Mixing
Tangent space of a set [closed]
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Let $P=leftlbrace (x,0):xinmathbb{R} wedge x<1rightrbrace$. What are the tangent space and
the orthogonal complement of the tangent space?
differential-geometry riemannian-geometry
asked Jan 14 at 19:34
Anemath2Anemath2
122
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closed as off-topic by Adrian Keister, David G. Stork, John Douma, Travis, Xander Henderson Jan 15 at 5:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Adrian Keister, David G. Stork, John Douma, Travis, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $P=leftlbrace (x,0):xinmathbb{R} wedge x<1rightrbrace$. What are the tangent space and
the orthogonal complement of the tangent space?
differential-geometry riemannian-geometry
asked Jan 14 at 19:34
Anemath2Anemath2
122
$endgroup$
closed as off-topic by Adrian Keister, David G. Stork, John Douma, Travis, Xander Henderson Jan 15 at 5:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Adrian Keister, David G. Stork, John Douma, Travis, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $P=leftlbrace (x,0):xinmathbb{R} wedge x<1rightrbrace$. What are the tangent space and
the orthogonal complement of the tangent space?
differential-geometry riemannian-geometry
asked Jan 14 at 19:34
Anemath2Anemath2
122
$endgroup$
Let $P=leftlbrace (x,0):xinmathbb{R} wedge x<1rightrbrace$. What are the tangent space and
the orthogonal complement of the tangent space?
differential-geometry riemannian-geometry
differential-geometry riemannian-geometry
asked Jan 14 at 19:34
Anemath2Anemath2
122
asked Jan 14 at 19:34
Anemath2Anemath2
122
asked Jan 14 at 19:34
Anemath2Anemath2
122
asked Jan 14 at 19:34
Anemath2Anemath2
122
asked Jan 14 at 19:34
Anemath2Anemath2
122
122
closed as off-topic by Adrian Keister, David G. Stork, John Douma, Travis, Xander Henderson Jan 15 at 5:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Adrian Keister, David G. Stork, John Douma, Travis, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Adrian Keister, David G. Stork, John Douma, Travis, Xander Henderson Jan 15 at 5:07
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Adrian Keister, David G. Stork, John Douma, Travis, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
An atlas for your 1-manifold M is simply ${(M,phi:=proj_1)}$ where proj_1 is the projection on first factor.
So the tangent space in a point $x_0$ is
$T_{x_0}M={afrac{partial}{partial phi}|_{x_0}: ain mathbb{R}}$
Now we can calculate the orthogonal complement of $T_{x_0}M$ in $T_{x_0}mathbb{R}^2$ with respect the flat metric.
The differential with respect the inclusion map $i$ is
$di_{x_0}(frac{partial}{partial phi}|_{x_0})(g)=frac{partial gcirc i}{partial phi}|_{x_0}=frac{partial g}{partial x}_{(x_0,0)}+0 frac{partial g}{partial y}$
for every germs $gin U_{(x_0,0)}$ on $mathbb{R}^2$
So
$di_{x_0}(frac{partial}{partial phi}|_{x_0})(g)=frac{partial }{partial x}|_{(x_0,0)}$
and the orthogonal complement is
${bfrac{partial}{partial y}|_{x_0}: bin mathbb{R}}$
with $x_0in M$
answered Jan 14 at 19:48
Federico FalluccaFederico Fallucca
2,034210
$endgroup$
$begingroup$
Soory, I do not understand who are the a and b.
$endgroup$
– Anemath2
Jan 14 at 20:18
$begingroup$
So the orthogonal complement at $x=2$ is $2partial_y$, even though the set is empty at $x=2$? Interesting.
$endgroup$
– Fly by Night
Jan 14 at 20:23
$begingroup$
I corrected it,sorry
$endgroup$
– Federico Fallucca
Jan 14 at 20:26
$begingroup$
@Anemath2 the tangent space must be a 1-dimensional vector space.. so they are simply coefficient
$endgroup$
– Federico Fallucca
Jan 14 at 20:27
$begingroup$
What is the problem? I corrected it
$endgroup$
– Federico Fallucca
Jan 14 at 20:32
|
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
An atlas for your 1-manifold M is simply ${(M,phi:=proj_1)}$ where proj_1 is the projection on first factor.
So the tangent space in a point $x_0$ is
$T_{x_0}M={afrac{partial}{partial phi}|_{x_0}: ain mathbb{R}}$
Now we can calculate the orthogonal complement of $T_{x_0}M$ in $T_{x_0}mathbb{R}^2$ with respect the flat metric.
