Show that there are at least two points on a manifold to which a vector is normal Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Differential Forms on Surfaces. Show that $Ncdot (nabla times V)eta=dphi$ on $x(D)$.Defining a metric in the tangent spaces $ T_xM $Vectorfields satisfying SO(3) algebra 2 dimensional?line integral of 3D vector fieldDifferential geometry of Barrett O NeilShow that $(x, y) in V times V $ is not a submanifold of the vector space $Vtimes V$Guillemin-Pollack Exercise 1.5.3: Normal IntersectionsSmooth function from regular surfacesProve there exists a outward unit normal field on the boundary this manifoldShow that if $f$ is a smooth function, $M$ is a manifold and $x$ is a local extremum of $f$ on $M$, then $D_f(x)(v) = 0$ in the tangent space.
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Show that there are at least two points on a manifold to which a vector is normal
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Differential Forms on Surfaces. Show that $Ncdot (nabla times V)eta=dphi$ on $x(D)$.Defining a metric in the tangent spaces $ T_xM $Vectorfields satisfying SO(3) algebra 2 dimensional?line integral of 3D vector fieldDifferential geometry of Barrett O NeilShow that $(x, y) in V times V $ is not a submanifold of the vector space $Vtimes V$Guillemin-Pollack Exercise 1.5.3: Normal IntersectionsSmooth function from regular surfacesProve there exists a outward unit normal field on the boundary this manifoldShow that if $f$ is a smooth function, $M$ is a manifold and $x$ is a local extremum of $f$ on $M$, then $D_f(x)(v) = 0$ in the tangent space.
$begingroup$
Let $M subset mathbb R^3$ be a $2$-dimensional manifold which is also a compact set. And let $v in mathbb R^3$ be a vector which satisfies $ ||v|| = 1$.
The task is to prove that there are at least two points $x,y in M$ such that $v$ is a normal vector to their tangent spaces.
There was given a hint to look at the extremum points of the function $f(x) = <x,v>$ which means $f(x) = x_1v_1 + x_2v_2 + x_3v_3$. Since the functions is continuous and $M$ is compact then the function get a minimum and maximum on $M$. Let's say the $x$ is the maximum and $y$ is a the minimum. Moreover, I noticed that $nabla f = (v_1,v_2,v_3) = v$.
However, I am not so sure how to continue from here. I am supposed to show that $v$ is normal to both $T_xM$ and $T_yM$ but I don't see how exactly. I still haven't used the fact that $M$ is a manifold so it probably uses that.
Help would be appreciated
calculus multivariable-calculus differential-geometry manifolds
edited Apr 8 at 21:49
Laz
969510
asked Apr 8 at 20:13
Gabi GGabi G
553210
$endgroup$
add a comment |
$begingroup$
Let $M subset mathbb R^3$ be a $2$-dimensional manifold which is also a compact set. And let $v in mathbb R^3$ be a vector which satisfies $ ||v|| = 1$.
The task is to prove that there are at least two points $x,y in M$ such that $v$ is a normal vector to their tangent spaces.
There was given a hint to look at the extremum points of the function $f(x) = <x,v>$ which means $f(x) = x_1v_1 + x_2v_2 + x_3v_3$. Since the functions is continuous and $M$ is compact then the function get a minimum and maximum on $M$. Let's say the $x$ is the maximum and $y$ is a the minimum. Moreover, I noticed that $nabla f = (v_1,v_2,v_3) = v$.
However, I am not so sure how to continue from here. I am supposed to show that $v$ is normal to both $T_xM$ and $T_yM$ but I don't see how exactly. I still haven't used the fact that $M$ is a manifold so it probably uses that.
Help would be appreciated
calculus multivariable-calculus differential-geometry manifolds
edited Apr 8 at 21:49
Laz
969510
asked Apr 8 at 20:13
Gabi GGabi G
553210
$endgroup$
2
$begingroup$
You really need to tell us what definition of a manifold contained in $BbbR^2$ you are using. (Some assumptions are needed to ensure the existence of the tangent spaces.) However, to see that a minimum/maximum, $x_0$ say, of your function $f(x)$ must be at a point where $v$ is normal to the tangent bundle, consider an arbitrary differentiable curve $gamma$ in $M$ with $gamma(0) = x_0$ and note that $0$ is a local minimum/maximum for $f circ gamma$ (hence $(f circ gamma)'(0) = 0$).
