An identity in terms of the Laplace-Beltrami operator

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Let $Gamma:eta=(eta_1(x_1,x_2),eta_2(x_1,x_2),eta_3(x_1,x_2))$ be a smooth surface in $mathbbR^3$ with induced megetric $g=(g_alpha,beta)$, where $g_alpha, beta=partial _alphaetacdot partial _ betaeta $. And let $ sqrtg=mathopmathrmdetnolimits (g _alpha,beta) $. I would like to ask for help in proving
begingather
displaystyle -Delta _geta=mathcalHn,
endgather
where $ mathcalH $ is twice the mean curvature of the surface $ Gamma $, $ n:=fracpartial _1eta times partial _2eta $ is the unit normal to $ Gamma $, $ Delta _g $ is the Laplace–Beltrami operator on $ Gamma $ in terms of $ g $, which can be precisely given as follows
begingather
displaystyle Delta _g :=sqrtg^-1partial _alpha[sqrtg g ^alpha betapartial _beta], g ^alpha beta=(g _alpha beta)^-1~(mathrmthe inverse of the matrix (g _alpha beta)).
endgather

I hope someone would be kind enough to give me a hand.

differential-geometry

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asked Jul 20 at 18:27

user562244

111

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Let $Gamma:eta=(eta_1(x_1,x_2),eta_2(x_1,x_2),eta_3(x_1,x_2))$ be a smooth surface in $mathbbR^3$ with induced megetric $g=(g_alpha,beta)$, where $g_alpha, beta=partial _alphaetacdot partial _ betaeta $. And let $ sqrtg=mathopmathrmdetnolimits (g _alpha,beta) $. I would like to ask for help in proving
begingather
displaystyle -Delta _geta=mathcalHn,
endgather
where $ mathcalH $ is twice the mean curvature of the surface $ Gamma $, $ n:=fracpartial _1eta times partial _2eta $ is the unit normal to $ Gamma $, $ Delta _g $ is the Laplace–Beltrami operator on $ Gamma $ in terms of $ g $, which can be precisely given as follows
begingather
displaystyle Delta _g :=sqrtg^-1partial _alpha[sqrtg g ^alpha betapartial _beta], g ^alpha beta=(g _alpha beta)^-1~(mathrmthe inverse of the matrix (g _alpha beta)).
endgather

I hope someone would be kind enough to give me a hand.

differential-geometry

share|cite|improve this question

asked Jul 20 at 18:27

user562244

111

add a comment | 

up vote
2
down vote

favorite

up vote
2
down vote

favorite

Let $Gamma:eta=(eta_1(x_1,x_2),eta_2(x_1,x_2),eta_3(x_1,x_2))$ be a smooth surface in $mathbbR^3$ with induced megetric $g=(g_alpha,beta)$, where $g_alpha, beta=partial _alphaetacdot partial _ betaeta $. And let $ sqrtg=mathopmathrmdetnolimits (g _alpha,beta) $. I would like to ask for help in proving
begingather
displaystyle -Delta _geta=mathcalHn,
endgather
where $ mathcalH $ is twice the mean curvature of the surface $ Gamma $, $ n:=fracpartial _1eta times partial _2eta $ is the unit normal to $ Gamma $, $ Delta _g $ is the Laplace–Beltrami operator on $ Gamma $ in terms of $ g $, which can be precisely given as follows
begingather
displaystyle Delta _g :=sqrtg^-1partial _alpha[sqrtg g ^alpha betapartial _beta], g ^alpha beta=(g _alpha beta)^-1~(mathrmthe inverse of the matrix (g _alpha beta)).
endgather

I hope someone would be kind enough to give me a hand.

differential-geometry

share|cite|improve this question

asked Jul 20 at 18:27

user562244

111

Let $Gamma:eta=(eta_1(x_1,x_2),eta_2(x_1,x_2),eta_3(x_1,x_2))$ be a smooth surface in $mathbbR^3$ with induced megetric $g=(g_alpha,beta)$, where $g_alpha, beta=partial _alphaetacdot partial _ betaeta $. And let $ sqrtg=mathopmathrmdetnolimits (g _alpha,beta) $. I would like to ask for help in proving
begingather
displaystyle -Delta _geta=mathcalHn,
endgather
where $ mathcalH $ is twice the mean curvature of the surface $ Gamma $, $ n:=fracpartial _1eta times partial _2eta $ is the unit normal to $ Gamma $, $ Delta _g $ is the Laplace–Beltrami operator on $ Gamma $ in terms of $ g $, which can be precisely given as follows
begingather
displaystyle Delta _g :=sqrtg^-1partial _alpha[sqrtg g ^alpha betapartial _beta], g ^alpha beta=(g _alpha beta)^-1~(mathrmthe inverse of the matrix (g _alpha beta)).
endgather

I hope someone would be kind enough to give me a hand.

differential-geometry

share|cite|improve this question

asked Jul 20 at 18:27

user562244

111

share|cite|improve this question

share|cite|improve this question

asked Jul 20 at 18:27

user562244

111

asked Jul 20 at 18:27

user562244

111

asked Jul 20 at 18:27

user562244

111

111

add a comment | 

add a comment | 

1 Answer
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Recall that the Hessian of a function $f$ on a Riemannian manifold $(M,g)$ is the covariant derivative of $df$ with respect to $nabla$, the Levi-Civita connection of $g$. Explicitly, for $X,Yin mathscrX(M)$ we have $$nabla^2f(X,Y)=X(df(Y))-df(nabla_XY).$$ Here $f$ could be multi-valued.

