On Manifolds: TOP, PDIFF, DIFF, PL

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I wonder why in geometric topology, only the following structure of manifolds are emphasized:



  • (TOP) topological manifolds

  • (PDIFF), for piecewise differentiable

  • (PL) piecewise-smooth manifolds

  • (DIFF) the smooth manifolds

Can we have more than the ones shown in the graph here: https://en.wikipedia.org/wiki/PDIFF with more subtle or refined structures?



For example, can we consider the affine structure, or Riemannian structure, etc? Or somewhere in between in the straight lines of the figure, can we have other refinement? other structures?



enter image description here




general-topology manifolds geometric-topology




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    I added a figure to clarify what you said (from the cited URL). Hope this is what you had in mind...
    – wonderich
    Jul 26 at 3:18














up vote
1
down vote

favorite
1












I wonder why in geometric topology, only the following structure of manifolds are emphasized:



  • (TOP) topological manifolds

  • (PDIFF), for piecewise differentiable

  • (PL) piecewise-smooth manifolds

  • (DIFF) the smooth manifolds

Can we have more than the ones shown in the graph here: https://en.wikipedia.org/wiki/PDIFF with more subtle or refined structures?



For example, can we consider the affine structure, or Riemannian structure, etc? Or somewhere in between in the straight lines of the figure, can we have other refinement? other structures?



enter image description here




general-topology manifolds geometric-topology




share|cite|improve this question

















  • 1




    I added a figure to clarify what you said (from the cited URL). Hope this is what you had in mind...
    – wonderich
    Jul 26 at 3:18












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





I wonder why in geometric topology, only the following structure of manifolds are emphasized:



  • (TOP) topological manifolds

  • (PDIFF), for piecewise differentiable

  • (PL) piecewise-smooth manifolds

  • (DIFF) the smooth manifolds

Can we have more than the ones shown in the graph here: https://en.wikipedia.org/wiki/PDIFF with more subtle or refined structures?



For example, can we consider the affine structure, or Riemannian structure, etc? Or somewhere in between in the straight lines of the figure, can we have other refinement? other structures?



enter image description here




general-topology manifolds geometric-topology




share|cite|improve this question













I wonder why in geometric topology, only the following structure of manifolds are emphasized:



  • (TOP) topological manifolds

  • (PDIFF), for piecewise differentiable

  • (PL) piecewise-smooth manifolds

  • (DIFF) the smooth manifolds

Can we have more than the ones shown in the graph here: https://en.wikipedia.org/wiki/PDIFF with more subtle or refined structures?



For example, can we consider the affine structure, or Riemannian structure, etc? Or somewhere in between in the straight lines of the figure, can we have other refinement? other structures?



enter image description here





general-topology manifolds geometric-topology





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 26 at 3:30
























asked Jul 26 at 3:09









annie heart

549616




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  • 1




    I added a figure to clarify what you said (from the cited URL). Hope this is what you had in mind...
    – wonderich
    Jul 26 at 3:18












  • 1




    I added a figure to clarify what you said (from the cited URL). Hope this is what you had in mind...
    – wonderich
    Jul 26 at 3:18







1




1




I added a figure to clarify what you said (from the cited URL). Hope this is what you had in mind...
– wonderich
Jul 26 at 3:18




I added a figure to clarify what you said (from the cited URL). Hope this is what you had in mind...
– wonderich
Jul 26 at 3:18










2 Answers
2






active

oldest

votes

















up vote
1
down vote













One quick way to get a lot of different structures on manifolds is by considering the transition maps given by local trivializations $f_alpha:U_alpha to mathbb R^n$.



