Computing the $2^textnd$ fundamental form of $mathbbR_+times M^nto mathbbR^n+p+1$ The Next CEO of Stack OverflowDerivative of a metric tensor along a curveEvaluating the second fundamental form for a curveProperty of normal coordinatesJacobi fields in polar coordinates.The Definition of the Second Fundamental FormProve without coordinates that covariant derivatives are “almost” related under isometric immersion?prove iff conditions on $(M_1,g_1)$, $(M_2,g_2)$ so that $big(M_1times M_2,g_1+g_2big)$ satisfies $textRic^0=0$Justifying “determinant” of second fundamental form using the definition of sectional curvatureCoordinates of the tensor of the second fundamental form for a submanifoldComputations with respect to a Riemannian metric
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Computing the $2^textnd$ fundamental form of $mathbbR_+times M^nto mathbbR^n+p+1$
The Next CEO of Stack OverflowDerivative of a metric tensor along a curveEvaluating the second fundamental form for a curveProperty of normal coordinatesJacobi fields in polar coordinates.The Definition of the Second Fundamental FormProve without coordinates that covariant derivatives are “almost” related under isometric immersion?prove iff conditions on $(M_1,g_1)$, $(M_2,g_2)$ so that $big(M_1times M_2,g_1+g_2big)$ satisfies $textRic^0=0$Justifying “determinant” of second fundamental form using the definition of sectional curvatureCoordinates of the tensor of the second fundamental form for a submanifoldComputations with respect to a Riemannian metric
$begingroup$
Let $f: M^nto mathbbS^n+p$ be an isometric immersion. The cone over $f$ is defined to be the immersion
beginalign*
F:mathbbR_+times M &to mathbbR^n+p+1\
(t,x)&mapsto tf(x)
endalign*
Compute the second fundamental form of $F$.
[where $mathbbR_+:=tinmathbbRmid t>0$]
I could verify that $F$ is an immersion by considering $f(x)=(f_1(x),...,f_n+p+1(x))inmathbbR^n+p+1$ with $sum_if_i^2=1$. That way, $langlefracpartial fpartial x_i,frangle=0$ for all $i$ and, since $f$ is an immersion, we have $fracpartial fpartial x_1,...,fracpartial fpartial x_n$ linearly independent, so $textrank(dF)=textrankleft(f,fracpartial fpartial x_1,...,fracpartial fpartial x_nright)=n+1$.
For the $2^textnd$ fundamental form, here's where I'm at: If $widetildenabla$, $nabla$ are Levi-Civita connections for $mathbbR^n+p+1$ and $mathbbR_+times M$ respectively, the second fundamental form is by definition $alpha(X,Y)=widetildenabla_widetildeXwidetildeY-nabla_XY$. By tensoriality and symmetry of $alpha$, we only need to compute $alphaleft(fracpartialpartial t,fracpartialpartial tright),alphaleft(fracpartialpartial t,fracpartialpartial x_iright)$ and $alphaleft(fracpartialpartial x_i,fracpartialpartial x_jright)$.
To compute $alphaleft(fracpartialpartial t,fracpartialpartial tright)$ for example, can see that $nabla_fracpartialpartial tfracpartialpartial t=0$, but I don't know how compute $widetildenabla_fracpartial Fpartial tfracpartial Fpartial t$. Obviously, $fracpartial Fpartial t(t,x)=f(x)$, but I can't figure out what $widetildenabla_f(x)f(x)$ even means.
Any suggestions?
riemannian-geometry smooth-manifolds riemannian-metric
$endgroup$
|
show 1 more comment
$begingroup$
Let $f: M^nto mathbbS^n+p$ be an isometric immersion. The cone over $f$ is defined to be the immersion
beginalign*
F:mathbbR_+times M &to mathbbR^n+p+1\
(t,x)&mapsto tf(x)
endalign*
Compute the second fundamental form of $F$.
[where $mathbbR_+:=tinmathbbRmid t>0$]
I could verify that $F$ is an immersion by considering $f(x)=(f_1(x),...,f_n+p+1(x))inmathbbR^n+p+1$ with $sum_if_i^2=1$. That way, $langlefracpartial fpartial x_i,frangle=0$ for all $i$ and, since $f$ is an immersion, we have $fracpartial fpartial x_1,...,fracpartial fpartial x_n$ linearly independent, so $textrank(dF)=textrankleft(f,fracpartial fpartial x_1,...,fracpartial fpartial x_nright)=n+1$.
For the $2^textnd$ fundamental form, here's where I'm at: If $widetildenabla$, $nabla$ are Levi-Civita connections for $mathbbR^n+p+1$ and $mathbbR_+times M$ respectively, the second fundamental form is by definition $alpha(X,Y)=widetildenabla_widetildeXwidetildeY-nabla_XY$. By tensoriality and symmetry of $alpha$, we only need to compute $alphaleft(fracpartialpartial t,fracpartialpartial tright),alphaleft(fracpartialpartial t,fracpartialpartial x_iright)$ and $alphaleft(fracpartialpartial x_i,fracpartialpartial x_jright)$.
To compute $alphaleft(fracpartialpartial t,fracpartialpartial tright)$ for example, can see that $nabla_fracpartialpartial tfracpartialpartial t=0$, but I don't know how compute $widetildenabla_fracpartial Fpartial tfracpartial Fpartial t$. Obviously, $fracpartial Fpartial t(t,x)=f(x)$, but I can't figure out what $widetildenabla_f(x)f(x)$ even means.
Any suggestions?
riemannian-geometry smooth-manifolds riemannian-metric
$endgroup$
$begingroup$
I wonder why this question received so little attention... is it because it's way too easy, way too hard or way too uninteresting?
$endgroup$
– rmdmc89
Mar 24 at 22:08
1
$begingroup$
This is a very interesting question, but it exhibits a few technical and conceptual difficulties, and the experts who can help seem to be busy with other things. A couple of remarks: 1) $widetildenabla$ should be the standard Euclidean connection (all Christoffels vahish); 2) the expression $widetildenabla_fracpartial Fpartial tfracpartial Fpartial t$ does not make sense. You would rather try to compute $widetildenabla_fracpartialpartial tfracpartialpartial t$ (without $F$) etc.
$endgroup$
– Yuri Vyatkin
Mar 25 at 19:42
1
$begingroup$
In fact, I would start with trying to understand the simplest possible case just to see how the answer could look like. So, what would be the answer in the case of $M = mathbbS^1$, and $p = n = 1$. where $f$ is the canonical immersion $ mathbbS^1 to mathbbS^2$ (as the equator)? The next step would be to figure out what is gonna happen when the immersion varies. After that it should become more clear, how this generalizes to higher dimensions.
