Group of matrices form a manifold or euclidean space

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There is a very interesting question How can a group of matrices for a manifold. From the answers it looks more like group of matrices form euclidean space than a general manifold. I understand that euclidean space is a manifold, but manifold is very general and has curvature. My question is what exactly makes a group of matrices a manifold but not simply a euclidean space.



  • I am not a mathematician so please correct me if there is anything wrong with the question or the way I posed it.





linear-algebra abstract-algebra manifolds euclidean-geometry linear-groups







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    Under addition, they can be Euclidean. However, all the interesting math happens multiplicatively, where they usually aren't flat.
    – Randall
    30 mins ago














up vote
2
down vote

favorite












There is a very interesting question How can a group of matrices for a manifold. From the answers it looks more like group of matrices form euclidean space than a general manifold. I understand that euclidean space is a manifold, but manifold is very general and has curvature. My question is what exactly makes a group of matrices a manifold but not simply a euclidean space.



  • I am not a mathematician so please correct me if there is anything wrong with the question or the way I posed it.





linear-algebra abstract-algebra manifolds euclidean-geometry linear-groups







share|cite|improve this question



















  • 1




    Under addition, they can be Euclidean. However, all the interesting math happens multiplicatively, where they usually aren't flat.
    – Randall
    30 mins ago












up vote
2
down vote

favorite









up vote
2
down vote

favorite











There is a very interesting question How can a group of matrices for a manifold. From the answers it looks more like group of matrices form euclidean space than a general manifold. I understand that euclidean space is a manifold, but manifold is very general and has curvature. My question is what exactly makes a group of matrices a manifold but not simply a euclidean space.



  • I am not a mathematician so please correct me if there is anything wrong with the question or the way I posed it.





linear-algebra abstract-algebra manifolds euclidean-geometry linear-groups







share|cite|improve this question















There is a very interesting question How can a group of matrices for a manifold. From the answers it looks more like group of matrices form euclidean space than a general manifold. I understand that euclidean space is a manifold, but manifold is very general and has curvature. My question is what exactly makes a group of matrices a manifold but not simply a euclidean space.



  • I am not a mathematician so please correct me if there is anything wrong with the question or the way I posed it.





linear-algebra abstract-algebra manifolds euclidean-geometry linear-groups




linear-algebra abstract-algebra manifolds euclidean-geometry linear-groups






share|cite|improve this question















share|cite|improve this question













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edited 23 mins ago









José Carlos Santos

133k17107196




133k17107196










asked 36 mins ago









Shew

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501411







  • 1




    Under addition, they can be Euclidean. However, all the interesting math happens multiplicatively, where they usually aren't flat.
    – Randall
    30 mins ago












  • 1




    Under addition, they can be Euclidean. However, all the interesting math happens multiplicatively, where they usually aren't flat.
    – Randall
    30 mins ago







1




1




Under addition, they can be Euclidean. However, all the interesting math happens multiplicatively, where they usually aren't flat.
– Randall
30 mins ago




Under addition, they can be Euclidean. However, all the interesting math happens multiplicatively, where they usually aren't flat.
– Randall
30 mins ago










2 Answers
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Take $SO(2,mathbbR)$, for instance. This is the group of the $2times2$ orthogonal matrices whose determinant is $1$. But then$$SO(2,mathbbR)=left,thetainmathbb Rright.$$This can be seen as a circle in $mathbbR^2$. Therefore, it is naturally a manifold, but in no way an Euclidean space.






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    In the question to which you refer the group of matrices is given a manifold structure by virtue of its embedding in a Euclidean space. So for example you can think of a torus as a two dimensional manifold (with interesting curvature) by using the differentiable structure it gets from the usual embedding as a curved surface in $3$-space. That doesn't make the torus a Euclidean space.






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      2 Answers
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      2 Answers
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      Take $SO(2,mathbbR)$, for instance. This is the group of the $2times2$ orthogonal matrices whose determinant is $1$. But then$$SO(2,mathbbR)=left,thetainmathbb Rright.$$This can be seen as a circle in $mathbbR^2$. Therefore, it is naturally a manifold, but in no way an Euclidean space.






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        up vote
        4
        down vote



        accepted










        Take $SO(2,mathbbR)$, for instance. This is the group of the $2times2$ orthogonal matrices whose determinant is $1$. But then$$SO(2,mathbbR)=left,thetainmathbb Rright.$$This can be seen as a circle in $mathbbR^2$. Therefore, it is naturally a manifold, but in no way an Euclidean space.






        share|cite|improve this answer






















          up vote
          4
          down vote



          accepted







          up vote
          4
          down vote



          accepted






          Take $SO(2,mathbbR)$, for instance. This is the group of the $2times2$ orthogonal matrices whose determinant is $1$. But then$$SO(2,mathbbR)=left,thetainmathbb Rright.$$This can be seen as a circle in $mathbbR^2$. Therefore, it is naturally a manifold, but in no way an Euclidean space.






          share|cite|improve this answer












          Take $SO(2,mathbbR)$, for instance. This is the group of the $2times2$ orthogonal matrices whose determinant is $1$. But then$$SO(2,mathbbR)=left,thetainmathbb Rright.$$This can be seen as a circle in $mathbbR^2$. Therefore, it is naturally a manifold, but in no way an Euclidean space.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 28 mins ago









          José Carlos Santos

          133k17107196




          133k17107196




















              up vote
              2
              down vote













              In the question to which you refer the group of matrices is given a manifold structure by virtue of its embedding in a Euclidean space. So for example you can think of a torus as a two dimensional manifold (with interesting curvature) by using the differentiable structure it gets from the usual embedding as a curved surface in $3$-space. That doesn't make the torus a Euclidean space.






              share|cite|improve this answer
























                up vote
                2
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                In the question to which you refer the group of matrices is given a manifold structure by virtue of its embedding in a Euclidean space. So for example you can think of a torus as a two dimensional manifold (with interesting curvature) by using the differentiable structure it gets from the usual embedding as a curved surface in $3$-space. That doesn't make the torus a Euclidean space.






                share|cite|improve this answer






















                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  In the question to which you refer the group of matrices is given a manifold structure by virtue of its embedding in a Euclidean space. So for example you can think of a torus as a two dimensional manifold (with interesting curvature) by using the differentiable structure it gets from the usual embedding as a curved surface in $3$-space. That doesn't make the torus a Euclidean space.






                  share|cite|improve this answer












                  In the question to which you refer the group of matrices is given a manifold structure by virtue of its embedding in a Euclidean space. So for example you can think of a torus as a two dimensional manifold (with interesting curvature) by using the differentiable structure it gets from the usual embedding as a curved surface in $3$-space. That doesn't make the torus a Euclidean space.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 32 mins ago









                  Ethan Bolker

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