Are the higher homotopy groups of a compact manifold finitely generated as $mathbb{Z}[pi_1]$-modules?
$begingroup$
Let $M$ be a compact manifold. The homology and cohomology groups of $M$ are necessarily finitely generated, as is the fundamental group. Serre proved that a simply connected finite CW complex has finitely generated homotopy groups, so if $M$ is simply connected, or has finite fundamental group, then the higher homotopy groups of $M$ are also finitely generated. Without this restriction on the fundamental group though, the conclusion no longer holds.
Let $M = T^4#mathbb{CP}^2$. The universal cover of $M$ is the connected sum of $mathbb{R}^4$ with a $mathbb{CP}^2$ at each point of the lattice $mathbb{Z}^4$; denote this manifold by $widetilde{M}$. Then we see that
$$pi_2(M) cong pi_2(widetilde{M}) cong H_2(widetilde{M}; mathbb{Z}) cong H_2(mathbb{R}^4; mathbb{Z})oplusbigoplus_{mathbb{Z}^4}H_2(mathbb{CP}^2; mathbb{Z}) cong bigoplus_{mathbb{Z}^4}mathbb{Z}$$
which is not a finitely generated group.
Recall that $pi_1(M)$ acts on $pi_n(M)$ by identifying $pi_1(M)$ with the deck transformations of $widetilde{M} to M$, so $pi_n(M)$ has the structure of a $mathbb{Z}[pi_1(M)]$-module. In the example above, the deck transformations act by translations, moving from one copy of $mathbb{CP}^2$ to another. So we see that $pi_2(M)$ is actually a free, rank one $mathbb{Z}[pi_1(M)]$-module - in particular, it is a finitely generated $mathbb{Z}[pi_1(M)]$-module. Is this always the case?
Let $M$ be a compact manifold. Is $pi_n(M)$ a finitely generated $mathbb{Z}[pi_1(M)]$-module for every $n geq 2$?
algebraic-topology higher-homotopy-groups
asked Dec 19 '18 at 18:49
Michael AlbaneseMichael Albanese
63.6k1599307
$endgroup$
add a comment |
$begingroup$
Let $M$ be a compact manifold. The homology and cohomology groups of $M$ are necessarily finitely generated, as is the fundamental group. Serre proved that a simply connected finite CW complex has finitely generated homotopy groups, so if $M$ is simply connected, or has finite fundamental group, then the higher homotopy groups of $M$ are also finitely generated. Without this restriction on the fundamental group though, the conclusion no longer holds.
Let $M = T^4#mathbb{CP}^2$. The universal cover of $M$ is the connected sum of $mathbb{R}^4$ with a $mathbb{CP}^2$ at each point of the lattice $mathbb{Z}^4$; denote this manifold by $widetilde{M}$. Then we see that
$$pi_2(M) cong pi_2(widetilde{M}) cong H_2(widetilde{M}; mathbb{Z}) cong H_2(mathbb{R}^4; mathbb{Z})oplusbigoplus_{mathbb{Z}^4}H_2(mathbb{CP}^2; mathbb{Z}) cong bigoplus_{mathbb{Z}^4}mathbb{Z}$$
which is not a finitely generated group.
Recall that $pi_1(M)$ acts on $pi_n(M)$ by identifying $pi_1(M)$ with the deck transformations of $widetilde{M} to M$, so $pi_n(M)$ has the structure of a $mathbb{Z}[pi_1(M)]$-module. In the example above, the deck transformations act by translations, moving from one copy of $mathbb{CP}^2$ to another. So we see that $pi_2(M)$ is actually a free, rank one $mathbb{Z}[pi_1(M)]$-module - in particular, it is a finitely generated $mathbb{Z}[pi_1(M)]$-module. Is this always the case?
Let $M$ be a compact manifold. Is $pi_n(M)$ a finitely generated $mathbb{Z}[pi_1(M)]$-module for every $n geq 2$?
algebraic-topology higher-homotopy-groups
asked Dec 19 '18 at 18:49
Michael AlbaneseMichael Albanese
63.6k1599307
$endgroup$
1
$begingroup$
This is false (I remember finding an article about an example somewhere between dimensions 6 or 8), but I don't know the reference. I'll try digging, but I figured it might help the search to know the result (assuming you trust my memory).
$endgroup$
– user98602
Dec 19 '18 at 19:48
1
$begingroup$
Remark: $pi_2 M$ is always finitely generated as a $pi_1$-module so long as the group $pi_1$ is Noetherian (for instance, finitely generated abelian). For one has the natural isomorphism $pi_2 M cong H_2 tilde M$ of $pi_1$-modules, and if one calculates the latter using a cellular chain complex corresponding to the universal cover of a cell decomposition of the compact manifold $M$, the chain groups themselves are finitely generated $pi_1$-modules. The subcomplex of cycles is finitely generated by the Noetherian hypothesis and the quotient is then clearly finitely generated.
$endgroup$
– user98602
Dec 20 '18 at 2:12
add a comment |
$begingroup$
Let $M$ be a compact manifold. The homology and cohomology groups of $M$ are necessarily finitely generated, as is the fundamental group. Serre proved that a simply connected finite CW complex has finitely generated homotopy groups, so if $M$ is simply connected, or has finite fundamental group, then the higher homotopy groups of $M$ are also finitely generated. Without this restriction on the fundamental group though, the conclusion no longer holds.
Let $M = T^4#mathbb{CP}^2$. The universal cover of $M$ is the connected sum of $mathbb{R}^4$ with a $mathbb{CP}^2$ at each point of the lattice $mathbb{Z}^4$; denote this manifold by $widetilde{M}$. Then we see that
$$pi_2(M) cong pi_2(widetilde{M}) cong H_2(widetilde{M}; mathbb{Z}) cong H_2(mathbb{R}^4; mathbb{Z})oplusbigoplus_{mathbb{Z}^4}H_2(mathbb{CP}^2; mathbb{Z}) cong bigoplus_{mathbb{Z}^4}mathbb{Z}$$
which is not a finitely generated group.