The differential with respect the inclusion map $i$ is
$di_{x_0}(frac{partial}{partial phi}|_{x_0})(g)=frac{partial gcirc i}{partial phi}|_{x_0}=frac{partial g}{partial x}_{(x_0,0)}+0 frac{partial g}{partial y}$
for every germs $gin U_{(x_0,0)}$ on $mathbb{R}^2$
So
$di_{x_0}(frac{partial}{partial phi}|_{x_0})(g)=frac{partial }{partial x}|_{(x_0,0)}$
and the orthogonal complement is
${bfrac{partial}{partial y}|_{x_0}: bin mathbb{R}}$
with $x_0in M$
answered Jan 14 at 19:48
Federico FalluccaFederico Fallucca
2,034210
$endgroup$
$begingroup$
Soory, I do not understand who are the a and b.
$endgroup$
– Anemath2
Jan 14 at 20:18
$begingroup$
So the orthogonal complement at $x=2$ is $2partial_y$, even though the set is empty at $x=2$? Interesting.
$endgroup$
– Fly by Night
Jan 14 at 20:23
$begingroup$
I corrected it,sorry
$endgroup$
– Federico Fallucca
Jan 14 at 20:26
$begingroup$
@Anemath2 the tangent space must be a 1-dimensional vector space.. so they are simply coefficient
$endgroup$
– Federico Fallucca
Jan 14 at 20:27
$begingroup$
What is the problem? I corrected it
$endgroup$
– Federico Fallucca
Jan 14 at 20:32
|
show 1 more comment
$begingroup$
An atlas for your 1-manifold M is simply ${(M,phi:=proj_1)}$ where proj_1 is the projection on first factor.
So the tangent space in a point $x_0$ is
$T_{x_0}M={afrac{partial}{partial phi}|_{x_0}: ain mathbb{R}}$
Now we can calculate the orthogonal complement of $T_{x_0}M$ in $T_{x_0}mathbb{R}^2$ with respect the flat metric.
The differential with respect the inclusion map $i$ is
$di_{x_0}(frac{partial}{partial phi}|_{x_0})(g)=frac{partial gcirc i}{partial phi}|_{x_0}=frac{partial g}{partial x}_{(x_0,0)}+0 frac{partial g}{partial y}$
for every germs $gin U_{(x_0,0)}$ on $mathbb{R}^2$
So
$di_{x_0}(frac{partial}{partial phi}|_{x_0})(g)=frac{partial }{partial x}|_{(x_0,0)}$
and the orthogonal complement is
${bfrac{partial}{partial y}|_{x_0}: bin mathbb{R}}$
with $x_0in M$
answered Jan 14 at 19:48
Federico FalluccaFederico Fallucca
2,034210
$endgroup$
$begingroup$
Soory, I do not understand who are the a and b.
$endgroup$
– Anemath2
Jan 14 at 20:18
$begingroup$
So the orthogonal complement at $x=2$ is $2partial_y$, even though the set is empty at $x=2$? Interesting.
$endgroup$
– Fly by Night
Jan 14 at 20:23
$begingroup$
I corrected it,sorry
$endgroup$
– Federico Fallucca
Jan 14 at 20:26
$begingroup$
@Anemath2 the tangent space must be a 1-dimensional vector space.. so they are simply coefficient
$endgroup$
– Federico Fallucca
Jan 14 at 20:27
$begingroup$
What is the problem? I corrected it
$endgroup$
– Federico Fallucca
Jan 14 at 20:32
|
show 1 more comment
$begingroup$
An atlas for your 1-manifold M is simply ${(M,phi:=proj_1)}$ where proj_1 is the projection on first factor.
So the tangent space in a point $x_0$ is
$T_{x_0}M={afrac{partial}{partial phi}|_{x_0}: ain mathbb{R}}$
Now we can calculate the orthogonal complement of $T_{x_0}M$ in $T_{x_0}mathbb{R}^2$ with respect the flat metric.
The differential with respect the inclusion map $i$ is
$di_{x_0}(frac{partial}{partial phi}|_{x_0})(g)=frac{partial gcirc i}{partial phi}|_{x_0}=frac{partial g}{partial x}_{(x_0,0)}+0 frac{partial g}{partial y}$
for every germs $gin U_{(x_0,0)}$ on $mathbb{R}^2$
So
$di_{x_0}(frac{partial}{partial phi}|_{x_0})(g)=frac{partial }{partial x}|_{(x_0,0)}$
and the orthogonal complement is
${bfrac{partial}{partial y}|_{x_0}: bin mathbb{R}}$
with $x_0in M$
answered Jan 14 at 19:48
Federico FalluccaFederico Fallucca
2,034210
$endgroup$
An atlas for your 1-manifold M is simply ${(M,phi:=proj_1)}$ where proj_1 is the projection on first factor.
