Does the integral cohomology ring determine the ring structure with other coefficients?












11












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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
















11












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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














11












11








11


3



$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










1 Answer
1






active

oldest

votes


















2





+50







$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

votes






active

oldest

votes









2





+50







$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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
















2





+50







$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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














2





+50







2





+50



2




+50



$begingroup$

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.






share|cite|improve this answer











$endgroup$



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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 14 at 16:20

























answered Jan 14 at 3:21









Moishe CohenMoishe Cohen

47.2k343108




47.2k343108












  • $begingroup$
    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










  • $begingroup$
    @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


















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張江高科駅

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張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
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