Mixing
Does the integral cohomology ring determine the ring structure with other coefficients?
$begingroup$
Let $A$ be an abelian group. By the universal coefficient theorem, the cohomology group of a manifold with coefficient $A$ is determined by the integral cohomology group. How about the ring structure?
Let $M_1$ and $M_2$ be manifolds with $H^*(M_1; mathbb{Z}) cong H^*(M_2; mathbb{Z})$ as a graded ring. Is it true that $H^*(M_1; R) cong H^*(M_2; R)$ for any commutative ring $R$?
I guess the answer is no. What would be an example?
algebraic-topology manifolds homology-cohomology
asked Dec 12 '18 at 9:45
HwangHwang
302112
$endgroup$
add a comment |
$begingroup$
Let $A$ be an abelian group. By the universal coefficient theorem, the cohomology group of a manifold with coefficient $A$ is determined by the integral cohomology group. How about the ring structure?
Let $M_1$ and $M_2$ be manifolds with $H^*(M_1; mathbb{Z}) cong H^*(M_2; mathbb{Z})$ as a graded ring. Is it true that $H^*(M_1; R) cong H^*(M_2; R)$ for any commutative ring $R$?
I guess the answer is no. What would be an example?
algebraic-topology manifolds homology-cohomology
asked Dec 12 '18 at 9:45
HwangHwang
302112
$endgroup$
7
$begingroup$
The answer for CW complexes is no: $mathbb RP^3$ and $mathbb RP^2vee S^3$ have isomorphic integral cohomology rings. I don't have manifold examples, though.
$endgroup$
– Justin Young
Dec 12 '18 at 15:32
1
$begingroup$
@JustinYoung Let $D^2$ be a two dimensional open disc. Is $(mathbb{RP}^2 times D^2) setminus { mathrm{point} }$ homotopy equivalent to $mathbb{RP}^2 vee S^3$?
$endgroup$
– David E Speyer
Jan 10 at 2:22
1
$begingroup$
Could you explain a little more? The complement of a point in $mathbb{R}mathbb{P}^2 times D^2$ is homotopy equivalent to $(mathbb{R}mathbb{P}^2 times S^1) cup S^1 times D^2$. How do I see that this is homotopy equivalent to $mathbb{R}mathbb{P}^2 vee S^3$?
$endgroup$
– Hwang
Jan 11 at 1:13
2
$begingroup$
Every finite (even countable) CW complex is homotopy-equivalent to a manifold (typically noncompact). Hence, you also get a manifold example.
$endgroup$
– Moishe Cohen
Jan 12 at 3:03
add a comment |
$begingroup$
Let $A$ be an abelian group. By the universal coefficient theorem, the cohomology group of a manifold with coefficient $A$ is determined by the integral cohomology group. How about the ring structure?
Let $M_1$ and $M_2$ be manifolds with $H^*(M_1; mathbb{Z}) cong H^*(M_2; mathbb{Z})$ as a graded ring. Is it true that $H^*(M_1; R) cong H^*(M_2; R)$ for any commutative ring $R$?
I guess the answer is no. What would be an example?
algebraic-topology manifolds homology-cohomology
asked Dec 12 '18 at 9:45
HwangHwang
302112
$endgroup$
Let $A$ be an abelian group. By the universal coefficient theorem, the cohomology group of a manifold with coefficient $A$ is determined by the integral cohomology group. How about the ring structure?
Let $M_1$ and $M_2$ be manifolds with $H^*(M_1; mathbb{Z}) cong H^*(M_2; mathbb{Z})$ as a graded ring. Is it true that $H^*(M_1; R) cong H^*(M_2; R)$ for any commutative ring $R$?
I guess the answer is no. What would be an example?
algebraic-topology manifolds homology-cohomology
algebraic-topology manifolds homology-cohomology
asked Dec 12 '18 at 9:45
HwangHwang
302112
asked Dec 12 '18 at 9:45
HwangHwang
302112
asked Dec 12 '18 at 9:45
HwangHwang
302112
asked Dec 12 '18 at 9:45
HwangHwang
302112
asked Dec 12 '18 at 9:45
HwangHwang
302112
302112
7
$begingroup$
The answer for CW complexes is no: $mathbb RP^3$ and $mathbb RP^2vee S^3$ have isomorphic integral cohomology rings. I don't have manifold examples, though.