Recall that $pi_1(M)$ acts on $pi_n(M)$ by identifying $pi_1(M)$ with the deck transformations of $widetilde{M} to M$, so $pi_n(M)$ has the structure of a $mathbb{Z}[pi_1(M)]$-module. In the example above, the deck transformations act by translations, moving from one copy of $mathbb{CP}^2$ to another. So we see that $pi_2(M)$ is actually a free, rank one $mathbb{Z}[pi_1(M)]$-module - in particular, it is a finitely generated $mathbb{Z}[pi_1(M)]$-module. Is this always the case?
Let $M$ be a compact manifold. Is $pi_n(M)$ a finitely generated $mathbb{Z}[pi_1(M)]$-module for every $n geq 2$?
algebraic-topology higher-homotopy-groups
asked Dec 19 '18 at 18:49
Michael AlbaneseMichael Albanese
63.6k1599307
$endgroup$
Let $M$ be a compact manifold. The homology and cohomology groups of $M$ are necessarily finitely generated, as is the fundamental group. Serre proved that a simply connected finite CW complex has finitely generated homotopy groups, so if $M$ is simply connected, or has finite fundamental group, then the higher homotopy groups of $M$ are also finitely generated. Without this restriction on the fundamental group though, the conclusion no longer holds.
Let $M = T^4#mathbb{CP}^2$. The universal cover of $M$ is the connected sum of $mathbb{R}^4$ with a $mathbb{CP}^2$ at each point of the lattice $mathbb{Z}^4$; denote this manifold by $widetilde{M}$. Then we see that
$$pi_2(M) cong pi_2(widetilde{M}) cong H_2(widetilde{M}; mathbb{Z}) cong H_2(mathbb{R}^4; mathbb{Z})oplusbigoplus_{mathbb{Z}^4}H_2(mathbb{CP}^2; mathbb{Z}) cong bigoplus_{mathbb{Z}^4}mathbb{Z}$$
which is not a finitely generated group.
Recall that $pi_1(M)$ acts on $pi_n(M)$ by identifying $pi_1(M)$ with the deck transformations of $widetilde{M} to M$, so $pi_n(M)$ has the structure of a $mathbb{Z}[pi_1(M)]$-module. In the example above, the deck transformations act by translations, moving from one copy of $mathbb{CP}^2$ to another. So we see that $pi_2(M)$ is actually a free, rank one $mathbb{Z}[pi_1(M)]$-module - in particular, it is a finitely generated $mathbb{Z}[pi_1(M)]$-module. Is this always the case?
Let $M$ be a compact manifold. Is $pi_n(M)$ a finitely generated $mathbb{Z}[pi_1(M)]$-module for every $n geq 2$?
algebraic-topology higher-homotopy-groups
algebraic-topology higher-homotopy-groups
asked Dec 19 '18 at 18:49
Michael AlbaneseMichael Albanese
63.6k1599307
asked Dec 19 '18 at 18:49
Michael AlbaneseMichael Albanese
63.6k1599307
asked Dec 19 '18 at 18:49
Michael AlbaneseMichael Albanese
63.6k1599307
asked Dec 19 '18 at 18:49
Michael AlbaneseMichael Albanese
63.6k1599307
asked Dec 19 '18 at 18:49
Michael AlbaneseMichael Albanese
63.6k1599307
63.6k1599307
1
$begingroup$
This is false (I remember finding an article about an example somewhere between dimensions 6 or 8), but I don't know the reference. I'll try digging, but I figured it might help the search to know the result (assuming you trust my memory).
$endgroup$
– user98602
Dec 19 '18 at 19:48
1
$begingroup$
Remark: $pi_2 M$ is always finitely generated as a $pi_1$-module so long as the group $pi_1$ is Noetherian (for instance, finitely generated abelian). For one has the natural isomorphism $pi_2 M cong H_2 tilde M$ of $pi_1$-modules, and if one calculates the latter using a cellular chain complex corresponding to the universal cover of a cell decomposition of the compact manifold $M$, the chain groups themselves are finitely generated $pi_1$-modules. The subcomplex of cycles is finitely generated by the Noetherian hypothesis and the quotient is then clearly finitely generated.
$endgroup$
– user98602
Dec 20 '18 at 2:12
add a comment |
1
$begingroup$
This is false (I remember finding an article about an example somewhere between dimensions 6 or 8), but I don't know the reference. I'll try digging, but I figured it might help the search to know the result (assuming you trust my memory).
$endgroup$
– user98602
Dec 19 '18 at 19:48
1
$begingroup$
Remark: $pi_2 M$ is always finitely generated as a $pi_1$-module so long as the group $pi_1$ is Noetherian (for instance, finitely generated abelian). For one has the natural isomorphism $pi_2 M cong H_2 tilde M$ of $pi_1$-modules, and if one calculates the latter using a cellular chain complex corresponding to the universal cover of a cell decomposition of the compact manifold $M$, the chain groups themselves are finitely generated $pi_1$-modules. The subcomplex of cycles is finitely generated by the Noetherian hypothesis and the quotient is then clearly finitely generated.
$endgroup$
– user98602
Dec 20 '18 at 2:12
1
1
$begingroup$
This is false (I remember finding an article about an example somewhere between dimensions 6 or 8), but I don't know the reference. I'll try digging, but I figured it might help the search to know the result (assuming you trust my memory).
$endgroup$
– user98602
Dec 19 '18 at 19:48
$begingroup$
This is false (I remember finding an article about an example somewhere between dimensions 6 or 8), but I don't know the reference. I'll try digging, but I figured it might help the search to know the result (assuming you trust my memory).