The procedure is by considering functions $f_alphaf_beta^-1:mathbb R^n to mathbb R^n$. If you want smooth a smooth structure you require these maps to be diffeomorphisms, if you want a complex structure you require them to be holomorphic, PL etc.






share|cite|improve this answer





















  • I don't know of a way to do this for riemannian manifolds, I think you really need the extra data of a tangent bundle for that.
    – Andres Mejia
    Jul 26 at 3:21











  • A Riemannian metric on $Bbb R^n$ is just a smooth map to positive definite matrices, and then the transition functions should pull back one of these smooth functions to the other, more or less. Everything can be phrased without an invocation of tangent bundles.
    – Mike Miller
    Jul 26 at 12:47











  • @MikeMiller isn’t that just basically invoking some bundle structure in disguise?
    – Andres Mejia
    Jul 26 at 12:58










  • No, that was a dumb comment. By that logic the whole answer is moot. Good point.
    – Andres Mejia
    Jul 26 at 12:59






  • 1




    I don't think it's a dumb comment, I agree with you that the tangent bundle is lurking in my point. (In fact it is rather interesting that the tangent bundle, a global concept, can be described entirely locally.)
    – Mike Miller
    Jul 26 at 13:20

















up vote
1
down vote













Other examples of structures on $n$-manifolds:



  • For a Euclidean structure, require your transition maps to be isometries of the Euclidean metric on $mathbb R^n$.

  • For a spherical structure, require your your local trivializations to take values in the $n$-sphere $mathbb S^n$, and your transition maps to be isometries of $mathbb S^n$, i.e elements of the orthogonal group $O(n+1)$.

In that last example, you can see that it is not necessary for the local trivializations to take values in $mathbb R^n$ itself, it is sufficient that they take values in some $n$-dimensional manifold.



  • For a hyperbolic structure, require your transition maps to be in hyperbolic $n$-space $mathbb H^n$, and your transition maps to be isometries of $mathbb H^n$.

In each of these three examples, restricting the local trivializations and the transition maps is used to define a certain kind of geometric manifold. The "model geometry" is a specific Riemannian manifold, and the "transition maps" are elements of the isometry group of that manifold. With that idea in mind, you can start making up your own examples of geometric manifolds.






share|cite|improve this answer

















  • 1




    I like the example of symplectic manifolds, which due to Darboux's theorem are the same as manifolds built up of open subsets of $Bbb R^2n$ with transition maps that have $textJac_x(varphi) in textSp_2n$.
    – Mike Miller
    Jul 26 at 13:36











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote













One quick way to get a lot of different structures on manifolds is by considering the transition maps given by local trivializations $f_alpha:U_alpha to mathbb R^n$.



The procedure is by considering functions $f_alphaf_beta^-1:mathbb R^n to mathbb R^n$. If you want smooth a smooth structure you require these maps to be diffeomorphisms, if you want a complex structure you require them to be holomorphic, PL etc.






share|cite|improve this answer





















  • I don't know of a way to do this for riemannian manifolds, I think you really need the extra data of a tangent bundle for that.
    – Andres Mejia
    Jul 26 at 3:21











  • A Riemannian metric on $Bbb R^n$ is just a smooth map to positive definite matrices, and then the transition functions should pull back one of these smooth functions to the other, more or less. Everything can be phrased without an invocation of tangent bundles.
    – Mike Miller
    Jul 26 at 12:47











  • @MikeMiller isn’t that just basically invoking some bundle structure in disguise?
    – Andres Mejia
    Jul 26 at 12:58










  • No, that was a dumb comment. By that logic the whole answer is moot. Good point.
    – Andres Mejia
    Jul 26 at 12:59






  • 1




    I don't think it's a dumb comment, I agree with you that the tangent bundle is lurking in my point. (In fact it is rather interesting that the tangent bundle, a global concept, can be described entirely locally.)
    – Mike Miller
    Jul 26 at 13:20














up vote
1
down vote













One quick way to get a lot of different structures on manifolds is by considering the transition maps given by local trivializations $f_alpha:U_alpha to mathbb R^n$.