So the tangent space in a point $x_0$ is
$T_{x_0}M={afrac{partial}{partial phi}|_{x_0}: ain mathbb{R}}$
Now we can calculate the orthogonal complement of $T_{x_0}M$ in $T_{x_0}mathbb{R}^2$ with respect the flat metric.
The differential with respect the inclusion map $i$ is
$di_{x_0}(frac{partial}{partial phi}|_{x_0})(g)=frac{partial gcirc i}{partial phi}|_{x_0}=frac{partial g}{partial x}_{(x_0,0)}+0 frac{partial g}{partial y}$
for every germs $gin U_{(x_0,0)}$ on $mathbb{R}^2$
So
$di_{x_0}(frac{partial}{partial phi}|_{x_0})(g)=frac{partial }{partial x}|_{(x_0,0)}$
and the orthogonal complement is
${bfrac{partial}{partial y}|_{x_0}: bin mathbb{R}}$
with $x_0in M$
answered Jan 14 at 19:48
Federico FalluccaFederico Fallucca
2,034210
edited Jan 14 at 20:39
answered Jan 14 at 19:48
Federico FalluccaFederico Fallucca
2,034210
answered Jan 14 at 19:48
Federico FalluccaFederico Fallucca
2,034210
answered Jan 14 at 19:48
Federico FalluccaFederico Fallucca
2,034210
2,034210
$begingroup$
Soory, I do not understand who are the a and b.
$endgroup$
– Anemath2
Jan 14 at 20:18
$begingroup$
So the orthogonal complement at $x=2$ is $2partial_y$, even though the set is empty at $x=2$? Interesting.
$endgroup$
– Fly by Night
Jan 14 at 20:23
$begingroup$
I corrected it,sorry
$endgroup$
– Federico Fallucca
Jan 14 at 20:26
$begingroup$
@Anemath2 the tangent space must be a 1-dimensional vector space.. so they are simply coefficient
$endgroup$
– Federico Fallucca
Jan 14 at 20:27
$begingroup$
What is the problem? I corrected it
$endgroup$
– Federico Fallucca
Jan 14 at 20:32
|
show 1 more comment
$begingroup$
Soory, I do not understand who are the a and b.
$endgroup$
– Anemath2
Jan 14 at 20:18
$begingroup$
So the orthogonal complement at $x=2$ is $2partial_y$, even though the set is empty at $x=2$? Interesting.
$endgroup$
– Fly by Night
Jan 14 at 20:23
$begingroup$
I corrected it,sorry
$endgroup$
– Federico Fallucca
Jan 14 at 20:26
$begingroup$
@Anemath2 the tangent space must be a 1-dimensional vector space.. so they are simply coefficient
$endgroup$
– Federico Fallucca
Jan 14 at 20:27
$begingroup$
What is the problem? I corrected it
$endgroup$
– Federico Fallucca
Jan 14 at 20:32
$begingroup$
Soory, I do not understand who are the a and b.
$endgroup$
– Anemath2
Jan 14 at 20:18
$begingroup$
Soory, I do not understand who are the a and b.
$endgroup$
– Anemath2
Jan 14 at 20:18
$begingroup$
So the orthogonal complement at $x=2$ is $2partial_y$, even though the set is empty at $x=2$? Interesting.
$endgroup$
– Fly by Night
Jan 14 at 20:23
$begingroup$
So the orthogonal complement at $x=2$ is $2partial_y$, even though the set is empty at $x=2$? Interesting.
$endgroup$
– Fly by Night
Jan 14 at 20:23
$begingroup$
I corrected it,sorry
$endgroup$
– Federico Fallucca
Jan 14 at 20:26
$begingroup$
I corrected it,sorry
$endgroup$
– Federico Fallucca
Jan 14 at 20:26
$begingroup$
@Anemath2 the tangent space must be a 1-dimensional vector space.. so they are simply coefficient
$endgroup$
– Federico Fallucca
Jan 14 at 20:27
$begingroup$
@Anemath2 the tangent space must be a 1-dimensional vector space.. so they are simply coefficient
$endgroup$
– Federico Fallucca
Jan 14 at 20:27
$begingroup$
What is the problem? I corrected it
$endgroup$
– Federico Fallucca
Jan 14 at 20:32
$begingroup$
What is the problem? I corrected it
$endgroup$
– Federico Fallucca
Jan 14 at 20:32
|
show 1 more comment
