Stably equivalent but not homotopy equivalent
up vote
15
down vote
favorite
What are some examples of (compact, say) manifolds $X$ and $Y$ that are stably equivalent, i.e. $Sigma^{infty}X_+simeqSigma^{infty}Y_+$, but are not homotopy equivalent?
at.algebraic-topology
asked Nov 23 at 17:33
user131711
1214
New contributor
user131711 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
up vote
15
down vote
favorite
What are some examples of (compact, say) manifolds $X$ and $Y$ that are stably equivalent, i.e. $Sigma^{infty}X_+simeqSigma^{infty}Y_+$, but are not homotopy equivalent?
at.algebraic-topology
asked Nov 23 at 17:33
user131711
1214
New contributor
user131711 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
11
Any pair of homology spheres of the same dimension. After one stabilization they are equivalent by Whitehead.
– Mike Miller
Nov 23 at 19:11
add a comment |
up vote
15
down vote
favorite
up vote
15
down vote
favorite
What are some examples of (compact, say) manifolds $X$ and $Y$ that are stably equivalent, i.e. $Sigma^{infty}X_+simeqSigma^{infty}Y_+$, but are not homotopy equivalent?
at.algebraic-topology
asked Nov 23 at 17:33
user131711
1214
New contributor
user131711 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What are some examples of (compact, say) manifolds $X$ and $Y$ that are stably equivalent, i.e. $Sigma^{infty}X_+simeqSigma^{infty}Y_+$, but are not homotopy equivalent?
at.algebraic-topology
at.algebraic-topology
asked Nov 23 at 17:33
user131711
1214
New contributor
user131711 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 23 at 17:33
user131711
1214
New contributor
user131711 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 23 at 17:33
user131711
1214
New contributor
user131711 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 23 at 17:33
user131711
1214
asked Nov 23 at 17:33
user131711
1214
1214
New contributor
user131711 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user131711 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user131711 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
11
Any pair of homology spheres of the same dimension. After one stabilization they are equivalent by Whitehead.
– Mike Miller
Nov 23 at 19:11
add a comment |
11
Any pair of homology spheres of the same dimension. After one stabilization they are equivalent by Whitehead.
– Mike Miller
Nov 23 at 19:11
11
11
Any pair of homology spheres of the same dimension. After one stabilization they are equivalent by Whitehead.
– Mike Miller
Nov 23 at 19:11
Any pair of homology spheres of the same dimension. After one stabilization they are equivalent by Whitehead.
– Mike Miller
Nov 23 at 19:11
add a comment |
2 Answers
2
active
oldest
votes
up vote
11
down vote
accepted
Maybe it is worth adding some simply-connected examples.
Every simply connected closed 4-manifold may be described as $X = D^4 cup_f (S^2 vee cdots vee S^2)$, where $f$ is a map $S^3 to S^2 vee cdots vee S^2$; $pi_3$ of this wedge is known to be generated by Hopf maps and Whitehead products of two factors, so we may represent such a map by a symmetric integer matrix $A$; this integer matrix may just as well be interpreted as the cup product pairing $H^2(X;Bbb Z) otimes H^2(X;Bbb Z) to H^4(X;Bbb Z) = Bbb Z$. The resulting homotopy type is determined up to isomorphism by $A$ up to isomorphism of symmetric bilinear forms over $Bbb Z$. See, for instance, Example 4.52 in Hatcher's algebraic topology book.
On the other hand, if $n > 2$, then $$pi_{n+1}(vee^k S^n) = (pi_{n+1} S^n)^k = (Bbb Z/2)^k.$$ The Whitehead product factors vanish. This follows by an inductive argument using the Hilton-Milnor theorem, as stated in this answer.
Furthermore, any map in $GL_k(Bbb Z)$ may be realized as an automorphism of $H_n(vee^k S^n)$ by some autoequivalence of $vee^k S^n$, and the map $GL_k(Bbb Z) to GL_k(Bbb Z/2)$ is surjective (check at the level of the generating set of elementary matrices). Because every two nonzero vectors in $(Bbb Z/2)^k$ are related by some matrix in $GL_k(Bbb Z/2)$, the homotopy type of $D^{n+2} cup_f (vee^k S^n)$ is determined entirely by whether or not $f$ is nontrivial.
Suspending the presentation given for $X$ takes the diagonals of the matrix $A$ mod 2, and so the homotopy type of the suspension is dictated by whether or not $X$ was spin (that is, whether or not its intersection form was even). If $X$ was spin, then $$Sigma X simeq S^5 vee^{b_2} S^3;$$ if $X$ was not spin, then $$Sigma X simeq Sigma Bbb{CP}^2 vee^{b_2-1} S^3.$$ In particular, the stable homotopy type is determined entirely by "even-ness" and rank of the intersection form.
answered Nov 23 at 23:46
Mike Miller
3,42452340
If I understood correctly, this gives examples of manifolds that are stably equivalent but have different cup product structures. Doesn't an equivalence of the spectra imply an isomorphism of the cohomology rings?