$endgroup$
– Yuri Vyatkin
Mar 25 at 19:48
$begingroup$
@YuriVyatkin, I'm confused with the expression $widetildenabla_fracpartialpartial tfracpartialpartial t$. As I understand it, $fracpartialpartial t$ makes sense in $mathbbR_+$, but to see it as in $mathbbR^n+p+1$ we need to consider the immersion $F$, right? That's why I thought $fracpartial Fpartial t=dFleft(fracpartialpartial tright)$ was correct
$endgroup$
– rmdmc89
Mar 25 at 20:13
1
$begingroup$
I posted my incomplete answer to show you the picture, from which using some imagination one can see that $fracpartial Fpartial t$ is a vector field along the cone. So, basically, you are on the right track, just the nature of the things has to be understood more clearly. Sorry, I don't have much time to finish this up, but I will do my best.
$endgroup$
– Yuri Vyatkin
Mar 26 at 9:47
|
show 1 more comment
$begingroup$
Let $f: M^nto mathbbS^n+p$ be an isometric immersion. The cone over $f$ is defined to be the immersion
beginalign*
F:mathbbR_+times M &to mathbbR^n+p+1\
(t,x)&mapsto tf(x)
endalign*
Compute the second fundamental form of $F$.
[where $mathbbR_+:=tinmathbbRmid t>0$]
I could verify that $F$ is an immersion by considering $f(x)=(f_1(x),...,f_n+p+1(x))inmathbbR^n+p+1$ with $sum_if_i^2=1$. That way, $langlefracpartial fpartial x_i,frangle=0$ for all $i$ and, since $f$ is an immersion, we have $fracpartial fpartial x_1,...,fracpartial fpartial x_n$ linearly independent, so $textrank(dF)=textrankleft(f,fracpartial fpartial x_1,...,fracpartial fpartial x_nright)=n+1$.
For the $2^textnd$ fundamental form, here's where I'm at: If $widetildenabla$, $nabla$ are Levi-Civita connections for $mathbbR^n+p+1$ and $mathbbR_+times M$ respectively, the second fundamental form is by definition $alpha(X,Y)=widetildenabla_widetildeXwidetildeY-nabla_XY$. By tensoriality and symmetry of $alpha$, we only need to compute $alphaleft(fracpartialpartial t,fracpartialpartial tright),alphaleft(fracpartialpartial t,fracpartialpartial x_iright)$ and $alphaleft(fracpartialpartial x_i,fracpartialpartial x_jright)$.
To compute $alphaleft(fracpartialpartial t,fracpartialpartial tright)$ for example, can see that $nabla_fracpartialpartial tfracpartialpartial t=0$, but I don't know how compute $widetildenabla_fracpartial Fpartial tfracpartial Fpartial t$. Obviously, $fracpartial Fpartial t(t,x)=f(x)$, but I can't figure out what $widetildenabla_f(x)f(x)$ even means.
Any suggestions?
riemannian-geometry smooth-manifolds riemannian-metric
$endgroup$
Let $f: M^nto mathbbS^n+p$ be an isometric immersion. The cone over $f$ is defined to be the immersion
beginalign*
F:mathbbR_+times M &to mathbbR^n+p+1\
(t,x)&mapsto tf(x)
endalign*
Compute the second fundamental form of $F$.
[where $mathbbR_+:=tinmathbbRmid t>0$]
I could verify that $F$ is an immersion by considering $f(x)=(f_1(x),...,f_n+p+1(x))inmathbbR^n+p+1$ with $sum_if_i^2=1$. That way, $langlefracpartial fpartial x_i,frangle=0$ for all $i$ and, since $f$ is an immersion, we have $fracpartial fpartial x_1,...,fracpartial fpartial x_n$ linearly independent, so $textrank(dF)=textrankleft(f,fracpartial fpartial x_1,...,fracpartial fpartial x_nright)=n+1$.
For the $2^textnd$ fundamental form, here's where I'm at: If $widetildenabla$, $nabla$ are Levi-Civita connections for $mathbbR^n+p+1$ and $mathbbR_+times M$ respectively, the second fundamental form is by definition $alpha(X,Y)=widetildenabla_widetildeXwidetildeY-nabla_XY$. By tensoriality and symmetry of $alpha$, we only need to compute $alphaleft(fracpartialpartial t,fracpartialpartial tright),alphaleft(fracpartialpartial t,fracpartialpartial x_iright)$ and $alphaleft(fracpartialpartial x_i,fracpartialpartial x_jright)$.
To compute $alphaleft(fracpartialpartial t,fracpartialpartial tright)$ for example, can see that $nabla_fracpartialpartial tfracpartialpartial t=0$, but I don't know how compute $widetildenabla_fracpartial Fpartial tfracpartial Fpartial t$. Obviously, $fracpartial Fpartial t(t,x)=f(x)$, but I can't figure out what $widetildenabla_f(x)f(x)$ even means.
Any suggestions?
riemannian-geometry smooth-manifolds riemannian-metric
riemannian-geometry smooth-manifolds riemannian-metric
edited Mar 25 at 20:35
rmdmc89
asked Mar 19 at 20:31
rmdmc89rmdmc89
2,2771923
2,2771923
$begingroup$
I wonder why this question received so little attention... is it because it's way too easy, way too hard or way too uninteresting?
$endgroup$
– rmdmc89
Mar 24 at 22:08
1
$begingroup$
This is a very interesting question, but it exhibits a few technical and conceptual difficulties, and the experts who can help seem to be busy with other things. A couple of remarks: 1) $widetildenabla$ should be the standard Euclidean connection (all Christoffels vahish); 2) the expression $widetildenabla_fracpartial Fpartial tfracpartial Fpartial t$ does not make sense. You would rather try to compute $widetildenabla_fracpartialpartial tfracpartialpartial t$ (without $F$) etc.
$endgroup$
– Yuri Vyatkin
Mar 25 at 19:42
1
$begingroup$
In fact, I would start with trying to understand the simplest possible case just to see how the answer could look like. So, what would be the answer in the case of $M = mathbbS^1$, and $p = n = 1$. where $f$ is the canonical immersion $ mathbbS^1 to mathbbS^2$ (as the equator)? The next step would be to figure out what is gonna happen when the immersion varies. After that it should become more clear, how this generalizes to higher dimensions.