In your case, $eta$ is an embedding of a surface into $mathbbR^3$ and in particular a multi-valued function on the surface. The differential $deta$ carries a vector tangent to the surface to its realization in $mathbbR^3$. Let $X$ and $Y$ be vector fields on the surface. Note that covariant derivation in $mathbbR^3$ is the usual differentiation. Hence, the above equation shows that, by definition, we have $$II(X,Y)=nabla^2eta(X,Y),$$where $II$ is my pour notation for the second fundamental form. In other words, the Hessian of $eta$ coincides with the second fundamental form. To obtain the desired equality, take the traces of both sides.

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answered Jul 20 at 22:01

Amitai Yuval

14.4k11026

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hi @Amitai, thank you for your answer. I am not that familiar with Riemann geometry, so would you please kindly recommend several readable references to me? And by the way, if the surface is given by $F(x_1,x_2,x_3)=0$, and $n=fracnabla Fnabla F$, can we get a similar identity? And what is the precise form? Thank you very much.
  – user562244
  Jul 20 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user562244 I can't think of any textbook in particular, but for basic terms such as Hessian and second fundamental form, I guess most of the textbooks will do. As for your other question, it seems to me more appropriate to post it separately.
  – Amitai Yuval
  Jul 20 at 23:01
  

  
  
  
  

add a comment | 

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1 Answer
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1 Answer
1

active

oldest

votes

active

oldest

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up vote
0
down vote

Recall that the Hessian of a function $f$ on a Riemannian manifold $(M,g)$ is the covariant derivative of $df$ with respect to $nabla$, the Levi-Civita connection of $g$. Explicitly, for $X,Yin mathscrX(M)$ we have $$nabla^2f(X,Y)=X(df(Y))-df(nabla_XY).$$ Here $f$ could be multi-valued.

In your case, $eta$ is an embedding of a surface into $mathbbR^3$ and in particular a multi-valued function on the surface. The differential $deta$ carries a vector tangent to the surface to its realization in $mathbbR^3$. Let $X$ and $Y$ be vector fields on the surface. Note that covariant derivation in $mathbbR^3$ is the usual differentiation. Hence, the above equation shows that, by definition, we have $$II(X,Y)=nabla^2eta(X,Y),$$where $II$ is my pour notation for the second fundamental form. In other words, the Hessian of $eta$ coincides with the second fundamental form. To obtain the desired equality, take the traces of both sides.

share|cite|improve this answer

answered Jul 20 at 22:01

Amitai Yuval

14.4k11026

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hi @Amitai, thank you for your answer. I am not that familiar with Riemann geometry, so would you please kindly recommend several readable references to me? And by the way, if the surface is given by $F(x_1,x_2,x_3)=0$, and $n=fracnabla Fnabla F$, can we get a similar identity? And what is the precise form? Thank you very much.
  – user562244
  Jul 20 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user562244 I can't think of any textbook in particular, but for basic terms such as Hessian and second fundamental form, I guess most of the textbooks will do. As for your other question, it seems to me more appropriate to post it separately.
  – Amitai Yuval
  Jul 20 at 23:01
  

  
  
  
  

add a comment | 

up vote
0
down vote

Recall that the Hessian of a function $f$ on a Riemannian manifold $(M,g)$ is the covariant derivative of $df$ with respect to $nabla$, the Levi-Civita connection of $g$. Explicitly, for $X,Yin mathscrX(M)$ we have $$nabla^2f(X,Y)=X(df(Y))-df(nabla_XY).$$ Here $f$ could be multi-valued.

In your case, $eta$ is an embedding of a surface into $mathbbR^3$ and in particular a multi-valued function on the surface. The differential $deta$ carries a vector tangent to the surface to its realization in $mathbbR^3$. Let $X$ and $Y$ be vector fields on the surface. Note that covariant derivation in $mathbbR^3$ is the usual differentiation. Hence, the above equation shows that, by definition, we have $$II(X,Y)=nabla^2eta(X,Y),$$where $II$ is my pour notation for the second fundamental form. In other words, the Hessian of $eta$ coincides with the second fundamental form. To obtain the desired equality, take the traces of both sides.

share|cite|improve this answer

answered Jul 20 at 22:01

Amitai Yuval

14.4k11026

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hi @Amitai, thank you for your answer. I am not that familiar with Riemann geometry, so would you please kindly recommend several readable references to me? And by the way, if the surface is given by $F(x_1,x_2,x_3)=0$, and $n=fracnabla Fnabla F$, can we get a similar identity? And what is the precise form? Thank you very much.
  – user562244
  Jul 20 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user562244 I can't think of any textbook in particular, but for basic terms such as Hessian and second fundamental form, I guess most of the textbooks will do. As for your other question, it seems to me more appropriate to post it separately.
  – Amitai Yuval
  Jul 20 at 23:01
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

Recall that the Hessian of a function $f$ on a Riemannian manifold $(M,g)$ is the covariant derivative of $df$ with respect to $nabla$, the Levi-Civita connection of $g$. Explicitly, for $X,Yin mathscrX(M)$ we have $$nabla^2f(X,Y)=X(df(Y))-df(nabla_XY).$$ Here $f$ could be multi-valued.