$endgroup$
– Rob Arthan
Apr 8 at 20:29
$begingroup$
$M$ is a manifold if for every $x in M$ there exists an open set $U subset R^3$ such that $M bigcap U$ is a graph of a smooth function
$endgroup$
– Gabi G
Apr 8 at 20:31
$begingroup$
That definition needs some more details, I think. How are you viewing $M cap U$ as the graph of a function?
$endgroup$
– Rob Arthan
Apr 8 at 21:56
add a comment |
$begingroup$
Let $M subset mathbb R^3$ be a $2$-dimensional manifold which is also a compact set. And let $v in mathbb R^3$ be a vector which satisfies $ ||v|| = 1$.
The task is to prove that there are at least two points $x,y in M$ such that $v$ is a normal vector to their tangent spaces.
There was given a hint to look at the extremum points of the function $f(x) = <x,v>$ which means $f(x) = x_1v_1 + x_2v_2 + x_3v_3$. Since the functions is continuous and $M$ is compact then the function get a minimum and maximum on $M$. Let's say the $x$ is the maximum and $y$ is a the minimum. Moreover, I noticed that $nabla f = (v_1,v_2,v_3) = v$.
However, I am not so sure how to continue from here. I am supposed to show that $v$ is normal to both $T_xM$ and $T_yM$ but I don't see how exactly. I still haven't used the fact that $M$ is a manifold so it probably uses that.
Help would be appreciated
calculus multivariable-calculus differential-geometry manifolds
edited Apr 8 at 21:49
Laz
969510
asked Apr 8 at 20:13
Gabi GGabi G
553210
$endgroup$
Let $M subset mathbb R^3$ be a $2$-dimensional manifold which is also a compact set. And let $v in mathbb R^3$ be a vector which satisfies $ ||v|| = 1$.
The task is to prove that there are at least two points $x,y in M$ such that $v$ is a normal vector to their tangent spaces.
There was given a hint to look at the extremum points of the function $f(x) = <x,v>$ which means $f(x) = x_1v_1 + x_2v_2 + x_3v_3$. Since the functions is continuous and $M$ is compact then the function get a minimum and maximum on $M$. Let's say the $x$ is the maximum and $y$ is a the minimum. Moreover, I noticed that $nabla f = (v_1,v_2,v_3) = v$.
However, I am not so sure how to continue from here. I am supposed to show that $v$ is normal to both $T_xM$ and $T_yM$ but I don't see how exactly. I still haven't used the fact that $M$ is a manifold so it probably uses that.
Help would be appreciated
calculus multivariable-calculus differential-geometry manifolds
calculus multivariable-calculus differential-geometry manifolds
edited Apr 8 at 21:49
Laz
969510
asked Apr 8 at 20:13
Gabi GGabi G
553210
edited Apr 8 at 21:49
Laz
969510
asked Apr 8 at 20:13
Gabi GGabi G
553210
edited Apr 8 at 21:49
Laz
969510
edited Apr 8 at 21:49
Laz
969510
edited Apr 8 at 21:49
Laz
969510
969510
asked Apr 8 at 20:13
Gabi GGabi G
553210
asked Apr 8 at 20:13
Gabi GGabi G
553210
asked Apr 8 at 20:13
Gabi GGabi G
553210
553210
2
$begingroup$
You really need to tell us what definition of a manifold contained in $BbbR^2$ you are using. (Some assumptions are needed to ensure the existence of the tangent spaces.) However, to see that a minimum/maximum, $x_0$ say, of your function $f(x)$ must be at a point where $v$ is normal to the tangent bundle, consider an arbitrary differentiable curve $gamma$ in $M$ with $gamma(0) = x_0$ and note that $0$ is a local minimum/maximum for $f circ gamma$ (hence $(f circ gamma)'(0) = 0$).
$endgroup$
– Rob Arthan
Apr 8 at 20:29
$begingroup$
$M$ is a manifold if for every $x in M$ there exists an open set $U subset R^3$ such that $M bigcap U$ is a graph of a smooth function
$endgroup$
– Gabi G
Apr 8 at 20:31
$begingroup$
That definition needs some more details, I think. How are you viewing $M cap U$ as the graph of a function?
$endgroup$
– Rob Arthan
Apr 8 at 21:56
add a comment |
2
$begingroup$
You really need to tell us what definition of a manifold contained in $BbbR^2$ you are using. (Some assumptions are needed to ensure the existence of the tangent spaces.) However, to see that a minimum/maximum, $x_0$ say, of your function $f(x)$ must be at a point where $v$ is normal to the tangent bundle, consider an arbitrary differentiable curve $gamma$ in $M$ with $gamma(0) = x_0$ and note that $0$ is a local minimum/maximum for $f circ gamma$ (hence $(f circ gamma)'(0) = 0$).