$endgroup$
– Yuri Vyatkin
Mar 25 at 19:48
$begingroup$
@YuriVyatkin, I'm confused with the expression $widetildenabla_fracpartialpartial tfracpartialpartial t$. As I understand it, $fracpartialpartial t$ makes sense in $mathbbR_+$, but to see it as in $mathbbR^n+p+1$ we need to consider the immersion $F$, right? That's why I thought $fracpartial Fpartial t=dFleft(fracpartialpartial tright)$ was correct
$endgroup$
– rmdmc89
Mar 25 at 20:13
1
$begingroup$
I posted my incomplete answer to show you the picture, from which using some imagination one can see that $fracpartial Fpartial t$ is a vector field along the cone. So, basically, you are on the right track, just the nature of the things has to be understood more clearly. Sorry, I don't have much time to finish this up, but I will do my best.
$endgroup$
– Yuri Vyatkin
Mar 26 at 9:47
|
show 1 more comment
$begingroup$
I wonder why this question received so little attention... is it because it's way too easy, way too hard or way too uninteresting?
$endgroup$
– rmdmc89
Mar 24 at 22:08
1
$begingroup$
This is a very interesting question, but it exhibits a few technical and conceptual difficulties, and the experts who can help seem to be busy with other things. A couple of remarks: 1) $widetildenabla$ should be the standard Euclidean connection (all Christoffels vahish); 2) the expression $widetildenabla_fracpartial Fpartial tfracpartial Fpartial t$ does not make sense. You would rather try to compute $widetildenabla_fracpartialpartial tfracpartialpartial t$ (without $F$) etc.
$endgroup$
– Yuri Vyatkin
Mar 25 at 19:42
1
$begingroup$
In fact, I would start with trying to understand the simplest possible case just to see how the answer could look like. So, what would be the answer in the case of $M = mathbbS^1$, and $p = n = 1$. where $f$ is the canonical immersion $ mathbbS^1 to mathbbS^2$ (as the equator)? The next step would be to figure out what is gonna happen when the immersion varies. After that it should become more clear, how this generalizes to higher dimensions.
$endgroup$
– Yuri Vyatkin
Mar 25 at 19:48
$begingroup$
@YuriVyatkin, I'm confused with the expression $widetildenabla_fracpartialpartial tfracpartialpartial t$. As I understand it, $fracpartialpartial t$ makes sense in $mathbbR_+$, but to see it as in $mathbbR^n+p+1$ we need to consider the immersion $F$, right? That's why I thought $fracpartial Fpartial t=dFleft(fracpartialpartial tright)$ was correct
$endgroup$
– rmdmc89
Mar 25 at 20:13
1
$begingroup$
I posted my incomplete answer to show you the picture, from which using some imagination one can see that $fracpartial Fpartial t$ is a vector field along the cone. So, basically, you are on the right track, just the nature of the things has to be understood more clearly. Sorry, I don't have much time to finish this up, but I will do my best.
$endgroup$
– Yuri Vyatkin
Mar 26 at 9:47
$begingroup$
I wonder why this question received so little attention... is it because it's way too easy, way too hard or way too uninteresting?
$endgroup$
– rmdmc89
Mar 24 at 22:08
$begingroup$
I wonder why this question received so little attention... is it because it's way too easy, way too hard or way too uninteresting?
$endgroup$
– rmdmc89
Mar 24 at 22:08
1
1
$begingroup$
This is a very interesting question, but it exhibits a few technical and conceptual difficulties, and the experts who can help seem to be busy with other things. A couple of remarks: 1) $widetildenabla$ should be the standard Euclidean connection (all Christoffels vahish); 2) the expression $widetildenabla_fracpartial Fpartial tfracpartial Fpartial t$ does not make sense. You would rather try to compute $widetildenabla_fracpartialpartial tfracpartialpartial t$ (without $F$) etc.
$endgroup$
– Yuri Vyatkin
Mar 25 at 19:42
$begingroup$
This is a very interesting question, but it exhibits a few technical and conceptual difficulties, and the experts who can help seem to be busy with other things. A couple of remarks: 1) $widetildenabla$ should be the standard Euclidean connection (all Christoffels vahish); 2) the expression $widetildenabla_fracpartial Fpartial tfracpartial Fpartial t$ does not make sense. You would rather try to compute $widetildenabla_fracpartialpartial tfracpartialpartial t$ (without $F$) etc.
$endgroup$
– Yuri Vyatkin
Mar 25 at 19:42
1
1
$begingroup$
In fact, I would start with trying to understand the simplest possible case just to see how the answer could look like. So, what would be the answer in the case of $M = mathbbS^1$, and $p = n = 1$. where $f$ is the canonical immersion $ mathbbS^1 to mathbbS^2$ (as the equator)? The next step would be to figure out what is gonna happen when the immersion varies. After that it should become more clear, how this generalizes to higher dimensions.
$endgroup$
– Yuri Vyatkin
Mar 25 at 19:48
$begingroup$
In fact, I would start with trying to understand the simplest possible case just to see how the answer could look like. So, what would be the answer in the case of $M = mathbbS^1$, and $p = n = 1$. where $f$ is the canonical immersion $ mathbbS^1 to mathbbS^2$ (as the equator)? The next step would be to figure out what is gonna happen when the immersion varies. After that it should become more clear, how this generalizes to higher dimensions.
$endgroup$
– Yuri Vyatkin
Mar 25 at 19:48
$begingroup$
@YuriVyatkin, I'm confused with the expression $widetildenabla_fracpartialpartial tfracpartialpartial t$. As I understand it, $fracpartialpartial t$ makes sense in $mathbbR_+$, but to see it as in $mathbbR^n+p+1$ we need to consider the immersion $F$, right? That's why I thought $fracpartial Fpartial t=dFleft(fracpartialpartial tright)$ was correct
$endgroup$
– rmdmc89
Mar 25 at 20:13
$begingroup$
@YuriVyatkin, I'm confused with the expression $widetildenabla_fracpartialpartial tfracpartialpartial t$. As I understand it, $fracpartialpartial t$ makes sense in $mathbbR_+$, but to see it as in $mathbbR^n+p+1$ we need to consider the immersion $F$, right? That's why I thought $fracpartial Fpartial t=dFleft(fracpartialpartial tright)$ was correct
$endgroup$
– rmdmc89
Mar 25 at 20:13
1
1
$begingroup$
I posted my incomplete answer to show you the picture, from which using some imagination one can see that $fracpartial Fpartial t$ is a vector field along the cone. So, basically, you are on the right track, just the nature of the things has to be understood more clearly. Sorry, I don't have much time to finish this up, but I will do my best.
