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$M$ is orientable if and only if $M setminus p$ is orientable.
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Let $M$ a manifold of class $C^infty$. Show that $M$ is orientable if and only if $M setminus p$ is orientable.
Comments:
($Rightarrow$) Let $omega: M longrightarrow Lambda^n(M)$ a orientation form of M then $omega: Msetminusp longrightarrow Lambda^n(Msetminusp)$ is a form which is never null then defines a orientation in $Msetminusp$.
I am having difficulty justifying the ($Leftarrow$).
geometry manifolds orientation
asked Jul 23 at 15:02
Croos
789415
add a comment |Â
up vote
2
down vote
favorite
Let $M$ a manifold of class $C^infty$. Show that $M$ is orientable if and only if $M setminus p$ is orientable.
Comments:
($Rightarrow$) Let $omega: M longrightarrow Lambda^n(M)$ a orientation form of M then $omega: Msetminusp longrightarrow Lambda^n(Msetminusp)$ is a form which is never null then defines a orientation in $Msetminusp$.
I am having difficulty justifying the ($Leftarrow$).
geometry manifolds orientation
asked Jul 23 at 15:02
Croos
789415
You should give a bit more context, explain the symbols, and so on. I can guess that $M$ is any manifold and $p in M$ a point, but still I'd prefer to hear that from you.
– Luke
Jul 23 at 15:18
4
For $Leftarrow$, in a chart at $p$ write $omega=gdx_1wedge...wedge dx_n$. The $g$ is always positive except possibly at $0$. Then take $omega'=g'dx_1wedge...wedge dx_n$, where $g'$ is equal to $1$ at $0$ and equal to $g$ outside a neighborhood of $0$.
– user577471
Jul 23 at 15:33
@HGLandcaster Do you say to define $g (p) = 1$?
– Croos
Jul 26 at 0:35
1
@Croos Yes, but doing only that wouldn't keep it smooth. It has to be adjusted in a whole neighborhood of $p$ to keep it smooth. Fortunately smooth functions are quite flexible.
– user577471
Jul 26 at 14:49
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $M$ a manifold of class $C^infty$. Show that $M$ is orientable if and only if $M setminus p$ is orientable.
Comments:
($Rightarrow$) Let $omega: M longrightarrow Lambda^n(M)$ a orientation form of M then $omega: Msetminusp longrightarrow Lambda^n(Msetminusp)$ is a form which is never null then defines a orientation in $Msetminusp$.
I am having difficulty justifying the ($Leftarrow$).
geometry manifolds orientation
asked Jul 23 at 15:02
Croos
789415
Let $M$ a manifold of class $C^infty$. Show that $M$ is orientable if and only if $M setminus p$ is orientable.
Comments:
($Rightarrow$) Let $omega: M longrightarrow Lambda^n(M)$ a orientation form of M then $omega: Msetminusp longrightarrow Lambda^n(Msetminusp)$ is a form which is never null then defines a orientation in $Msetminusp$.
I am having difficulty justifying the ($Leftarrow$).
geometry manifolds orientation
asked Jul 23 at 15:02
Croos
789415
edited Jul 25 at 23:59
asked Jul 23 at 15:02
Croos
789415
asked Jul 23 at 15:02
Croos
789415
asked Jul 23 at 15:02
Croos
789415
789415
You should give a bit more context, explain the symbols, and so on. I can guess that $M$ is any manifold and $p in M$ a point, but still I'd prefer to hear that from you.
– Luke
Jul 23 at 15:18
4
For $Leftarrow$, in a chart at $p$ write $omega=gdx_1wedge...wedge dx_n$. The $g$ is always positive except possibly at $0$. Then take $omega'=g'dx_1wedge...wedge dx_n$, where $g'$ is equal to $1$ at $0$ and equal to $g$ outside a neighborhood of $0$.
– user577471
Jul 23 at 15:33
@HGLandcaster Do you say to define $g (p) = 1$?
