Construct compact space with some homology group not finitely generated
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I would like to construct a compact space $X$ that has a non finitely generated singular homology group $H_n (X)$ for some $n$. I thought about taking a countable wedge sum of 1-spheres, but this space is not compact. Another idea would be the Hawaiian earring which should be compact and have an infinitely generated homology group, but this is quite difficult to calculate.
algebraic-topology homology-cohomology
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I would like to construct a compact space $X$ that has a non finitely generated singular homology group $H_n (X)$ for some $n$. I thought about taking a countable wedge sum of 1-spheres, but this space is not compact. Another idea would be the Hawaiian earring which should be compact and have an infinitely generated homology group, but this is quite difficult to calculate.
algebraic-topology homology-cohomology
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This thread might interest you: math.stackexchange.com/questions/2512580/…
– Aleksandar Milivojevic
4 hours ago
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up vote
2
down vote
favorite
up vote
2
down vote
favorite
I would like to construct a compact space $X$ that has a non finitely generated singular homology group $H_n (X)$ for some $n$. I thought about taking a countable wedge sum of 1-spheres, but this space is not compact. Another idea would be the Hawaiian earring which should be compact and have an infinitely generated homology group, but this is quite difficult to calculate.
algebraic-topology homology-cohomology
I would like to construct a compact space $X$ that has a non finitely generated singular homology group $H_n (X)$ for some $n$. I thought about taking a countable wedge sum of 1-spheres, but this space is not compact. Another idea would be the Hawaiian earring which should be compact and have an infinitely generated homology group, but this is quite difficult to calculate.
algebraic-topology homology-cohomology
algebraic-topology homology-cohomology
asked 5 hours ago
CHwC
229111
229111
1
This thread might interest you: math.stackexchange.com/questions/2512580/…
– Aleksandar Milivojevic
4 hours ago
add a comment |
1
This thread might interest you: math.stackexchange.com/questions/2512580/…
– Aleksandar Milivojevic
4 hours ago
1
1
This thread might interest you: math.stackexchange.com/questions/2512580/…
– Aleksandar Milivojevic
4 hours ago
This thread might interest you: math.stackexchange.com/questions/2512580/…
– Aleksandar Milivojevic
4 hours ago
add a comment |
2 Answers
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The Hawaiian earring works just fine. You don't have to explicitly compute its homology to show it is not finitely generated. Specifically let $X$ be the Hawaiian earring space. Note that for any $ninmathbb{N}$, $X$ retracts onto a wedge of $n$ circles (just take $n$ of the circles that make up $X$, and map all the rest of the circles to the point where the circles meet). This implies that $H_1(X)$ has $mathbb{Z}^n$ as a direct summand for all $ninmathbb{Z}$. This implies $H_1(X)$ is not finitely generated.
For an even easier example, you could take $X$ to be any infinite compact totally disconnected space (say, ${0}cup{1/n:ninmathbb{Z}_+}$, or a Cantor set). Then $H_0(X)$ is not finitely generated, since it is freely generated by the path-components of $X$ and there are infinitely many path-components.
how do you conclude that Z^n has to be a direct summand and the step where you say that it is freely generated by the path-components?
– CHwC
4 hours ago
okay the second point is simply the definition of singular homology, but what about the first?
– CHwC
3 hours ago
The retraction gives you a surjection of $H_1$ onto $mathbb{Z}^n$. Since $mathbb{Z}^n$ is free abelian, this surjects splits, and the claim follows by the splitting lemma en.m.wikipedia.org/wiki/Splitting_lemma
– Aleksandar Milivojevic
2 hours ago
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3
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The Cantor set $C subset [0,1]$ has uncountably many path components (in fact, each single point subset is a path component of $C$). Hence $H_0(C)$ is a free abelian group with uncountably many generators.
Taking the suspension $Sigma$ and noting that $tilde{H}_{n+1}(Sigma X) approx tilde{H}_n(X)$, where $tilde{H}_*$ denotes reduced homology, you can construct examples for all $H_i$ with $i ge 0$. Recall that $tilde{H}_i = H_i$ for $i > 0$.
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
The Hawaiian earring works just fine. You don't have to explicitly compute its homology to show it is not finitely generated. Specifically let $X$ be the Hawaiian earring space. Note that for any $ninmathbb{N}$, $X$ retracts onto a wedge of $n$ circles (just take $n$ of the circles that make up $X$, and map all the rest of the circles to the point where the circles meet). This implies that $H_1(X)$ has $mathbb{Z}^n$ as a direct summand for all $ninmathbb{Z}$. This implies $H_1(X)$ is not finitely generated.
