Why is the exponential map clearly the identity map for $mathbb R^n$?












1












$begingroup$


Suppose that $M$ is a smooth Riemannian manifold, $qin M$. There exists an $epsilon$ such that $exp_q:B_{epsilon}(0)rightarrow M$ is a diffeomorphism.
In DoCarmo’s Riemannian Geometry book, it has been written that for $M=mathbb R^n$, this map $exp_q$ is the identity map, while $exp_q(v)=q+v$. So what does DoCarmo mean?






riemannian-geometry smooth-manifolds







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  • $begingroup$
    What's your definition of the tangent space $T_qmathbb{R}^n$?
    $endgroup$
    – Dante Grevino
    Dec 17 '18 at 1:12










  • $begingroup$
    I identify it with $mathbb R^n$ itself. Why?
    $endgroup$
    – User12239
    Dec 17 '18 at 1:13






  • 1




    $begingroup$
    Some people think in $T_qmathbb{R}^n$ as the affine space $q + mathbb{R}^n$ with the linear structure $(q+v)+(q+w)=q+(v+w)$ and $lambda (q+v)=q+lambda v$ for every $v,winmathbb{R}^n$ and $lambdainmathbb{R}$.
    $endgroup$
    – Dante Grevino
    Dec 17 '18 at 1:17










  • $begingroup$
    Thanks, I got what he meant now @DanteGrevino
    $endgroup$
    – User12239
    Dec 17 '18 at 1:19
















1












$begingroup$


Suppose that $M$ is a smooth Riemannian manifold, $qin M$. There exists an $epsilon$ such that $exp_q:B_{epsilon}(0)rightarrow M$ is a diffeomorphism.
In DoCarmo’s Riemannian Geometry book, it has been written that for $M=mathbb R^n$, this map $exp_q$ is the identity map, while $exp_q(v)=q+v$. So what does DoCarmo mean?






riemannian-geometry smooth-manifolds







share|cite|improve this question











$endgroup$












  • $begingroup$
    What's your definition of the tangent space $T_qmathbb{R}^n$?
    $endgroup$
    – Dante Grevino
    Dec 17 '18 at 1:12










  • $begingroup$
    I identify it with $mathbb R^n$ itself. Why?
    $endgroup$
    – User12239
    Dec 17 '18 at 1:13






  • 1




    $begingroup$
    Some people think in $T_qmathbb{R}^n$ as the affine space $q + mathbb{R}^n$ with the linear structure $(q+v)+(q+w)=q+(v+w)$ and $lambda (q+v)=q+lambda v$ for every $v,winmathbb{R}^n$ and $lambdainmathbb{R}$.
    $endgroup$
    – Dante Grevino
    Dec 17 '18 at 1:17










  • $begingroup$
    Thanks, I got what he meant now @DanteGrevino
    $endgroup$
    – User12239
    Dec 17 '18 at 1:19














1












1








1





$begingroup$


Suppose that $M$ is a smooth Riemannian manifold, $qin M$. There exists an $epsilon$ such that $exp_q:B_{epsilon}(0)rightarrow M$ is a diffeomorphism.
In DoCarmo’s Riemannian Geometry book, it has been written that for $M=mathbb R^n$, this map $exp_q$ is the identity map, while $exp_q(v)=q+v$. So what does DoCarmo mean?






riemannian-geometry smooth-manifolds







share|cite|improve this question











$endgroup$




Suppose that $M$ is a smooth Riemannian manifold, $qin M$. There exists an $epsilon$ such that $exp_q:B_{epsilon}(0)rightarrow M$ is a diffeomorphism.
In DoCarmo’s Riemannian Geometry book, it has been written that for $M=mathbb R^n$, this map $exp_q$ is the identity map, while $exp_q(v)=q+v$. So what does DoCarmo mean?






riemannian-geometry smooth-manifolds




riemannian-geometry smooth-manifolds






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 17 '18 at 1:09









Batominovski

33k33293




33k33293










asked Dec 17 '18 at 1:04









User12239User12239

453216




453216












  • $begingroup$
    What's your definition of the tangent space $T_qmathbb{R}^n$?
    $endgroup$
    – Dante Grevino
    Dec 17 '18 at 1:12










  • $begingroup$
    I identify it with $mathbb R^n$ itself. Why?
    $endgroup$
    – User12239
    Dec 17 '18 at 1:13






  • 1




    $begingroup$
    Some people think in $T_qmathbb{R}^n$ as the affine space $q + mathbb{R}^n$ with the linear structure $(q+v)+(q+w)=q+(v+w)$ and $lambda (q+v)=q+lambda v$ for every $v,winmathbb{R}^n$ and $lambdainmathbb{R}$.
    $endgroup$
    – Dante Grevino
    Dec 17 '18 at 1:17










  • $begingroup$
    Thanks, I got what he meant now @DanteGrevino
    $endgroup$
    – User12239
    Dec 17 '18 at 1:19


















  • $begingroup$
    What's your definition of the tangent space $T_qmathbb{R}^n$?
    $endgroup$
    – Dante Grevino
    Dec 17 '18 at 1:12










  • $begingroup$
    I identify it with $mathbb R^n$ itself. Why?
    $endgroup$
    – User12239
    Dec 17 '18 at 1:13






  • 1




    $begingroup$
    Some people think in $T_qmathbb{R}^n$ as the affine space $q + mathbb{R}^n$ with the linear structure $(q+v)+(q+w)=q+(v+w)$ and $lambda (q+v)=q+lambda v$ for every $v,winmathbb{R}^n$ and $lambdainmathbb{R}$.
    $endgroup$
    – Dante Grevino
    Dec 17 '18 at 1:17










  • $begingroup$
    Thanks, I got what he meant now @DanteGrevino
    $endgroup$
    – User12239
    Dec 17 '18 at 1:19
















$begingroup$
What's your definition of the tangent space $T_qmathbb{R}^n$?
$endgroup$
– Dante Grevino
Dec 17 '18 at 1:12




$begingroup$
What's your definition of the tangent space $T_qmathbb{R}^n$?
$endgroup$
– Dante Grevino
Dec 17 '18 at 1:12












$begingroup$
I identify it with $mathbb R^n$ itself. Why?
$endgroup$
– User12239
Dec 17 '18 at 1:13




$begingroup$
I identify it with $mathbb R^n$ itself. Why?
$endgroup$
– User12239
Dec 17 '18 at 1:13




1




1




$begingroup$
Some people think in $T_qmathbb{R}^n$ as the affine space $q + mathbb{R}^n$ with the linear structure $(q+v)+(q+w)=q+(v+w)$ and $lambda (q+v)=q+lambda v$ for every $v,winmathbb{R}^n$ and $lambdainmathbb{R}$.
$endgroup$
– Dante Grevino
Dec 17 '18 at 1:17




$begingroup$
Some people think in $T_qmathbb{R}^n$ as the affine space $q + mathbb{R}^n$ with the linear structure $(q+v)+(q+w)=q+(v+w)$ and $lambda (q+v)=q+lambda v$ for every $v,winmathbb{R}^n$ and $lambdainmathbb{R}$.
$endgroup$
– Dante Grevino
Dec 17 '18 at 1:17












$begingroup$
Thanks, I got what he meant now @DanteGrevino
$endgroup$
– User12239
Dec 17 '18 at 1:19




$begingroup$
Thanks, I got what he meant now @DanteGrevino
$endgroup$
– User12239
Dec 17 '18 at 1:19










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