Homogeneous polynomial on unit sphere.
Suppose $ F $ is a homogeneous polynomial function of degree $ m $ on Euclidean space $ mathbb{R}^{n+1} $. Restrict $ F $ to the unit sphere $ S^n $ and we get a function on $ S^n $ denoted by $ f $, prove:
$(1)$ $ |nabla_{S^n}f|^2=|nabla_{mathbb{R}^{n+1}}F|^2-m^2f^2 $;
$(2)$ $ Delta_{S^n}f=Delta_{mathbb{R}^{n+1}}F-m(m-1)f-mnf .$
Where $ nabla_{S^n}, nabla_{mathbb{R}^{n+1}} $ are gradients on $ S^n $ and $ mathbb{R}^{n+1} $ respectively, $ Delta_{S^n}, Delta_{mathbb{R}^{n+1}} $ are Laplace operators on $ S^n $ and $ mathbb{R}^{n+1} $ respectively.
Hint: Use Euler's homogeneous function theorem $ langle (nabla_{mathbb{R}^{n+1}}F)_z, z rangle=mF(z) $.
The question above comes from my Riemannian geometry textbook. Can someone give me a hint about how to deal with $ nabla_{S^n}f $ ? I can't find any direct relation between $ nabla_{S^n}f $ and $ nabla_{mathbb{R}^{n+1}}F $. Though we can use local coordinates to expand $ nabla_{S^n}f $ and compute directly, I am still looking forward to a more elegant way to do this.
differential-geometry riemannian-geometry
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Suppose $ F $ is a homogeneous polynomial function of degree $ m $ on Euclidean space $ mathbb{R}^{n+1} $. Restrict $ F $ to the unit sphere $ S^n $ and we get a function on $ S^n $ denoted by $ f $, prove:
$(1)$ $ |nabla_{S^n}f|^2=|nabla_{mathbb{R}^{n+1}}F|^2-m^2f^2 $;
$(2)$ $ Delta_{S^n}f=Delta_{mathbb{R}^{n+1}}F-m(m-1)f-mnf .$
Where $ nabla_{S^n}, nabla_{mathbb{R}^{n+1}} $ are gradients on $ S^n $ and $ mathbb{R}^{n+1} $ respectively, $ Delta_{S^n}, Delta_{mathbb{R}^{n+1}} $ are Laplace operators on $ S^n $ and $ mathbb{R}^{n+1} $ respectively.
Hint: Use Euler's homogeneous function theorem $ langle (nabla_{mathbb{R}^{n+1}}F)_z, z rangle=mF(z) $.
The question above comes from my Riemannian geometry textbook. Can someone give me a hint about how to deal with $ nabla_{S^n}f $ ? I can't find any direct relation between $ nabla_{S^n}f $ and $ nabla_{mathbb{R}^{n+1}}F $. Though we can use local coordinates to expand $ nabla_{S^n}f $ and compute directly, I am still looking forward to a more elegant way to do this.
differential-geometry riemannian-geometry
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Suppose $ F $ is a homogeneous polynomial function of degree $ m $ on Euclidean space $ mathbb{R}^{n+1} $. Restrict $ F $ to the unit sphere $ S^n $ and we get a function on $ S^n $ denoted by $ f $, prove:
$(1)$ $ |nabla_{S^n}f|^2=|nabla_{mathbb{R}^{n+1}}F|^2-m^2f^2 $;
$(2)$ $ Delta_{S^n}f=Delta_{mathbb{R}^{n+1}}F-m(m-1)f-mnf .$
Where $ nabla_{S^n}, nabla_{mathbb{R}^{n+1}} $ are gradients on $ S^n $ and $ mathbb{R}^{n+1} $ respectively, $ Delta_{S^n}, Delta_{mathbb{R}^{n+1}} $ are Laplace operators on $ S^n $ and $ mathbb{R}^{n+1} $ respectively.
Hint: Use Euler's homogeneous function theorem $ langle (nabla_{mathbb{R}^{n+1}}F)_z, z rangle=mF(z) $.