$endgroup$
– user98602
Dec 19 '18 at 19:48
1
1
$begingroup$
Remark: $pi_2 M$ is always finitely generated as a $pi_1$-module so long as the group $pi_1$ is Noetherian (for instance, finitely generated abelian). For one has the natural isomorphism $pi_2 M cong H_2 tilde M$ of $pi_1$-modules, and if one calculates the latter using a cellular chain complex corresponding to the universal cover of a cell decomposition of the compact manifold $M$, the chain groups themselves are finitely generated $pi_1$-modules. The subcomplex of cycles is finitely generated by the Noetherian hypothesis and the quotient is then clearly finitely generated.
$endgroup$
– user98602
Dec 20 '18 at 2:12
$begingroup$
Remark: $pi_2 M$ is always finitely generated as a $pi_1$-module so long as the group $pi_1$ is Noetherian (for instance, finitely generated abelian). For one has the natural isomorphism $pi_2 M cong H_2 tilde M$ of $pi_1$-modules, and if one calculates the latter using a cellular chain complex corresponding to the universal cover of a cell decomposition of the compact manifold $M$, the chain groups themselves are finitely generated $pi_1$-modules. The subcomplex of cycles is finitely generated by the Noetherian hypothesis and the quotient is then clearly finitely generated.
$endgroup$
– user98602
Dec 20 '18 at 2:12
add a comment |
1 Answer
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Not necessarily. Couldn't find that reference I mentioned, though.
$require{AMScd}$
The CW-complex model for this phenomenon can be taken as $pi_3(S^1 vee S^2)$. Then this is the same as $pi_3(vee_{Bbb Z} S^2)$; in the case of finitely many wedge factors, this may be identified as the space of symmetric matrices, and so this group may be identified as finitely supported symmetric matrices drawn on the grid $Bbb Z times Bbb Z$. The action of $pi_1$ is the right-shift on $Bbb Z$, but quotienting by this action, the group is still isomorphic to $Bbb Z^infty$, the isomorphism given by taking the sum over each line $x + y = n$ for $n geq 0$. The extra terms come from Whitehead products of the two factors (for instance, if $iota_2$ is the identity map on the 2-sphere, the bracket $[iota_2, iota_2] in pi_3(S^2vee S^2)$ is the map whose adjunction space gives $S^2 times S^2$, and this is distinct from the spaces you would get from $pi_3 S^2 oplus pi_3 S^2$.)
If all you want is a compact manifold, possibly with boundary, we are essentially finished: any CW complex of dimension $k$ embeds in $Bbb R^N$ for large $N$ so that it has a regular neighborhood (here taken to mean codimension 0 compact manifold with boundary containing the image of the CW complex which deformation retracts onto the complex). But in particular you could just take a copy of $S^2 times S^1$ and delete a small open ball.
The rest of this answer assumes you really care about closed manifolds.
Following your intuition to take the $S^1 vee S^2$ idea and make it into a question about tori instead (most relevantly: a place where the universal cover splits as a comprehensible infinite wedge of the same space, in a way which is geometrically clear), here we should instead take $M = (S^1 times S^2) # (S^1 times S^2)$. You may identify the universal cover as what it would look like to take the boundary of a regular neighborhood of the Cayley graph of $F_2$ in $Bbb R^4$; precisely, at each 4-valent vertex, we add in a copy of $(S^3 setminus 4B^3)$, the four-times punctured 3-sphere, and along each edge we add in a copy of $S^2 times [0,1]$.
Now define a CW complex $X$, whose 1-skeleton is the Cayley graph, and 2-skeleton additionally adds two 2-spheres at each vertex. (That's not a typo: two at each.)
There is a continuous map $X to widetilde{M}$ given by embedding the Cayley graph as the universal cover of some appropriate wedge of two circles in $M$, with the wedge point lying on the connected sum sphere. In each $S^3 setminus 3B^4$, at a vertex of the Cayley graph, we wedge on the two "diagonal $S^2$s", one which has the north and east necks on one side, and the other of which has the north and west necks on one side.
I claim that this map $X to widetilde{M}$ is a homology equivalence and hence a homotopy equivalence by Whitehead's theorem.
Here is a version of this statement that can be proved by induction. Let $X_0$ a wedge of two 2-spheres, thought of as lying above the vertex of the Cayley graph corresponding to the identity. Let $X_i$ be the subspace of $X$ obtained from $X_{i-1}$ by adding a new edge to the Cayley graph (and the corresponding 2-spheres), and $widetilde{M}_i$ the subspace built "above this part of the Cayley graph". (Assume I have chosen my enumeration so that this exhausts $X$.)
The space $widetilde{M}_i$ is the $2i+2$-punctured 3-sphere, and hence has $H_2(widetilde{M}_i;Bbb Z) cong Bbb Z^{2i+1}$. The map $X_i to widetilde{M}_i$ is injective on $H_2$, and any of the boundary spheres of $widetilde{M}_i$ generate the cokernel of this map.
This follows from the Mayer-Vietoris sequence, adding one new 4-punctured $S^3$ at a time (containing a new pair of 2-spheres in it); the base case is clear. Under the inductive assumption and writing $P$ for the 4-punctured 3-sphere, the Mayer-Vietoris sequence gives
begin{CD}
0 @>>> H_2(*) @>>> H_2(X_i) oplus H_2(S^2 vee S^2) @>>> H_2(X_{i+1}) @>>> 0\
@. @VVV @VVV @VVV \
0 @>>> H_2(S^2) @>>> H_2(widetilde{M}_i) oplus H_2(P) @>>> H_2(widetilde{M}_{i+1}) @>>> 0.
end{CD}
Clearly by inductive assumption the middle map is injective, and so the only map which is not clearly injective is the rightmost; its injectivity follows from one of the four-lemmas. The cokernel of $$H_2(X_i) oplus H_2(S^2 vee S^2) to H_2(widetilde{M}_i) oplus H_2(P)$$ is isomorphic to $Bbb Z oplus Bbb Z$, both terms generated by the image of $H_2(S^2)$.