$endgroup$
– Rob Arthan
Apr 8 at 20:29
$begingroup$
$M$ is a manifold if for every $x in M$ there exists an open set $U subset R^3$ such that $M bigcap U$ is a graph of a smooth function
$endgroup$
– Gabi G
Apr 8 at 20:31
$begingroup$
That definition needs some more details, I think. How are you viewing $M cap U$ as the graph of a function?
$endgroup$
– Rob Arthan
Apr 8 at 21:56
2
2
$begingroup$
You really need to tell us what definition of a manifold contained in $BbbR^2$ you are using. (Some assumptions are needed to ensure the existence of the tangent spaces.) However, to see that a minimum/maximum, $x_0$ say, of your function $f(x)$ must be at a point where $v$ is normal to the tangent bundle, consider an arbitrary differentiable curve $gamma$ in $M$ with $gamma(0) = x_0$ and note that $0$ is a local minimum/maximum for $f circ gamma$ (hence $(f circ gamma)'(0) = 0$).
$endgroup$
– Rob Arthan
Apr 8 at 20:29
$begingroup$
You really need to tell us what definition of a manifold contained in $BbbR^2$ you are using. (Some assumptions are needed to ensure the existence of the tangent spaces.) However, to see that a minimum/maximum, $x_0$ say, of your function $f(x)$ must be at a point where $v$ is normal to the tangent bundle, consider an arbitrary differentiable curve $gamma$ in $M$ with $gamma(0) = x_0$ and note that $0$ is a local minimum/maximum for $f circ gamma$ (hence $(f circ gamma)'(0) = 0$).
$endgroup$
– Rob Arthan
Apr 8 at 20:29
$begingroup$
$M$ is a manifold if for every $x in M$ there exists an open set $U subset R^3$ such that $M bigcap U$ is a graph of a smooth function
$endgroup$
– Gabi G
Apr 8 at 20:31
$begingroup$
$M$ is a manifold if for every $x in M$ there exists an open set $U subset R^3$ such that $M bigcap U$ is a graph of a smooth function
$endgroup$
– Gabi G
Apr 8 at 20:31
$begingroup$
That definition needs some more details, I think. How are you viewing $M cap U$ as the graph of a function?
$endgroup$
– Rob Arthan
Apr 8 at 21:56
$begingroup$
That definition needs some more details, I think. How are you viewing $M cap U$ as the graph of a function?
$endgroup$
– Rob Arthan
Apr 8 at 21:56
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Following the hint you propose, $f(x)=langle x, vrangle$ has a maximum and a minimum. If they both coincide, it turns out that your $2$-manifold is a compact subset of a plane in $mathbbR^3$, which would imply that it is a surface with boundary, contradicting your definition of submanifold as something that is locally a graph of a function with open domain in $mathbbR^2$ (Invariance of Domain implicitly used).
Then if $x_0$ is the maximum of $f$ and $gamma: (-epsilon, epsilon)rightarrow M$ is a curve with $gamma(0)=x_0$, $t=0$ is a critical point of $fcirc gamma$ (this is just a directional derivative of $f$ at $x_0$). This means that $0=nabla f(x_0)cdot gamma'(0)=vcdot gamma'(0)$. This proves that $vperp T_x_0M$, since $T_x_0M$ is exactly the plane containing all vectors tangent to curves through $x_0$.
The exact same argument proves that $vperp T_y_0M$.
answered Apr 8 at 21:15
LazLaz
969510
$endgroup$
$begingroup$
So the solution is working with the definition of tangent space. Thanks!
$endgroup$
– Gabi G
Apr 8 at 21:35
$begingroup$
Exactly, this is one of the many definitions of tangent space. This one is very useful, because it is very geometric, but it requieres our manifold to be sitting naturally inside some $mathbbR^n$. Your welcome!
$endgroup$
– Laz
Apr 8 at 21:37
1
$begingroup$
@Laz: the question only makes sense for a manifold sitting inside $BbbR^3$ (it's about a relationship between a vector in $BbbR^3$ and the tangent spaces, (which only makes sense if the tangent spaces are identified with planes in $BbbR^3$). (And it was me who provided the hint and not the OP $ddotsmile$.)