The question above comes from my Riemannian geometry textbook. Can someone give me a hint about how to deal with $ nabla_{S^n}f $ ? I can't find any direct relation between $ nabla_{S^n}f $ and $ nabla_{mathbb{R}^{n+1}}F $. Though we can use local coordinates to expand $ nabla_{S^n}f $ and compute directly, I am still looking forward to a more elegant way to do this.
differential-geometry riemannian-geometry
Suppose $ F $ is a homogeneous polynomial function of degree $ m $ on Euclidean space $ mathbb{R}^{n+1} $. Restrict $ F $ to the unit sphere $ S^n $ and we get a function on $ S^n $ denoted by $ f $, prove:
$(1)$ $ |nabla_{S^n}f|^2=|nabla_{mathbb{R}^{n+1}}F|^2-m^2f^2 $;
$(2)$ $ Delta_{S^n}f=Delta_{mathbb{R}^{n+1}}F-m(m-1)f-mnf .$
Where $ nabla_{S^n}, nabla_{mathbb{R}^{n+1}} $ are gradients on $ S^n $ and $ mathbb{R}^{n+1} $ respectively, $ Delta_{S^n}, Delta_{mathbb{R}^{n+1}} $ are Laplace operators on $ S^n $ and $ mathbb{R}^{n+1} $ respectively.
Hint: Use Euler's homogeneous function theorem $ langle (nabla_{mathbb{R}^{n+1}}F)_z, z rangle=mF(z) $.
The question above comes from my Riemannian geometry textbook. Can someone give me a hint about how to deal with $ nabla_{S^n}f $ ? I can't find any direct relation between $ nabla_{S^n}f $ and $ nabla_{mathbb{R}^{n+1}}F $. Though we can use local coordinates to expand $ nabla_{S^n}f $ and compute directly, I am still looking forward to a more elegant way to do this.
differential-geometry riemannian-geometry
differential-geometry riemannian-geometry
asked Nov 21 '18 at 13:23
Philip
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Assume we have a Riemannian manifold $(M,g)$ and a submanifold $N subseteq M$. Denote by $h$ the induced Riemannian metric on $N$ (the pullback of $g$ under the inclusion map $i colon N hookrightarrow M$). Given a smooth function $F colon M rightarrow mathbb{R}$, we can consider $f = F|_{N}$ which is a smooth function on $N$. Given $p in N$, what is the relation between $(nabla_M F)(p)$ and $(nabla_N f)(p)$?
Denote by $P colon T_pM rightarrow T_p N$ the orthogonal projection onto $N$. Then $(nabla_N f)(p) = P((nabla_M F)(p))$ because
$$ df|_p(v) = d(F circ i)|_p(v) = dF|_p(di|_p(v)) = left< (nabla_M F)(p), di|_p(v) right>_g = left< di|_p left( P((nabla_M F)(p)) right), di|_p(v) right>_g = left< P((nabla_M F)(p)), v right>_h $$
for all $v in T_pN$.
In your case, $M = mathbb{R}^{n+1}$ with the standard Euclidean metric and $N = S^n$ has codimension one. In addition, given $p in S^n$, we have the orthogonal direct sum decomposition
$$ T_p(M) = T_p(S^{n}) oplus operatorname{span}_{mathbb{R}} { p } $$
where we think of $p$ both as a point in $mathbb{R}^{n+1}$ and a tangent vector at $T_p(mathbb{R}^{n+1})$. Hence,
$$ P(v) = v - left<v, p right>p $$
and so
$$ (nabla_{S^n} f)(p) = (nabla_{mathbb{R}^{n+1}} F)(p) - left< (nabla_{mathbb{R}^{n+1}} F)(p), p right> p, \
| (nabla_{mathbb{R}^{n+1}} F)(p) |^2 = | (nabla_{S^n} f)(p) |^2 + left| left< nabla_{mathbb{R}^{n+1}} F)(p), p right> right|^2 = | (nabla_{S^n} f)(p) |^2 + m^2 |f(p)|^2.$$
This handles the first part. I'll leave the second part to you.