For an even easier example, you could take $X$ to be any infinite compact totally disconnected space (say, ${0}cup{1/n:ninmathbb{Z}_+}$, or a Cantor set). Then $H_0(X)$ is not finitely generated, since it is freely generated by the path-components of $X$ and there are infinitely many path-components.
how do you conclude that Z^n has to be a direct summand and the step where you say that it is freely generated by the path-components?
– CHwC
4 hours ago
okay the second point is simply the definition of singular homology, but what about the first?
– CHwC
3 hours ago
The retraction gives you a surjection of $H_1$ onto $mathbb{Z}^n$. Since $mathbb{Z}^n$ is free abelian, this surjects splits, and the claim follows by the splitting lemma en.m.wikipedia.org/wiki/Splitting_lemma
– Aleksandar Milivojevic
2 hours ago
add a comment |
up vote
3
down vote
The Hawaiian earring works just fine. You don't have to explicitly compute its homology to show it is not finitely generated. Specifically let $X$ be the Hawaiian earring space. Note that for any $ninmathbb{N}$, $X$ retracts onto a wedge of $n$ circles (just take $n$ of the circles that make up $X$, and map all the rest of the circles to the point where the circles meet). This implies that $H_1(X)$ has $mathbb{Z}^n$ as a direct summand for all $ninmathbb{Z}$. This implies $H_1(X)$ is not finitely generated.
For an even easier example, you could take $X$ to be any infinite compact totally disconnected space (say, ${0}cup{1/n:ninmathbb{Z}_+}$, or a Cantor set). Then $H_0(X)$ is not finitely generated, since it is freely generated by the path-components of $X$ and there are infinitely many path-components.
how do you conclude that Z^n has to be a direct summand and the step where you say that it is freely generated by the path-components?
– CHwC
4 hours ago
okay the second point is simply the definition of singular homology, but what about the first?
– CHwC
3 hours ago
The retraction gives you a surjection of $H_1$ onto $mathbb{Z}^n$. Since $mathbb{Z}^n$ is free abelian, this surjects splits, and the claim follows by the splitting lemma en.m.wikipedia.org/wiki/Splitting_lemma
– Aleksandar Milivojevic
2 hours ago
add a comment |
up vote
3
down vote
up vote
3
down vote
The Hawaiian earring works just fine. You don't have to explicitly compute its homology to show it is not finitely generated. Specifically let $X$ be the Hawaiian earring space. Note that for any $ninmathbb{N}$, $X$ retracts onto a wedge of $n$ circles (just take $n$ of the circles that make up $X$, and map all the rest of the circles to the point where the circles meet). This implies that $H_1(X)$ has $mathbb{Z}^n$ as a direct summand for all $ninmathbb{Z}$. This implies $H_1(X)$ is not finitely generated.
For an even easier example, you could take $X$ to be any infinite compact totally disconnected space (say, ${0}cup{1/n:ninmathbb{Z}_+}$, or a Cantor set). Then $H_0(X)$ is not finitely generated, since it is freely generated by the path-components of $X$ and there are infinitely many path-components.
The Hawaiian earring works just fine. You don't have to explicitly compute its homology to show it is not finitely generated. Specifically let $X$ be the Hawaiian earring space. Note that for any $ninmathbb{N}$, $X$ retracts onto a wedge of $n$ circles (just take $n$ of the circles that make up $X$, and map all the rest of the circles to the point where the circles meet). This implies that $H_1(X)$ has $mathbb{Z}^n$ as a direct summand for all $ninmathbb{Z}$. This implies $H_1(X)$ is not finitely generated.
For an even easier example, you could take $X$ to be any infinite compact totally disconnected space (say, ${0}cup{1/n:ninmathbb{Z}_+}$, or a Cantor set). Then $H_0(X)$ is not finitely generated, since it is freely generated by the path-components of $X$ and there are infinitely many path-components.
answered 4 hours ago
Eric Wofsey
176k12202326
176k12202326
how do you conclude that Z^n has to be a direct summand and the step where you say that it is freely generated by the path-components?
– CHwC
4 hours ago
okay the second point is simply the definition of singular homology, but what about the first?
