Tangent space for the differentiable manifold $S^1$
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Since $S^1$ is a compact 1-dimensional regular submanifold in $mathbb{R}^2$ (it's $S^1 = f^{-1}(1)$ for $f : mathbb{R}^2 to mathbb{R}$ given as $f(x,y) = x^2+y^2$), we can find the tangent space for $S^1$ in (1,0) as $$T_{(1,0)}(S^1) = Ker(d(f)_{(1,0)}).$$ Intuitively $T_{(1,0)}(S^1)$ is the plane ${(1,y): y in mathbb{R}}.$
But we get $$Ker(d(f)_{(1,0)}) = Ker((2x,2y)_{(1,0)}) = Ker((2,0)) = {(0,y) : y in mathbb{R}}.$$
What it is wrong? Thanks in advance!
differential-geometry tangent-spaces
asked Nov 18 at 19:17
user540275
597
add a comment |
up vote
1
down vote
favorite
Since $S^1$ is a compact 1-dimensional regular submanifold in $mathbb{R}^2$ (it's $S^1 = f^{-1}(1)$ for $f : mathbb{R}^2 to mathbb{R}$ given as $f(x,y) = x^2+y^2$), we can find the tangent space for $S^1$ in (1,0) as $$T_{(1,0)}(S^1) = Ker(d(f)_{(1,0)}).$$ Intuitively $T_{(1,0)}(S^1)$ is the plane ${(1,y): y in mathbb{R}}.$
But we get $$Ker(d(f)_{(1,0)}) = Ker((2x,2y)_{(1,0)}) = Ker((2,0)) = {(0,y) : y in mathbb{R}}.$$
What it is wrong? Thanks in advance!
differential-geometry tangent-spaces
asked Nov 18 at 19:17
user540275
597
1
You are calculating the tangent space as a subspace of $mathbb{R}^2$. So, it passes through the origin. You can then translate it to the point $(1, 0)$.
– Joe Johnson 126
Nov 18 at 19:29
2
Your result is a subspace of tangent space $T_{(1,0)}R^2$. Nothing wrong with your calculation. The result represent the components of tangent vectors. That is the elements of the kernel should be in the form $0 partial_x + spartial_y$ for $sin R$. Which is exactly what you wanted (up to the identification).
– Kelvin Lois
Nov 18 at 19:29
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Since $S^1$ is a compact 1-dimensional regular submanifold in $mathbb{R}^2$ (it's $S^1 = f^{-1}(1)$ for $f : mathbb{R}^2 to mathbb{R}$ given as $f(x,y) = x^2+y^2$), we can find the tangent space for $S^1$ in (1,0) as $$T_{(1,0)}(S^1) = Ker(d(f)_{(1,0)}).$$ Intuitively $T_{(1,0)}(S^1)$ is the plane ${(1,y): y in mathbb{R}}.$
But we get $$Ker(d(f)_{(1,0)}) = Ker((2x,2y)_{(1,0)}) = Ker((2,0)) = {(0,y) : y in mathbb{R}}.$$
What it is wrong? Thanks in advance!
differential-geometry tangent-spaces
asked Nov 18 at 19:17
user540275
597
Since $S^1$ is a compact 1-dimensional regular submanifold in $mathbb{R}^2$ (it's $S^1 = f^{-1}(1)$ for $f : mathbb{R}^2 to mathbb{R}$ given as $f(x,y) = x^2+y^2$), we can find the tangent space for $S^1$ in (1,0) as $$T_{(1,0)}(S^1) = Ker(d(f)_{(1,0)}).$$ Intuitively $T_{(1,0)}(S^1)$ is the plane ${(1,y): y in mathbb{R}}.$
But we get $$Ker(d(f)_{(1,0)}) = Ker((2x,2y)_{(1,0)}) = Ker((2,0)) = {(0,y) : y in mathbb{R}}.$$
What it is wrong? Thanks in advance!
differential-geometry tangent-spaces
differential-geometry tangent-spaces
asked Nov 18 at 19:17
user540275
597
asked Nov 18 at 19:17
user540275
597
asked Nov 18 at 19:17
user540275
597
asked Nov 18 at 19:17
user540275
597
asked Nov 18 at 19:17
user540275
597
597
1
You are calculating the tangent space as a subspace of $mathbb{R}^2$. So, it passes through the origin. You can then translate it to the point $(1, 0)$.
– Joe Johnson 126
Nov 18 at 19:29
2
Your result is a subspace of tangent space $T_{(1,0)}R^2$. Nothing wrong with your calculation. The result represent the components of tangent vectors. That is the elements of the kernel should be in the form $0 partial_x + spartial_y$ for $sin R$. Which is exactly what you wanted (up to the identification).
– Kelvin Lois
Nov 18 at 19:29
add a comment |
1
You are calculating the tangent space as a subspace of $mathbb{R}^2$. So, it passes through the origin. You can then translate it to the point $(1, 0)$.
– Joe Johnson 126
Nov 18 at 19:29
2
Your result is a subspace of tangent space $T_{(1,0)}R^2$. Nothing wrong with your calculation. The result represent the components of tangent vectors. That is the elements of the kernel should be in the form $0 partial_x + spartial_y$ for $sin R$. Which is exactly what you wanted (up to the identification).
– Kelvin Lois
Nov 18 at 19:29
1
1
You are calculating the tangent space as a subspace of $mathbb{R}^2$. So, it passes through the origin. You can then translate it to the point $(1, 0)$.
– Joe Johnson 126
Nov 18 at 19:29
You are calculating the tangent space as a subspace of $mathbb{R}^2$. So, it passes through the origin. You can then translate it to the point $(1, 0)$.
– Joe Johnson 126
Nov 18 at 19:29
2
2
Your result is a subspace of tangent space $T_{(1,0)}R^2$. Nothing wrong with your calculation. The result represent the components of tangent vectors. That is the elements of the kernel should be in the form $0 partial_x + spartial_y$ for $sin R$. Which is exactly what you wanted (up to the identification).
– Kelvin Lois
Nov 18 at 19:29
Your result is a subspace of tangent space $T_{(1,0)}R^2$. Nothing wrong with your calculation. The result represent the components of tangent vectors. That is the elements of the kernel should be in the form $0 partial_x + spartial_y$ for $sin R$. Which is exactly what you wanted (up to the identification).
– Kelvin Lois
Nov 18 at 19:29
add a comment |
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1
You are calculating the tangent space as a subspace of $mathbb{R}^2$. So, it passes through the origin. You can then translate it to the point $(1, 0)$.
– Joe Johnson 126
Nov 18 at 19:29
2
Your result is a subspace of tangent space $T_{(1,0)}R^2$. Nothing wrong with your calculation. The result represent the components of tangent vectors. That is the elements of the kernel should be in the form $0 partial_x + spartial_y$ for $sin R$. Which is exactly what you wanted (up to the identification).
– Kelvin Lois
Nov 18 at 19:29