Topology of the torus
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Theorem: There is no open covering $T^k=U_1cup...cup U_k$ of the $k$-torus such that the map $$H_1(V,mathbb{Z}) rightarrow H_1(T^k,mathbb{Z}) $$ has rank at most $(i-1)$ for every component $V$ of $U_i$.
McMullen states this in his 2004 article on Minkowski's conjecture without proof and without any reference.
I have no idea how to prove it. Can anyone suggest a proof or give any reference?
general-topology homology-cohomology covering-spaces dimension-theory arithmetic-topology
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Theorem: There is no open covering $T^k=U_1cup...cup U_k$ of the $k$-torus such that the map $$H_1(V,mathbb{Z}) rightarrow H_1(T^k,mathbb{Z}) $$ has rank at most $(i-1)$ for every component $V$ of $U_i$.
McMullen states this in his 2004 article on Minkowski's conjecture without proof and without any reference.
I have no idea how to prove it. Can anyone suggest a proof or give any reference?
general-topology homology-cohomology covering-spaces dimension-theory arithmetic-topology
add a comment |
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Theorem: There is no open covering $T^k=U_1cup...cup U_k$ of the $k$-torus such that the map $$H_1(V,mathbb{Z}) rightarrow H_1(T^k,mathbb{Z}) $$ has rank at most $(i-1)$ for every component $V$ of $U_i$.
McMullen states this in his 2004 article on Minkowski's conjecture without proof and without any reference.
I have no idea how to prove it. Can anyone suggest a proof or give any reference?
general-topology homology-cohomology covering-spaces dimension-theory arithmetic-topology
Theorem: There is no open covering $T^k=U_1cup...cup U_k$ of the $k$-torus such that the map $$H_1(V,mathbb{Z}) rightarrow H_1(T^k,mathbb{Z}) $$ has rank at most $(i-1)$ for every component $V$ of $U_i$.
McMullen states this in his 2004 article on Minkowski's conjecture without proof and without any reference.
I have no idea how to prove it. Can anyone suggest a proof or give any reference?
general-topology homology-cohomology covering-spaces dimension-theory arithmetic-topology
general-topology homology-cohomology covering-spaces dimension-theory arithmetic-topology
edited Nov 5 '17 at 4:33
asked Nov 5 '17 at 3:54
Vladimir Kondratiev
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3 Answers
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This follows from a more general result of McMullen:
Definition: The order of a cover $mathfrak{U}$ is the greatest integer $n$ so that the intersection of $n+1$ elements is nontrivial.
Loosely stated theorem: Let $mathfrak{U}$ be an open cover of the $n$-torus. Suppose for that all components $V$ of the intersection $U_1 capdotscap U_k$ with $k leq n$, we have that the induced map by inclusion on homology has at most rank $(n-k)$, then $mathfrak{U}$ has order at least $n$.
The basic ingredient for the proof is the Cech DeRham complex, reference for which can be found here, on page 6.
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In Mc Mullen paper "Minkowski's conjecture, well rounded lattices and topological dimension" it is referred as Corollary 2.2, and it is proved at p.5, as a trivial consequence of Theorem 2.1, which is the "Loosely stated theorem" as pointed out by Andres Mejia.
Sincerely, to me, is not clear how that theorem implies the corollary you are looking for, so I cannot help you with that.
add a comment |
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I think the statement follows from the theorem cited above in the following way:
Suppose there was a cover $(U_i)_{ileq k}$ such that $$H_1(V)rightarrow H_1(T^k)$$ has rank at most $i-1$ for each component $V$ of $U_i$. Let $Isubseteq lbrace 1,dots,krbrace$, $#I=l$. Then $j=min Ileq k-(l-1)$. Let $Vsubseteqbigcap_I U_i$ be a component of the corresponding $l$-intersection. Then there is another component $Wsubseteq U_j$ with $Vsubseteq W$. It follows $$mathrm{rk}(H_1(V)rightarrow H_1(T^k))leqmathrm{rk}(H_1(W)rightarrow H_1(T^k))leq j-1leq k-l.$$
Therefore, the order of $(U_i)_{ileq k}$ is at least $k$, which is impossible.
New contributor
This doesn't improve this answer whatsoever.