$endgroup$
– Yuri Vyatkin
Mar 26 at 9:47
$begingroup$
I posted my incomplete answer to show you the picture, from which using some imagination one can see that $fracpartial Fpartial t$ is a vector field along the cone. So, basically, you are on the right track, just the nature of the things has to be understood more clearly. Sorry, I don't have much time to finish this up, but I will do my best.
$endgroup$
– Yuri Vyatkin
Mar 26 at 9:47
|
show 1 more comment
1 Answer
1
active
oldest
votes
$begingroup$
The question is purely local, so we may choose a coordinate chart $varphi colon U subset mathbbR^n to M$ around some point $x in M$. We use this chart to indentify points in $U$ and $varphi(U)$, and thus we regard $f$ locally as $varphi circ f$, so that the mapping $f colon U to mathbbR^n+p+1$ is a parametrization of its image, and
$$
langle f, f rangle = 1
$$
The mapping
$F colon mathbbR_+ times U to mathbbR^n+p+1 colon (t, x) mapsto t f(x)$
is also a local parametrization of its image, which is a piece of the cone $C = F(mathbbR_+ times U)$.
We calculate at the point $(t, x) in C$.
Here is a picture showing schematically what is going on:
Corresponding to the chart $varphi$, there are local coordinates $x^i$ in $U$, and let $t$ be a local coordinate in $mathbbR_+$, so that $ t, x^i, i=1, dots, n $ form a coordinate system on $mathbbR_+ times U$.
Let us introduce the following notation: $partial_i := fracpartialpartial x^i$, $partial_t := fracpartialpartial x^i$, $f_i := partial_i f$, $F_i := partial_i F$, $F_t := partial_t F$, $i = 1, dots, n$. Since $F = t f$, we have:
$$
F_i = tf_i text and F_t = f
$$
Notice that $F_i$ and $F_t$ are vector fields along the cone $C$, and
$$
F_i|_(t,x) = t pmatrix
f^1_i(x)\
vdots\
f^n+p+1_i(x)
text and
F_t|_(t,x) = pmatrix
f^1(x)\
vdots\
f^n+p+1(x)
$$
In order to differentiate $F_i$ and $F_t$ in $mathbbR^n+p+1$ we need some local coordinates there and some extensions $widetildeF_i$ and $widetildeF_t$ of those fields, so we just take the slice coordinates in $mathbbR_+ times U times mathbbR^p$
$$
y^0 = t, y^1 = x^1, dots, y^n = x^n, y^n+1 = y^n+1, dots, y^n+p = y^n+p
$$
(where $y^n+1, dots, y^n+p$ are the standard coordinates in $mathbbR^p$), and define
$$
widetildeF(t, x^1, dots, x^n, *, dots, *) := F(t, x^1, dots, x^n)
$$
so that $widetildeF_i = partial_i widetildeF$ and $widetildeF_t = partial_i widetildeF$.
Since in $mathbbR^n+p+1$ we have the standard Euclidean metric and the standard Euclidean connection (all Christoffel symbols vanish identically),
we have
$$
widetildenabla_i widetildeF_j = partial_i F_j \
widetildenabla_t widetildeF_i = partial_t F_i \
widetildenabla_t widetildeF_t = partial_t F_t
$$
In order to compute the s.f.f. $alpha(X,Y)=widetildenabla_widetildeXwidetildeY-nabla_XY$ it remains to understand what $nabla_XY$ is here.
It is known that $nabla_XY = (widetildenabla_widetildeXwidetildeY)^top = Pi(widetildenabla_widetildeXwidetildeY)$, where the tangential projection operator is given by
$$
Pi colon v mapsto frac1F_t langle v, F_t rangle F_t + sum_i=1^n frac1 langle v, F_i rangle F_i
$$
This is almost it, because now we can write
$$
nabla_i F_j = Pi(widetildenabla_i widetildeF_j) = frac1F_t langle widetildenabla_i widetildeF_j, F_t rangle F_t + sum_k=1^n frac1F_k langle widetildenabla_i widetildeF_j, F_k rangle F_k \
nabla_t F_i = Pi(widetildenabla_t widetildeF_i) = frac1F_t langle widetildenabla_t widetildeF_i, F_t rangle F_t + sum_k=1^n frac1F_k langle widetildenabla_t widetildeF_i, F_k rangle F_k \
nabla_t F_t = Pi(widetildenabla_t widetildeF_t) = frac1F_t langle widetildenabla_t widetildeF_t, F_t rangle F_t + sum_k=1^n frac1F_k langle widetildenabla_t widetildeF_t, F_k rangle F_k
$$
and collecting all the pieces it is now straightforward to obtain the final expressions.
$endgroup$
$begingroup$
Geometrically, I think it's reasonable expect $alphaleft(fracpartialpartial t,fracpartialpartial tright)=0$. First, it's obvious that $nabla_fracpartialpartial tfracpartialpartial t=0$. Second, we can see that $widetildenabla_fracpartialpartial tfracpartialpartial t=0$ because we can take the geodesic $gamma(t)=(1+t)f(x)$ in $mathbbR^n+p+1$, for which $gamma'(0)=f(x)=fracpartial Fpartial t(t,x)$, therefore $widetildenabla_gamma'gamma'=0$.
$endgroup$
– rmdmc89
Mar 27 at 0:56
$begingroup$
But for $alphaleft(fracpartialpartial t,fracpartialpartial x_iright)$ and $alphaleft(fracpartialpartial x_i,fracpartialpartial x_jright)$, I'm still pretty lost. I only know that in general they can't be all zero, since in that case the immersion would be totally geodesic, which is not necessarily the case.
$endgroup$
– rmdmc89
Mar 27 at 1:00
1
$begingroup$
@rmdmc89 I completely rewrote the answer trying to make it more clear. See if this helps, please.
$endgroup$
– Yuri Vyatkin
yesterday
$begingroup$
Thank you so much for your patience, Yuri
$endgroup$
– rmdmc89
13 hours ago
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$begingroup$
The question is purely local, so we may choose a coordinate chart $varphi colon U subset mathbbR^n to M$ around some point $x in M$. We use this chart to indentify points in $U$ and $varphi(U)$, and thus we regard $f$ locally as $varphi circ f$, so that the mapping $f colon U to mathbbR^n+p+1$ is a parametrization of its image, and
$$
langle f, f rangle = 1
$$
The mapping
$F colon mathbbR_+ times U to mathbbR^n+p+1 colon (t, x) mapsto t f(x)$
is also a local parametrization of its image, which is a piece of the cone $C = F(mathbbR_+ times U)$.
