Mixing
Topology of the torus
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
asked Nov 5 '17 at 3:54
Vladimir Kondratiev
164
add a comment |
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
asked Nov 5 '17 at 3:54
Vladimir Kondratiev
164
add a comment |
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
asked Nov 5 '17 at 3:54
Vladimir Kondratiev
164
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
asked Nov 5 '17 at 3:54
Vladimir Kondratiev
164
asked Nov 5 '17 at 3:54
Vladimir Kondratiev
164
edited Nov 5 '17 at 4:33
asked Nov 5 '17 at 3:54
Vladimir Kondratiev
164
asked Nov 5 '17 at 3:54
Vladimir Kondratiev
164
asked Nov 5 '17 at 3:54
Vladimir Kondratiev
164
164
add a comment |
add a comment |
3 Answers
3
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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.
answered Nov 5 '17 at 18:58
Andres Mejia
16k21546
add a comment |
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
add a comment |
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.
answered Nov 20 '18 at 10:35
Sundance Kid
1
This doesn't improve this answer whatsoever.
– José Carlos Santos
Nov 20 '18 at 10:55
add a comment |
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3 Answers
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oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
16k21546
add a comment |
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
16k21546
add a comment |
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
16k21546
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
16k21546
answered Nov 5 '17 at 18:58
Andres Mejia
16k21546
answered Nov 5 '17 at 18:58
Andres Mejia
16k21546
answered Nov 5 '17 at 18:58
Andres Mejia
16k21546
16k21546
add a comment |
add a comment |
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
add a comment |
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
add a comment |
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
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
answered Nov 6 '17 at 17:11
usr1123
11
answered Nov 6 '17 at 17:11
usr1123
11
answered Nov 6 '17 at 17:11
usr1123
11
11
add a comment |
add a comment |
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.
answered Nov 20 '18 at 10:35
Sundance Kid
1
This doesn't improve this answer whatsoever.
– José Carlos Santos
Nov 20 '18 at 10:55
add a comment |
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.
answered Nov 20 '18 at 10:35
Sundance Kid
1
This doesn't improve this answer whatsoever.
– José Carlos Santos
Nov 20 '18 at 10:55
add a comment |
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.
answered Nov 20 '18 at 10:35
Sundance Kid
1
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.
answered Nov 20 '18 at 10:35
Sundance Kid
1
answered Nov 20 '18 at 10:35
Sundance Kid
1
answered Nov 20 '18 at 10:35
Sundance Kid
1
answered Nov 20 '18 at 10:35
Sundance Kid
1
1
This doesn't improve this answer whatsoever.
– José Carlos Santos
Nov 20 '18 at 10:55
add a comment |
This doesn't improve this answer whatsoever.
– José Carlos Santos
Nov 20 '18 at 10:55
This doesn't improve this answer whatsoever.
– José Carlos Santos
Nov 20 '18 at 10:55
This doesn't improve this answer whatsoever.
– José Carlos Santos
Nov 20 '18 at 10:55
add a comment |
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