In particular, because naturality and exactness identifies $text{Im}(H_2(X_{i+1}))$ with the image of the composite $$H_2(X_i) oplus H_2(S^2 vee S^2) to H_2(widetilde{M}_i) oplus H_2 P to H_2(widetilde{M}_{i+1}),$$ and the final map in the bottom row quotients by the diagonal factor of $H_2(S^2)$ in $H_2(widetilde{M}_i) oplus H_2(P)$, we see that we may identify the cokernel with the class given by $H_2(S^2) to H_2(widetilde{M}_{i+1})$. This is the right dimension-count, but not the right class; but one may use the existing 2-spheres in the image to find a homologous cycle given as a union of the boundary 2-sphere and 2-spheres in the image of $X_i$.
This proves the inductive claim.
One therefore has at the level of the colimit that $H_2(X) to H_2(widetilde{M})$ is an isomorphism.
Therefore, we see that $widetilde M = vee_{F_2 times F_2} S^2$, with the natural translation action of $pi_1$. Now as before we have $pi_3(vee_{F_2 times F_2} S^2)$ equal to symmetric, finitely supported integer functions on $(F_2 times F_2) times (F_2 times F_2)$, and the translation action acts diagonally; the quotient by the $pi_1$-action is free on the set $(F_2 times F_2)/(x sim x^{-1})$ of "upper-diagonal lines".
edited Dec 20 '18 at 5:37
Michael Albanese
63.6k1599307
$endgroup$
$begingroup$
The first couple versions of this answer claimed the same result for different complexes $X$, and they were wrong. I think the kinks have all been ironed out now.
$endgroup$
– user98602
Dec 20 '18 at 1:58
add a comment |
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Not necessarily. Couldn't find that reference I mentioned, though.
$require{AMScd}$
The CW-complex model for this phenomenon can be taken as $pi_3(S^1 vee S^2)$. Then this is the same as $pi_3(vee_{Bbb Z} S^2)$; in the case of finitely many wedge factors, this may be identified as the space of symmetric matrices, and so this group may be identified as finitely supported symmetric matrices drawn on the grid $Bbb Z times Bbb Z$. The action of $pi_1$ is the right-shift on $Bbb Z$, but quotienting by this action, the group is still isomorphic to $Bbb Z^infty$, the isomorphism given by taking the sum over each line $x + y = n$ for $n geq 0$. The extra terms come from Whitehead products of the two factors (for instance, if $iota_2$ is the identity map on the 2-sphere, the bracket $[iota_2, iota_2] in pi_3(S^2vee S^2)$ is the map whose adjunction space gives $S^2 times S^2$, and this is distinct from the spaces you would get from $pi_3 S^2 oplus pi_3 S^2$.)
If all you want is a compact manifold, possibly with boundary, we are essentially finished: any CW complex of dimension $k$ embeds in $Bbb R^N$ for large $N$ so that it has a regular neighborhood (here taken to mean codimension 0 compact manifold with boundary containing the image of the CW complex which deformation retracts onto the complex). But in particular you could just take a copy of $S^2 times S^1$ and delete a small open ball.
The rest of this answer assumes you really care about closed manifolds.
Following your intuition to take the $S^1 vee S^2$ idea and make it into a question about tori instead (most relevantly: a place where the universal cover splits as a comprehensible infinite wedge of the same space, in a way which is geometrically clear), here we should instead take $M = (S^1 times S^2) # (S^1 times S^2)$. You may identify the universal cover as what it would look like to take the boundary of a regular neighborhood of the Cayley graph of $F_2$ in $Bbb R^4$; precisely, at each 4-valent vertex, we add in a copy of $(S^3 setminus 4B^3)$, the four-times punctured 3-sphere, and along each edge we add in a copy of $S^2 times [0,1]$.
Now define a CW complex $X$, whose 1-skeleton is the Cayley graph, and 2-skeleton additionally adds two 2-spheres at each vertex. (That's not a typo: two at each.)
There is a continuous map $X to widetilde{M}$ given by embedding the Cayley graph as the universal cover of some appropriate wedge of two circles in $M$, with the wedge point lying on the connected sum sphere. In each $S^3 setminus 3B^4$, at a vertex of the Cayley graph, we wedge on the two "diagonal $S^2$s", one which has the north and east necks on one side, and the other of which has the north and west necks on one side.
I claim that this map $X to widetilde{M}$ is a homology equivalence and hence a homotopy equivalence by Whitehead's theorem.
Here is a version of this statement that can be proved by induction. Let $X_0$ a wedge of two 2-spheres, thought of as lying above the vertex of the Cayley graph corresponding to the identity. Let $X_i$ be the subspace of $X$ obtained from $X_{i-1}$ by adding a new edge to the Cayley graph (and the corresponding 2-spheres), and $widetilde{M}_i$ the subspace built "above this part of the Cayley graph". (Assume I have chosen my enumeration so that this exhausts $X$.)
The space $widetilde{M}_i$ is the $2i+2$-punctured 3-sphere, and hence has $H_2(widetilde{M}_i;Bbb Z) cong Bbb Z^{2i+1}$. The map $X_i to widetilde{M}_i$ is injective on $H_2$, and any of the boundary spheres of $widetilde{M}_i$ generate the cokernel of this map.