The procedure is by considering functions $f_alphaf_beta^-1:mathbb R^n to mathbb R^n$. If you want smooth a smooth structure you require these maps to be diffeomorphisms, if you want a complex structure you require them to be holomorphic, PL etc.






share|cite|improve this answer





















  • I don't know of a way to do this for riemannian manifolds, I think you really need the extra data of a tangent bundle for that.
    – Andres Mejia
    Jul 26 at 3:21











  • A Riemannian metric on $Bbb R^n$ is just a smooth map to positive definite matrices, and then the transition functions should pull back one of these smooth functions to the other, more or less. Everything can be phrased without an invocation of tangent bundles.
    – Mike Miller
    Jul 26 at 12:47











  • @MikeMiller isn’t that just basically invoking some bundle structure in disguise?
    – Andres Mejia
    Jul 26 at 12:58










  • No, that was a dumb comment. By that logic the whole answer is moot. Good point.
    – Andres Mejia
    Jul 26 at 12:59






  • 1




    I don't think it's a dumb comment, I agree with you that the tangent bundle is lurking in my point. (In fact it is rather interesting that the tangent bundle, a global concept, can be described entirely locally.)
    – Mike Miller
    Jul 26 at 13:20












up vote
1
down vote










up vote
1
down vote









One quick way to get a lot of different structures on manifolds is by considering the transition maps given by local trivializations $f_alpha:U_alpha to mathbb R^n$.



The procedure is by considering functions $f_alphaf_beta^-1:mathbb R^n to mathbb R^n$. If you want smooth a smooth structure you require these maps to be diffeomorphisms, if you want a complex structure you require them to be holomorphic, PL etc.






share|cite|improve this answer













One quick way to get a lot of different structures on manifolds is by considering the transition maps given by local trivializations $f_alpha:U_alpha to mathbb R^n$.



The procedure is by considering functions $f_alphaf_beta^-1:mathbb R^n to mathbb R^n$. If you want smooth a smooth structure you require these maps to be diffeomorphisms, if you want a complex structure you require them to be holomorphic, PL etc.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 26 at 3:19









Andres Mejia

14.5k11243




14.5k11243











  • I don't know of a way to do this for riemannian manifolds, I think you really need the extra data of a tangent bundle for that.
    – Andres Mejia
    Jul 26 at 3:21











  • A Riemannian metric on $Bbb R^n$ is just a smooth map to positive definite matrices, and then the transition functions should pull back one of these smooth functions to the other, more or less. Everything can be phrased without an invocation of tangent bundles.
    – Mike Miller
    Jul 26 at 12:47











  • @MikeMiller isn’t that just basically invoking some bundle structure in disguise?
    – Andres Mejia
    Jul 26 at 12:58










  • No, that was a dumb comment. By that logic the whole answer is moot. Good point.
    – Andres Mejia
    Jul 26 at 12:59






  • 1




    I don't think it's a dumb comment, I agree with you that the tangent bundle is lurking in my point. (In fact it is rather interesting that the tangent bundle, a global concept, can be described entirely locally.)
    – Mike Miller
    Jul 26 at 13:20
















  • I don't know of a way to do this for riemannian manifolds, I think you really need the extra data of a tangent bundle for that.
    – Andres Mejia
    Jul 26 at 3:21











  • A Riemannian metric on $Bbb R^n$ is just a smooth map to positive definite matrices, and then the transition functions should pull back one of these smooth functions to the other, more or less. Everything can be phrased without an invocation of tangent bundles.
    – Mike Miller
    Jul 26 at 12:47











  • @MikeMiller isn’t that just basically invoking some bundle structure in disguise?
    – Andres Mejia
    Jul 26 at 12:58










  • No, that was a dumb comment. By that logic the whole answer is moot. Good point.
    – Andres Mejia
    Jul 26 at 12:59






  • 1




    I don't think it's a dumb comment, I agree with you that the tangent bundle is lurking in my point. (In fact it is rather interesting that the tangent bundle, a global concept, can be described entirely locally.)
    – Mike Miller
    Jul 26 at 13:20















