Tangent space of $S^1$

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Compute $T_t(S^1)$ in $(cos(t),sin(t))$ with parametrization $f(t)=(cos(t),sin(t)).$



I have this:



I know, $T_tf:T_tmathbbRto T_f(t)S^1$ now $T_tf(left.fracddsright|_s=0(t+sK))=left.fracddsright|_s=0 f(t+sK)=K(-sin(t),cos(t))$.



Therefore $T_tf(K)=K(-sin(t),cos(t))$.



Now that I have the tangent function. How do I get $T_t(S^1)$?




tangent-spaces




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  • There are more rigorous ways to do this, but as you can see from drawing a picture, the tangent space is simply spanned by $(-sin(t),cos(t))$.
    – Cave Johnson
    Jul 30 at 1:35










  • From what I did, we can conclude that: $T_cos(t),sin(t)S^1=leftK(-sin(t),cos(t):Kin T_tmathbbRright$ ?
    – eraldcoil
    Jul 30 at 1:44















up vote
0
down vote

favorite












Compute $T_t(S^1)$ in $(cos(t),sin(t))$ with parametrization $f(t)=(cos(t),sin(t)).$



I have this:



I know, $T_tf:T_tmathbbRto T_f(t)S^1$ now $T_tf(left.fracddsright|_s=0(t+sK))=left.fracddsright|_s=0 f(t+sK)=K(-sin(t),cos(t))$.



Therefore $T_tf(K)=K(-sin(t),cos(t))$.



Now that I have the tangent function. How do I get $T_t(S^1)$?




tangent-spaces




share|cite|improve this question



















  • There are more rigorous ways to do this, but as you can see from drawing a picture, the tangent space is simply spanned by $(-sin(t),cos(t))$.
    – Cave Johnson
    Jul 30 at 1:35










  • From what I did, we can conclude that: $T_cos(t),sin(t)S^1=leftK(-sin(t),cos(t):Kin T_tmathbbRright$ ?
    – eraldcoil
    Jul 30 at 1:44













up vote
0
down vote

favorite









up vote
0
down vote

favorite











Compute $T_t(S^1)$ in $(cos(t),sin(t))$ with parametrization $f(t)=(cos(t),sin(t)).$



I have this:



I know, $T_tf:T_tmathbbRto T_f(t)S^1$ now $T_tf(left.fracddsright|_s=0(t+sK))=left.fracddsright|_s=0 f(t+sK)=K(-sin(t),cos(t))$.



Therefore $T_tf(K)=K(-sin(t),cos(t))$.



Now that I have the tangent function. How do I get $T_t(S^1)$?




tangent-spaces




share|cite|improve this question











Compute $T_t(S^1)$ in $(cos(t),sin(t))$ with parametrization $f(t)=(cos(t),sin(t)).$



I have this:



I know, $T_tf:T_tmathbbRto T_f(t)S^1$ now $T_tf(left.fracddsright|_s=0(t+sK))=left.fracddsright|_s=0 f(t+sK)=K(-sin(t),cos(t))$.



Therefore $T_tf(K)=K(-sin(t),cos(t))$.



Now that I have the tangent function. How do I get $T_t(S^1)$?





tangent-spaces





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 30 at 1:28









eraldcoil

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  • There are more rigorous ways to do this, but as you can see from drawing a picture, the tangent space is simply spanned by $(-sin(t),cos(t))$.
    – Cave Johnson
    Jul 30 at 1:35










  • From what I did, we can conclude that: $T_cos(t),sin(t)S^1=leftK(-sin(t),cos(t):Kin T_tmathbbRright$ ?
    – eraldcoil
    Jul 30 at 1:44

















  • There are more rigorous ways to do this, but as you can see from drawing a picture, the tangent space is simply spanned by $(-sin(t),cos(t))$.
    – Cave Johnson
    Jul 30 at 1:35










  • From what I did, we can conclude that: $T_cos(t),sin(t)S^1=leftK(-sin(t),cos(t):Kin T_tmathbbRright$ ?
    – eraldcoil
    Jul 30 at 1:44
















There are more rigorous ways to do this, but as you can see from drawing a picture, the tangent space is simply spanned by $(-sin(t),cos(t))$.
– Cave Johnson
Jul 30 at 1:35




There are more rigorous ways to do this, but as you can see from drawing a picture, the tangent space is simply spanned by $(-sin(t),cos(t))$.
– Cave Johnson
Jul 30 at 1:35












From what I did, we can conclude that: $T_cos(t),sin(t)S^1=leftK(-sin(t),cos(t):Kin T_tmathbbRright$ ?
– eraldcoil
Jul 30 at 1:44





From what I did, we can conclude that: $T_cos(t),sin(t)S^1=leftK(-sin(t),cos(t):Kin T_tmathbbRright$ ?
– eraldcoil
Jul 30 at 1:44
















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