– user131711
Nov 24 at 18:22
1
@user131711 No. In fact, the cup product is destroyed after one stabilization: $tilde H^*(Sigma X)$ has trivial cup products. (This is a straightforward exercise using the relative cup product and the fact that $Sigma X$ has a cover by two contractible pieces.) All that lives is the Steenrod algebra action. This is visible at the level of spectra: $H^*(X; Bbb Z/2) = [Sigma^infty X, HBbb Z/2]$, and the Steenrod algebra is $[HBbb Z/2, HBbb Z/2]$, which acts by postcomposition.
– Mike Miller
Nov 24 at 18:29
The Steenrod algebra action is how one checks that $Sigma Bbb{CP}^2$ and $S^5$ are not stably equivalent.
– Mike Miller
Nov 24 at 18:32
Correct me if I am wrong, but the cup product on $Sigma X_+$ is given using the diagonal $Sigma^{infty}(Sigma X_+)rightarrowSigma^{infty}(Sigma X_+)wedge Sigma^{infty}(Sigma X_+)$. Since $S^1rightarrow S^1wedge S^1$ is nullhomotopic, there is no cup product. On the other hand, the cup product on $X$ is induced by compositions $Sigma^{infty}X_+rightarrowSigma^{infty}X_+wedgeSigma^{infty}X_+rightarrow HRwedge HRrightarrow HR$. Shouldn't an equivalence of the spectra give an identification of the diagonal maps, and so the same cup products?
– user131711
Nov 24 at 18:38
@user131711 Clearly there must be an error by the simple fact that if $Sigma X sim Sigma Y$, then $X$ and $Y$ are stable homotopy equivalent (by definition), and as above you may kill the cup product. In particular, suspending $D^4 cup_{2 in pi_3 S^2} S^2$ you get $S^5 vee S^3$. I think the error is this: The definition of the first diagonal map relies on the space $X$ as opposed to its stable homotopy type; suspension spectra have diagonal maps but that structure depends on $X$. You have therefore defined the cohomology ring of a space. (Spectra do not have natural diagonals.)
– Mike Miller
Nov 24 at 18:49
|
show 1 more comment
up vote
18
down vote
The easiest examples are given by complements of open neighborhoods of distinct knots in $S^n$.
More generally, Spanier-Whitehead duality says that if $X$ is a compact simplicial complex and if $f,g:X rightarrow S^n$ are embeddings, then $S^n - Im(f)$ is stabily equivalent to $S^n - Im(g)$.
See my notes here for an elementary discussion.
answered Nov 23 at 19:01
Andy Putman
30.9k5132211
[Dold, Albrecht, A simple proof of the Jordan-Alexander complement theorem. Amer. Math. Monthly 100 (1993), no. 9, 856–857] has a six line elementary (but sneaky) proof that if A, a closed subset of R^m is homeomorphic to B, a closed subset of R^n, then the complement of A x 0 in R^{m+n} is homeomorphic to the complement of 0 x B is R^{m+n}.
– Nicholas Kuhn
yesterday
@NicholasKuhn: Wow, that's a really beautiful proof! Thanks for sharing it.
– Andy Putman
yesterday
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
11
down vote
accepted
Maybe it is worth adding some simply-connected examples.
Every simply connected closed 4-manifold may be described as $X = D^4 cup_f (S^2 vee cdots vee S^2)$, where $f$ is a map $S^3 to S^2 vee cdots vee S^2$; $pi_3$ of this wedge is known to be generated by Hopf maps and Whitehead products of two factors, so we may represent such a map by a symmetric integer matrix $A$; this integer matrix may just as well be interpreted as the cup product pairing $H^2(X;Bbb Z) otimes H^2(X;Bbb Z) to H^4(X;Bbb Z) = Bbb Z$. The resulting homotopy type is determined up to isomorphism by $A$ up to isomorphism of symmetric bilinear forms over $Bbb Z$. See, for instance, Example 4.52 in Hatcher's algebraic topology book.
On the other hand, if $n > 2$, then $$pi_{n+1}(vee^k S^n) = (pi_{n+1} S^n)^k = (Bbb Z/2)^k.$$ The Whitehead product factors vanish. This follows by an inductive argument using the Hilton-Milnor theorem, as stated in this answer.