– CHwC
3 hours ago
The retraction gives you a surjection of $H_1$ onto $mathbb{Z}^n$. Since $mathbb{Z}^n$ is free abelian, this surjects splits, and the claim follows by the splitting lemma en.m.wikipedia.org/wiki/Splitting_lemma
– Aleksandar Milivojevic
2 hours ago
add a comment |
how do you conclude that Z^n has to be a direct summand and the step where you say that it is freely generated by the path-components?
– CHwC
4 hours ago
okay the second point is simply the definition of singular homology, but what about the first?
– CHwC
3 hours ago
The retraction gives you a surjection of $H_1$ onto $mathbb{Z}^n$. Since $mathbb{Z}^n$ is free abelian, this surjects splits, and the claim follows by the splitting lemma en.m.wikipedia.org/wiki/Splitting_lemma
– Aleksandar Milivojevic
2 hours ago
how do you conclude that Z^n has to be a direct summand and the step where you say that it is freely generated by the path-components?
– CHwC
4 hours ago
how do you conclude that Z^n has to be a direct summand and the step where you say that it is freely generated by the path-components?
– CHwC
4 hours ago
okay the second point is simply the definition of singular homology, but what about the first?
– CHwC
3 hours ago
okay the second point is simply the definition of singular homology, but what about the first?
– CHwC
3 hours ago
The retraction gives you a surjection of $H_1$ onto $mathbb{Z}^n$. Since $mathbb{Z}^n$ is free abelian, this surjects splits, and the claim follows by the splitting lemma en.m.wikipedia.org/wiki/Splitting_lemma
– Aleksandar Milivojevic
2 hours ago
The retraction gives you a surjection of $H_1$ onto $mathbb{Z}^n$. Since $mathbb{Z}^n$ is free abelian, this surjects splits, and the claim follows by the splitting lemma en.m.wikipedia.org/wiki/Splitting_lemma
– Aleksandar Milivojevic
2 hours ago
add a comment |
up vote
3
down vote
The Cantor set $C subset [0,1]$ has uncountably many path components (in fact, each single point subset is a path component of $C$). Hence $H_0(C)$ is a free abelian group with uncountably many generators.
Taking the suspension $Sigma$ and noting that $tilde{H}_{n+1}(Sigma X) approx tilde{H}_n(X)$, where $tilde{H}_*$ denotes reduced homology, you can construct examples for all $H_i$ with $i ge 0$. Recall that $tilde{H}_i = H_i$ for $i > 0$.
add a comment |
up vote
3
down vote
The Cantor set $C subset [0,1]$ has uncountably many path components (in fact, each single point subset is a path component of $C$). Hence $H_0(C)$ is a free abelian group with uncountably many generators.
Taking the suspension $Sigma$ and noting that $tilde{H}_{n+1}(Sigma X) approx tilde{H}_n(X)$, where $tilde{H}_*$ denotes reduced homology, you can construct examples for all $H_i$ with $i ge 0$. Recall that $tilde{H}_i = H_i$ for $i > 0$.
add a comment |
up vote
3
down vote
up vote
3
down vote
The Cantor set $C subset [0,1]$ has uncountably many path components (in fact, each single point subset is a path component of $C$). Hence $H_0(C)$ is a free abelian group with uncountably many generators.
Taking the suspension $Sigma$ and noting that $tilde{H}_{n+1}(Sigma X) approx tilde{H}_n(X)$, where $tilde{H}_*$ denotes reduced homology, you can construct examples for all $H_i$ with $i ge 0$. Recall that $tilde{H}_i = H_i$ for $i > 0$.
The Cantor set $C subset [0,1]$ has uncountably many path components (in fact, each single point subset is a path component of $C$). Hence $H_0(C)$ is a free abelian group with uncountably many generators.
Taking the suspension $Sigma$ and noting that $tilde{H}_{n+1}(Sigma X) approx tilde{H}_n(X)$, where $tilde{H}_*$ denotes reduced homology, you can construct examples for all $H_i$ with $i ge 0$. Recall that $tilde{H}_i = H_i$ for $i > 0$.
edited 1 hour ago
answered 3 hours ago
Paul Frost
7,8841527
7,8841527
add a comment |
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This thread might interest you: math.stackexchange.com/questions/2512580/…
– Aleksandar Milivojevic
4 hours ago