps of sets (of course
it wouldn't make sense to say order preserving as $X$ doesn't come
with an order).
I'm wondering about a description of $F^{bullet}(X)$. For example if $X = {0,1}$ then there are 2 0-simplices, may as well call them $[0] and [1]$ and 2 1-simplices $[0, 1]$ and $[1,0]$ glued together to form a copy of $S^1$.
Edit: as pointed out in Goodwillie's answer, this is not the end of the story, there are way more higher dimensional non-degenerate simplices.
Is there an analogous description when $X = {0, 1, 2}$?
A closely related question is whether there's a right adjoint to the
forgetful functor from the simplex category $Delta$ (finite ordered
sets) to, say, finite (unordered) sets -- and if so what is it.
Example where such simplicial sets arise: given a map of topological spaces $f: X
to Y$ we can always form a
simplicial object $mathcal{S}^{bullet}(f)$ with
$$
mathcal{S}^{n} = prodnolimits_{X}^{n} = underbrace{X times_{Y}
cdots times_{Y} X}_{ntext{ times }}
$$
with face and degeneracy maps given by projections and diagonals
respectively. Taking connected components gives a simplicial set.
When $Y$ is the union $bigcup_{i=1}^{N} H_{i}$ of the coordinate
hyperplanes in $mathbb{C}^{N}$ and $f: X=coprod_{i=1}^{N} H_{i} to
bigcup_{i=1}^{N} H_{i}=Y$ is the obvious map, I believe the simplicial
set we get is $F^{bullet}({1, dots, n})$.
ag.algebraic-geometry at.algebraic-topology ct.category-theory simplicial-stuff hyperplane-arrangements
asked 5 hours ago
cgodfreycgodfrey
35819
$endgroup$
Let $X$ be a (finite if you want) set and form the simplicial set $F^{bullet}(X)$ with
$$
F^{n}(X) = mathrm{Hom}_{mathrm{set}} ([n], X)
$$
where the right hand side denotes arbitrary maps of sets (of course
it wouldn't make sense to say order preserving as $X$ doesn't come
with an order).
I'm wondering about a description of $F^{bullet}(X)$. For example if $X = {0,1}$ then there are 2 0-simplices, may as well call them $[0] and [1]$ and 2 1-simplices $[0, 1]$ and $[1,0]$ glued together to form a copy of $S^1$.
Edit: as pointed out in Goodwillie's answer, this is not the end of the story, there are way more higher dimensional non-degenerate simplices.
Is there an analogous description when $X = {0, 1, 2}$?
A closely related question is whether there's a right adjoint to the
forgetful functor from the simplex category $Delta$ (finite ordered
sets) to, say, finite (unordered) sets -- and if so what is it.
Example where such simplicial sets arise: given a map of topological spaces $f: X
to Y$ we can always form a
simplicial object $mathcal{S}^{bullet}(f)$ with
$$
mathcal{S}^{n} = prodnolimits_{X}^{n} = underbrace{X times_{Y}
cdots times_{Y} X}_{ntext{ times }}
$$
with face and degeneracy maps given by projections and diagonals
respectively. Taking connected components gives a simplicial set.
When $Y$ is the union $bigcup_{i=1}^{N} H_{i}$ of the coordinate
hyperplanes in $mathbb{C}^{N}$ and $f: X=coprod_{i=1}^{N} H_{i} to
bigcup_{i=1}^{N} H_{i}=Y$ is the obvious map, I believe the simplicial
set we get is $F^{bullet}({1, dots, n})$.
ag.algebraic-geometry at.algebraic-topology ct.category-theory simplicial-stuff hyperplane-arrangements
ag.algebraic-geometry at.algebraic-topology ct.category-theory simplicial-stuff hyperplane-arrangements
asked 5 hours ago
cgodfreycgodfrey
35819
asked 5 hours ago
cgodfreycgodfrey
35819
edited 4 hours ago
cgodfrey
asked 5 hours ago
cgodfreycgodfrey
35819
asked 5 hours ago
cgodfreycgodfrey
35819
asked 5 hours ago
cgodfreycgodfrey
35819
35819
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You are overlooking some nondegenerate simplices. For example, when $X={0,1}$ there are the $2$-cells $[0,1,0]$ and $[1,0,1]$. In fact, the thing you call $F^bullet(X)$ is infinite dimensional if $X$ has more than one element.
It is contractible whenever $X$ is non-empty; this can be seen by identifying it with the nerve of a category, a category equivalent to the point category with one morphism.
If $X=G$ has a group structure then $F^bullet(G)$ is often called $EG$; it is a contractible space with free $G$-action.
answered 4 hours ago
Tom GoodwillieTom Goodwillie
40.5k3111201
$endgroup$
$begingroup$
Ah! Absolutely overlooked those, many thanks. Quick follow up: as an intermediate step, can we view $F^bullet(X)$ as the nerve of the category with objects the points $x in X$ and with a unique morphism $x to y$ for every 2 points $x, y in X$ (this is at least the case when $X= G$ a group and we build $EG$ as the nerve of the Cayley graph I think...)? Then including a point $x in X$ with its identity morphism would give an equivalence of categories ...
$endgroup$
– cgodfrey
4 hours ago
add a comment |
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1 Answer
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$begingroup$
You are overlooking some nondegenerate simplices. For example, when $X={0,1}$ there are the $2$-cells $[0,1,0]$ and $[1,0,1]$. In fact, the thing you call $F^bullet(X)$ is infinite dimensional if $X$ has more than one element.