Furthermore, any map in $GL_k(Bbb Z)$ may be realized as an automorphism of $H_n(vee^k S^n)$ by some autoequivalence of $vee^k S^n$, and the map $GL_k(Bbb Z) to GL_k(Bbb Z/2)$ is surjective (check at the level of the generating set of elementary matrices). Because every two nonzero vectors in $(Bbb Z/2)^k$ are related by some matrix in $GL_k(Bbb Z/2)$, the homotopy type of $D^{n+2} cup_f (vee^k S^n)$ is determined entirely by whether or not $f$ is nontrivial.
Suspending the presentation given for $X$ takes the diagonals of the matrix $A$ mod 2, and so the homotopy type of the suspension is dictated by whether or not $X$ was spin (that is, whether or not its intersection form was even). If $X$ was spin, then $$Sigma X simeq S^5 vee^{b_2} S^3;$$ if $X$ was not spin, then $$Sigma X simeq Sigma Bbb{CP}^2 vee^{b_2-1} S^3.$$ In particular, the stable homotopy type is determined entirely by "even-ness" and rank of the intersection form.
answered Nov 23 at 23:46
Mike Miller
3,42452340
If I understood correctly, this gives examples of manifolds that are stably equivalent but have different cup product structures. Doesn't an equivalence of the spectra imply an isomorphism of the cohomology rings?
– user131711
Nov 24 at 18:22
1
@user131711 No. In fact, the cup product is destroyed after one stabilization: $tilde H^*(Sigma X)$ has trivial cup products. (This is a straightforward exercise using the relative cup product and the fact that $Sigma X$ has a cover by two contractible pieces.) All that lives is the Steenrod algebra action. This is visible at the level of spectra: $H^*(X; Bbb Z/2) = [Sigma^infty X, HBbb Z/2]$, and the Steenrod algebra is $[HBbb Z/2, HBbb Z/2]$, which acts by postcomposition.
– Mike Miller
Nov 24 at 18:29
The Steenrod algebra action is how one checks that $Sigma Bbb{CP}^2$ and $S^5$ are not stably equivalent.
– Mike Miller
Nov 24 at 18:32
Correct me if I am wrong, but the cup product on $Sigma X_+$ is given using the diagonal $Sigma^{infty}(Sigma X_+)rightarrowSigma^{infty}(Sigma X_+)wedge Sigma^{infty}(Sigma X_+)$. Since $S^1rightarrow S^1wedge S^1$ is nullhomotopic, there is no cup product. On the other hand, the cup product on $X$ is induced by compositions $Sigma^{infty}X_+rightarrowSigma^{infty}X_+wedgeSigma^{infty}X_+rightarrow HRwedge HRrightarrow HR$. Shouldn't an equivalence of the spectra give an identification of the diagonal maps, and so the same cup products?
– user131711
Nov 24 at 18:38
@user131711 Clearly there must be an error by the simple fact that if $Sigma X sim Sigma Y$, then $X$ and $Y$ are stable homotopy equivalent (by definition), and as above you may kill the cup product. In particular, suspending $D^4 cup_{2 in pi_3 S^2} S^2$ you get $S^5 vee S^3$. I think the error is this: The definition of the first diagonal map relies on the space $X$ as opposed to its stable homotopy type; suspension spectra have diagonal maps but that structure depends on $X$. You have therefore defined the cohomology ring of a space. (Spectra do not have natural diagonals.)
– Mike Miller
Nov 24 at 18:49
|
show 1 more comment
up vote
11
down vote
accepted
Maybe it is worth adding some simply-connected examples.
Every simply connected closed 4-manifold may be described as $X = D^4 cup_f (S^2 vee cdots vee S^2)$, where $f$ is a map $S^3 to S^2 vee cdots vee S^2$; $pi_3$ of this wedge is known to be generated by Hopf maps and Whitehead products of two factors, so we may represent such a map by a symmetric integer matrix $A$; this integer matrix may just as well be interpreted as the cup product pairing $H^2(X;Bbb Z) otimes H^2(X;Bbb Z) to H^4(X;Bbb Z) = Bbb Z$. The resulting homotopy type is determined up to isomorphism by $A$ up to isomorphism of symmetric bilinear forms over $Bbb Z$. See, for instance, Example 4.52 in Hatcher's algebraic topology book.
On the other hand, if $n > 2$, then $$pi_{n+1}(vee^k S^n) = (pi_{n+1} S^n)^k = (Bbb Z/2)^k.$$ The Whitehead product factors vanish. This follows by an inductive argument using the Hilton-Milnor theorem, as stated in this answer.