$endgroup$
– Rob Arthan
Apr 8 at 22:06
$begingroup$
Of course @RobArthan, I was refering to the definition of tangent space in the general setting.
$endgroup$
– Laz
Apr 8 at 22:08
$begingroup$
@Laz: sure. The problem could have been phrased in terms of an immersion or embedding of an abstract differentiable manifold in $BbbR^3$, which is why I asked the OP to say more about the definitions being used. Anyway, the OP is happy with your answer, so let's go our way rejoicing.
$endgroup$
– Rob Arthan
Apr 8 at 22:15
|
show 1 more comment
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1 Answer
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1 Answer
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$begingroup$
Following the hint you propose, $f(x)=langle x, vrangle$ has a maximum and a minimum. If they both coincide, it turns out that your $2$-manifold is a compact subset of a plane in $mathbbR^3$, which would imply that it is a surface with boundary, contradicting your definition of submanifold as something that is locally a graph of a function with open domain in $mathbbR^2$ (Invariance of Domain implicitly used).
Then if $x_0$ is the maximum of $f$ and $gamma: (-epsilon, epsilon)rightarrow M$ is a curve with $gamma(0)=x_0$, $t=0$ is a critical point of $fcirc gamma$ (this is just a directional derivative of $f$ at $x_0$). This means that $0=nabla f(x_0)cdot gamma'(0)=vcdot gamma'(0)$. This proves that $vperp T_x_0M$, since $T_x_0M$ is exactly the plane containing all vectors tangent to curves through $x_0$.
The exact same argument proves that $vperp T_y_0M$.
answered Apr 8 at 21:15
LazLaz
969510
$endgroup$
$begingroup$
So the solution is working with the definition of tangent space. Thanks!
$endgroup$
– Gabi G
Apr 8 at 21:35
$begingroup$
Exactly, this is one of the many definitions of tangent space. This one is very useful, because it is very geometric, but it requieres our manifold to be sitting naturally inside some $mathbbR^n$. Your welcome!
$endgroup$
– Laz
Apr 8 at 21:37
1
$begingroup$
@Laz: the question only makes sense for a manifold sitting inside $BbbR^3$ (it's about a relationship between a vector in $BbbR^3$ and the tangent spaces, (which only makes sense if the tangent spaces are identified with planes in $BbbR^3$). (And it was me who provided the hint and not the OP $ddotsmile$.)
$endgroup$
– Rob Arthan
Apr 8 at 22:06
$begingroup$
Of course @RobArthan, I was refering to the definition of tangent space in the general setting.
$endgroup$
– Laz
Apr 8 at 22:08
$begingroup$
@Laz: sure. The problem could have been phrased in terms of an immersion or embedding of an abstract differentiable manifold in $BbbR^3$, which is why I asked the OP to say more about the definitions being used. Anyway, the OP is happy with your answer, so let's go our way rejoicing.
$endgroup$
– Rob Arthan
Apr 8 at 22:15
|
show 1 more comment
$begingroup$
Following the hint you propose, $f(x)=langle x, vrangle$ has a maximum and a minimum. If they both coincide, it turns out that your $2$-manifold is a compact subset of a plane in $mathbbR^3$, which would imply that it is a surface with boundary, contradicting your definition of submanifold as something that is locally a graph of a function with open domain in $mathbbR^2$ (Invariance of Domain implicitly used).
Then if $x_0$ is the maximum of $f$ and $gamma: (-epsilon, epsilon)rightarrow M$ is a curve with $gamma(0)=x_0$, $t=0$ is a critical point of $fcirc gamma$ (this is just a directional derivative of $f$ at $x_0$). This means that $0=nabla f(x_0)cdot gamma'(0)=vcdot gamma'(0)$. This proves that $vperp T_x_0M$, since $T_x_0M$ is exactly the plane containing all vectors tangent to curves through $x_0$.
The exact same argument proves that $vperp T_y_0M$.
answered Apr 8 at 21:15
LazLaz
969510
$endgroup$
$begingroup$
So the solution is working with the definition of tangent space. Thanks!
$endgroup$
– Gabi G
Apr 8 at 21:35
$begingroup$
Exactly, this is one of the many definitions of tangent space. This one is very useful, because it is very geometric, but it requieres our manifold to be sitting naturally inside some $mathbbR^n$. Your welcome!