– José Carlos Santos
Nov 20 at 10:55
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
This follows from a more general result of McMullen:
Definition: The order of a cover $mathfrak{U}$ is the greatest integer $n$ so that the intersection of $n+1$ elements is nontrivial.
Loosely stated theorem: Let $mathfrak{U}$ be an open cover of the $n$-torus. Suppose for that all components $V$ of the intersection $U_1 capdotscap U_k$ with $k leq n$, we have that the induced map by inclusion on homology has at most rank $(n-k)$, then $mathfrak{U}$ has order at least $n$.
The basic ingredient for the proof is the Cech DeRham complex, reference for which can be found here, on page 6.
add a comment |
up vote
0
down vote
This follows from a more general result of McMullen:
Definition: The order of a cover $mathfrak{U}$ is the greatest integer $n$ so that the intersection of $n+1$ elements is nontrivial.
Loosely stated theorem: Let $mathfrak{U}$ be an open cover of the $n$-torus. Suppose for that all components $V$ of the intersection $U_1 capdotscap U_k$ with $k leq n$, we have that the induced map by inclusion on homology has at most rank $(n-k)$, then $mathfrak{U}$ has order at least $n$.
The basic ingredient for the proof is the Cech DeRham complex, reference for which can be found here, on page 6.
add a comment |
up vote
0
down vote
up vote
0
down vote
This follows from a more general result of McMullen:
Definition: The order of a cover $mathfrak{U}$ is the greatest integer $n$ so that the intersection of $n+1$ elements is nontrivial.
Loosely stated theorem: Let $mathfrak{U}$ be an open cover of the $n$-torus. Suppose for that all components $V$ of the intersection $U_1 capdotscap U_k$ with $k leq n$, we have that the induced map by inclusion on homology has at most rank $(n-k)$, then $mathfrak{U}$ has order at least $n$.
The basic ingredient for the proof is the Cech DeRham complex, reference for which can be found here, on page 6.
This follows from a more general result of McMullen:
Definition: The order of a cover $mathfrak{U}$ is the greatest integer $n$ so that the intersection of $n+1$ elements is nontrivial.
Loosely stated theorem: Let $mathfrak{U}$ be an open cover of the $n$-torus. Suppose for that all components $V$ of the intersection $U_1 capdotscap U_k$ with $k leq n$, we have that the induced map by inclusion on homology has at most rank $(n-k)$, then $mathfrak{U}$ has order at least $n$.
The basic ingredient for the proof is the Cech DeRham complex, reference for which can be found here, on page 6.
answered Nov 5 '17 at 18:58
Andres Mejia
15.9k21445
15.9k21445
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In Mc Mullen paper "Minkowski's conjecture, well rounded lattices and topological dimension" it is referred as Corollary 2.2, and it is proved at p.5, as a trivial consequence of Theorem 2.1, which is the "Loosely stated theorem" as pointed out by Andres Mejia.
Sincerely, to me, is not clear how that theorem implies the corollary you are looking for, so I cannot help you with that.
add a comment |
up vote
0
down vote
In Mc Mullen paper "Minkowski's conjecture, well rounded lattices and topological dimension" it is referred as Corollary 2.2, and it is proved at p.5, as a trivial consequence of Theorem 2.1, which is the "Loosely stated theorem" as pointed out by Andres Mejia.
Sincerely, to me, is not clear how that theorem implies the corollary you are looking for, so I cannot help you with that.
add a comment |
up vote
0
down vote
up vote
0
down vote
In Mc Mullen paper "Minkowski's conjecture, well rounded lattices and topological dimension" it is referred as Corollary 2.2, and it is proved at p.5, as a trivial consequence of Theorem 2.1, which is the "Loosely stated theorem" as pointed out by Andres Mejia.
Sincerely, to me, is not clear how that theorem implies the corollary you are looking for, so I cannot help you with that.
In Mc Mullen paper "Minkowski's conjecture, well rounded lattices and topological dimension" it is referred as Corollary 2.2, and it is proved at p.5, as a trivial consequence of Theorem 2.1, which is the "Loosely stated theorem" as pointed out by Andres Mejia.