Furthermore, any map in $GL_k(Bbb Z)$ may be realized as an automorphism of $H_n(vee^k S^n)$ by some autoequivalence of $vee^k S^n$, and the map $GL_k(Bbb Z) to GL_k(Bbb Z/2)$ is surjective (check at the level of the generating set of elementary matrices). Because every two nonzero vectors in $(Bbb Z/2)^k$ are related by some matrix in $GL_k(Bbb Z/2)$, the homotopy type of $D^{n+2} cup_f (vee^k S^n)$ is determined entirely by whether or not $f$ is nontrivial.
Suspending the presentation given for $X$ takes the diagonals of the matrix $A$ mod 2, and so the homotopy type of the suspension is dictated by whether or not $X$ was spin (that is, whether or not its intersection form was even). If $X$ was spin, then $$Sigma X simeq S^5 vee^{b_2} S^3;$$ if $X$ was not spin, then $$Sigma X simeq Sigma Bbb{CP}^2 vee^{b_2-1} S^3.$$ In particular, the stable homotopy type is determined entirely by "even-ness" and rank of the intersection form.
answered Nov 23 at 23:46
Mike Miller
3,42452340
If I understood correctly, this gives examples of manifolds that are stably equivalent but have different cup product structures. Doesn't an equivalence of the spectra imply an isomorphism of the cohomology rings?
– user131711
Nov 24 at 18:22
1
@user131711 No. In fact, the cup product is destroyed after one stabilization: $tilde H^*(Sigma X)$ has trivial cup products. (This is a straightforward exercise using the relative cup product and the fact that $Sigma X$ has a cover by two contractible pieces.) All that lives is the Steenrod algebra action. This is visible at the level of spectra: $H^*(X; Bbb Z/2) = [Sigma^infty X, HBbb Z/2]$, and the Steenrod algebra is $[HBbb Z/2, HBbb Z/2]$, which acts by postcomposition.
– Mike Miller
Nov 24 at 18:29
The Steenrod algebra action is how one checks that $Sigma Bbb{CP}^2$ and $S^5$ are not stably equivalent.
– Mike Miller
Nov 24 at 18:32
Correct me if I am wrong, but the cup product on $Sigma X_+$ is given using the diagonal $Sigma^{infty}(Sigma X_+)rightarrowSigma^{infty}(Sigma X_+)wedge Sigma^{infty}(Sigma X_+)$. Since $S^1rightarrow S^1wedge S^1$ is nullhomotopic, there is no cup product. On the other hand, the cup product on $X$ is induced by compositions $Sigma^{infty}X_+rightarrowSigma^{infty}X_+wedgeSigma^{infty}X_+rightarrow HRwedge HRrightarrow HR$. Shouldn't an equivalence of the spectra give an identification of the diagonal maps, and so the same cup products?
– user131711
Nov 24 at 18:38
@user131711 Clearly there must be an error by the simple fact that if $Sigma X sim Sigma Y$, then $X$ and $Y$ are stable homotopy equivalent (by definition), and as above you may kill the cup product. In particular, suspending $D^4 cup_{2 in pi_3 S^2} S^2$ you get $S^5 vee S^3$. I think the error is this: The definition of the first diagonal map relies on the space $X$ as opposed to its stable homotopy type; suspension spectra have diagonal maps but that structure depends on $X$. You have therefore defined the cohomology ring of a space. (Spectra do not have natural diagonals.)
– Mike Miller
Nov 24 at 18:49
|
show 1 more comment
up vote
11
down vote
accepted
up vote
11
down vote
accepted
Maybe it is worth adding some simply-connected examples.
Every simply connected closed 4-manifold may be described as $X = D^4 cup_f (S^2 vee cdots vee S^2)$, where $f$ is a map $S^3 to S^2 vee cdots vee S^2$; $pi_3$ of this wedge is known to be generated by Hopf maps and Whitehead products of two factors, so we may represent such a map by a symmetric integer matrix $A$; this integer matrix may just as well be interpreted as the cup product pairing $H^2(X;Bbb Z) otimes H^2(X;Bbb Z) to H^4(X;Bbb Z) = Bbb Z$. The resulting homotopy type is determined up to isomorphism by $A$ up to isomorphism of symmetric bilinear forms over $Bbb Z$. See, for instance, Example 4.52 in Hatcher's algebraic topology book.
On the other hand, if $n > 2$, then $$pi_{n+1}(vee^k S^n) = (pi_{n+1} S^n)^k = (Bbb Z/2)^k.$$ The Whitehead product factors vanish. This follows by an inductive argument using the Hilton-Milnor theorem, as stated in this answer.