This follows from the Mayer-Vietoris sequence, adding one new 4-punctured $S^3$ at a time (containing a new pair of 2-spheres in it); the base case is clear. Under the inductive assumption and writing $P$ for the 4-punctured 3-sphere, the Mayer-Vietoris sequence gives
begin{CD}
0 @>>> H_2(*) @>>> H_2(X_i) oplus H_2(S^2 vee S^2) @>>> H_2(X_{i+1}) @>>> 0\
@. @VVV @VVV @VVV \
0 @>>> H_2(S^2) @>>> H_2(widetilde{M}_i) oplus H_2(P) @>>> H_2(widetilde{M}_{i+1}) @>>> 0.
end{CD}
Clearly by inductive assumption the middle map is injective, and so the only map which is not clearly injective is the rightmost; its injectivity follows from one of the four-lemmas. The cokernel of $$H_2(X_i) oplus H_2(S^2 vee S^2) to H_2(widetilde{M}_i) oplus H_2(P)$$ is isomorphic to $Bbb Z oplus Bbb Z$, both terms generated by the image of $H_2(S^2)$.
In particular, because naturality and exactness identifies $text{Im}(H_2(X_{i+1}))$ with the image of the composite $$H_2(X_i) oplus H_2(S^2 vee S^2) to H_2(widetilde{M}_i) oplus H_2 P to H_2(widetilde{M}_{i+1}),$$ and the final map in the bottom row quotients by the diagonal factor of $H_2(S^2)$ in $H_2(widetilde{M}_i) oplus H_2(P)$, we see that we may identify the cokernel with the class given by $H_2(S^2) to H_2(widetilde{M}_{i+1})$. This is the right dimension-count, but not the right class; but one may use the existing 2-spheres in the image to find a homologous cycle given as a union of the boundary 2-sphere and 2-spheres in the image of $X_i$.
This proves the inductive claim.
One therefore has at the level of the colimit that $H_2(X) to H_2(widetilde{M})$ is an isomorphism.
Therefore, we see that $widetilde M = vee_{F_2 times F_2} S^2$, with the natural translation action of $pi_1$. Now as before we have $pi_3(vee_{F_2 times F_2} S^2)$ equal to symmetric, finitely supported integer functions on $(F_2 times F_2) times (F_2 times F_2)$, and the translation action acts diagonally; the quotient by the $pi_1$-action is free on the set $(F_2 times F_2)/(x sim x^{-1})$ of "upper-diagonal lines".
edited Dec 20 '18 at 5:37
Michael Albanese
63.6k1599307
$endgroup$
$begingroup$
The first couple versions of this answer claimed the same result for different complexes $X$, and they were wrong. I think the kinks have all been ironed out now.
$endgroup$
– user98602
Dec 20 '18 at 1:58
add a comment |
$begingroup$
Not necessarily. Couldn't find that reference I mentioned, though.
$require{AMScd}$
The CW-complex model for this phenomenon can be taken as $pi_3(S^1 vee S^2)$. Then this is the same as $pi_3(vee_{Bbb Z} S^2)$; in the case of finitely many wedge factors, this may be identified as the space of symmetric matrices, and so this group may be identified as finitely supported symmetric matrices drawn on the grid $Bbb Z times Bbb Z$. The action of $pi_1$ is the right-shift on $Bbb Z$, but quotienting by this action, the group is still isomorphic to $Bbb Z^infty$, the isomorphism given by taking the sum over each line $x + y = n$ for $n geq 0$. The extra terms come from Whitehead products of the two factors (for instance, if $iota_2$ is the identity map on the 2-sphere, the bracket $[iota_2, iota_2] in pi_3(S^2vee S^2)$ is the map whose adjunction space gives $S^2 times S^2$, and this is distinct from the spaces you would get from $pi_3 S^2 oplus pi_3 S^2$.)
If all you want is a compact manifold, possibly with boundary, we are essentially finished: any CW complex of dimension $k$ embeds in $Bbb R^N$ for large $N$ so that it has a regular neighborhood (here taken to mean codimension 0 compact manifold with boundary containing the image of the CW complex which deformation retracts onto the complex). But in particular you could just take a copy of $S^2 times S^1$ and delete a small open ball.
The rest of this answer assumes you really care about closed manifolds.
Following your intuition to take the $S^1 vee S^2$ idea and make it into a question about tori instead (most relevantly: a place where the universal cover splits as a comprehensible infinite wedge of the same space, in a way which is geometrically clear), here we should instead take $M = (S^1 times S^2) # (S^1 times S^2)$. You may identify the universal cover as what it would look like to take the boundary of a regular neighborhood of the Cayley graph of $F_2$ in $Bbb R^4$; precisely, at each 4-valent vertex, we add in a copy of $(S^3 setminus 4B^3)$, the four-times punctured 3-sphere, and along each edge we add in a copy of $S^2 times [0,1]$.
Now define a CW complex $X$, whose 1-skeleton is the Cayley graph, and 2-skeleton additionally adds two 2-spheres at each vertex. (That's not a typo: two at each.)
There is a continuous map $X to widetilde{M}$ given by embedding the Cayley graph as the universal cover of some appropriate wedge of two circles in $M$, with the wedge point lying on the connected sum sphere. In each $S^3 setminus 3B^4$, at a vertex of the Cayley graph, we wedge on the two "diagonal $S^2$s", one which has the north and east necks on one side, and the other of which has the north and west necks on one side.
I claim that this map $X to widetilde{M}$ is a homology equivalence and hence a homotopy equivalence by Whitehead's theorem.
Here is a version of this statement that can be proved by induction. Let $X_0$ a wedge of two 2-spheres, thought of as lying above the vertex of the Cayley graph corresponding to the identity. Let $X_i$ be the subspace of $X$ obtained from $X_{i-1}$ by adding a new edge to the Cayley graph (and the corresponding 2-spheres), and $widetilde{M}_i$ the subspace built "above this part of the Cayley graph". (Assume I have chosen my enumeration so that this exhausts $X$.)