We calculate at the point $(t, x) in C$.
Here is a picture showing schematically what is going on:
Corresponding to the chart $varphi$, there are local coordinates $x^i$ in $U$, and let $t$ be a local coordinate in $mathbbR_+$, so that $ t, x^i, i=1, dots, n $ form a coordinate system on $mathbbR_+ times U$.
Let us introduce the following notation: $partial_i := fracpartialpartial x^i$, $partial_t := fracpartialpartial x^i$, $f_i := partial_i f$, $F_i := partial_i F$, $F_t := partial_t F$, $i = 1, dots, n$. Since $F = t f$, we have:
$$
F_i = tf_i text and F_t = f
$$
Notice that $F_i$ and $F_t$ are vector fields along the cone $C$, and
$$
F_i|_(t,x) = t pmatrix
f^1_i(x)\
vdots\
f^n+p+1_i(x)
text and
F_t|_(t,x) = pmatrix
f^1(x)\
vdots\
f^n+p+1(x)
$$
In order to differentiate $F_i$ and $F_t$ in $mathbbR^n+p+1$ we need some local coordinates there and some extensions $widetildeF_i$ and $widetildeF_t$ of those fields, so we just take the slice coordinates in $mathbbR_+ times U times mathbbR^p$
$$
y^0 = t, y^1 = x^1, dots, y^n = x^n, y^n+1 = y^n+1, dots, y^n+p = y^n+p
$$
(where $y^n+1, dots, y^n+p$ are the standard coordinates in $mathbbR^p$), and define
$$
widetildeF(t, x^1, dots, x^n, *, dots, *) := F(t, x^1, dots, x^n)
$$
so that $widetildeF_i = partial_i widetildeF$ and $widetildeF_t = partial_i widetildeF$.
Since in $mathbbR^n+p+1$ we have the standard Euclidean metric and the standard Euclidean connection (all Christoffel symbols vanish identically),
we have
$$
widetildenabla_i widetildeF_j = partial_i F_j \
widetildenabla_t widetildeF_i = partial_t F_i \
widetildenabla_t widetildeF_t = partial_t F_t
$$
In order to compute the s.f.f. $alpha(X,Y)=widetildenabla_widetildeXwidetildeY-nabla_XY$ it remains to understand what $nabla_XY$ is here.
It is known that $nabla_XY = (widetildenabla_widetildeXwidetildeY)^top = Pi(widetildenabla_widetildeXwidetildeY)$, where the tangential projection operator is given by
$$
Pi colon v mapsto frac1F_t langle v, F_t rangle F_t + sum_i=1^n frac1 langle v, F_i rangle F_i
$$
This is almost it, because now we can write
$$
nabla_i F_j = Pi(widetildenabla_i widetildeF_j) = frac1F_t langle widetildenabla_i widetildeF_j, F_t rangle F_t + sum_k=1^n frac1F_k langle widetildenabla_i widetildeF_j, F_k rangle F_k \
nabla_t F_i = Pi(widetildenabla_t widetildeF_i) = frac1F_t langle widetildenabla_t widetildeF_i, F_t rangle F_t + sum_k=1^n frac1F_k langle widetildenabla_t widetildeF_i, F_k rangle F_k \
nabla_t F_t = Pi(widetildenabla_t widetildeF_t) = frac1F_t langle widetildenabla_t widetildeF_t, F_t rangle F_t + sum_k=1^n frac1F_k langle widetildenabla_t widetildeF_t, F_k rangle F_k
$$
and collecting all the pieces it is now straightforward to obtain the final expressions.
$endgroup$
$begingroup$
Geometrically, I think it's reasonable expect $alphaleft(fracpartialpartial t,fracpartialpartial tright)=0$. First, it's obvious that $nabla_fracpartialpartial tfracpartialpartial t=0$. Second, we can see that $widetildenabla_fracpartialpartial tfracpartialpartial t=0$ because we can take the geodesic $gamma(t)=(1+t)f(x)$ in $mathbbR^n+p+1$, for which $gamma'(0)=f(x)=fracpartial Fpartial t(t,x)$, therefore $widetildenabla_gamma'gamma'=0$.
$endgroup$
– rmdmc89
Mar 27 at 0:56
$begingroup$
But for $alphaleft(fracpartialpartial t,fracpartialpartial x_iright)$ and $alphaleft(fracpartialpartial x_i,fracpartialpartial x_jright)$, I'm still pretty lost. I only know that in general they can't be all zero, since in that case the immersion would be totally geodesic, which is not necessarily the case.
$endgroup$
– rmdmc89
Mar 27 at 1:00
1
$begingroup$
@rmdmc89 I completely rewrote the answer trying to make it more clear. See if this helps, please.
$endgroup$
– Yuri Vyatkin
yesterday
$begingroup$
Thank you so much for your patience, Yuri
$endgroup$
– rmdmc89
13 hours ago
add a comment |
$begingroup$
The question is purely local, so we may choose a coordinate chart $varphi colon U subset mathbbR^n to M$ around some point $x in M$. We use this chart to indentify points in $U$ and $varphi(U)$, and thus we regard $f$ locally as $varphi circ f$, so that the mapping $f colon U to mathbbR^n+p+1$ is a parametrization of its image, and
$$
langle f, f rangle = 1
$$
The mapping
$F colon mathbbR_+ times U to mathbbR^n+p+1 colon (t, x) mapsto t f(x)$
is also a local parametrization of its image, which is a piece of the cone $C = F(mathbbR_+ times U)$.
We calculate at the point $(t, x) in C$.
Here is a picture showing schematically what is going on:
Corresponding to the chart $varphi$, there are local coordinates $x^i$ in $U$, and let $t$ be a local coordinate in $mathbbR_+$, so that $ t, x^i, i=1, dots, n $ form a coordinate system on $mathbbR_+ times U$.