I don't know of a way to do this for riemannian manifolds, I think you really need the extra data of a tangent bundle for that.
– Andres Mejia
Jul 26 at 3:21





I don't know of a way to do this for riemannian manifolds, I think you really need the extra data of a tangent bundle for that.
– Andres Mejia
Jul 26 at 3:21













A Riemannian metric on $Bbb R^n$ is just a smooth map to positive definite matrices, and then the transition functions should pull back one of these smooth functions to the other, more or less. Everything can be phrased without an invocation of tangent bundles.
– Mike Miller
Jul 26 at 12:47





A Riemannian metric on $Bbb R^n$ is just a smooth map to positive definite matrices, and then the transition functions should pull back one of these smooth functions to the other, more or less. Everything can be phrased without an invocation of tangent bundles.
– Mike Miller
Jul 26 at 12:47













@MikeMiller isn’t that just basically invoking some bundle structure in disguise?
– Andres Mejia
Jul 26 at 12:58




@MikeMiller isn’t that just basically invoking some bundle structure in disguise?
– Andres Mejia
Jul 26 at 12:58












No, that was a dumb comment. By that logic the whole answer is moot. Good point.
– Andres Mejia
Jul 26 at 12:59




No, that was a dumb comment. By that logic the whole answer is moot. Good point.
– Andres Mejia
Jul 26 at 12:59




1




1




I don't think it's a dumb comment, I agree with you that the tangent bundle is lurking in my point. (In fact it is rather interesting that the tangent bundle, a global concept, can be described entirely locally.)
– Mike Miller
Jul 26 at 13:20




I don't think it's a dumb comment, I agree with you that the tangent bundle is lurking in my point. (In fact it is rather interesting that the tangent bundle, a global concept, can be described entirely locally.)
– Mike Miller
Jul 26 at 13:20










up vote
1
down vote













Other examples of structures on $n$-manifolds:



  • For a Euclidean structure, require your transition maps to be isometries of the Euclidean metric on $mathbb R^n$.

  • For a spherical structure, require your your local trivializations to take values in the $n$-sphere $mathbb S^n$, and your transition maps to be isometries of $mathbb S^n$, i.e elements of the orthogonal group $O(n+1)$.

In that last example, you can see that it is not necessary for the local trivializations to take values in $mathbb R^n$ itself, it is sufficient that they take values in some $n$-dimensional manifold.



  • For a hyperbolic structure, require your transition maps to be in hyperbolic $n$-space $mathbb H^n$, and your transition maps to be isometries of $mathbb H^n$.

In each of these three examples, restricting the local trivializations and the transition maps is used to define a certain kind of geometric manifold. The "model geometry" is a specific Riemannian manifold, and the "transition maps" are elements of the isometry group of that manifold. With that idea in mind, you can start making up your own examples of geometric manifolds.






share|cite|improve this answer

















  • 1




    I like the example of symplectic manifolds, which due to Darboux's theorem are the same as manifolds built up of open subsets of $Bbb R^2n$ with transition maps that have $textJac_x(varphi) in textSp_2n$.
    – Mike Miller
    Jul 26 at 13:36















up vote
1
down vote













Other examples of structures on $n$-manifolds:



  • For a Euclidean structure, require your transition maps to be isometries of the Euclidean metric on $mathbb R^n$.

  • For a spherical structure, require your your local trivializations to take values in the $n$-sphere $mathbb S^n$, and your transition maps to be isometries of $mathbb S^n$, i.e elements of the orthogonal group $O(n+1)$.

In that last example, you can see that it is not necessary for the local trivializations to take values in $mathbb R^n$ itself, it is sufficient that they take values in some $n$-dimensional manifold.



  • For a hyperbolic structure, require your transition maps to be in hyperbolic $n$-space $mathbb H^n$, and your transition maps to be isometries of $mathbb H^n$.