The space $widetilde{M}_i$ is the $2i+2$-punctured 3-sphere, and hence has $H_2(widetilde{M}_i;Bbb Z) cong Bbb Z^{2i+1}$. The map $X_i to widetilde{M}_i$ is injective on $H_2$, and any of the boundary spheres of $widetilde{M}_i$ generate the cokernel of this map.
This follows from the Mayer-Vietoris sequence, adding one new 4-punctured $S^3$ at a time (containing a new pair of 2-spheres in it); the base case is clear. Under the inductive assumption and writing $P$ for the 4-punctured 3-sphere, the Mayer-Vietoris sequence gives
begin{CD}
0 @>>> H_2(*) @>>> H_2(X_i) oplus H_2(S^2 vee S^2) @>>> H_2(X_{i+1}) @>>> 0\
@. @VVV @VVV @VVV \
0 @>>> H_2(S^2) @>>> H_2(widetilde{M}_i) oplus H_2(P) @>>> H_2(widetilde{M}_{i+1}) @>>> 0.
end{CD}
Clearly by inductive assumption the middle map is injective, and so the only map which is not clearly injective is the rightmost; its injectivity follows from one of the four-lemmas. The cokernel of $$H_2(X_i) oplus H_2(S^2 vee S^2) to H_2(widetilde{M}_i) oplus H_2(P)$$ is isomorphic to $Bbb Z oplus Bbb Z$, both terms generated by the image of $H_2(S^2)$.
In particular, because naturality and exactness identifies $text{Im}(H_2(X_{i+1}))$ with the image of the composite $$H_2(X_i) oplus H_2(S^2 vee S^2) to H_2(widetilde{M}_i) oplus H_2 P to H_2(widetilde{M}_{i+1}),$$ and the final map in the bottom row quotients by the diagonal factor of $H_2(S^2)$ in $H_2(widetilde{M}_i) oplus H_2(P)$, we see that we may identify the cokernel with the class given by $H_2(S^2) to H_2(widetilde{M}_{i+1})$. This is the right dimension-count, but not the right class; but one may use the existing 2-spheres in the image to find a homologous cycle given as a union of the boundary 2-sphere and 2-spheres in the image of $X_i$.
This proves the inductive claim.
One therefore has at the level of the colimit that $H_2(X) to H_2(widetilde{M})$ is an isomorphism.
Therefore, we see that $widetilde M = vee_{F_2 times F_2} S^2$, with the natural translation action of $pi_1$. Now as before we have $pi_3(vee_{F_2 times F_2} S^2)$ equal to symmetric, finitely supported integer functions on $(F_2 times F_2) times (F_2 times F_2)$, and the translation action acts diagonally; the quotient by the $pi_1$-action is free on the set $(F_2 times F_2)/(x sim x^{-1})$ of "upper-diagonal lines".
edited Dec 20 '18 at 5:37
Michael Albanese
63.6k1599307
$endgroup$
$begingroup$
The first couple versions of this answer claimed the same result for different complexes $X$, and they were wrong. I think the kinks have all been ironed out now.
$endgroup$
– user98602
Dec 20 '18 at 1:58
add a comment |
$begingroup$
Not necessarily. Couldn't find that reference I mentioned, though.
$require{AMScd}$
The CW-complex model for this phenomenon can be taken as $pi_3(S^1 vee S^2)$. Then this is the same as $pi_3(vee_{Bbb Z} S^2)$; in the case of finitely many wedge factors, this may be identified as the space of symmetric matrices, and so this group may be identified as finitely supported symmetric matrices drawn on the grid $Bbb Z times Bbb Z$. The action of $pi_1$ is the right-shift on $Bbb Z$, but quotienting by this action, the group is still isomorphic to $Bbb Z^infty$, the isomorphism given by taking the sum over each line $x + y = n$ for $n geq 0$. The extra terms come from Whitehead products of the two factors (for instance, if $iota_2$ is the identity map on the 2-sphere, the bracket $[iota_2, iota_2] in pi_3(S^2vee S^2)$ is the map whose adjunction space gives $S^2 times S^2$, and this is distinct from the spaces you would get from $pi_3 S^2 oplus pi_3 S^2$.)
If all you want is a compact manifold, possibly with boundary, we are essentially finished: any CW complex of dimension $k$ embeds in $Bbb R^N$ for large $N$ so that it has a regular neighborhood (here taken to mean codimension 0 compact manifold with boundary containing the image of the CW complex which deformation retracts onto the complex). But in particular you could just take a copy of $S^2 times S^1$ and delete a small open ball.
The rest of this answer assumes you really care about closed manifolds.
Following your intuition to take the $S^1 vee S^2$ idea and make it into a question about tori instead (most relevantly: a place where the universal cover splits as a comprehensible infinite wedge of the same space, in a way which is geometrically clear), here we should instead take $M = (S^1 times S^2) # (S^1 times S^2)$. You may identify the universal cover as what it would look like to take the boundary of a regular neighborhood of the Cayley graph of $F_2$ in $Bbb R^4$; precisely, at each 4-valent vertex, we add in a copy of $(S^3 setminus 4B^3)$, the four-times punctured 3-sphere, and along each edge we add in a copy of $S^2 times [0,1]$.
Now define a CW complex $X$, whose 1-skeleton is the Cayley graph, and 2-skeleton additionally adds two 2-spheres at each vertex. (That's not a typo: two at each.)
There is a continuous map $X to widetilde{M}$ given by embedding the Cayley graph as the universal cover of some appropriate wedge of two circles in $M$, with the wedge point lying on the connected sum sphere. In each $S^3 setminus 3B^4$, at a vertex of the Cayley graph, we wedge on the two "diagonal $S^2$s", one which has the north and east necks on one side, and the other of which has the north and west necks on one side.
I claim that this map $X to widetilde{M}$ is a homology equivalence and hence a homotopy equivalence by Whitehead's theorem.