Let us introduce the following notation: $partial_i := fracpartialpartial x^i$, $partial_t := fracpartialpartial x^i$, $f_i := partial_i f$, $F_i := partial_i F$, $F_t := partial_t F$, $i = 1, dots, n$. Since $F = t f$, we have:
$$
F_i = tf_i text and F_t = f
$$
Notice that $F_i$ and $F_t$ are vector fields along the cone $C$, and
$$
F_i|_(t,x) = t pmatrix
f^1_i(x)\
vdots\
f^n+p+1_i(x)
text and
F_t|_(t,x) = pmatrix
f^1(x)\
vdots\
f^n+p+1(x)
$$
In order to differentiate $F_i$ and $F_t$ in $mathbbR^n+p+1$ we need some local coordinates there and some extensions $widetildeF_i$ and $widetildeF_t$ of those fields, so we just take the slice coordinates in $mathbbR_+ times U times mathbbR^p$
$$
y^0 = t, y^1 = x^1, dots, y^n = x^n, y^n+1 = y^n+1, dots, y^n+p = y^n+p
$$
(where $y^n+1, dots, y^n+p$ are the standard coordinates in $mathbbR^p$), and define
$$
widetildeF(t, x^1, dots, x^n, *, dots, *) := F(t, x^1, dots, x^n)
$$
so that $widetildeF_i = partial_i widetildeF$ and $widetildeF_t = partial_i widetildeF$.
Since in $mathbbR^n+p+1$ we have the standard Euclidean metric and the standard Euclidean connection (all Christoffel symbols vanish identically),
we have
$$
widetildenabla_i widetildeF_j = partial_i F_j \
widetildenabla_t widetildeF_i = partial_t F_i \
widetildenabla_t widetildeF_t = partial_t F_t
$$
In order to compute the s.f.f. $alpha(X,Y)=widetildenabla_widetildeXwidetildeY-nabla_XY$ it remains to understand what $nabla_XY$ is here.
It is known that $nabla_XY = (widetildenabla_widetildeXwidetildeY)^top = Pi(widetildenabla_widetildeXwidetildeY)$, where the tangential projection operator is given by
$$
Pi colon v mapsto frac1F_t langle v, F_t rangle F_t + sum_i=1^n frac1 langle v, F_i rangle F_i
$$
This is almost it, because now we can write
$$
nabla_i F_j = Pi(widetildenabla_i widetildeF_j) = frac1F_t langle widetildenabla_i widetildeF_j, F_t rangle F_t + sum_k=1^n frac1F_k langle widetildenabla_i widetildeF_j, F_k rangle F_k \
nabla_t F_i = Pi(widetildenabla_t widetildeF_i) = frac1F_t langle widetildenabla_t widetildeF_i, F_t rangle F_t + sum_k=1^n frac1F_k langle widetildenabla_t widetildeF_i, F_k rangle F_k \
nabla_t F_t = Pi(widetildenabla_t widetildeF_t) = frac1F_t langle widetildenabla_t widetildeF_t, F_t rangle F_t + sum_k=1^n frac1F_k langle widetildenabla_t widetildeF_t, F_k rangle F_k
$$
and collecting all the pieces it is now straightforward to obtain the final expressions.
$endgroup$
$begingroup$
Geometrically, I think it's reasonable expect $alphaleft(fracpartialpartial t,fracpartialpartial tright)=0$. First, it's obvious that $nabla_fracpartialpartial tfracpartialpartial t=0$. Second, we can see that $widetildenabla_fracpartialpartial tfracpartialpartial t=0$ because we can take the geodesic $gamma(t)=(1+t)f(x)$ in $mathbbR^n+p+1$, for which $gamma'(0)=f(x)=fracpartial Fpartial t(t,x)$, therefore $widetildenabla_gamma'gamma'=0$.
$endgroup$
– rmdmc89
Mar 27 at 0:56
$begingroup$
But for $alphaleft(fracpartialpartial t,fracpartialpartial x_iright)$ and $alphaleft(fracpartialpartial x_i,fracpartialpartial x_jright)$, I'm still pretty lost. I only know that in general they can't be all zero, since in that case the immersion would be totally geodesic, which is not necessarily the case.
$endgroup$
– rmdmc89
Mar 27 at 1:00
1
$begingroup$
@rmdmc89 I completely rewrote the answer trying to make it more clear. See if this helps, please.
$endgroup$
– Yuri Vyatkin
yesterday
$begingroup$
Thank you so much for your patience, Yuri
$endgroup$
– rmdmc89
13 hours ago
add a comment |
$begingroup$
The question is purely local, so we may choose a coordinate chart $varphi colon U subset mathbbR^n to M$ around some point $x in M$. We use this chart to indentify points in $U$ and $varphi(U)$, and thus we regard $f$ locally as $varphi circ f$, so that the mapping $f colon U to mathbbR^n+p+1$ is a parametrization of its image, and
$$
langle f, f rangle = 1
$$
The mapping
$F colon mathbbR_+ times U to mathbbR^n+p+1 colon (t, x) mapsto t f(x)$
is also a local parametrization of its image, which is a piece of the cone $C = F(mathbbR_+ times U)$.
We calculate at the point $(t, x) in C$.
Here is a picture showing schematically what is going on:
Corresponding to the chart $varphi$, there are local coordinates $x^i$ in $U$, and let $t$ be a local coordinate in $mathbbR_+$, so that $ t, x^i, i=1, dots, n $ form a coordinate system on $mathbbR_+ times U$.
Let us introduce the following notation: $partial_i := fracpartialpartial x^i$, $partial_t := fracpartialpartial x^i$, $f_i := partial_i f$, $F_i := partial_i F$, $F_t := partial_t F$, $i = 1, dots, n$. Since $F = t f$, we have:
$$
F_i = tf_i text and F_t = f
$$
Notice that $F_i$ and $F_t$ are vector fields along the cone $C$, and
$$
F_i|_(t,x) = t pmatrix
f^1_i(x)\
vdots\
f^n+p+1_i(x)
text and
F_t|_(t,x) = pmatrix
f^1(x)\
vdots\
f^n+p+1(x)
$$
In order to differentiate $F_i$ and $F_t$ in $mathbbR^n+p+1$ we need some local coordinates there and some extensions $widetildeF_i$ and $widetildeF_t$ of those fields, so we just take the slice coordinates in $mathbbR_+ times U times mathbbR^p$
$$
y^0 = t, y^1 = x^1, dots, y^n = x^n, y^n+1 = y^n+1, dots, y^n+p = y^n+p
$$
(where $y^n+1, dots, y^n+p$ are the standard coordinates in $mathbbR^p$), and define
$$
widetildeF(t, x^1, dots, x^n, *, dots, *) := F(t, x^1, dots, x^n)
$$
so that $widetildeF_i = partial_i widetildeF$ and $widetildeF_t = partial_i widetildeF$.