In your case, $eta$ is an embedding of a surface into $mathbbR^3$ and in particular a multi-valued function on the surface. The differential $deta$ carries a vector tangent to the surface to its realization in $mathbbR^3$. Let $X$ and $Y$ be vector fields on the surface. Note that covariant derivation in $mathbbR^3$ is the usual differentiation. Hence, the above equation shows that, by definition, we have $$II(X,Y)=nabla^2eta(X,Y),$$where $II$ is my pour notation for the second fundamental form. In other words, the Hessian of $eta$ coincides with the second fundamental form. To obtain the desired equality, take the traces of both sides.

share|cite|improve this answer

answered Jul 20 at 22:01

Amitai Yuval

14.4k11026

Recall that the Hessian of a function $f$ on a Riemannian manifold $(M,g)$ is the covariant derivative of $df$ with respect to $nabla$, the Levi-Civita connection of $g$. Explicitly, for $X,Yin mathscrX(M)$ we have $$nabla^2f(X,Y)=X(df(Y))-df(nabla_XY).$$ Here $f$ could be multi-valued.

In your case, $eta$ is an embedding of a surface into $mathbbR^3$ and in particular a multi-valued function on the surface. The differential $deta$ carries a vector tangent to the surface to its realization in $mathbbR^3$. Let $X$ and $Y$ be vector fields on the surface. Note that covariant derivation in $mathbbR^3$ is the usual differentiation. Hence, the above equation shows that, by definition, we have $$II(X,Y)=nabla^2eta(X,Y),$$where $II$ is my pour notation for the second fundamental form. In other words, the Hessian of $eta$ coincides with the second fundamental form. To obtain the desired equality, take the traces of both sides.

share|cite|improve this answer

answered Jul 20 at 22:01

Amitai Yuval

14.4k11026

share|cite|improve this answer

share|cite|improve this answer

answered Jul 20 at 22:01

Amitai Yuval

14.4k11026

answered Jul 20 at 22:01

Amitai Yuval

14.4k11026

answered Jul 20 at 22:01

Amitai Yuval

14.4k11026

14.4k11026

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hi @Amitai, thank you for your answer. I am not that familiar with Riemann geometry, so would you please kindly recommend several readable references to me? And by the way, if the surface is given by $F(x_1,x_2,x_3)=0$, and $n=fracnabla Fnabla F$, can we get a similar identity? And what is the precise form? Thank you very much.
  – user562244
  Jul 20 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user562244 I can't think of any textbook in particular, but for basic terms such as Hessian and second fundamental form, I guess most of the textbooks will do. As for your other question, it seems to me more appropriate to post it separately.
  – Amitai Yuval
  Jul 20 at 23:01
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Hi @Amitai, thank you for your answer. I am not that familiar with Riemann geometry, so would you please kindly recommend several readable references to me? And by the way, if the surface is given by $F(x_1,x_2,x_3)=0$, and $n=fracnabla Fnabla F$, can we get a similar identity? And what is the precise form? Thank you very much.
  – user562244
  Jul 20 at 22:30
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  @user562244 I can't think of any textbook in particular, but for basic terms such as Hessian and second fundamental form, I guess most of the textbooks will do. As for your other question, it seems to me more appropriate to post it separately.
  – Amitai Yuval
  Jul 20 at 23:01
  

  
  
  
  

Hi @Amitai, thank you for your answer. I am not that familiar with Riemann geometry, so would you please kindly recommend several readable references to me? And by the way, if the surface is given by $F(x_1,x_2,x_3)=0$, and $n=fracnabla Fnabla F$, can we get a similar identity? And what is the precise form? Thank you very much.
– user562244
Jul 20 at 22:30

Hi @Amitai, thank you for your answer. I am not that familiar with Riemann geometry, so would you please kindly recommend several readable references to me? And by the way, if the surface is given by $F(x_1,x_2,x_3)=0$, and $n=fracnabla Fnabla F$, can we get a similar identity? And what is the precise form? Thank you very much.
– user562244
Jul 20 at 22:30

@user562244 I can't think of any textbook in particular, but for basic terms such as Hessian and second fundamental form, I guess most of the textbooks will do. As for your other question, it seems to me more appropriate to post it separately.
– Amitai Yuval
Jul 20 at 23:01

@user562244 I can't think of any textbook in particular, but for basic terms such as Hessian and second fundamental form, I guess most of the textbooks will do. As for your other question, it seems to me more appropriate to post it separately.
– Amitai Yuval
Jul 20 at 23:01

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