Can different topological spaces have the same square?
$begingroup$
For topological spaces $X$ and $Y$, is it possible that $X times X$ and $Y times Y$ are homeomorphic, but $X$ and $Y$ are not homeomorphic?
(I poked around with finite spaces, and manifolds, and the Cantor set, without seeing any examples.)
This was inspired by Existence of topological space which has no “square-root” but whose “cube” has a “square-root”. Cartesian product makes the proper class of topological spaces (up to homeomorphism) into a large abelian monoid, so here's a bonus question: what is known about the structure of this monoid? For example, the dogbone space shows that it is not cancellative. Does it have torsion in the sense that sometimes $X^n notcong X$ but $X^{n + 1} cong X$?
general-topology
asked Dec 31 '18 at 3:02
Hew WolffHew Wolff
2,255716
$endgroup$
add a comment |
$begingroup$
For topological spaces $X$ and $Y$, is it possible that $X times X$ and $Y times Y$ are homeomorphic, but $X$ and $Y$ are not homeomorphic?
(I poked around with finite spaces, and manifolds, and the Cantor set, without seeing any examples.)
This was inspired by Existence of topological space which has no “square-root” but whose “cube” has a “square-root”. Cartesian product makes the proper class of topological spaces (up to homeomorphism) into a large abelian monoid, so here's a bonus question: what is known about the structure of this monoid? For example, the dogbone space shows that it is not cancellative. Does it have torsion in the sense that sometimes $X^n notcong X$ but $X^{n + 1} cong X$?
general-topology
asked Dec 31 '18 at 3:02
Hew WolffHew Wolff
2,255716
$endgroup$
3
$begingroup$
It's not possible for finite spaces, by my argument at mathoverflow.net/questions/269542/….
$endgroup$
– Eric Wofsey
Dec 31 '18 at 3:15
1
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Maybe take a look at the Whitehead manifold. See: numdam.org/article/BSMF_1960__88__131_0.pdf and en.wikipedia.org/wiki/Whitehead_manifold
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– Alvin Jin
Dec 31 '18 at 3:31
2
$begingroup$
The existence of such spaces $X, Y$ is what's used in Mike Miller 's example, see his answer in the linked question. But Whitehead manifold would also do the job.
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– Moishe Cohen
Dec 31 '18 at 3:35
2
$begingroup$
As for the "torsion" examples, see references in mathoverflow.net/questions/10128/when-is-a-isomorphic-to-a3
$endgroup$
– Moishe Cohen
Dec 31 '18 at 4:13
$begingroup$
@MoisheCohen Interesting, for OP's sake I will observe that there exists a functor from groups to spaces, given by constructing a simplicial complex called "$BG$" whose $n$-simplices are labelled by sequences $[g_1, cdots, g_n]$ of elements of $G$ (and the gluings involve multiplications), and $B(G times H) = BG times BH$. In particular, functoriality means that if $G$ and $H$ are homeomorphic, so are $BG$ and $BH$, and conversely (if $BG cong BH$ then $G cong H$ by looking at fundamental groups). So if $A cong A^3 neq A^2$, then $BA cong (BA)^3 neq (BA)^2$.
$endgroup$
– Mike Miller
Dec 31 '18 at 4:23
add a comment |
$begingroup$
For topological spaces $X$ and $Y$, is it possible that $X times X$ and $Y times Y$ are homeomorphic, but $X$ and $Y$ are not homeomorphic?
(I poked around with finite spaces, and manifolds, and the Cantor set, without seeing any examples.)
This was inspired by Existence of topological space which has no “square-root” but whose “cube” has a “square-root”. Cartesian product makes the proper class of topological spaces (up to homeomorphism) into a large abelian monoid, so here's a bonus question: what is known about the structure of this monoid? For example, the dogbone space shows that it is not cancellative. Does it have torsion in the sense that sometimes $X^n notcong X$ but $X^{n + 1} cong X$?
general-topology
asked Dec 31 '18 at 3:02
Hew WolffHew Wolff
2,255716
$endgroup$
For topological spaces $X$ and $Y$, is it possible that $X times X$ and $Y times Y$ are homeomorphic, but $X$ and $Y$ are not homeomorphic?
(I poked around with finite spaces, and manifolds, and the Cantor set, without seeing any examples.)