Since in $mathbbR^n+p+1$ we have the standard Euclidean metric and the standard Euclidean connection (all Christoffel symbols vanish identically),
we have
$$
widetildenabla_i widetildeF_j = partial_i F_j \
widetildenabla_t widetildeF_i = partial_t F_i \
widetildenabla_t widetildeF_t = partial_t F_t
$$
In order to compute the s.f.f. $alpha(X,Y)=widetildenabla_widetildeXwidetildeY-nabla_XY$ it remains to understand what $nabla_XY$ is here.
It is known that $nabla_XY = (widetildenabla_widetildeXwidetildeY)^top = Pi(widetildenabla_widetildeXwidetildeY)$, where the tangential projection operator is given by
$$
Pi colon v mapsto frac1F_t langle v, F_t rangle F_t + sum_i=1^n frac1 langle v, F_i rangle F_i
$$
This is almost it, because now we can write
$$
nabla_i F_j = Pi(widetildenabla_i widetildeF_j) = frac1F_t langle widetildenabla_i widetildeF_j, F_t rangle F_t + sum_k=1^n frac1F_k langle widetildenabla_i widetildeF_j, F_k rangle F_k \
nabla_t F_i = Pi(widetildenabla_t widetildeF_i) = frac1F_t langle widetildenabla_t widetildeF_i, F_t rangle F_t + sum_k=1^n frac1F_k langle widetildenabla_t widetildeF_i, F_k rangle F_k \
nabla_t F_t = Pi(widetildenabla_t widetildeF_t) = frac1F_t langle widetildenabla_t widetildeF_t, F_t rangle F_t + sum_k=1^n frac1F_k langle widetildenabla_t widetildeF_t, F_k rangle F_k
$$
and collecting all the pieces it is now straightforward to obtain the final expressions.
$endgroup$
The question is purely local, so we may choose a coordinate chart $varphi colon U subset mathbbR^n to M$ around some point $x in M$. We use this chart to indentify points in $U$ and $varphi(U)$, and thus we regard $f$ locally as $varphi circ f$, so that the mapping $f colon U to mathbbR^n+p+1$ is a parametrization of its image, and
$$
langle f, f rangle = 1
$$
The mapping
$F colon mathbbR_+ times U to mathbbR^n+p+1 colon (t, x) mapsto t f(x)$
is also a local parametrization of its image, which is a piece of the cone $C = F(mathbbR_+ times U)$.
We calculate at the point $(t, x) in C$.
Here is a picture showing schematically what is going on:
Corresponding to the chart $varphi$, there are local coordinates $x^i$ in $U$, and let $t$ be a local coordinate in $mathbbR_+$, so that $ t, x^i, i=1, dots, n $ form a coordinate system on $mathbbR_+ times U$.
Let us introduce the following notation: $partial_i := fracpartialpartial x^i$, $partial_t := fracpartialpartial x^i$, $f_i := partial_i f$, $F_i := partial_i F$, $F_t := partial_t F$, $i = 1, dots, n$. Since $F = t f$, we have:
$$
F_i = tf_i text and F_t = f
$$
Notice that $F_i$ and $F_t$ are vector fields along the cone $C$, and
$$
F_i|_(t,x) = t pmatrix
f^1_i(x)\
vdots\
f^n+p+1_i(x)
text and
F_t|_(t,x) = pmatrix
f^1(x)\
vdots\
f^n+p+1(x)
$$
In order to differentiate $F_i$ and $F_t$ in $mathbbR^n+p+1$ we need some local coordinates there and some extensions $widetildeF_i$ and $widetildeF_t$ of those fields, so we just take the slice coordinates in $mathbbR_+ times U times mathbbR^p$
$$
y^0 = t, y^1 = x^1, dots, y^n = x^n, y^n+1 = y^n+1, dots, y^n+p = y^n+p
$$
(where $y^n+1, dots, y^n+p$ are the standard coordinates in $mathbbR^p$), and define
$$
widetildeF(t, x^1, dots, x^n, *, dots, *) := F(t, x^1, dots, x^n)
$$
so that $widetildeF_i = partial_i widetildeF$ and $widetildeF_t = partial_i widetildeF$.
Since in $mathbbR^n+p+1$ we have the standard Euclidean metric and the standard Euclidean connection (all Christoffel symbols vanish identically),
we have
$$
widetildenabla_i widetildeF_j = partial_i F_j \
widetildenabla_t widetildeF_i = partial_t F_i \
widetildenabla_t widetildeF_t = partial_t F_t
$$
In order to compute the s.f.f. $alpha(X,Y)=widetildenabla_widetildeXwidetildeY-nabla_XY$ it remains to understand what $nabla_XY$ is here.
It is known that $nabla_XY = (widetildenabla_widetildeXwidetildeY)^top = Pi(widetildenabla_widetildeXwidetildeY)$, where the tangential projection operator is given by
$$
Pi colon v mapsto frac1F_t langle v, F_t rangle F_t + sum_i=1^n frac1 langle v, F_i rangle F_i
$$
This is almost it, because now we can write
$$
nabla_i F_j = Pi(widetildenabla_i widetildeF_j) = frac1F_t langle widetildenabla_i widetildeF_j, F_t rangle F_t + sum_k=1^n frac1F_k langle widetildenabla_i widetildeF_j, F_k rangle F_k \
nabla_t F_i = Pi(widetildenabla_t widetildeF_i) = frac1F_t langle widetildenabla_t widetildeF_i, F_t rangle F_t + sum_k=1^n frac1F_k langle widetildenabla_t widetildeF_i, F_k rangle F_k \
nabla_t F_t = Pi(widetildenabla_t widetildeF_t) = frac1F_t langle widetildenabla_t widetildeF_t, F_t rangle F_t + sum_k=1^n frac1F_k langle widetildenabla_t widetildeF_t, F_k rangle F_k
$$
and collecting all the pieces it is now straightforward to obtain the final expressions.
edited 15 hours ago
answered Mar 26 at 9:42
Yuri VyatkinYuri Vyatkin
7,76322243
7,76322243
$begingroup$
Geometrically, I think it's reasonable expect $alphaleft(fracpartialpartial t,fracpartialpartial tright)=0$. First, it's obvious that $nabla_fracpartialpartial tfracpartialpartial t=0$. Second, we can see that $widetildenabla_fracpartialpartial tfracpartialpartial t=0$ because we can take the geodesic $gamma(t)=(1+t)f(x)$ in $mathbbR^n+p+1$, for which $gamma'(0)=f(x)=fracpartial Fpartial t(t,x)$, therefore $widetildenabla_gamma'gamma'=0$.