$endgroup$
– Laz
Apr 8 at 21:37
1
$begingroup$
@Laz: the question only makes sense for a manifold sitting inside $BbbR^3$ (it's about a relationship between a vector in $BbbR^3$ and the tangent spaces, (which only makes sense if the tangent spaces are identified with planes in $BbbR^3$). (And it was me who provided the hint and not the OP $ddotsmile$.)
$endgroup$
– Rob Arthan
Apr 8 at 22:06
$begingroup$
Of course @RobArthan, I was refering to the definition of tangent space in the general setting.
$endgroup$
– Laz
Apr 8 at 22:08
$begingroup$
@Laz: sure. The problem could have been phrased in terms of an immersion or embedding of an abstract differentiable manifold in $BbbR^3$, which is why I asked the OP to say more about the definitions being used. Anyway, the OP is happy with your answer, so let's go our way rejoicing.
$endgroup$
– Rob Arthan
Apr 8 at 22:15
|
show 1 more comment
$begingroup$
Following the hint you propose, $f(x)=langle x, vrangle$ has a maximum and a minimum. If they both coincide, it turns out that your $2$-manifold is a compact subset of a plane in $mathbbR^3$, which would imply that it is a surface with boundary, contradicting your definition of submanifold as something that is locally a graph of a function with open domain in $mathbbR^2$ (Invariance of Domain implicitly used).
Then if $x_0$ is the maximum of $f$ and $gamma: (-epsilon, epsilon)rightarrow M$ is a curve with $gamma(0)=x_0$, $t=0$ is a critical point of $fcirc gamma$ (this is just a directional derivative of $f$ at $x_0$). This means that $0=nabla f(x_0)cdot gamma'(0)=vcdot gamma'(0)$. This proves that $vperp T_x_0M$, since $T_x_0M$ is exactly the plane containing all vectors tangent to curves through $x_0$.
The exact same argument proves that $vperp T_y_0M$.
answered Apr 8 at 21:15
LazLaz
969510
$endgroup$
Following the hint you propose, $f(x)=langle x, vrangle$ has a maximum and a minimum. If they both coincide, it turns out that your $2$-manifold is a compact subset of a plane in $mathbbR^3$, which would imply that it is a surface with boundary, contradicting your definition of submanifold as something that is locally a graph of a function with open domain in $mathbbR^2$ (Invariance of Domain implicitly used).
Then if $x_0$ is the maximum of $f$ and $gamma: (-epsilon, epsilon)rightarrow M$ is a curve with $gamma(0)=x_0$, $t=0$ is a critical point of $fcirc gamma$ (this is just a directional derivative of $f$ at $x_0$). This means that $0=nabla f(x_0)cdot gamma'(0)=vcdot gamma'(0)$. This proves that $vperp T_x_0M$, since $T_x_0M$ is exactly the plane containing all vectors tangent to curves through $x_0$.
The exact same argument proves that $vperp T_y_0M$.
answered Apr 8 at 21:15
LazLaz
969510
edited Apr 8 at 21:21
answered Apr 8 at 21:15
LazLaz
969510
answered Apr 8 at 21:15
LazLaz
969510
answered Apr 8 at 21:15
LazLaz
969510
969510
$begingroup$
So the solution is working with the definition of tangent space. Thanks!
$endgroup$
– Gabi G
Apr 8 at 21:35
$begingroup$
Exactly, this is one of the many definitions of tangent space. This one is very useful, because it is very geometric, but it requieres our manifold to be sitting naturally inside some $mathbbR^n$. Your welcome!
$endgroup$
– Laz
Apr 8 at 21:37
1
$begingroup$
@Laz: the question only makes sense for a manifold sitting inside $BbbR^3$ (it's about a relationship between a vector in $BbbR^3$ and the tangent spaces, (which only makes sense if the tangent spaces are identified with planes in $BbbR^3$). (And it was me who provided the hint and not the OP $ddotsmile$.)
$endgroup$
– Rob Arthan
Apr 8 at 22:06
$begingroup$
Of course @RobArthan, I was refering to the definition of tangent space in the general setting.
$endgroup$
– Laz
Apr 8 at 22:08
$begingroup$
@Laz: sure. The problem could have been phrased in terms of an immersion or embedding of an abstract differentiable manifold in $BbbR^3$, which is why I asked the OP to say more about the definitions being used. Anyway, the OP is happy with your answer, so let's go our way rejoicing.
$endgroup$
– Rob Arthan
Apr 8 at 22:15
|
show 1 more comment
$begingroup$
So the solution is working with the definition of tangent space. Thanks!