$endgroup$
– Justin Young
Dec 12 '18 at 15:32
1
$begingroup$
@JustinYoung Let $D^2$ be a two dimensional open disc. Is $(mathbb{RP}^2 times D^2) setminus { mathrm{point} }$ homotopy equivalent to $mathbb{RP}^2 vee S^3$?
$endgroup$
– David E Speyer
Jan 10 at 2:22
1
$begingroup$
Could you explain a little more? The complement of a point in $mathbb{R}mathbb{P}^2 times D^2$ is homotopy equivalent to $(mathbb{R}mathbb{P}^2 times S^1) cup S^1 times D^2$. How do I see that this is homotopy equivalent to $mathbb{R}mathbb{P}^2 vee S^3$?
$endgroup$
– Hwang
Jan 11 at 1:13
2
$begingroup$
Every finite (even countable) CW complex is homotopy-equivalent to a manifold (typically noncompact). Hence, you also get a manifold example.
$endgroup$
– Moishe Cohen
Jan 12 at 3:03
add a comment |
7
$begingroup$
The answer for CW complexes is no: $mathbb RP^3$ and $mathbb RP^2vee S^3$ have isomorphic integral cohomology rings. I don't have manifold examples, though.
$endgroup$
– Justin Young
Dec 12 '18 at 15:32
1
$begingroup$
@JustinYoung Let $D^2$ be a two dimensional open disc. Is $(mathbb{RP}^2 times D^2) setminus { mathrm{point} }$ homotopy equivalent to $mathbb{RP}^2 vee S^3$?
$endgroup$
– David E Speyer
Jan 10 at 2:22
1
$begingroup$
Could you explain a little more? The complement of a point in $mathbb{R}mathbb{P}^2 times D^2$ is homotopy equivalent to $(mathbb{R}mathbb{P}^2 times S^1) cup S^1 times D^2$. How do I see that this is homotopy equivalent to $mathbb{R}mathbb{P}^2 vee S^3$?
$endgroup$
– Hwang
Jan 11 at 1:13
2
$begingroup$
Every finite (even countable) CW complex is homotopy-equivalent to a manifold (typically noncompact). Hence, you also get a manifold example.
$endgroup$
– Moishe Cohen
Jan 12 at 3:03
7
7
$begingroup$
The answer for CW complexes is no: $mathbb RP^3$ and $mathbb RP^2vee S^3$ have isomorphic integral cohomology rings. I don't have manifold examples, though.
$endgroup$
– Justin Young
Dec 12 '18 at 15:32
$begingroup$
The answer for CW complexes is no: $mathbb RP^3$ and $mathbb RP^2vee S^3$ have isomorphic integral cohomology rings. I don't have manifold examples, though.
$endgroup$
– Justin Young
Dec 12 '18 at 15:32
1
1
$begingroup$
@JustinYoung Let $D^2$ be a two dimensional open disc. Is $(mathbb{RP}^2 times D^2) setminus { mathrm{point} }$ homotopy equivalent to $mathbb{RP}^2 vee S^3$?
$endgroup$
– David E Speyer
Jan 10 at 2:22
$begingroup$
@JustinYoung Let $D^2$ be a two dimensional open disc. Is $(mathbb{RP}^2 times D^2) setminus { mathrm{point} }$ homotopy equivalent to $mathbb{RP}^2 vee S^3$?
$endgroup$
– David E Speyer
Jan 10 at 2:22
1
1
$begingroup$
Could you explain a little more? The complement of a point in $mathbb{R}mathbb{P}^2 times D^2$ is homotopy equivalent to $(mathbb{R}mathbb{P}^2 times S^1) cup S^1 times D^2$. How do I see that this is homotopy equivalent to $mathbb{R}mathbb{P}^2 vee S^3$?
$endgroup$
– Hwang
Jan 11 at 1:13
$begingroup$
Could you explain a little more? The complement of a point in $mathbb{R}mathbb{P}^2 times D^2$ is homotopy equivalent to $(mathbb{R}mathbb{P}^2 times S^1) cup S^1 times D^2$. How do I see that this is homotopy equivalent to $mathbb{R}mathbb{P}^2 vee S^3$?
$endgroup$
– Hwang
Jan 11 at 1:13
2
2
$begingroup$
Every finite (even countable) CW complex is homotopy-equivalent to a manifold (typically noncompact). Hence, you also get a manifold example.
$endgroup$
– Moishe Cohen
Jan 12 at 3:03
$begingroup$
Every finite (even countable) CW complex is homotopy-equivalent to a manifold (typically noncompact). Hence, you also get a manifold example.
