Mixing
Exponential map is orientation preserving
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
Let us define the map $f:mathbbRto mathbbS^1$ by $xto e^2pi i x$. I want to prove that this map is an orientation-preserving map and also it's restriction on the unit interval has an orientation-preserving inverse.
What I know:
$f: Mto N$ is an orientation preserving if the determinant of the derivative map on the tangent space level is positive. So here, we need to show $Df(p):T_pmathbbRto T_f(p)mathbbS^1$ has a positive determinant. But here $Df(p)$ is just a real number so we need to show it is a positive number. Consider a curve $gamma(t)=p+tv$ where $(p,v)in T_pmathbbR$. $f(gamma(t))=e^2pi i(p+tv)$. But here I am having difficulty as the derivative will be a $2times 1$ matrix.
differential-geometry orientation
asked Aug 2 at 12:00
I am pi
1889
add a comment |Â
up vote
0
down vote
favorite
Let us define the map $f:mathbbRto mathbbS^1$ by $xto e^2pi i x$. I want to prove that this map is an orientation-preserving map and also it's restriction on the unit interval has an orientation-preserving inverse.
What I know:
$f: Mto N$ is an orientation preserving if the determinant of the derivative map on the tangent space level is positive. So here, we need to show $Df(p):T_pmathbbRto T_f(p)mathbbS^1$ has a positive determinant. But here $Df(p)$ is just a real number so we need to show it is a positive number. Consider a curve $gamma(t)=p+tv$ where $(p,v)in T_pmathbbR$. $f(gamma(t))=e^2pi i(p+tv)$. But here I am having difficulty as the derivative will be a $2times 1$ matrix.
differential-geometry orientation
asked Aug 2 at 12:00
I am pi
1889
In the definition of orientation-preserving map I know there is an assumption that source and target manifolds have the same dimension (otherwise you encounter the problem that the derivative is not a square matrix).
– Paweł Czyż
Aug 4 at 22:07
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let us define the map $f:mathbbRto mathbbS^1$ by $xto e^2pi i x$. I want to prove that this map is an orientation-preserving map and also it's restriction on the unit interval has an orientation-preserving inverse.
What I know:
$f: Mto N$ is an orientation preserving if the determinant of the derivative map on the tangent space level is positive. So here, we need to show $Df(p):T_pmathbbRto T_f(p)mathbbS^1$ has a positive determinant. But here $Df(p)$ is just a real number so we need to show it is a positive number. Consider a curve $gamma(t)=p+tv$ where $(p,v)in T_pmathbbR$. $f(gamma(t))=e^2pi i(p+tv)$. But here I am having difficulty as the derivative will be a $2times 1$ matrix.
differential-geometry orientation
asked Aug 2 at 12:00
I am pi
1889
Let us define the map $f:mathbbRto mathbbS^1$ by $xto e^2pi i x$. I want to prove that this map is an orientation-preserving map and also it's restriction on the unit interval has an orientation-preserving inverse.
What I know:
$f: Mto N$ is an orientation preserving if the determinant of the derivative map on the tangent space level is positive. So here, we need to show $Df(p):T_pmathbbRto T_f(p)mathbbS^1$ has a positive determinant. But here $Df(p)$ is just a real number so we need to show it is a positive number. Consider a curve $gamma(t)=p+tv$ where $(p,v)in T_pmathbbR$. $f(gamma(t))=e^2pi i(p+tv)$. But here I am having difficulty as the derivative will be a $2times 1$ matrix.
differential-geometry orientation
asked Aug 2 at 12:00
I am pi
1889
asked Aug 2 at 12:00
I am pi
1889
asked Aug 2 at 12:00
I am pi
1889
asked Aug 2 at 12:00
I am pi
1889
1889
In the definition of orientation-preserving map I know there is an assumption that source and target manifolds have the same dimension (otherwise you encounter the problem that the derivative is not a square matrix).
– Paweł Czyż
Aug 4 at 22:07
add a comment |Â
In the definition of orientation-preserving map I know there is an assumption that source and target manifolds have the same dimension (otherwise you encounter the problem that the derivative is not a square matrix).
– Paweł Czyż
Aug 4 at 22:07
In the definition of orientation-preserving map I know there is an assumption that source and target manifolds have the same dimension (otherwise you encounter the problem that the derivative is not a square matrix).
– Paweł Czyż
Aug 4 at 22:07
In the definition of orientation-preserving map I know there is an assumption that source and target manifolds have the same dimension (otherwise you encounter the problem that the derivative is not a square matrix).
– Paweł Czyż
Aug 4 at 22:07
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2870001%2fexponential-map-is-orientation-preserving%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
In the definition of orientation-preserving map I know there is an assumption that source and target manifolds have the same dimension (otherwise you encounter the problem that the derivative is not a square matrix).
– Paweł Czyż
Aug 4 at 22:07