Sincerely, to me, is not clear how that theorem implies the corollary you are looking for, so I cannot help you with that.
answered Nov 6 '17 at 17:11
usr1123
11
11
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add a comment |
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0
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I think the statement follows from the theorem cited above in the following way:
Suppose there was a cover $(U_i)_{ileq k}$ such that $$H_1(V)rightarrow H_1(T^k)$$ has rank at most $i-1$ for each component $V$ of $U_i$. Let $Isubseteq lbrace 1,dots,krbrace$, $#I=l$. Then $j=min Ileq k-(l-1)$. Let $Vsubseteqbigcap_I U_i$ be a component of the corresponding $l$-intersection. Then there is another component $Wsubseteq U_j$ with $Vsubseteq W$. It follows $$mathrm{rk}(H_1(V)rightarrow H_1(T^k))leqmathrm{rk}(H_1(W)rightarrow H_1(T^k))leq j-1leq k-l.$$
Therefore, the order of $(U_i)_{ileq k}$ is at least $k$, which is impossible.
New contributor
This doesn't improve this answer whatsoever.
– José Carlos Santos
Nov 20 at 10:55
add a comment |
up vote
0
down vote
I think the statement follows from the theorem cited above in the following way:
Suppose there was a cover $(U_i)_{ileq k}$ such that $$H_1(V)rightarrow H_1(T^k)$$ has rank at most $i-1$ for each component $V$ of $U_i$. Let $Isubseteq lbrace 1,dots,krbrace$, $#I=l$. Then $j=min Ileq k-(l-1)$. Let $Vsubseteqbigcap_I U_i$ be a component of the corresponding $l$-intersection. Then there is another component $Wsubseteq U_j$ with $Vsubseteq W$. It follows $$mathrm{rk}(H_1(V)rightarrow H_1(T^k))leqmathrm{rk}(H_1(W)rightarrow H_1(T^k))leq j-1leq k-l.$$
Therefore, the order of $(U_i)_{ileq k}$ is at least $k$, which is impossible.
New contributor
This doesn't improve this answer whatsoever.
– José Carlos Santos
Nov 20 at 10:55
add a comment |
up vote
0
down vote
up vote
0
down vote
I think the statement follows from the theorem cited above in the following way:
Suppose there was a cover $(U_i)_{ileq k}$ such that $$H_1(V)rightarrow H_1(T^k)$$ has rank at most $i-1$ for each component $V$ of $U_i$. Let $Isubseteq lbrace 1,dots,krbrace$, $#I=l$. Then $j=min Ileq k-(l-1)$. Let $Vsubseteqbigcap_I U_i$ be a component of the corresponding $l$-intersection. Then there is another component $Wsubseteq U_j$ with $Vsubseteq W$. It follows $$mathrm{rk}(H_1(V)rightarrow H_1(T^k))leqmathrm{rk}(H_1(W)rightarrow H_1(T^k))leq j-1leq k-l.$$
Therefore, the order of $(U_i)_{ileq k}$ is at least $k$, which is impossible.
New contributor
I think the statement follows from the theorem cited above in the following way:
Suppose there was a cover $(U_i)_{ileq k}$ such that $$H_1(V)rightarrow H_1(T^k)$$ has rank at most $i-1$ for each component $V$ of $U_i$. Let $Isubseteq lbrace 1,dots,krbrace$, $#I=l$. Then $j=min Ileq k-(l-1)$. Let $Vsubseteqbigcap_I U_i$ be a component of the corresponding $l$-intersection. Then there is another component $Wsubseteq U_j$ with $Vsubseteq W$. It follows $$mathrm{rk}(H_1(V)rightarrow H_1(T^k))leqmathrm{rk}(H_1(W)rightarrow H_1(T^k))leq j-1leq k-l.$$
Therefore, the order of $(U_i)_{ileq k}$ is at least $k$, which is impossible.
New contributor
New contributor
answered Nov 20 at 10:35
Sundance Kid
1
1
New contributor
New contributor
This doesn't improve this answer whatsoever.
– José Carlos Santos
Nov 20 at 10:55
add a comment |
This doesn't improve this answer whatsoever.
– José Carlos Santos
Nov 20 at 10:55
This doesn't improve this answer whatsoever.
– José Carlos Santos
Nov 20 at 10:55
This doesn't improve this answer whatsoever.
– José Carlos Santos
Nov 20 at 10:55
add a comment |
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