This was inspired by Existence of topological space which has no “square-root” but whose “cube” has a “square-root”. Cartesian product makes the proper class of topological spaces (up to homeomorphism) into a large abelian monoid, so here's a bonus question: what is known about the structure of this monoid? For example, the dogbone space shows that it is not cancellative. Does it have torsion in the sense that sometimes $X^n notcong X$ but $X^{n + 1} cong X$?
general-topology
general-topology
asked Dec 31 '18 at 3:02
Hew WolffHew Wolff
2,255716
asked Dec 31 '18 at 3:02
Hew WolffHew Wolff
2,255716
asked Dec 31 '18 at 3:02
Hew WolffHew Wolff
2,255716
asked Dec 31 '18 at 3:02
Hew WolffHew Wolff
2,255716
asked Dec 31 '18 at 3:02
Hew WolffHew Wolff
2,255716
2,255716
3
$begingroup$
It's not possible for finite spaces, by my argument at mathoverflow.net/questions/269542/….
$endgroup$
– Eric Wofsey
Dec 31 '18 at 3:15
1
$begingroup$
Maybe take a look at the Whitehead manifold. See: numdam.org/article/BSMF_1960__88__131_0.pdf and en.wikipedia.org/wiki/Whitehead_manifold
$endgroup$
– Alvin Jin
Dec 31 '18 at 3:31
2
$begingroup$
The existence of such spaces $X, Y$ is what's used in Mike Miller 's example, see his answer in the linked question. But Whitehead manifold would also do the job.
$endgroup$
– Moishe Cohen
Dec 31 '18 at 3:35
2
$begingroup$
As for the "torsion" examples, see references in mathoverflow.net/questions/10128/when-is-a-isomorphic-to-a3
$endgroup$
– Moishe Cohen
Dec 31 '18 at 4:13
$begingroup$
@MoisheCohen Interesting, for OP's sake I will observe that there exists a functor from groups to spaces, given by constructing a simplicial complex called "$BG$" whose $n$-simplices are labelled by sequences $[g_1, cdots, g_n]$ of elements of $G$ (and the gluings involve multiplications), and $B(G times H) = BG times BH$. In particular, functoriality means that if $G$ and $H$ are homeomorphic, so are $BG$ and $BH$, and conversely (if $BG cong BH$ then $G cong H$ by looking at fundamental groups). So if $A cong A^3 neq A^2$, then $BA cong (BA)^3 neq (BA)^2$.
$endgroup$
– Mike Miller
Dec 31 '18 at 4:23
add a comment |
3
$begingroup$
It's not possible for finite spaces, by my argument at mathoverflow.net/questions/269542/….
$endgroup$
– Eric Wofsey
Dec 31 '18 at 3:15
1
$begingroup$
Maybe take a look at the Whitehead manifold. See: numdam.org/article/BSMF_1960__88__131_0.pdf and en.wikipedia.org/wiki/Whitehead_manifold
$endgroup$
– Alvin Jin
Dec 31 '18 at 3:31
2
$begingroup$
The existence of such spaces $X, Y$ is what's used in Mike Miller 's example, see his answer in the linked question. But Whitehead manifold would also do the job.
$endgroup$
– Moishe Cohen
Dec 31 '18 at 3:35
2
$begingroup$
As for the "torsion" examples, see references in mathoverflow.net/questions/10128/when-is-a-isomorphic-to-a3
$endgroup$
– Moishe Cohen
Dec 31 '18 at 4:13
$begingroup$
@MoisheCohen Interesting, for OP's sake I will observe that there exists a functor from groups to spaces, given by constructing a simplicial complex called "$BG$" whose $n$-simplices are labelled by sequences $[g_1, cdots, g_n]$ of elements of $G$ (and the gluings involve multiplications), and $B(G times H) = BG times BH$. In particular, functoriality means that if $G$ and $H$ are homeomorphic, so are $BG$ and $BH$, and conversely (if $BG cong BH$ then $G cong H$ by looking at fundamental groups). So if $A cong A^3 neq A^2$, then $BA cong (BA)^3 neq (BA)^2$.
$endgroup$
– Mike Miller
Dec 31 '18 at 4:23
3
3
$begingroup$
It's not possible for finite spaces, by my argument at mathoverflow.net/questions/269542/….
$endgroup$
– Eric Wofsey
Dec 31 '18 at 3:15
$begingroup$
It's not possible for finite spaces, by my argument at mathoverflow.net/questions/269542/….
$endgroup$
– Eric Wofsey
Dec 31 '18 at 3:15
1
1
$begingroup$
Maybe take a look at the Whitehead manifold. See: numdam.org/article/BSMF_1960__88__131_0.pdf and en.wikipedia.org/wiki/Whitehead_manifold
$endgroup$
– Alvin Jin
Dec 31 '18 at 3:31
$begingroup$
Maybe take a look at the Whitehead manifold. See: numdam.org/article/BSMF_1960__88__131_0.pdf and en.wikipedia.org/wiki/Whitehead_manifold
$endgroup$
– Alvin Jin
Dec 31 '18 at 3:31
2
2
$begingroup$
The existence of such spaces $X, Y$ is what's used in Mike Miller 's example, see his answer in the linked question. But Whitehead manifold would also do the job.