$endgroup$
– rmdmc89
Mar 27 at 0:56
$begingroup$
But for $alphaleft(fracpartialpartial t,fracpartialpartial x_iright)$ and $alphaleft(fracpartialpartial x_i,fracpartialpartial x_jright)$, I'm still pretty lost. I only know that in general they can't be all zero, since in that case the immersion would be totally geodesic, which is not necessarily the case.
$endgroup$
– rmdmc89
Mar 27 at 1:00
1
$begingroup$
@rmdmc89 I completely rewrote the answer trying to make it more clear. See if this helps, please.
$endgroup$
– Yuri Vyatkin
yesterday
$begingroup$
Thank you so much for your patience, Yuri
$endgroup$
– rmdmc89
13 hours ago
add a comment |
$begingroup$
Geometrically, I think it's reasonable expect $alphaleft(fracpartialpartial t,fracpartialpartial tright)=0$. First, it's obvious that $nabla_fracpartialpartial tfracpartialpartial t=0$. Second, we can see that $widetildenabla_fracpartialpartial tfracpartialpartial t=0$ because we can take the geodesic $gamma(t)=(1+t)f(x)$ in $mathbbR^n+p+1$, for which $gamma'(0)=f(x)=fracpartial Fpartial t(t,x)$, therefore $widetildenabla_gamma'gamma'=0$.
$endgroup$
– rmdmc89
Mar 27 at 0:56
$begingroup$
But for $alphaleft(fracpartialpartial t,fracpartialpartial x_iright)$ and $alphaleft(fracpartialpartial x_i,fracpartialpartial x_jright)$, I'm still pretty lost. I only know that in general they can't be all zero, since in that case the immersion would be totally geodesic, which is not necessarily the case.
$endgroup$
– rmdmc89
Mar 27 at 1:00
1
$begingroup$
@rmdmc89 I completely rewrote the answer trying to make it more clear. See if this helps, please.
$endgroup$
– Yuri Vyatkin
yesterday
$begingroup$
Thank you so much for your patience, Yuri
$endgroup$
– rmdmc89
13 hours ago
$begingroup$
Geometrically, I think it's reasonable expect $alphaleft(fracpartialpartial t,fracpartialpartial tright)=0$. First, it's obvious that $nabla_fracpartialpartial tfracpartialpartial t=0$. Second, we can see that $widetildenabla_fracpartialpartial tfracpartialpartial t=0$ because we can take the geodesic $gamma(t)=(1+t)f(x)$ in $mathbbR^n+p+1$, for which $gamma'(0)=f(x)=fracpartial Fpartial t(t,x)$, therefore $widetildenabla_gamma'gamma'=0$.
$endgroup$
– rmdmc89
Mar 27 at 0:56
$begingroup$
Geometrically, I think it's reasonable expect $alphaleft(fracpartialpartial t,fracpartialpartial tright)=0$. First, it's obvious that $nabla_fracpartialpartial tfracpartialpartial t=0$. Second, we can see that $widetildenabla_fracpartialpartial tfracpartialpartial t=0$ because we can take the geodesic $gamma(t)=(1+t)f(x)$ in $mathbbR^n+p+1$, for which $gamma'(0)=f(x)=fracpartial Fpartial t(t,x)$, therefore $widetildenabla_gamma'gamma'=0$.
$endgroup$
– rmdmc89
Mar 27 at 0:56
$begingroup$
But for $alphaleft(fracpartialpartial t,fracpartialpartial x_iright)$ and $alphaleft(fracpartialpartial x_i,fracpartialpartial x_jright)$, I'm still pretty lost. I only know that in general they can't be all zero, since in that case the immersion would be totally geodesic, which is not necessarily the case.
$endgroup$
– rmdmc89
Mar 27 at 1:00
$begingroup$
But for $alphaleft(fracpartialpartial t,fracpartialpartial x_iright)$ and $alphaleft(fracpartialpartial x_i,fracpartialpartial x_jright)$, I'm still pretty lost. I only know that in general they can't be all zero, since in that case the immersion would be totally geodesic, which is not necessarily the case.
$endgroup$
– rmdmc89
Mar 27 at 1:00
1
1
$begingroup$
@rmdmc89 I completely rewrote the answer trying to make it more clear. See if this helps, please.
$endgroup$
– Yuri Vyatkin
yesterday
$begingroup$
@rmdmc89 I completely rewrote the answer trying to make it more clear. See if this helps, please.
$endgroup$
– Yuri Vyatkin
yesterday
$begingroup$
Thank you so much for your patience, Yuri
$endgroup$
– rmdmc89
13 hours ago
$begingroup$
Thank you so much for your patience, Yuri
$endgroup$
– rmdmc89
13 hours ago
add a comment |
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$begingroup$
I wonder why this question received so little attention... is it because it's way too easy, way too hard or way too uninteresting?
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– rmdmc89
Mar 24 at 22:08
1
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This is a very interesting question, but it exhibits a few technical and conceptual difficulties, and the experts who can help seem to be busy with other things. A couple of remarks: 1) $widetildenabla$ should be the standard Euclidean connection (all Christoffels vahish); 2) the expression $widetildenabla_fracpartial Fpartial tfracpartial Fpartial t$ does not make sense. You would rather try to compute $widetildenabla_fracpartialpartial tfracpartialpartial t$ (without $F$) etc.
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– Yuri Vyatkin
Mar 25 at 19:42
1
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In fact, I would start with trying to understand the simplest possible case just to see how the answer could look like. So, what would be the answer in the case of $M = mathbbS^1$, and $p = n = 1$. where $f$ is the canonical immersion $ mathbbS^1 to mathbbS^2$ (as the equator)? The next step would be to figure out what is gonna happen when the immersion varies. After that it should become more clear, how this generalizes to higher dimensions.
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– Yuri Vyatkin
Mar 25 at 19:48
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@YuriVyatkin, I'm confused with the expression $widetildenabla_fracpartialpartial tfracpartialpartial t$. As I understand it, $fracpartialpartial t$ makes sense in $mathbbR_+$, but to see it as in $mathbbR^n+p+1$ we need to consider the immersion $F$, right? That's why I thought $fracpartial Fpartial t=dFleft(fracpartialpartial tright)$ was correct
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– rmdmc89
Mar 25 at 20:13
1
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I posted my incomplete answer to show you the picture, from which using some imagination one can see that $fracpartial Fpartial t$ is a vector field along the cone. So, basically, you are on the right track, just the nature of the things has to be understood more clearly. Sorry, I don't have much time to finish this up, but I will do my best.
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– Yuri Vyatkin
Mar 26 at 9:47