Here is a version of this statement that can be proved by induction. Let $X_0$ a wedge of two 2-spheres, thought of as lying above the vertex of the Cayley graph corresponding to the identity. Let $X_i$ be the subspace of $X$ obtained from $X_{i-1}$ by adding a new edge to the Cayley graph (and the corresponding 2-spheres), and $widetilde{M}_i$ the subspace built "above this part of the Cayley graph". (Assume I have chosen my enumeration so that this exhausts $X$.)
The space $widetilde{M}_i$ is the $2i+2$-punctured 3-sphere, and hence has $H_2(widetilde{M}_i;Bbb Z) cong Bbb Z^{2i+1}$. The map $X_i to widetilde{M}_i$ is injective on $H_2$, and any of the boundary spheres of $widetilde{M}_i$ generate the cokernel of this map.
This follows from the Mayer-Vietoris sequence, adding one new 4-punctured $S^3$ at a time (containing a new pair of 2-spheres in it); the base case is clear. Under the inductive assumption and writing $P$ for the 4-punctured 3-sphere, the Mayer-Vietoris sequence gives
begin{CD}
0 @>>> H_2(*) @>>> H_2(X_i) oplus H_2(S^2 vee S^2) @>>> H_2(X_{i+1}) @>>> 0\
@. @VVV @VVV @VVV \
0 @>>> H_2(S^2) @>>> H_2(widetilde{M}_i) oplus H_2(P) @>>> H_2(widetilde{M}_{i+1}) @>>> 0.
end{CD}
Clearly by inductive assumption the middle map is injective, and so the only map which is not clearly injective is the rightmost; its injectivity follows from one of the four-lemmas. The cokernel of $$H_2(X_i) oplus H_2(S^2 vee S^2) to H_2(widetilde{M}_i) oplus H_2(P)$$ is isomorphic to $Bbb Z oplus Bbb Z$, both terms generated by the image of $H_2(S^2)$.
In particular, because naturality and exactness identifies $text{Im}(H_2(X_{i+1}))$ with the image of the composite $$H_2(X_i) oplus H_2(S^2 vee S^2) to H_2(widetilde{M}_i) oplus H_2 P to H_2(widetilde{M}_{i+1}),$$ and the final map in the bottom row quotients by the diagonal factor of $H_2(S^2)$ in $H_2(widetilde{M}_i) oplus H_2(P)$, we see that we may identify the cokernel with the class given by $H_2(S^2) to H_2(widetilde{M}_{i+1})$. This is the right dimension-count, but not the right class; but one may use the existing 2-spheres in the image to find a homologous cycle given as a union of the boundary 2-sphere and 2-spheres in the image of $X_i$.
This proves the inductive claim.
One therefore has at the level of the colimit that $H_2(X) to H_2(widetilde{M})$ is an isomorphism.
Therefore, we see that $widetilde M = vee_{F_2 times F_2} S^2$, with the natural translation action of $pi_1$. Now as before we have $pi_3(vee_{F_2 times F_2} S^2)$ equal to symmetric, finitely supported integer functions on $(F_2 times F_2) times (F_2 times F_2)$, and the translation action acts diagonally; the quotient by the $pi_1$-action is free on the set $(F_2 times F_2)/(x sim x^{-1})$ of "upper-diagonal lines".
edited Dec 20 '18 at 5:37
Michael Albanese
63.6k1599307
$endgroup$
Not necessarily. Couldn't find that reference I mentioned, though.
$require{AMScd}$
The CW-complex model for this phenomenon can be taken as $pi_3(S^1 vee S^2)$. Then this is the same as $pi_3(vee_{Bbb Z} S^2)$; in the case of finitely many wedge factors, this may be identified as the space of symmetric matrices, and so this group may be identified as finitely supported symmetric matrices drawn on the grid $Bbb Z times Bbb Z$. The action of $pi_1$ is the right-shift on $Bbb Z$, but quotienting by this action, the group is still isomorphic to $Bbb Z^infty$, the isomorphism given by taking the sum over each line $x + y = n$ for $n geq 0$. The extra terms come from Whitehead products of the two factors (for instance, if $iota_2$ is the identity map on the 2-sphere, the bracket $[iota_2, iota_2] in pi_3(S^2vee S^2)$ is the map whose adjunction space gives $S^2 times S^2$, and this is distinct from the spaces you would get from $pi_3 S^2 oplus pi_3 S^2$.)
If all you want is a compact manifold, possibly with boundary, we are essentially finished: any CW complex of dimension $k$ embeds in $Bbb R^N$ for large $N$ so that it has a regular neighborhood (here taken to mean codimension 0 compact manifold with boundary containing the image of the CW complex which deformation retracts onto the complex). But in particular you could just take a copy of $S^2 times S^1$ and delete a small open ball.
The rest of this answer assumes you really care about closed manifolds.
Following your intuition to take the $S^1 vee S^2$ idea and make it into a question about tori instead (most relevantly: a place where the universal cover splits as a comprehensible infinite wedge of the same space, in a way which is geometrically clear), here we should instead take $M = (S^1 times S^2) # (S^1 times S^2)$. You may identify the universal cover as what it would look like to take the boundary of a regular neighborhood of the Cayley graph of $F_2$ in $Bbb R^4$; precisely, at each 4-valent vertex, we add in a copy of $(S^3 setminus 4B^3)$, the four-times punctured 3-sphere, and along each edge we add in a copy of $S^2 times [0,1]$.
Now define a CW complex $X$, whose 1-skeleton is the Cayley graph, and 2-skeleton additionally adds two 2-spheres at each vertex. (That's not a typo: two at each.)