$endgroup$
– Moishe Cohen
Dec 31 '18 at 3:35
$begingroup$
The existence of such spaces $X, Y$ is what's used in Mike Miller 's example, see his answer in the linked question. But Whitehead manifold would also do the job.
$endgroup$
– Moishe Cohen
Dec 31 '18 at 3:35
2
2
$begingroup$
As for the "torsion" examples, see references in mathoverflow.net/questions/10128/when-is-a-isomorphic-to-a3
$endgroup$
– Moishe Cohen
Dec 31 '18 at 4:13
$begingroup$
As for the "torsion" examples, see references in mathoverflow.net/questions/10128/when-is-a-isomorphic-to-a3
$endgroup$
– Moishe Cohen
Dec 31 '18 at 4:13
$begingroup$
@MoisheCohen Interesting, for OP's sake I will observe that there exists a functor from groups to spaces, given by constructing a simplicial complex called "$BG$" whose $n$-simplices are labelled by sequences $[g_1, cdots, g_n]$ of elements of $G$ (and the gluings involve multiplications), and $B(G times H) = BG times BH$. In particular, functoriality means that if $G$ and $H$ are homeomorphic, so are $BG$ and $BH$, and conversely (if $BG cong BH$ then $G cong H$ by looking at fundamental groups). So if $A cong A^3 neq A^2$, then $BA cong (BA)^3 neq (BA)^2$.
$endgroup$
– Mike Miller
Dec 31 '18 at 4:23
$begingroup$
@MoisheCohen Interesting, for OP's sake I will observe that there exists a functor from groups to spaces, given by constructing a simplicial complex called "$BG$" whose $n$-simplices are labelled by sequences $[g_1, cdots, g_n]$ of elements of $G$ (and the gluings involve multiplications), and $B(G times H) = BG times BH$. In particular, functoriality means that if $G$ and $H$ are homeomorphic, so are $BG$ and $BH$, and conversely (if $BG cong BH$ then $G cong H$ by looking at fundamental groups). So if $A cong A^3 neq A^2$, then $BA cong (BA)^3 neq (BA)^2$.
$endgroup$
– Mike Miller
Dec 31 '18 at 4:23
add a comment |
1 Answer
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$begingroup$
Copying part of my answer to this Math Overflow question:
Whether two (a) topological spaces, (b) metric spaces, or (c) groups $A$ and $B$, with isomorphic squares, are necessarily isomorphic, was Problem 77 (of Ulam) in the Scottish book. The following information is from the commentaries on Problem 77, by Mauldin for (a) and (b) and by Kaplansky for (c), in The Scottish Book: Mathematics from the Scottish Café, R. Daniel Mauldin, ed., 1981.
Mauldin gives a bibliography of 17 items for parts (a) and (b). The first counterexample for part (a), as well as a positive answer for two-dimensional compact manifolds, was given by R. H. Fox, On a problem of S. Ulam concerning Cartesian products, Fund. Math. 34 (1947), 278-287. The first counterexample to part (b) was given by G. Fournier, On a problem of S. Ulam, Proc. Amer. Math. Soc. 29 (1971), 622. Mauldin writes (in 1981):
"However, it is open whether there is an affirmative solution to (b) in the case that $A$ and $B$ are complete metric spaces. In fact, part (b) is open in the case where $A$ and $B$ are assumed to be compact."
answered Dec 31 '18 at 3:43
bofbof
51.5k558120
$endgroup$
$begingroup$
I deleted my silly comment, which parsed (b) as "metrizable".
$endgroup$
– Mike Miller
Dec 31 '18 at 7:59
$begingroup$
For completeness: I see the Fox paper at matwbn.icm.edu.pl/ksiazki/fm/fm34/fm34128.pdf, and another example from Marjanović and Vučemilović at dml.cz/bitstream/handle/10338.dmlcz/106394/….
$endgroup$
– Hew Wolff
Jan 1 at 18:04
add a comment |
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$begingroup$
Copying part of my answer to this Math Overflow question:
Whether two (a) topological spaces, (b) metric spaces, or (c) groups $A$ and $B$, with isomorphic squares, are necessarily isomorphic, was Problem 77 (of Ulam) in the Scottish book. The following information is from the commentaries on Problem 77, by Mauldin for (a) and (b) and by Kaplansky for (c), in The Scottish Book: Mathematics from the Scottish Café, R. Daniel Mauldin, ed., 1981.