Furthermore, any map in $GL_k(Bbb Z)$ may be realized as an automorphism of $H_n(vee^k S^n)$ by some autoequivalence of $vee^k S^n$, and the map $GL_k(Bbb Z) to GL_k(Bbb Z/2)$ is surjective (check at the level of the generating set of elementary matrices). Because every two nonzero vectors in $(Bbb Z/2)^k$ are related by some matrix in $GL_k(Bbb Z/2)$, the homotopy type of $D^{n+2} cup_f (vee^k S^n)$ is determined entirely by whether or not $f$ is nontrivial.
Suspending the presentation given for $X$ takes the diagonals of the matrix $A$ mod 2, and so the homotopy type of the suspension is dictated by whether or not $X$ was spin (that is, whether or not its intersection form was even). If $X$ was spin, then $$Sigma X simeq S^5 vee^{b_2} S^3;$$ if $X$ was not spin, then $$Sigma X simeq Sigma Bbb{CP}^2 vee^{b_2-1} S^3.$$ In particular, the stable homotopy type is determined entirely by "even-ness" and rank of the intersection form.
answered Nov 23 at 23:46
Mike Miller
3,42452340
Maybe it is worth adding some simply-connected examples.
Every simply connected closed 4-manifold may be described as $X = D^4 cup_f (S^2 vee cdots vee S^2)$, where $f$ is a map $S^3 to S^2 vee cdots vee S^2$; $pi_3$ of this wedge is known to be generated by Hopf maps and Whitehead products of two factors, so we may represent such a map by a symmetric integer matrix $A$; this integer matrix may just as well be interpreted as the cup product pairing $H^2(X;Bbb Z) otimes H^2(X;Bbb Z) to H^4(X;Bbb Z) = Bbb Z$. The resulting homotopy type is determined up to isomorphism by $A$ up to isomorphism of symmetric bilinear forms over $Bbb Z$. See, for instance, Example 4.52 in Hatcher's algebraic topology book.
On the other hand, if $n > 2$, then $$pi_{n+1}(vee^k S^n) = (pi_{n+1} S^n)^k = (Bbb Z/2)^k.$$ The Whitehead product factors vanish. This follows by an inductive argument using the Hilton-Milnor theorem, as stated in this answer.
Furthermore, any map in $GL_k(Bbb Z)$ may be realized as an automorphism of $H_n(vee^k S^n)$ by some autoequivalence of $vee^k S^n$, and the map $GL_k(Bbb Z) to GL_k(Bbb Z/2)$ is surjective (check at the level of the generating set of elementary matrices). Because every two nonzero vectors in $(Bbb Z/2)^k$ are related by some matrix in $GL_k(Bbb Z/2)$, the homotopy type of $D^{n+2} cup_f (vee^k S^n)$ is determined entirely by whether or not $f$ is nontrivial.
Suspending the presentation given for $X$ takes the diagonals of the matrix $A$ mod 2, and so the homotopy type of the suspension is dictated by whether or not $X$ was spin (that is, whether or not its intersection form was even). If $X$ was spin, then $$Sigma X simeq S^5 vee^{b_2} S^3;$$ if $X$ was not spin, then $$Sigma X simeq Sigma Bbb{CP}^2 vee^{b_2-1} S^3.$$ In particular, the stable homotopy type is determined entirely by "even-ness" and rank of the intersection form.
answered Nov 23 at 23:46
Mike Miller
3,42452340
edited Nov 24 at 6:56
answered Nov 23 at 23:46
Mike Miller
3,42452340
answered Nov 23 at 23:46
Mike Miller
3,42452340
answered Nov 23 at 23:46
Mike Miller
3,42452340
3,42452340
If I understood correctly, this gives examples of manifolds that are stably equivalent but have different cup product structures. Doesn't an equivalence of the spectra imply an isomorphism of the cohomology rings?
– user131711
Nov 24 at 18:22
1
@user131711 No. In fact, the cup product is destroyed after one stabilization: $tilde H^*(Sigma X)$ has trivial cup products. (This is a straightforward exercise using the relative cup product and the fact that $Sigma X$ has a cover by two contractible pieces.) All that lives is the Steenrod algebra action. This is visible at the level of spectra: $H^*(X; Bbb Z/2) = [Sigma^infty X, HBbb Z/2]$, and the Steenrod algebra is $[HBbb Z/2, HBbb Z/2]$, which acts by postcomposition.
– Mike Miller
Nov 24 at 18:29
The Steenrod algebra action is how one checks that $Sigma Bbb{CP}^2$ and $S^5$ are not stably equivalent.