There is a continuous map $X to widetilde{M}$ given by embedding the Cayley graph as the universal cover of some appropriate wedge of two circles in $M$, with the wedge point lying on the connected sum sphere. In each $S^3 setminus 3B^4$, at a vertex of the Cayley graph, we wedge on the two "diagonal $S^2$s", one which has the north and east necks on one side, and the other of which has the north and west necks on one side.
I claim that this map $X to widetilde{M}$ is a homology equivalence and hence a homotopy equivalence by Whitehead's theorem.
Here is a version of this statement that can be proved by induction. Let $X_0$ a wedge of two 2-spheres, thought of as lying above the vertex of the Cayley graph corresponding to the identity. Let $X_i$ be the subspace of $X$ obtained from $X_{i-1}$ by adding a new edge to the Cayley graph (and the corresponding 2-spheres), and $widetilde{M}_i$ the subspace built "above this part of the Cayley graph". (Assume I have chosen my enumeration so that this exhausts $X$.)
The space $widetilde{M}_i$ is the $2i+2$-punctured 3-sphere, and hence has $H_2(widetilde{M}_i;Bbb Z) cong Bbb Z^{2i+1}$. The map $X_i to widetilde{M}_i$ is injective on $H_2$, and any of the boundary spheres of $widetilde{M}_i$ generate the cokernel of this map.
This follows from the Mayer-Vietoris sequence, adding one new 4-punctured $S^3$ at a time (containing a new pair of 2-spheres in it); the base case is clear. Under the inductive assumption and writing $P$ for the 4-punctured 3-sphere, the Mayer-Vietoris sequence gives
begin{CD}
0 @>>> H_2(*) @>>> H_2(X_i) oplus H_2(S^2 vee S^2) @>>> H_2(X_{i+1}) @>>> 0\
@. @VVV @VVV @VVV \
0 @>>> H_2(S^2) @>>> H_2(widetilde{M}_i) oplus H_2(P) @>>> H_2(widetilde{M}_{i+1}) @>>> 0.
end{CD}
Clearly by inductive assumption the middle map is injective, and so the only map which is not clearly injective is the rightmost; its injectivity follows from one of the four-lemmas. The cokernel of $$H_2(X_i) oplus H_2(S^2 vee S^2) to H_2(widetilde{M}_i) oplus H_2(P)$$ is isomorphic to $Bbb Z oplus Bbb Z$, both terms generated by the image of $H_2(S^2)$.
In particular, because naturality and exactness identifies $text{Im}(H_2(X_{i+1}))$ with the image of the composite $$H_2(X_i) oplus H_2(S^2 vee S^2) to H_2(widetilde{M}_i) oplus H_2 P to H_2(widetilde{M}_{i+1}),$$ and the final map in the bottom row quotients by the diagonal factor of $H_2(S^2)$ in $H_2(widetilde{M}_i) oplus H_2(P)$, we see that we may identify the cokernel with the class given by $H_2(S^2) to H_2(widetilde{M}_{i+1})$. This is the right dimension-count, but not the right class; but one may use the existing 2-spheres in the image to find a homologous cycle given as a union of the boundary 2-sphere and 2-spheres in the image of $X_i$.
This proves the inductive claim.
One therefore has at the level of the colimit that $H_2(X) to H_2(widetilde{M})$ is an isomorphism.
Therefore, we see that $widetilde M = vee_{F_2 times F_2} S^2$, with the natural translation action of $pi_1$. Now as before we have $pi_3(vee_{F_2 times F_2} S^2)$ equal to symmetric, finitely supported integer functions on $(F_2 times F_2) times (F_2 times F_2)$, and the translation action acts diagonally; the quotient by the $pi_1$-action is free on the set $(F_2 times F_2)/(x sim x^{-1})$ of "upper-diagonal lines".
edited Dec 20 '18 at 5:37
Michael Albanese
63.6k1599307
edited Dec 20 '18 at 5:37
Michael Albanese
63.6k1599307
edited Dec 20 '18 at 5:37
Michael Albanese
63.6k1599307
edited Dec 20 '18 at 5:37
Michael Albanese
63.6k1599307
63.6k1599307
answered Dec 19 '18 at 20:32
user98602
$begingroup$
The first couple versions of this answer claimed the same result for different complexes $X$, and they were wrong. I think the kinks have all been ironed out now.
$endgroup$
– user98602
Dec 20 '18 at 1:58
add a comment |
$begingroup$
The first couple versions of this answer claimed the same result for different complexes $X$, and they were wrong. I think the kinks have all been ironed out now.
$endgroup$
– user98602
Dec 20 '18 at 1:58
$begingroup$
The first couple versions of this answer claimed the same result for different complexes $X$, and they were wrong. I think the kinks have all been ironed out now.
$endgroup$
– user98602
Dec 20 '18 at 1:58
$begingroup$
The first couple versions of this answer claimed the same result for different complexes $X$, and they were wrong. I think the kinks have all been ironed out now.
$endgroup$
– user98602
Dec 20 '18 at 1:58
add a comment |
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$begingroup$
This is false (I remember finding an article about an example somewhere between dimensions 6 or 8), but I don't know the reference. I'll try digging, but I figured it might help the search to know the result (assuming you trust my memory).
$endgroup$
– user98602
Dec 19 '18 at 19:48
1
$begingroup$
Remark: $pi_2 M$ is always finitely generated as a $pi_1$-module so long as the group $pi_1$ is Noetherian (for instance, finitely generated abelian). For one has the natural isomorphism $pi_2 M cong H_2 tilde M$ of $pi_1$-modules, and if one calculates the latter using a cellular chain complex corresponding to the universal cover of a cell decomposition of the compact manifold $M$, the chain groups themselves are finitely generated $pi_1$-modules. The subcomplex of cycles is finitely generated by the Noetherian hypothesis and the quotient is then clearly finitely generated.
$endgroup$
– user98602
Dec 20 '18 at 2:12