Mauldin gives a bibliography of 17 items for parts (a) and (b). The first counterexample for part (a), as well as a positive answer for two-dimensional compact manifolds, was given by R. H. Fox, On a problem of S. Ulam concerning Cartesian products, Fund. Math. 34 (1947), 278-287. The first counterexample to part (b) was given by G. Fournier, On a problem of S. Ulam, Proc. Amer. Math. Soc. 29 (1971), 622. Mauldin writes (in 1981):
"However, it is open whether there is an affirmative solution to (b) in the case that $A$ and $B$ are complete metric spaces. In fact, part (b) is open in the case where $A$ and $B$ are assumed to be compact."
answered Dec 31 '18 at 3:43
bofbof
51.5k558120
$endgroup$
$begingroup$
I deleted my silly comment, which parsed (b) as "metrizable".
$endgroup$
– Mike Miller
Dec 31 '18 at 7:59
$begingroup$
For completeness: I see the Fox paper at matwbn.icm.edu.pl/ksiazki/fm/fm34/fm34128.pdf, and another example from Marjanović and Vučemilović at dml.cz/bitstream/handle/10338.dmlcz/106394/….
$endgroup$
– Hew Wolff
Jan 1 at 18:04
add a comment |
$begingroup$
Copying part of my answer to this Math Overflow question:
Whether two (a) topological spaces, (b) metric spaces, or (c) groups $A$ and $B$, with isomorphic squares, are necessarily isomorphic, was Problem 77 (of Ulam) in the Scottish book. The following information is from the commentaries on Problem 77, by Mauldin for (a) and (b) and by Kaplansky for (c), in The Scottish Book: Mathematics from the Scottish Café, R. Daniel Mauldin, ed., 1981.
Mauldin gives a bibliography of 17 items for parts (a) and (b). The first counterexample for part (a), as well as a positive answer for two-dimensional compact manifolds, was given by R. H. Fox, On a problem of S. Ulam concerning Cartesian products, Fund. Math. 34 (1947), 278-287. The first counterexample to part (b) was given by G. Fournier, On a problem of S. Ulam, Proc. Amer. Math. Soc. 29 (1971), 622. Mauldin writes (in 1981):
"However, it is open whether there is an affirmative solution to (b) in the case that $A$ and $B$ are complete metric spaces. In fact, part (b) is open in the case where $A$ and $B$ are assumed to be compact."
answered Dec 31 '18 at 3:43
bofbof
51.5k558120
$endgroup$
$begingroup$
I deleted my silly comment, which parsed (b) as "metrizable".
$endgroup$
– Mike Miller
Dec 31 '18 at 7:59
$begingroup$
For completeness: I see the Fox paper at matwbn.icm.edu.pl/ksiazki/fm/fm34/fm34128.pdf, and another example from Marjanović and Vučemilović at dml.cz/bitstream/handle/10338.dmlcz/106394/….
$endgroup$
– Hew Wolff
Jan 1 at 18:04
add a comment |
$begingroup$
Copying part of my answer to this Math Overflow question:
Whether two (a) topological spaces, (b) metric spaces, or (c) groups $A$ and $B$, with isomorphic squares, are necessarily isomorphic, was Problem 77 (of Ulam) in the Scottish book. The following information is from the commentaries on Problem 77, by Mauldin for (a) and (b) and by Kaplansky for (c), in The Scottish Book: Mathematics from the Scottish Café, R. Daniel Mauldin, ed., 1981.
Mauldin gives a bibliography of 17 items for parts (a) and (b). The first counterexample for part (a), as well as a positive answer for two-dimensional compact manifolds, was given by R. H. Fox, On a problem of S. Ulam concerning Cartesian products, Fund. Math. 34 (1947), 278-287. The first counterexample to part (b) was given by G. Fournier, On a problem of S. Ulam, Proc. Amer. Math. Soc. 29 (1971), 622. Mauldin writes (in 1981):
"However, it is open whether there is an affirmative solution to (b) in the case that $A$ and $B$ are complete metric spaces. In fact, part (b) is open in the case where $A$ and $B$ are assumed to be compact."
answered Dec 31 '18 at 3:43
bofbof
51.5k558120
$endgroup$
Copying part of my answer to this Math Overflow question:
Whether two (a) topological spaces, (b) metric spaces, or (c) groups $A$ and $B$, with isomorphic squares, are necessarily isomorphic, was Problem 77 (of Ulam) in the Scottish book. The following information is from the commentaries on Problem 77, by Mauldin for (a) and (b) and by Kaplansky for (c), in The Scottish Book: Mathematics from the Scottish Café, R. Daniel Mauldin, ed., 1981.