– Mike Miller
Nov 24 at 18:32
Correct me if I am wrong, but the cup product on $Sigma X_+$ is given using the diagonal $Sigma^{infty}(Sigma X_+)rightarrowSigma^{infty}(Sigma X_+)wedge Sigma^{infty}(Sigma X_+)$. Since $S^1rightarrow S^1wedge S^1$ is nullhomotopic, there is no cup product. On the other hand, the cup product on $X$ is induced by compositions $Sigma^{infty}X_+rightarrowSigma^{infty}X_+wedgeSigma^{infty}X_+rightarrow HRwedge HRrightarrow HR$. Shouldn't an equivalence of the spectra give an identification of the diagonal maps, and so the same cup products?
– user131711
Nov 24 at 18:38
@user131711 Clearly there must be an error by the simple fact that if $Sigma X sim Sigma Y$, then $X$ and $Y$ are stable homotopy equivalent (by definition), and as above you may kill the cup product. In particular, suspending $D^4 cup_{2 in pi_3 S^2} S^2$ you get $S^5 vee S^3$. I think the error is this: The definition of the first diagonal map relies on the space $X$ as opposed to its stable homotopy type; suspension spectra have diagonal maps but that structure depends on $X$. You have therefore defined the cohomology ring of a space. (Spectra do not have natural diagonals.)
– Mike Miller
Nov 24 at 18:49
|
show 1 more comment
If I understood correctly, this gives examples of manifolds that are stably equivalent but have different cup product structures. Doesn't an equivalence of the spectra imply an isomorphism of the cohomology rings?
– user131711
Nov 24 at 18:22
1
@user131711 No. In fact, the cup product is destroyed after one stabilization: $tilde H^*(Sigma X)$ has trivial cup products. (This is a straightforward exercise using the relative cup product and the fact that $Sigma X$ has a cover by two contractible pieces.) All that lives is the Steenrod algebra action. This is visible at the level of spectra: $H^*(X; Bbb Z/2) = [Sigma^infty X, HBbb Z/2]$, and the Steenrod algebra is $[HBbb Z/2, HBbb Z/2]$, which acts by postcomposition.
– Mike Miller
Nov 24 at 18:29
The Steenrod algebra action is how one checks that $Sigma Bbb{CP}^2$ and $S^5$ are not stably equivalent.
– Mike Miller
Nov 24 at 18:32
Correct me if I am wrong, but the cup product on $Sigma X_+$ is given using the diagonal $Sigma^{infty}(Sigma X_+)rightarrowSigma^{infty}(Sigma X_+)wedge Sigma^{infty}(Sigma X_+)$. Since $S^1rightarrow S^1wedge S^1$ is nullhomotopic, there is no cup product. On the other hand, the cup product on $X$ is induced by compositions $Sigma^{infty}X_+rightarrowSigma^{infty}X_+wedgeSigma^{infty}X_+rightarrow HRwedge HRrightarrow HR$. Shouldn't an equivalence of the spectra give an identification of the diagonal maps, and so the same cup products?
– user131711
Nov 24 at 18:38
@user131711 Clearly there must be an error by the simple fact that if $Sigma X sim Sigma Y$, then $X$ and $Y$ are stable homotopy equivalent (by definition), and as above you may kill the cup product. In particular, suspending $D^4 cup_{2 in pi_3 S^2} S^2$ you get $S^5 vee S^3$. I think the error is this: The definition of the first diagonal map relies on the space $X$ as opposed to its stable homotopy type; suspension spectra have diagonal maps but that structure depends on $X$. You have therefore defined the cohomology ring of a space. (Spectra do not have natural diagonals.)
– Mike Miller
Nov 24 at 18:49
If I understood correctly, this gives examples of manifolds that are stably equivalent but have different cup product structures. Doesn't an equivalence of the spectra imply an isomorphism of the cohomology rings?
– user131711
Nov 24 at 18:22
If I understood correctly, this gives examples of manifolds that are stably equivalent but have different cup product structures. Doesn't an equivalence of the spectra imply an isomorphism of the cohomology rings?
– user131711
Nov 24 at 18:22
1
1
@user131711 No. In fact, the cup product is destroyed after one stabilization: $tilde H^*(Sigma X)$ has trivial cup products. (This is a straightforward exercise using the relative cup product and the fact that $Sigma X$ has a cover by two contractible pieces.) All that lives is the Steenrod algebra action. This is visible at the level of spectra: $H^*(X; Bbb Z/2) = [Sigma^infty X, HBbb Z/2]$, and the Steenrod algebra is $[HBbb Z/2, HBbb Z/2]$, which acts by postcomposition.
