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?










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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?










    share|cite|improve this question


























      up vote
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      down vote

      favorite
      1









      up vote
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      down vote

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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?










      share|cite|improve this question
















      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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      edited Nov 5 '17 at 4:33

























      asked Nov 5 '17 at 3:54









      Vladimir Kondratiev

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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.






          share|cite|improve this answer




























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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.






            share|cite|improve this answer




























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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.






              share|cite|improve this answer








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              • This doesn't improve this answer whatsoever.
                – José Carlos Santos
                Nov 20 at 10:55











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              3 Answers
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              active

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              3 Answers
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              active

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              active

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              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.






              share|cite|improve this answer

























                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.






                share|cite|improve this answer























                  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.






                  share|cite|improve this answer












                  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.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 5 '17 at 18:58









                  Andres Mejia

                  15.9k21445




                  15.9k21445






















                      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.






                      share|cite|improve this answer

























                        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.






                        share|cite|improve this answer























                          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.






                          share|cite|improve this answer












                          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.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 6 '17 at 17:11









                          usr1123

                          11




                          11






















                              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.






                              share|cite|improve this answer








                              New contributor




                              Sundance Kid is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                              Check out our Code of Conduct.


















                              • This doesn't improve this answer whatsoever.
                                – José Carlos Santos
                                Nov 20 at 10:55















                              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.






                              share|cite|improve this answer








                              New contributor




                              Sundance Kid is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                              Check out our Code of Conduct.


















                              • This doesn't improve this answer whatsoever.
                                – José Carlos Santos
                                Nov 20 at 10:55













                              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.






                              share|cite|improve this answer








                              New contributor




                              Sundance Kid is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                              Check out our Code of Conduct.









                              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.







                              share|cite|improve this answer








                              New contributor




                              Sundance Kid is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                              Check out our Code of Conduct.









                              share|cite|improve this answer



                              share|cite|improve this answer






                              New contributor




                              Sundance Kid is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                              Check out our Code of Conduct.









                              answered Nov 20 at 10:35









                              Sundance Kid

                              1




                              1




                              New contributor




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                              New contributor





                              Sundance Kid is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                              Check out our Code of Conduct.






                              Sundance Kid is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                              Check out our Code of Conduct.












                              • 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




                              This doesn't improve this answer whatsoever.
                              – José Carlos Santos
                              Nov 20 at 10:55


















                               

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