Mauldin gives a bibliography of 17 items for parts (a) and (b). The first counterexample for part (a), as well as a positive answer for two-dimensional compact manifolds, was given by R. H. Fox, On a problem of S. Ulam concerning Cartesian products, Fund. Math. 34 (1947), 278-287. The first counterexample to part (b) was given by G. Fournier, On a problem of S. Ulam, Proc. Amer. Math. Soc. 29 (1971), 622. Mauldin writes (in 1981):
"However, it is open whether there is an affirmative solution to (b) in the case that $A$ and $B$ are complete metric spaces. In fact, part (b) is open in the case where $A$ and $B$ are assumed to be compact."
answered Dec 31 '18 at 3:43
bofbof
51.5k558120
answered Dec 31 '18 at 3:43
bofbof
51.5k558120
answered Dec 31 '18 at 3:43
bofbof
51.5k558120
answered Dec 31 '18 at 3:43
bofbof
51.5k558120
51.5k558120
$begingroup$
I deleted my silly comment, which parsed (b) as "metrizable".
$endgroup$
– Mike Miller
Dec 31 '18 at 7:59
$begingroup$
For completeness: I see the Fox paper at matwbn.icm.edu.pl/ksiazki/fm/fm34/fm34128.pdf, and another example from Marjanović and Vučemilović at dml.cz/bitstream/handle/10338.dmlcz/106394/….
$endgroup$
– Hew Wolff
Jan 1 at 18:04
add a comment |
$begingroup$
I deleted my silly comment, which parsed (b) as "metrizable".
$endgroup$
– Mike Miller
Dec 31 '18 at 7:59
$begingroup$
For completeness: I see the Fox paper at matwbn.icm.edu.pl/ksiazki/fm/fm34/fm34128.pdf, and another example from Marjanović and Vučemilović at dml.cz/bitstream/handle/10338.dmlcz/106394/….
$endgroup$
– Hew Wolff
Jan 1 at 18:04
$begingroup$
I deleted my silly comment, which parsed (b) as "metrizable".
$endgroup$
– Mike Miller
Dec 31 '18 at 7:59
$begingroup$
I deleted my silly comment, which parsed (b) as "metrizable".
$endgroup$
– Mike Miller
Dec 31 '18 at 7:59
$begingroup$
For completeness: I see the Fox paper at matwbn.icm.edu.pl/ksiazki/fm/fm34/fm34128.pdf, and another example from Marjanović and Vučemilović at dml.cz/bitstream/handle/10338.dmlcz/106394/….
$endgroup$
– Hew Wolff
Jan 1 at 18:04
$begingroup$
For completeness: I see the Fox paper at matwbn.icm.edu.pl/ksiazki/fm/fm34/fm34128.pdf, and another example from Marjanović and Vučemilović at dml.cz/bitstream/handle/10338.dmlcz/106394/….
$endgroup$
– Hew Wolff
Jan 1 at 18:04
add a comment |
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3
$begingroup$
It's not possible for finite spaces, by my argument at mathoverflow.net/questions/269542/….
$endgroup$
– Eric Wofsey
Dec 31 '18 at 3:15
1
$begingroup$
Maybe take a look at the Whitehead manifold. See: numdam.org/article/BSMF_1960__88__131_0.pdf and en.wikipedia.org/wiki/Whitehead_manifold
$endgroup$
– Alvin Jin
Dec 31 '18 at 3:31
2
$begingroup$
The existence of such spaces $X, Y$ is what's used in Mike Miller 's example, see his answer in the linked question. But Whitehead manifold would also do the job.
$endgroup$
– Moishe Cohen
Dec 31 '18 at 3:35
2
$begingroup$
As for the "torsion" examples, see references in mathoverflow.net/questions/10128/when-is-a-isomorphic-to-a3
$endgroup$
– Moishe Cohen
Dec 31 '18 at 4:13
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@MoisheCohen Interesting, for OP's sake I will observe that there exists a functor from groups to spaces, given by constructing a simplicial complex called "$BG$" whose $n$-simplices are labelled by sequences $[g_1, cdots, g_n]$ of elements of $G$ (and the gluings involve multiplications), and $B(G times H) = BG times BH$. In particular, functoriality means that if $G$ and $H$ are homeomorphic, so are $BG$ and $BH$, and conversely (if $BG cong BH$ then $G cong H$ by looking at fundamental groups). So if $A cong A^3 neq A^2$, then $BA cong (BA)^3 neq (BA)^2$.
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– Mike Miller
Dec 31 '18 at 4:23