How to parameterize a manifold? The 2019 Stack Overflow Developer Survey Results Are InAssumptions required for an implicitely defined surface/manifold to have a specified dimensionBasic Manifold QuestionDefinition of a parallelizable manifoldShowing a function is a manifoldProblem defining a smooth m-manifold via a smooth atlasProof that this is a smooth manifoldProve that the $n-sphere$ is a manifoldProve that the unit circle, $S^1:=(x,y)in mathbbR^2: x^2+y^2=1$, Is a one-dimensional manifoldUnitary matrices form a manifoldExtension of the smoothness of a manifold to its interior
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How to parameterize a manifold?
The 2019 Stack Overflow Developer Survey Results Are InAssumptions required for an implicitely defined surface/manifold to have a specified dimensionBasic Manifold QuestionDefinition of a parallelizable manifoldShowing a function is a manifoldProblem defining a smooth m-manifold via a smooth atlasProof that this is a smooth manifoldProve that the $n-sphere$ is a manifoldProve that the unit circle, $S^1:=(x,y)in mathbbR^2: x^2+y^2=1$, Is a one-dimensional manifoldUnitary matrices form a manifoldExtension of the smoothness of a manifold to its interior
$begingroup$
If I have $M$ be a set in $mathbbR^3$ defined by $x-y^2+z^2=0$ and $y^2+z^2<4$, how would I find a function from $mathbbR^d rightarrow mathbbR^3$ ($d$ is the dimension of the manifold)that parameterizes $M$?
manifolds smooth-manifolds
asked Apr 6 at 20:43
hoya2021hoya2021
385
$endgroup$
add a comment |
$begingroup$
If I have $M$ be a set in $mathbbR^3$ defined by $x-y^2+z^2=0$ and $y^2+z^2<4$, how would I find a function from $mathbbR^d rightarrow mathbbR^3$ ($d$ is the dimension of the manifold)that parameterizes $M$?
manifolds smooth-manifolds
asked Apr 6 at 20:43
hoya2021hoya2021
385
$endgroup$
$begingroup$
How about $(y,z) mapsto (y^2-z^2, y, z)$?
$endgroup$
– MisterRiemann
Apr 6 at 20:48
$begingroup$
Can you help me understand how you came to this?
$endgroup$
– hoya2021
Apr 6 at 20:51
1
$begingroup$
It follows directly from your equation: $$ x-y^2+z^2 = 0 iff x = y^2-z^2, $$ which tells you how the $x$-coordinate depends on the other two.
$endgroup$
– MisterRiemann
Apr 6 at 20:52
add a comment |
$begingroup$
If I have $M$ be a set in $mathbbR^3$ defined by $x-y^2+z^2=0$ and $y^2+z^2<4$, how would I find a function from $mathbbR^d rightarrow mathbbR^3$ ($d$ is the dimension of the manifold)that parameterizes $M$?
manifolds smooth-manifolds
asked Apr 6 at 20:43
hoya2021hoya2021
385
$endgroup$
If I have $M$ be a set in $mathbbR^3$ defined by $x-y^2+z^2=0$ and $y^2+z^2<4$, how would I find a function from $mathbbR^d rightarrow mathbbR^3$ ($d$ is the dimension of the manifold)that parameterizes $M$?
manifolds smooth-manifolds
manifolds smooth-manifolds
asked Apr 6 at 20:43
hoya2021hoya2021
385
asked Apr 6 at 20:43
hoya2021hoya2021
385
asked Apr 6 at 20:43
hoya2021hoya2021
385
asked Apr 6 at 20:43
hoya2021hoya2021
385
asked Apr 6 at 20:43
hoya2021hoya2021
385
385
$begingroup$
How about $(y,z) mapsto (y^2-z^2, y, z)$?
$endgroup$
– MisterRiemann
Apr 6 at 20:48
$begingroup$
Can you help me understand how you came to this?
$endgroup$
– hoya2021
Apr 6 at 20:51
1
$begingroup$
It follows directly from your equation: $$ x-y^2+z^2 = 0 iff x = y^2-z^2, $$ which tells you how the $x$-coordinate depends on the other two.
$endgroup$
– MisterRiemann
Apr 6 at 20:52
add a comment |
$begingroup$
How about $(y,z) mapsto (y^2-z^2, y, z)$?
$endgroup$
– MisterRiemann
Apr 6 at 20:48
$begingroup$
Can you help me understand how you came to this?
$endgroup$
– hoya2021
Apr 6 at 20:51
1
$begingroup$
It follows directly from your equation: $$ x-y^2+z^2 = 0 iff x = y^2-z^2, $$ which tells you how the $x$-coordinate depends on the other two.
$endgroup$
– MisterRiemann
Apr 6 at 20:52
$begingroup$
How about $(y,z) mapsto (y^2-z^2, y, z)$?
$endgroup$
– MisterRiemann
Apr 6 at 20:48
$begingroup$
How about $(y,z) mapsto (y^2-z^2, y, z)$?
$endgroup$
– MisterRiemann
Apr 6 at 20:48
$begingroup$
Can you help me understand how you came to this?
$endgroup$
– hoya2021
Apr 6 at 20:51
$begingroup$
Can you help me understand how you came to this?
$endgroup$
– hoya2021
Apr 6 at 20:51
1
1
$begingroup$
It follows directly from your equation: $$ x-y^2+z^2 = 0 iff x = y^2-z^2, $$ which tells you how the $x$-coordinate depends on the other two.
$endgroup$
– MisterRiemann
Apr 6 at 20:52
$begingroup$
It follows directly from your equation: $$ x-y^2+z^2 = 0 iff x = y^2-z^2, $$ which tells you how the $x$-coordinate depends on the other two.
$endgroup$
– MisterRiemann
Apr 6 at 20:52
add a comment |
0
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$begingroup$
How about $(y,z) mapsto (y^2-z^2, y, z)$?
$endgroup$
– MisterRiemann
Apr 6 at 20:48
$begingroup$
Can you help me understand how you came to this?
$endgroup$
– hoya2021
Apr 6 at 20:51
1
$begingroup$
It follows directly from your equation: $$ x-y^2+z^2 = 0 iff x = y^2-z^2, $$ which tells you how the $x$-coordinate depends on the other two.
$endgroup$
– MisterRiemann
Apr 6 at 20:52