Stably equivalent but not homotopy equivalent
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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
1315
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up vote
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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
1315
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
1315
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
1315
asked Nov 23 at 17:33
user131711
1315
asked Nov 23 at 17:33
user131711
1315
asked Nov 23 at 17:33
user131711
1315
asked Nov 23 at 17:33
user131711
1315
1315
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
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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
|
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18
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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
31.1k5132212
[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
Nov 28 at 22:02
@NicholasKuhn: Wow, that's a really beautiful proof! Thanks for sharing it.
– Andy Putman
Nov 28 at 22:16
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2 Answers
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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
31.1k5132212
[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
Nov 28 at 22:02
@NicholasKuhn: Wow, that's a really beautiful proof! Thanks for sharing it.
– Andy Putman
Nov 28 at 22:16
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
31.1k5132212
[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
Nov 28 at 22:02
@NicholasKuhn: Wow, that's a really beautiful proof! Thanks for sharing it.
– Andy Putman
Nov 28 at 22:16
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
31.1k5132212
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
31.1k5132212
edited Nov 23 at 19:02
answered Nov 23 at 19:01
Andy Putman
31.1k5132212
answered Nov 23 at 19:01
Andy Putman
31.1k5132212
answered Nov 23 at 19:01
Andy Putman
31.1k5132212
31.1k5132212
[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
Nov 28 at 22:02
@NicholasKuhn: Wow, that's a really beautiful proof! Thanks for sharing it.
– Andy Putman
Nov 28 at 22:16
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
Nov 28 at 22:02
@NicholasKuhn: Wow, that's a really beautiful proof! Thanks for sharing it.
– Andy Putman
Nov 28 at 22:16
[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
Nov 28 at 22:02
[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
Nov 28 at 22:02
@NicholasKuhn: Wow, that's a really beautiful proof! Thanks for sharing it.
– Andy Putman
Nov 28 at 22:16
@NicholasKuhn: Wow, that's a really beautiful proof! Thanks for sharing it.
– Andy Putman
Nov 28 at 22:16
add a comment |
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Any pair of homology spheres of the same dimension. After one stabilization they are equivalent by Whitehead.
– Mike Miller
Nov 23 at 19:11