– Croos
Jul 26 at 0:35
1
@Croos Yes, but doing only that wouldn't keep it smooth. It has to be adjusted in a whole neighborhood of $p$ to keep it smooth. Fortunately smooth functions are quite flexible.
– user577471
Jul 26 at 14:49
add a comment |Â
You should give a bit more context, explain the symbols, and so on. I can guess that $M$ is any manifold and $p in M$ a point, but still I'd prefer to hear that from you.
– Luke
Jul 23 at 15:18
4
For $Leftarrow$, in a chart at $p$ write $omega=gdx_1wedge...wedge dx_n$. The $g$ is always positive except possibly at $0$. Then take $omega'=g'dx_1wedge...wedge dx_n$, where $g'$ is equal to $1$ at $0$ and equal to $g$ outside a neighborhood of $0$.
– user577471
Jul 23 at 15:33
@HGLandcaster Do you say to define $g (p) = 1$?
– Croos
Jul 26 at 0:35
1
@Croos Yes, but doing only that wouldn't keep it smooth. It has to be adjusted in a whole neighborhood of $p$ to keep it smooth. Fortunately smooth functions are quite flexible.
– user577471
Jul 26 at 14:49
You should give a bit more context, explain the symbols, and so on. I can guess that $M$ is any manifold and $p in M$ a point, but still I'd prefer to hear that from you.
– Luke
Jul 23 at 15:18
You should give a bit more context, explain the symbols, and so on. I can guess that $M$ is any manifold and $p in M$ a point, but still I'd prefer to hear that from you.
– Luke
Jul 23 at 15:18
4
4
For $Leftarrow$, in a chart at $p$ write $omega=gdx_1wedge...wedge dx_n$. The $g$ is always positive except possibly at $0$. Then take $omega'=g'dx_1wedge...wedge dx_n$, where $g'$ is equal to $1$ at $0$ and equal to $g$ outside a neighborhood of $0$.
– user577471
Jul 23 at 15:33
For $Leftarrow$, in a chart at $p$ write $omega=gdx_1wedge...wedge dx_n$. The $g$ is always positive except possibly at $0$. Then take $omega'=g'dx_1wedge...wedge dx_n$, where $g'$ is equal to $1$ at $0$ and equal to $g$ outside a neighborhood of $0$.
– user577471
Jul 23 at 15:33
@HGLandcaster Do you say to define $g (p) = 1$?
– Croos
Jul 26 at 0:35
@HGLandcaster Do you say to define $g (p) = 1$?
– Croos
Jul 26 at 0:35
1
1
@Croos Yes, but doing only that wouldn't keep it smooth. It has to be adjusted in a whole neighborhood of $p$ to keep it smooth. Fortunately smooth functions are quite flexible.
– user577471
Jul 26 at 14:49
@Croos Yes, but doing only that wouldn't keep it smooth. It has to be adjusted in a whole neighborhood of $p$ to keep it smooth. Fortunately smooth functions are quite flexible.
– user577471
Jul 26 at 14:49
add a comment |Â
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You should give a bit more context, explain the symbols, and so on. I can guess that $M$ is any manifold and $p in M$ a point, but still I'd prefer to hear that from you.
– Luke
Jul 23 at 15:18
4
For $Leftarrow$, in a chart at $p$ write $omega=gdx_1wedge...wedge dx_n$. The $g$ is always positive except possibly at $0$. Then take $omega'=g'dx_1wedge...wedge dx_n$, where $g'$ is equal to $1$ at $0$ and equal to $g$ outside a neighborhood of $0$.
– user577471
Jul 23 at 15:33
@HGLandcaster Do you say to define $g (p) = 1$?
– Croos
Jul 26 at 0:35
1
@Croos Yes, but doing only that wouldn't keep it smooth. It has to be adjusted in a whole neighborhood of $p$ to keep it smooth. Fortunately smooth functions are quite flexible.
– user577471
Jul 26 at 14:49