$endgroup$
– Gabi G
Apr 8 at 21:35
$begingroup$
Exactly, this is one of the many definitions of tangent space. This one is very useful, because it is very geometric, but it requieres our manifold to be sitting naturally inside some $mathbbR^n$. Your welcome!
$endgroup$
– Laz
Apr 8 at 21:37
1
$begingroup$
@Laz: the question only makes sense for a manifold sitting inside $BbbR^3$ (it's about a relationship between a vector in $BbbR^3$ and the tangent spaces, (which only makes sense if the tangent spaces are identified with planes in $BbbR^3$). (And it was me who provided the hint and not the OP $ddotsmile$.)
$endgroup$
– Rob Arthan
Apr 8 at 22:06
$begingroup$
Of course @RobArthan, I was refering to the definition of tangent space in the general setting.
$endgroup$
– Laz
Apr 8 at 22:08
$begingroup$
@Laz: sure. The problem could have been phrased in terms of an immersion or embedding of an abstract differentiable manifold in $BbbR^3$, which is why I asked the OP to say more about the definitions being used. Anyway, the OP is happy with your answer, so let's go our way rejoicing.
$endgroup$
– Rob Arthan
Apr 8 at 22:15
$begingroup$
So the solution is working with the definition of tangent space. Thanks!
$endgroup$
– Gabi G
Apr 8 at 21:35
$begingroup$
So the solution is working with the definition of tangent space. Thanks!
$endgroup$
– Gabi G
Apr 8 at 21:35
$begingroup$
Exactly, this is one of the many definitions of tangent space. This one is very useful, because it is very geometric, but it requieres our manifold to be sitting naturally inside some $mathbbR^n$. Your welcome!
$endgroup$
– Laz
Apr 8 at 21:37
$begingroup$
Exactly, this is one of the many definitions of tangent space. This one is very useful, because it is very geometric, but it requieres our manifold to be sitting naturally inside some $mathbbR^n$. Your welcome!
$endgroup$
– Laz
Apr 8 at 21:37
1
1
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@Laz: the question only makes sense for a manifold sitting inside $BbbR^3$ (it's about a relationship between a vector in $BbbR^3$ and the tangent spaces, (which only makes sense if the tangent spaces are identified with planes in $BbbR^3$). (And it was me who provided the hint and not the OP $ddotsmile$.)
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– Rob Arthan
Apr 8 at 22:06
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@Laz: the question only makes sense for a manifold sitting inside $BbbR^3$ (it's about a relationship between a vector in $BbbR^3$ and the tangent spaces, (which only makes sense if the tangent spaces are identified with planes in $BbbR^3$). (And it was me who provided the hint and not the OP $ddotsmile$.)
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– Rob Arthan
Apr 8 at 22:06
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Of course @RobArthan, I was refering to the definition of tangent space in the general setting.
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– Laz
Apr 8 at 22:08
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Of course @RobArthan, I was refering to the definition of tangent space in the general setting.
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– Laz
Apr 8 at 22:08
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@Laz: sure. The problem could have been phrased in terms of an immersion or embedding of an abstract differentiable manifold in $BbbR^3$, which is why I asked the OP to say more about the definitions being used. Anyway, the OP is happy with your answer, so let's go our way rejoicing.
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– Rob Arthan
Apr 8 at 22:15
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@Laz: sure. The problem could have been phrased in terms of an immersion or embedding of an abstract differentiable manifold in $BbbR^3$, which is why I asked the OP to say more about the definitions being used. Anyway, the OP is happy with your answer, so let's go our way rejoicing.
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– Rob Arthan
Apr 8 at 22:15
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You really need to tell us what definition of a manifold contained in $BbbR^2$ you are using. (Some assumptions are needed to ensure the existence of the tangent spaces.) However, to see that a minimum/maximum, $x_0$ say, of your function $f(x)$ must be at a point where $v$ is normal to the tangent bundle, consider an arbitrary differentiable curve $gamma$ in $M$ with $gamma(0) = x_0$ and note that $0$ is a local minimum/maximum for $f circ gamma$ (hence $(f circ gamma)'(0) = 0$).
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– Rob Arthan
Apr 8 at 20:29
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$M$ is a manifold if for every $x in M$ there exists an open set $U subset R^3$ such that $M bigcap U$ is a graph of a smooth function
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– Gabi G
Apr 8 at 20:31
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That definition needs some more details, I think. How are you viewing $M cap U$ as the graph of a function?
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– Rob Arthan
Apr 8 at 21:56