– Mike Miller
Nov 24 at 18:29
@user131711 No. In fact, the cup product is destroyed after one stabilization: $tilde H^*(Sigma X)$ has trivial cup products. (This is a straightforward exercise using the relative cup product and the fact that $Sigma X$ has a cover by two contractible pieces.) All that lives is the Steenrod algebra action. This is visible at the level of spectra: $H^*(X; Bbb Z/2) = [Sigma^infty X, HBbb Z/2]$, and the Steenrod algebra is $[HBbb Z/2, HBbb Z/2]$, which acts by postcomposition.
– Mike Miller
Nov 24 at 18:29
The Steenrod algebra action is how one checks that $Sigma Bbb{CP}^2$ and $S^5$ are not stably equivalent.
– Mike Miller
Nov 24 at 18:32
The Steenrod algebra action is how one checks that $Sigma Bbb{CP}^2$ and $S^5$ are not stably equivalent.
– Mike Miller
Nov 24 at 18:32
Correct me if I am wrong, but the cup product on $Sigma X_+$ is given using the diagonal $Sigma^{infty}(Sigma X_+)rightarrowSigma^{infty}(Sigma X_+)wedge Sigma^{infty}(Sigma X_+)$. Since $S^1rightarrow S^1wedge S^1$ is nullhomotopic, there is no cup product. On the other hand, the cup product on $X$ is induced by compositions $Sigma^{infty}X_+rightarrowSigma^{infty}X_+wedgeSigma^{infty}X_+rightarrow HRwedge HRrightarrow HR$. Shouldn't an equivalence of the spectra give an identification of the diagonal maps, and so the same cup products?
– user131711
Nov 24 at 18:38
Correct me if I am wrong, but the cup product on $Sigma X_+$ is given using the diagonal $Sigma^{infty}(Sigma X_+)rightarrowSigma^{infty}(Sigma X_+)wedge Sigma^{infty}(Sigma X_+)$. Since $S^1rightarrow S^1wedge S^1$ is nullhomotopic, there is no cup product. On the other hand, the cup product on $X$ is induced by compositions $Sigma^{infty}X_+rightarrowSigma^{infty}X_+wedgeSigma^{infty}X_+rightarrow HRwedge HRrightarrow HR$. Shouldn't an equivalence of the spectra give an identification of the diagonal maps, and so the same cup products?
– user131711
Nov 24 at 18:38
@user131711 Clearly there must be an error by the simple fact that if $Sigma X sim Sigma Y$, then $X$ and $Y$ are stable homotopy equivalent (by definition), and as above you may kill the cup product. In particular, suspending $D^4 cup_{2 in pi_3 S^2} S^2$ you get $S^5 vee S^3$. I think the error is this: The definition of the first diagonal map relies on the space $X$ as opposed to its stable homotopy type; suspension spectra have diagonal maps but that structure depends on $X$. You have therefore defined the cohomology ring of a space. (Spectra do not have natural diagonals.)
– Mike Miller
Nov 24 at 18:49
@user131711 Clearly there must be an error by the simple fact that if $Sigma X sim Sigma Y$, then $X$ and $Y$ are stable homotopy equivalent (by definition), and as above you may kill the cup product. In particular, suspending $D^4 cup_{2 in pi_3 S^2} S^2$ you get $S^5 vee S^3$. I think the error is this: The definition of the first diagonal map relies on the space $X$ as opposed to its stable homotopy type; suspension spectra have diagonal maps but that structure depends on $X$. You have therefore defined the cohomology ring of a space. (Spectra do not have natural diagonals.)
– Mike Miller
Nov 24 at 18:49
|
show 1 more comment
up vote
18
down vote
The easiest examples are given by complements of open neighborhoods of distinct knots in $S^n$.
More generally, Spanier-Whitehead duality says that if $X$ is a compact simplicial complex and if $f,g:X rightarrow S^n$ are embeddings, then $S^n - Im(f)$ is stabily equivalent to $S^n - Im(g)$.
See my notes here for an elementary discussion.
answered Nov 23 at 19:01
Andy Putman
30.9k5132211
[Dold, Albrecht, A simple proof of the Jordan-Alexander complement theorem. Amer. Math. Monthly 100 (1993), no. 9, 856–857] has a six line elementary (but sneaky) proof that if A, a closed subset of R^m is homeomorphic to B, a closed subset of R^n, then the complement of A x 0 in R^{m+n} is homeomorphic to the complement of 0 x B is R^{m+n}.
– Nicholas Kuhn
yesterday
@NicholasKuhn: Wow, that's a really beautiful proof! Thanks for sharing it.
– Andy Putman
yesterday
add a comment |
up vote
18
down vote
The easiest examples are given by complements of open neighborhoods of distinct knots in $S^n$.
More generally, Spanier-Whitehead duality says that if $X$ is a compact simplicial complex and if $f,g:X rightarrow S^n$ are embeddings, then $S^n - Im(f)$ is stabily equivalent to $S^n - Im(g)$.