In each of these three examples, restricting the local trivializations and the transition maps is used to define a certain kind of geometric manifold. The "model geometry" is a specific Riemannian manifold, and the "transition maps" are elements of the isometry group of that manifold. With that idea in mind, you can start making up your own examples of geometric manifolds.






share|cite|improve this answer

















  • 1




    I like the example of symplectic manifolds, which due to Darboux's theorem are the same as manifolds built up of open subsets of $Bbb R^2n$ with transition maps that have $textJac_x(varphi) in textSp_2n$.
    – Mike Miller
    Jul 26 at 13:36













up vote
1
down vote










up vote
1
down vote









Other examples of structures on $n$-manifolds:



  • For a Euclidean structure, require your transition maps to be isometries of the Euclidean metric on $mathbb R^n$.

  • For a spherical structure, require your your local trivializations to take values in the $n$-sphere $mathbb S^n$, and your transition maps to be isometries of $mathbb S^n$, i.e elements of the orthogonal group $O(n+1)$.

In that last example, you can see that it is not necessary for the local trivializations to take values in $mathbb R^n$ itself, it is sufficient that they take values in some $n$-dimensional manifold.



  • For a hyperbolic structure, require your transition maps to be in hyperbolic $n$-space $mathbb H^n$, and your transition maps to be isometries of $mathbb H^n$.

In each of these three examples, restricting the local trivializations and the transition maps is used to define a certain kind of geometric manifold. The "model geometry" is a specific Riemannian manifold, and the "transition maps" are elements of the isometry group of that manifold. With that idea in mind, you can start making up your own examples of geometric manifolds.






share|cite|improve this answer













Other examples of structures on $n$-manifolds:



  • For a Euclidean structure, require your transition maps to be isometries of the Euclidean metric on $mathbb R^n$.

  • For a spherical structure, require your your local trivializations to take values in the $n$-sphere $mathbb S^n$, and your transition maps to be isometries of $mathbb S^n$, i.e elements of the orthogonal group $O(n+1)$.

In that last example, you can see that it is not necessary for the local trivializations to take values in $mathbb R^n$ itself, it is sufficient that they take values in some $n$-dimensional manifold.



  • For a hyperbolic structure, require your transition maps to be in hyperbolic $n$-space $mathbb H^n$, and your transition maps to be isometries of $mathbb H^n$.

In each of these three examples, restricting the local trivializations and the transition maps is used to define a certain kind of geometric manifold. The "model geometry" is a specific Riemannian manifold, and the "transition maps" are elements of the isometry group of that manifold. With that idea in mind, you can start making up your own examples of geometric manifolds.







share|cite|improve this answer













share|cite|improve this answer



share|cite|improve this answer











answered Jul 26 at 12:30









Lee Mosher

45.4k33478




45.4k33478







  • 1




    I like the example of symplectic manifolds, which due to Darboux's theorem are the same as manifolds built up of open subsets of $Bbb R^2n$ with transition maps that have $textJac_x(varphi) in textSp_2n$.
    – Mike Miller
    Jul 26 at 13:36













  • 1




    I like the example of symplectic manifolds, which due to Darboux's theorem are the same as manifolds built up of open subsets of $Bbb R^2n$ with transition maps that have $textJac_x(varphi) in textSp_2n$.
    – Mike Miller
    Jul 26 at 13:36








1




1




I like the example of symplectic manifolds, which due to Darboux's theorem are the same as manifolds built up of open subsets of $Bbb R^2n$ with transition maps that have $textJac_x(varphi) in textSp_2n$.
– Mike Miller
Jul 26 at 13:36





I like the example of symplectic manifolds, which due to Darboux's theorem are the same as manifolds built up of open subsets of $Bbb R^2n$ with transition maps that have $textJac_x(varphi) in textSp_2n$.
– Mike Miller
Jul 26 at 13:36













 

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