$endgroup$
– Moishe Cohen
Jan 12 at 3:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Lemma. Every finite CW complex is homotopy-equivalent to a smooth manifold .
Proof. (Rourke and Sanderson's book "Introduction to Piecewise-Linear Topology" will contain all the necessary background information.) Every finite CW complex is homotopy-equivalent to the underlying space of a finite simplicial complex. Every finite simplicial complex is a subcomplex of the face-complex of a simplex. Thus, every finite simplicial complex embeds in $R^n$. (One can actually do a bit better: Each finite $k$-dimensional simplicial complex embeds in $R^{2k+1}$, this is a version of Whitney's embedding theorem in the PL category.) Now, let $Xsubset R^n$ be a finite simplicial complex. Let $N(X)$ denote the regular neighborhood of $X$ in $R^n$. (The regular neighborhood of a subcomplex $X$ in a simplicial complex $Y$ is defined as follows. Take the 2nd barycentric subdivision $Y''$ of $Y$ and let $N(X)$ denote the subcomplex of $Y''$ consisting of simplices having nonempty intersection with $X''$.)
Then $N(R)$ is homotopy-equivalent to $|X|$. (This is a general fact about regular neighborhoods of subcomplexes in simplicial complexes.) The regular neighborhood is a codimension zero PL manifold with boundary in $R^n$. Hence, the interior of $N(X)$ is an open subset in $R^n$ homotopy-equivalent to $|X|$. qed
Remark. One can do a bit better: If $X$ is a $k$-dimensional countable CW complex, then $X$ is homotopy-equivalent to a smooth $2k$-dimensional manifold. But you do not need this.
In view of this lemma, since you already have finite CW complexes answering your question ($RP^3$ and $RP^2vee S^3$), you also have manifolds answering your question.
answered Jan 14 at 3:21
Moishe CohenMoishe Cohen
46.6k342108
$endgroup$
$begingroup$
It would be nice to have a closed example too, which requires a little more work (which as you see by my deleted answer I wasn't willing to do).
$endgroup$
– Mike Miller
Jan 14 at 23:18
$begingroup$
@MikeMiller: True. One can try to double the manifold in my answer along the boundary. I do not have time for the computation now...
$endgroup$
– Moishe Cohen
Jan 14 at 23:53
add a comment |
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$begingroup$
Lemma. Every finite CW complex is homotopy-equivalent to a smooth manifold .
Proof. (Rourke and Sanderson's book "Introduction to Piecewise-Linear Topology" will contain all the necessary background information.) Every finite CW complex is homotopy-equivalent to the underlying space of a finite simplicial complex. Every finite simplicial complex is a subcomplex of the face-complex of a simplex. Thus, every finite simplicial complex embeds in $R^n$. (One can actually do a bit better: Each finite $k$-dimensional simplicial complex embeds in $R^{2k+1}$, this is a version of Whitney's embedding theorem in the PL category.) Now, let $Xsubset R^n$ be a finite simplicial complex. Let $N(X)$ denote the regular neighborhood of $X$ in $R^n$. (The regular neighborhood of a subcomplex $X$ in a simplicial complex $Y$ is defined as follows. Take the 2nd barycentric subdivision $Y''$ of $Y$ and let $N(X)$ denote the subcomplex of $Y''$ consisting of simplices having nonempty intersection with $X''$.)
Then $N(R)$ is homotopy-equivalent to $|X|$. (This is a general fact about regular neighborhoods of subcomplexes in simplicial complexes.) The regular neighborhood is a codimension zero PL manifold with boundary in $R^n$. Hence, the interior of $N(X)$ is an open subset in $R^n$ homotopy-equivalent to $|X|$. qed
Remark. One can do a bit better: If $X$ is a $k$-dimensional countable CW complex, then $X$ is homotopy-equivalent to a smooth $2k$-dimensional manifold. But you do not need this.
In view of this lemma, since you already have finite CW complexes answering your question ($RP^3$ and $RP^2vee S^3$), you also have manifolds answering your question.
answered Jan 14 at 3:21
Moishe CohenMoishe Cohen
46.6k342108
$endgroup$
$begingroup$
It would be nice to have a closed example too, which requires a little more work (which as you see by my deleted answer I wasn't willing to do).
$endgroup$
– Mike Miller
Jan 14 at 23:18
$begingroup$
@MikeMiller: True. One can try to double the manifold in my answer along the boundary. I do not have time for the computation now...