See my notes here for an elementary discussion.
answered Nov 23 at 19:01
Andy Putman
30.9k5132211
[Dold, Albrecht, A simple proof of the Jordan-Alexander complement theorem. Amer. Math. Monthly 100 (1993), no. 9, 856–857] has a six line elementary (but sneaky) proof that if A, a closed subset of R^m is homeomorphic to B, a closed subset of R^n, then the complement of A x 0 in R^{m+n} is homeomorphic to the complement of 0 x B is R^{m+n}.
– Nicholas Kuhn
yesterday
@NicholasKuhn: Wow, that's a really beautiful proof! Thanks for sharing it.
– Andy Putman
yesterday
add a comment |
up vote
18
down vote
up vote
18
down vote
The easiest examples are given by complements of open neighborhoods of distinct knots in $S^n$.
More generally, Spanier-Whitehead duality says that if $X$ is a compact simplicial complex and if $f,g:X rightarrow S^n$ are embeddings, then $S^n - Im(f)$ is stabily equivalent to $S^n - Im(g)$.
See my notes here for an elementary discussion.
answered Nov 23 at 19:01
Andy Putman
30.9k5132211
The easiest examples are given by complements of open neighborhoods of distinct knots in $S^n$.
More generally, Spanier-Whitehead duality says that if $X$ is a compact simplicial complex and if $f,g:X rightarrow S^n$ are embeddings, then $S^n - Im(f)$ is stabily equivalent to $S^n - Im(g)$.
See my notes here for an elementary discussion.
answered Nov 23 at 19:01
Andy Putman
30.9k5132211
edited Nov 23 at 19:02
answered Nov 23 at 19:01
Andy Putman
30.9k5132211
answered Nov 23 at 19:01
Andy Putman
30.9k5132211
answered Nov 23 at 19:01
Andy Putman
30.9k5132211
30.9k5132211
[Dold, Albrecht, A simple proof of the Jordan-Alexander complement theorem. Amer. Math. Monthly 100 (1993), no. 9, 856–857] has a six line elementary (but sneaky) proof that if A, a closed subset of R^m is homeomorphic to B, a closed subset of R^n, then the complement of A x 0 in R^{m+n} is homeomorphic to the complement of 0 x B is R^{m+n}.
– Nicholas Kuhn
yesterday
@NicholasKuhn: Wow, that's a really beautiful proof! Thanks for sharing it.
– Andy Putman
yesterday
add a comment |
[Dold, Albrecht, A simple proof of the Jordan-Alexander complement theorem. Amer. Math. Monthly 100 (1993), no. 9, 856–857] has a six line elementary (but sneaky) proof that if A, a closed subset of R^m is homeomorphic to B, a closed subset of R^n, then the complement of A x 0 in R^{m+n} is homeomorphic to the complement of 0 x B is R^{m+n}.
– Nicholas Kuhn
yesterday
@NicholasKuhn: Wow, that's a really beautiful proof! Thanks for sharing it.
– Andy Putman
yesterday
[Dold, Albrecht, A simple proof of the Jordan-Alexander complement theorem. Amer. Math. Monthly 100 (1993), no. 9, 856–857] has a six line elementary (but sneaky) proof that if A, a closed subset of R^m is homeomorphic to B, a closed subset of R^n, then the complement of A x 0 in R^{m+n} is homeomorphic to the complement of 0 x B is R^{m+n}.
– Nicholas Kuhn
yesterday
[Dold, Albrecht, A simple proof of the Jordan-Alexander complement theorem. Amer. Math. Monthly 100 (1993), no. 9, 856–857] has a six line elementary (but sneaky) proof that if A, a closed subset of R^m is homeomorphic to B, a closed subset of R^n, then the complement of A x 0 in R^{m+n} is homeomorphic to the complement of 0 x B is R^{m+n}.
– Nicholas Kuhn
yesterday
@NicholasKuhn: Wow, that's a really beautiful proof! Thanks for sharing it.
– Andy Putman
yesterday
@NicholasKuhn: Wow, that's a really beautiful proof! Thanks for sharing it.
– Andy Putman
yesterday
add a comment |
user131711 is a new contributor. Be nice, and check out our Code of Conduct.
user131711 is a new contributor. Be nice, and check out our Code of Conduct.
user131711 is a new contributor. Be nice, and check out our Code of Conduct.
user131711 is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to MathOverflow!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f316046%2fstably-equivalent-but-not-homotopy-equivalent%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
11
Any pair of homology spheres of the same dimension. After one stabilization they are equivalent by Whitehead.
– Mike Miller
Nov 23 at 19:11