Tangent space of smooth manifold $M={(x,x^3,e^{x-1}) : x in Bbb{R}}$ at $(1,1,1)$
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What's the tangent space of $M={(x,x^3,e^{x-1}): x in Bbb{R}}$ at the point $(1,1,1)$, where $M$ is a manifold of smoothness $C^infty$.
I know how to find the tangent space of a manifold in the form that gives an implicit function such as $M={(x,y,z) in Bbb{R}^3: x^2+y^2-z^2=1}$. The tangent space of $M$ in this case $= ker(mbox{dg}(x))$ at the given point which as $2x+zy-2z=0$.
Can anyone help with the question that only the coordinate was given? Any hint would be helpful. :)
manifolds smooth-manifolds tangent-spaces
edited Dec 8 '18 at 19:23
Brahadeesh
6,24242361
asked Mar 12 '18 at 15:38
geggeg
112
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add a comment |
$begingroup$
What's the tangent space of $M={(x,x^3,e^{x-1}): x in Bbb{R}}$ at the point $(1,1,1)$, where $M$ is a manifold of smoothness $C^infty$.
I know how to find the tangent space of a manifold in the form that gives an implicit function such as $M={(x,y,z) in Bbb{R}^3: x^2+y^2-z^2=1}$. The tangent space of $M$ in this case $= ker(mbox{dg}(x))$ at the given point which as $2x+zy-2z=0$.
Can anyone help with the question that only the coordinate was given? Any hint would be helpful. :)
manifolds smooth-manifolds tangent-spaces
edited Dec 8 '18 at 19:23
Brahadeesh
6,24242361
asked Mar 12 '18 at 15:38
geggeg
112
$endgroup$
$begingroup$
Isn't your manifold just a one dimensional curve and the tangent space should be the tangent line at that point?
$endgroup$
– mastrok
Mar 12 '18 at 17:35
add a comment |
$begingroup$
What's the tangent space of $M={(x,x^3,e^{x-1}): x in Bbb{R}}$ at the point $(1,1,1)$, where $M$ is a manifold of smoothness $C^infty$.
I know how to find the tangent space of a manifold in the form that gives an implicit function such as $M={(x,y,z) in Bbb{R}^3: x^2+y^2-z^2=1}$. The tangent space of $M$ in this case $= ker(mbox{dg}(x))$ at the given point which as $2x+zy-2z=0$.
Can anyone help with the question that only the coordinate was given? Any hint would be helpful. :)
manifolds smooth-manifolds tangent-spaces
edited Dec 8 '18 at 19:23
Brahadeesh
6,24242361
asked Mar 12 '18 at 15:38
geggeg
112
$endgroup$
What's the tangent space of $M={(x,x^3,e^{x-1}): x in Bbb{R}}$ at the point $(1,1,1)$, where $M$ is a manifold of smoothness $C^infty$.
I know how to find the tangent space of a manifold in the form that gives an implicit function such as $M={(x,y,z) in Bbb{R}^3: x^2+y^2-z^2=1}$. The tangent space of $M$ in this case $= ker(mbox{dg}(x))$ at the given point which as $2x+zy-2z=0$.
Can anyone help with the question that only the coordinate was given? Any hint would be helpful. :)
manifolds smooth-manifolds tangent-spaces
manifolds smooth-manifolds tangent-spaces
edited Dec 8 '18 at 19:23
Brahadeesh
6,24242361
asked Mar 12 '18 at 15:38
geggeg
112
edited Dec 8 '18 at 19:23
Brahadeesh
6,24242361
asked Mar 12 '18 at 15:38
geggeg
112
edited Dec 8 '18 at 19:23
Brahadeesh
6,24242361
edited Dec 8 '18 at 19:23
Brahadeesh
6,24242361
edited Dec 8 '18 at 19:23
Brahadeesh
6,24242361
6,24242361
asked Mar 12 '18 at 15:38
geggeg
112
asked Mar 12 '18 at 15:38
geggeg
112
asked Mar 12 '18 at 15:38
geggeg
112
112
$begingroup$
Isn't your manifold just a one dimensional curve and the tangent space should be the tangent line at that point?
$endgroup$
– mastrok
Mar 12 '18 at 17:35
add a comment |
$begingroup$
Isn't your manifold just a one dimensional curve and the tangent space should be the tangent line at that point?
$endgroup$
– mastrok
Mar 12 '18 at 17:35
$begingroup$
Isn't your manifold just a one dimensional curve and the tangent space should be the tangent line at that point?
$endgroup$
– mastrok
Mar 12 '18 at 17:35
$begingroup$
Isn't your manifold just a one dimensional curve and the tangent space should be the tangent line at that point?
$endgroup$
– mastrok
Mar 12 '18 at 17:35
add a comment |
1 Answer
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From the comment above by @mastrok.
Your manifold is just a one-dimensional curve, so the tangent space should be the tangent line at that point. So, it is the set of points of the form $${ (p,v) : p = (1,1,1) text{ and } v = (lambda, 3lambda, lambda), lambda in Bbb{R} }.$$
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1 Answer
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1 Answer
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$begingroup$
From the comment above by @mastrok.
Your manifold is just a one-dimensional curve, so the tangent space should be the tangent line at that point. So, it is the set of points of the form $${ (p,v) : p = (1,1,1) text{ and } v = (lambda, 3lambda, lambda), lambda in Bbb{R} }.$$
$endgroup$
add a comment |
$begingroup$
From the comment above by @mastrok.
Your manifold is just a one-dimensional curve, so the tangent space should be the tangent line at that point. So, it is the set of points of the form $${ (p,v) : p = (1,1,1) text{ and } v = (lambda, 3lambda, lambda), lambda in Bbb{R} }.$$
$endgroup$
add a comment |
$begingroup$
From the comment above by @mastrok.
Your manifold is just a one-dimensional curve, so the tangent space should be the tangent line at that point. So, it is the set of points of the form $${ (p,v) : p = (1,1,1) text{ and } v = (lambda, 3lambda, lambda), lambda in Bbb{R} }.$$
$endgroup$
From the comment above by @mastrok.
Your manifold is just a one-dimensional curve, so the tangent space should be the tangent line at that point. So, it is the set of points of the form $${ (p,v) : p = (1,1,1) text{ and } v = (lambda, 3lambda, lambda), lambda in Bbb{R} }.$$
answered Dec 8 '18 at 19:21
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Brahadeesh
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$begingroup$
Isn't your manifold just a one dimensional curve and the tangent space should be the tangent line at that point?
$endgroup$
– mastrok
Mar 12 '18 at 17:35