$endgroup$
– Moishe Cohen
Jan 14 at 23:53
add a comment |
$begingroup$
Lemma. Every finite CW complex is homotopy-equivalent to a smooth manifold .
Proof. (Rourke and Sanderson's book "Introduction to Piecewise-Linear Topology" will contain all the necessary background information.) Every finite CW complex is homotopy-equivalent to the underlying space of a finite simplicial complex. Every finite simplicial complex is a subcomplex of the face-complex of a simplex. Thus, every finite simplicial complex embeds in $R^n$. (One can actually do a bit better: Each finite $k$-dimensional simplicial complex embeds in $R^{2k+1}$, this is a version of Whitney's embedding theorem in the PL category.) Now, let $Xsubset R^n$ be a finite simplicial complex. Let $N(X)$ denote the regular neighborhood of $X$ in $R^n$. (The regular neighborhood of a subcomplex $X$ in a simplicial complex $Y$ is defined as follows. Take the 2nd barycentric subdivision $Y''$ of $Y$ and let $N(X)$ denote the subcomplex of $Y''$ consisting of simplices having nonempty intersection with $X''$.)
Then $N(R)$ is homotopy-equivalent to $|X|$. (This is a general fact about regular neighborhoods of subcomplexes in simplicial complexes.) The regular neighborhood is a codimension zero PL manifold with boundary in $R^n$. Hence, the interior of $N(X)$ is an open subset in $R^n$ homotopy-equivalent to $|X|$. qed
Remark. One can do a bit better: If $X$ is a $k$-dimensional countable CW complex, then $X$ is homotopy-equivalent to a smooth $2k$-dimensional manifold. But you do not need this.
In view of this lemma, since you already have finite CW complexes answering your question ($RP^3$ and $RP^2vee S^3$), you also have manifolds answering your question.
answered Jan 14 at 3:21
Moishe CohenMoishe Cohen
46.6k342108
$endgroup$
$begingroup$
It would be nice to have a closed example too, which requires a little more work (which as you see by my deleted answer I wasn't willing to do).
$endgroup$
– Mike Miller
Jan 14 at 23:18
$begingroup$
@MikeMiller: True. One can try to double the manifold in my answer along the boundary. I do not have time for the computation now...
$endgroup$
– Moishe Cohen
Jan 14 at 23:53
add a comment |
$begingroup$
Lemma. Every finite CW complex is homotopy-equivalent to a smooth manifold .
Proof. (Rourke and Sanderson's book "Introduction to Piecewise-Linear Topology" will contain all the necessary background information.) Every finite CW complex is homotopy-equivalent to the underlying space of a finite simplicial complex. Every finite simplicial complex is a subcomplex of the face-complex of a simplex. Thus, every finite simplicial complex embeds in $R^n$. (One can actually do a bit better: Each finite $k$-dimensional simplicial complex embeds in $R^{2k+1}$, this is a version of Whitney's embedding theorem in the PL category.) Now, let $Xsubset R^n$ be a finite simplicial complex. Let $N(X)$ denote the regular neighborhood of $X$ in $R^n$. (The regular neighborhood of a subcomplex $X$ in a simplicial complex $Y$ is defined as follows. Take the 2nd barycentric subdivision $Y''$ of $Y$ and let $N(X)$ denote the subcomplex of $Y''$ consisting of simplices having nonempty intersection with $X''$.)
Then $N(R)$ is homotopy-equivalent to $|X|$. (This is a general fact about regular neighborhoods of subcomplexes in simplicial complexes.) The regular neighborhood is a codimension zero PL manifold with boundary in $R^n$. Hence, the interior of $N(X)$ is an open subset in $R^n$ homotopy-equivalent to $|X|$. qed
Remark. One can do a bit better: If $X$ is a $k$-dimensional countable CW complex, then $X$ is homotopy-equivalent to a smooth $2k$-dimensional manifold. But you do not need this.
In view of this lemma, since you already have finite CW complexes answering your question ($RP^3$ and $RP^2vee S^3$), you also have manifolds answering your question.
answered Jan 14 at 3:21
Moishe CohenMoishe Cohen
46.6k342108
$endgroup$
Lemma. Every finite CW complex is homotopy-equivalent to a smooth manifold .