It is contractible whenever $X$ is non-empty; this can be seen by identifying it with the nerve of a category, a category equivalent to the point category with one morphism.
If $X=G$ has a group structure then $F^bullet(G)$ is often called $EG$; it is a contractible space with free $G$-action.
answered 4 hours ago
Tom GoodwillieTom Goodwillie
40.5k3111201
$endgroup$
$begingroup$
Ah! Absolutely overlooked those, many thanks. Quick follow up: as an intermediate step, can we view $F^bullet(X)$ as the nerve of the category with objects the points $x in X$ and with a unique morphism $x to y$ for every 2 points $x, y in X$ (this is at least the case when $X= G$ a group and we build $EG$ as the nerve of the Cayley graph I think...)? Then including a point $x in X$ with its identity morphism would give an equivalence of categories ...
$endgroup$
– cgodfrey
4 hours ago
add a comment |
$begingroup$
You are overlooking some nondegenerate simplices. For example, when $X={0,1}$ there are the $2$-cells $[0,1,0]$ and $[1,0,1]$. In fact, the thing you call $F^bullet(X)$ is infinite dimensional if $X$ has more than one element.
It is contractible whenever $X$ is non-empty; this can be seen by identifying it with the nerve of a category, a category equivalent to the point category with one morphism.
If $X=G$ has a group structure then $F^bullet(G)$ is often called $EG$; it is a contractible space with free $G$-action.
answered 4 hours ago
Tom GoodwillieTom Goodwillie
40.5k3111201
$endgroup$
$begingroup$
Ah! Absolutely overlooked those, many thanks. Quick follow up: as an intermediate step, can we view $F^bullet(X)$ as the nerve of the category with objects the points $x in X$ and with a unique morphism $x to y$ for every 2 points $x, y in X$ (this is at least the case when $X= G$ a group and we build $EG$ as the nerve of the Cayley graph I think...)? Then including a point $x in X$ with its identity morphism would give an equivalence of categories ...
$endgroup$
– cgodfrey
4 hours ago
add a comment |
$begingroup$
You are overlooking some nondegenerate simplices. For example, when $X={0,1}$ there are the $2$-cells $[0,1,0]$ and $[1,0,1]$. In fact, the thing you call $F^bullet(X)$ is infinite dimensional if $X$ has more than one element.
It is contractible whenever $X$ is non-empty; this can be seen by identifying it with the nerve of a category, a category equivalent to the point category with one morphism.
If $X=G$ has a group structure then $F^bullet(G)$ is often called $EG$; it is a contractible space with free $G$-action.
answered 4 hours ago
Tom GoodwillieTom Goodwillie
40.5k3111201
$endgroup$
You are overlooking some nondegenerate simplices. For example, when $X={0,1}$ there are the $2$-cells $[0,1,0]$ and $[1,0,1]$. In fact, the thing you call $F^bullet(X)$ is infinite dimensional if $X$ has more than one element.
It is contractible whenever $X$ is non-empty; this can be seen by identifying it with the nerve of a category, a category equivalent to the point category with one morphism.
If $X=G$ has a group structure then $F^bullet(G)$ is often called $EG$; it is a contractible space with free $G$-action.
answered 4 hours ago
Tom GoodwillieTom Goodwillie
40.5k3111201
answered 4 hours ago
Tom GoodwillieTom Goodwillie
40.5k3111201
answered 4 hours ago
Tom GoodwillieTom Goodwillie
40.5k3111201
answered 4 hours ago
Tom GoodwillieTom Goodwillie
40.5k3111201
40.5k3111201
$begingroup$
Ah! Absolutely overlooked those, many thanks. Quick follow up: as an intermediate step, can we view $F^bullet(X)$ as the nerve of the category with objects the points $x in X$ and with a unique morphism $x to y$ for every 2 points $x, y in X$ (this is at least the case when $X= G$ a group and we build $EG$ as the nerve of the Cayley graph I think...)? Then including a point $x in X$ with its identity morphism would give an equivalence of categories ...
$endgroup$
– cgodfrey
4 hours ago
add a comment |
$begingroup$
Ah! Absolutely overlooked those, many thanks. Quick follow up: as an intermediate step, can we view $F^bullet(X)$ as the nerve of the category with objects the points $x in X$ and with a unique morphism $x to y$ for every 2 points $x, y in X$ (this is at least the case when $X= G$ a group and we build $EG$ as the nerve of the Cayley graph I think...)? Then including a point $x in X$ with its identity morphism would give an equivalence of categories ...
$endgroup$
– cgodfrey
4 hours ago
$begingroup$
Ah! Absolutely overlooked those, many thanks. Quick follow up: as an intermediate step, can we view $F^bullet(X)$ as the nerve of the category with objects the points $x in X$ and with a unique morphism $x to y$ for every 2 points $x, y in X$ (this is at least the case when $X= G$ a group and we build $EG$ as the nerve of the Cayley graph I think...)? Then including a point $x in X$ with its identity morphism would give an equivalence of categories ...
$endgroup$
– cgodfrey
4 hours ago
$begingroup$
Ah! Absolutely overlooked those, many thanks. Quick follow up: as an intermediate step, can we view $F^bullet(X)$ as the nerve of the category with objects the points $x in X$ and with a unique morphism $x to y$ for every 2 points $x, y in X$ (this is at least the case when $X= G$ a group and we build $EG$ as the nerve of the Cayley graph I think...)? Then including a point $x in X$ with its identity morphism would give an equivalence of categories ...
$endgroup$
– cgodfrey
4 hours ago
add a comment |
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