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Assume we have a Riemannian manifold $(M,g)$ and a submanifold $N subseteq M$. Denote by $h$ the induced Riemannian metric on $N$ (the pullback of $g$ under the inclusion map $i colon N hookrightarrow M$). Given a smooth function $F colon M rightarrow mathbb{R}$, we can consider $f = F|_{N}$ which is a smooth function on $N$. Given $p in N$, what is the relation between $(nabla_M F)(p)$ and $(nabla_N f)(p)$?
Denote by $P colon T_pM rightarrow T_p N$ the orthogonal projection onto $N$. Then $(nabla_N f)(p) = P((nabla_M F)(p))$ because
$$ df|_p(v) = d(F circ i)|_p(v) = dF|_p(di|_p(v)) = left< (nabla_M F)(p), di|_p(v) right>_g = left< di|_p left( P((nabla_M F)(p)) right), di|_p(v) right>_g = left< P((nabla_M F)(p)), v right>_h $$
for all $v in T_pN$.
In your case, $M = mathbb{R}^{n+1}$ with the standard Euclidean metric and $N = S^n$ has codimension one. In addition, given $p in S^n$, we have the orthogonal direct sum decomposition
$$ T_p(M) = T_p(S^{n}) oplus operatorname{span}_{mathbb{R}} { p } $$
where we think of $p$ both as a point in $mathbb{R}^{n+1}$ and a tangent vector at $T_p(mathbb{R}^{n+1})$. Hence,
$$ P(v) = v - left<v, p right>p $$
and so
$$ (nabla_{S^n} f)(p) = (nabla_{mathbb{R}^{n+1}} F)(p) - left< (nabla_{mathbb{R}^{n+1}} F)(p), p right> p, \
| (nabla_{mathbb{R}^{n+1}} F)(p) |^2 = | (nabla_{S^n} f)(p) |^2 + left| left< nabla_{mathbb{R}^{n+1}} F)(p), p right> right|^2 = | (nabla_{S^n} f)(p) |^2 + m^2 |f(p)|^2.$$
This handles the first part. I'll leave the second part to you.
add a comment |
Assume we have a Riemannian manifold $(M,g)$ and a submanifold $N subseteq M$. Denote by $h$ the induced Riemannian metric on $N$ (the pullback of $g$ under the inclusion map $i colon N hookrightarrow M$). Given a smooth function $F colon M rightarrow mathbb{R}$, we can consider $f = F|_{N}$ which is a smooth function on $N$. Given $p in N$, what is the relation between $(nabla_M F)(p)$ and $(nabla_N f)(p)$?
Denote by $P colon T_pM rightarrow T_p N$ the orthogonal projection onto $N$. Then $(nabla_N f)(p) = P((nabla_M F)(p))$ because
$$ df|_p(v) = d(F circ i)|_p(v) = dF|_p(di|_p(v)) = left< (nabla_M F)(p), di|_p(v) right>_g = left< di|_p left( P((nabla_M F)(p)) right), di|_p(v) right>_g = left< P((nabla_M F)(p)), v right>_h $$
for all $v in T_pN$.
In your case, $M = mathbb{R}^{n+1}$ with the standard Euclidean metric and $N = S^n$ has codimension one. In addition, given $p in S^n$, we have the orthogonal direct sum decomposition
$$ T_p(M) = T_p(S^{n}) oplus operatorname{span}_{mathbb{R}} { p } $$
where we think of $p$ both as a point in $mathbb{R}^{n+1}$ and a tangent vector at $T_p(mathbb{R}^{n+1})$. Hence,
$$ P(v) = v - left<v, p right>p $$
and so
$$ (nabla_{S^n} f)(p) = (nabla_{mathbb{R}^{n+1}} F)(p) - left< (nabla_{mathbb{R}^{n+1}} F)(p), p right> p, \
| (nabla_{mathbb{R}^{n+1}} F)(p) |^2 = | (nabla_{S^n} f)(p) |^2 + left| left< nabla_{mathbb{R}^{n+1}} F)(p), p right> right|^2 = | (nabla_{S^n} f)(p) |^2 + m^2 |f(p)|^2.$$
This handles the first part. I'll leave the second part to you.
add a comment |
Assume we have a Riemannian manifold $(M,g)$ and a submanifold $N subseteq M$. Denote by $h$ the induced Riemannian metric on $N$ (the pullback of $g$ under the inclusion map $i colon N hookrightarrow M$). Given a smooth function $F colon M rightarrow mathbb{R}$, we can consider $f = F|_{N}$ which is a smooth function on $N$. Given $p in N$, what is the relation between $(nabla_M F)(p)$ and $(nabla_N f)(p)$?
