Homogeneous polynomial on unit sphere.












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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.










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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.










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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.










      share|cite|improve this question














      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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      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.






            share|cite|improve this answer


























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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.






              share|cite|improve this answer
























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                1








                1






                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.






                share|cite|improve this answer












                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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                answered Nov 21 '18 at 15:22









                levap

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