Cocompact & discrete lattice












2












$begingroup$


I don't understand a step in the proof of the following proposition:




Let $Lambda$ be a subgroup of a real vector space $V$ of finite dimension. Then $Lambda$ is a full lattice if and only if $Lambda$ is discrete and $V/Lambda$ is compact.



Proof.(the part I don't understand) Assume $Lambda$ is discrete and $V/Lambda$ is compact, let $W$ be the subspace of $V$ spanned by $Lambda$, then the real vector space $V/W$ can't have positive dimension since $V/Lambda$ is compact.




Why is $V/W$ a trivial vector space?






number-theory algebraic-number-theory vector-lattices







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$endgroup$

















    2












    $begingroup$


    I don't understand a step in the proof of the following proposition:




    Let $Lambda$ be a subgroup of a real vector space $V$ of finite dimension. Then $Lambda$ is a full lattice if and only if $Lambda$ is discrete and $V/Lambda$ is compact.



    Proof.(the part I don't understand) Assume $Lambda$ is discrete and $V/Lambda$ is compact, let $W$ be the subspace of $V$ spanned by $Lambda$, then the real vector space $V/W$ can't have positive dimension since $V/Lambda$ is compact.




    Why is $V/W$ a trivial vector space?






    number-theory algebraic-number-theory vector-lattices







    share|cite|improve this question









    $endgroup$















      2












      2








      2





      $begingroup$


      I don't understand a step in the proof of the following proposition:




      Let $Lambda$ be a subgroup of a real vector space $V$ of finite dimension. Then $Lambda$ is a full lattice if and only if $Lambda$ is discrete and $V/Lambda$ is compact.



      Proof.(the part I don't understand) Assume $Lambda$ is discrete and $V/Lambda$ is compact, let $W$ be the subspace of $V$ spanned by $Lambda$, then the real vector space $V/W$ can't have positive dimension since $V/Lambda$ is compact.




      Why is $V/W$ a trivial vector space?






      number-theory algebraic-number-theory vector-lattices







      share|cite|improve this question









      $endgroup$




      I don't understand a step in the proof of the following proposition:




      Let $Lambda$ be a subgroup of a real vector space $V$ of finite dimension. Then $Lambda$ is a full lattice if and only if $Lambda$ is discrete and $V/Lambda$ is compact.



      Proof.(the part I don't understand) Assume $Lambda$ is discrete and $V/Lambda$ is compact, let $W$ be the subspace of $V$ spanned by $Lambda$, then the real vector space $V/W$ can't have positive dimension since $V/Lambda$ is compact.




      Why is $V/W$ a trivial vector space?






      number-theory algebraic-number-theory vector-lattices




      number-theory algebraic-number-theory vector-lattices






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 14 at 14:16









      CYCCYC

      972711




      972711






















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          $begingroup$

          The first thing going on here is a sequence of equivalencies: $Lambda$ does not span $V iff W$ is a proper subspace of $V iff V/W$ has positive dimension $iff V/W$ is a nontrivial vector space.



          But if all of that happens then $V/W$ is noncompact. The inclusion $Lambda subset W$ induces a surjective continuous function $V / Lambda mapsto V / W$, and therefore $V / Lambda$ is noncompact, a contradiction.






          share|cite|improve this answer









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            1 Answer
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            active

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            $begingroup$

            The first thing going on here is a sequence of equivalencies: $Lambda$ does not span $V iff W$ is a proper subspace of $V iff V/W$ has positive dimension $iff V/W$ is a nontrivial vector space.



            But if all of that happens then $V/W$ is noncompact. The inclusion $Lambda subset W$ induces a surjective continuous function $V / Lambda mapsto V / W$, and therefore $V / Lambda$ is noncompact, a contradiction.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              The first thing going on here is a sequence of equivalencies: $Lambda$ does not span $V iff W$ is a proper subspace of $V iff V/W$ has positive dimension $iff V/W$ is a nontrivial vector space.



              But if all of that happens then $V/W$ is noncompact. The inclusion $Lambda subset W$ induces a surjective continuous function $V / Lambda mapsto V / W$, and therefore $V / Lambda$ is noncompact, a contradiction.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                The first thing going on here is a sequence of equivalencies: $Lambda$ does not span $V iff W$ is a proper subspace of $V iff V/W$ has positive dimension $iff V/W$ is a nontrivial vector space.



                But if all of that happens then $V/W$ is noncompact. The inclusion $Lambda subset W$ induces a surjective continuous function $V / Lambda mapsto V / W$, and therefore $V / Lambda$ is noncompact, a contradiction.






                share|cite|improve this answer









                $endgroup$



                The first thing going on here is a sequence of equivalencies: $Lambda$ does not span $V iff W$ is a proper subspace of $V iff V/W$ has positive dimension $iff V/W$ is a nontrivial vector space.



                But if all of that happens then $V/W$ is noncompact. The inclusion $Lambda subset W$ induces a surjective continuous function $V / Lambda mapsto V / W$, and therefore $V / Lambda$ is noncompact, a contradiction.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 14 at 14:31









                Lee MosherLee Mosher

                49.4k33686




                49.4k33686






























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