Proof. (Rourke and Sanderson's book "Introduction to Piecewise-Linear Topology" will contain all the necessary background information.) Every finite CW complex is homotopy-equivalent to the underlying space of a finite simplicial complex. Every finite simplicial complex is a subcomplex of the face-complex of a simplex. Thus, every finite simplicial complex embeds in $R^n$. (One can actually do a bit better: Each finite $k$-dimensional simplicial complex embeds in $R^{2k+1}$, this is a version of Whitney's embedding theorem in the PL category.) Now, let $Xsubset R^n$ be a finite simplicial complex. Let $N(X)$ denote the regular neighborhood of $X$ in $R^n$. (The regular neighborhood of a subcomplex $X$ in a simplicial complex $Y$ is defined as follows. Take the 2nd barycentric subdivision $Y''$ of $Y$ and let $N(X)$ denote the subcomplex of $Y''$ consisting of simplices having nonempty intersection with $X''$.)
Then $N(R)$ is homotopy-equivalent to $|X|$. (This is a general fact about regular neighborhoods of subcomplexes in simplicial complexes.) The regular neighborhood is a codimension zero PL manifold with boundary in $R^n$. Hence, the interior of $N(X)$ is an open subset in $R^n$ homotopy-equivalent to $|X|$. qed
Remark. One can do a bit better: If $X$ is a $k$-dimensional countable CW complex, then $X$ is homotopy-equivalent to a smooth $2k$-dimensional manifold. But you do not need this.
In view of this lemma, since you already have finite CW complexes answering your question ($RP^3$ and $RP^2vee S^3$), you also have manifolds answering your question.
answered Jan 14 at 3:21
Moishe CohenMoishe Cohen
46.6k342108
edited Jan 14 at 16:20
answered Jan 14 at 3:21
Moishe CohenMoishe Cohen
46.6k342108
answered Jan 14 at 3:21
Moishe CohenMoishe Cohen
46.6k342108
answered Jan 14 at 3:21
Moishe CohenMoishe Cohen
46.6k342108
46.6k342108
$begingroup$
It would be nice to have a closed example too, which requires a little more work (which as you see by my deleted answer I wasn't willing to do).
$endgroup$
– Mike Miller
Jan 14 at 23:18
$begingroup$
@MikeMiller: True. One can try to double the manifold in my answer along the boundary. I do not have time for the computation now...
$endgroup$
– Moishe Cohen
Jan 14 at 23:53
add a comment |
$begingroup$
It would be nice to have a closed example too, which requires a little more work (which as you see by my deleted answer I wasn't willing to do).
$endgroup$
– Mike Miller
Jan 14 at 23:18
$begingroup$
@MikeMiller: True. One can try to double the manifold in my answer along the boundary. I do not have time for the computation now...
$endgroup$
– Moishe Cohen
Jan 14 at 23:53
$begingroup$
It would be nice to have a closed example too, which requires a little more work (which as you see by my deleted answer I wasn't willing to do).
$endgroup$
– Mike Miller
Jan 14 at 23:18
$begingroup$
It would be nice to have a closed example too, which requires a little more work (which as you see by my deleted answer I wasn't willing to do).
$endgroup$
– Mike Miller
Jan 14 at 23:18
$begingroup$
@MikeMiller: True. One can try to double the manifold in my answer along the boundary. I do not have time for the computation now...
$endgroup$
– Moishe Cohen
Jan 14 at 23:53
$begingroup$
@MikeMiller: True. One can try to double the manifold in my answer along the boundary. I do not have time for the computation now...
$endgroup$
– Moishe Cohen
Jan 14 at 23:53
add a comment |
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$begingroup$
The answer for CW complexes is no: $mathbb RP^3$ and $mathbb RP^2vee S^3$ have isomorphic integral cohomology rings. I don't have manifold examples, though.
$endgroup$
– Justin Young
Dec 12 '18 at 15:32
1
$begingroup$
@JustinYoung Let $D^2$ be a two dimensional open disc. Is $(mathbb{RP}^2 times D^2) setminus { mathrm{point} }$ homotopy equivalent to $mathbb{RP}^2 vee S^3$?
$endgroup$
– David E Speyer
Jan 10 at 2:22
1
$begingroup$
Could you explain a little more? The complement of a point in $mathbb{R}mathbb{P}^2 times D^2$ is homotopy equivalent to $(mathbb{R}mathbb{P}^2 times S^1) cup S^1 times D^2$. How do I see that this is homotopy equivalent to $mathbb{R}mathbb{P}^2 vee S^3$?
$endgroup$
– Hwang
Jan 11 at 1:13
2
$begingroup$
Every finite (even countable) CW complex is homotopy-equivalent to a manifold (typically noncompact). Hence, you also get a manifold example.
$endgroup$
– Moishe Cohen
Jan 12 at 3:03