Denote by $P colon T_pM rightarrow T_p N$ the orthogonal projection onto $N$. Then $(nabla_N f)(p) = P((nabla_M F)(p))$ because
$$ df|_p(v) = d(F circ i)|_p(v) = dF|_p(di|_p(v)) = left< (nabla_M F)(p), di|_p(v) right>_g = left< di|_p left( P((nabla_M F)(p)) right), di|_p(v) right>_g = left< P((nabla_M F)(p)), v right>_h $$
for all $v in T_pN$.
In your case, $M = mathbb{R}^{n+1}$ with the standard Euclidean metric and $N = S^n$ has codimension one. In addition, given $p in S^n$, we have the orthogonal direct sum decomposition
$$ T_p(M) = T_p(S^{n}) oplus operatorname{span}_{mathbb{R}} { p } $$
where we think of $p$ both as a point in $mathbb{R}^{n+1}$ and a tangent vector at $T_p(mathbb{R}^{n+1})$. Hence,
$$ P(v) = v - left<v, p right>p $$
and so
$$ (nabla_{S^n} f)(p) = (nabla_{mathbb{R}^{n+1}} F)(p) - left< (nabla_{mathbb{R}^{n+1}} F)(p), p right> p, \
| (nabla_{mathbb{R}^{n+1}} F)(p) |^2 = | (nabla_{S^n} f)(p) |^2 + left| left< nabla_{mathbb{R}^{n+1}} F)(p), p right> right|^2 = | (nabla_{S^n} f)(p) |^2 + m^2 |f(p)|^2.$$
This handles the first part. I'll leave the second part to you.
Assume we have a Riemannian manifold $(M,g)$ and a submanifold $N subseteq M$. Denote by $h$ the induced Riemannian metric on $N$ (the pullback of $g$ under the inclusion map $i colon N hookrightarrow M$). Given a smooth function $F colon M rightarrow mathbb{R}$, we can consider $f = F|_{N}$ which is a smooth function on $N$. Given $p in N$, what is the relation between $(nabla_M F)(p)$ and $(nabla_N f)(p)$?
Denote by $P colon T_pM rightarrow T_p N$ the orthogonal projection onto $N$. Then $(nabla_N f)(p) = P((nabla_M F)(p))$ because
$$ df|_p(v) = d(F circ i)|_p(v) = dF|_p(di|_p(v)) = left< (nabla_M F)(p), di|_p(v) right>_g = left< di|_p left( P((nabla_M F)(p)) right), di|_p(v) right>_g = left< P((nabla_M F)(p)), v right>_h $$
for all $v in T_pN$.
In your case, $M = mathbb{R}^{n+1}$ with the standard Euclidean metric and $N = S^n$ has codimension one. In addition, given $p in S^n$, we have the orthogonal direct sum decomposition
$$ T_p(M) = T_p(S^{n}) oplus operatorname{span}_{mathbb{R}} { p } $$
where we think of $p$ both as a point in $mathbb{R}^{n+1}$ and a tangent vector at $T_p(mathbb{R}^{n+1})$. Hence,
$$ P(v) = v - left<v, p right>p $$
and so
$$ (nabla_{S^n} f)(p) = (nabla_{mathbb{R}^{n+1}} F)(p) - left< (nabla_{mathbb{R}^{n+1}} F)(p), p right> p, \
| (nabla_{mathbb{R}^{n+1}} F)(p) |^2 = | (nabla_{S^n} f)(p) |^2 + left| left< nabla_{mathbb{R}^{n+1}} F)(p), p right> right|^2 = | (nabla_{S^n} f)(p) |^2 + m^2 |f(p)|^2.$$
This handles the first part. I'll leave the second part to you.
answered Nov 21 '18 at 15:22
levap
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