Trace-free part of the second fundamental form
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I found this notion on this part of a survey by S. Brendle:
What exactly is the definition (without using basis) of the “trace-free part of the second fundamental formâ€�
For me, the second fundamental form of a hypersurface $Sigma$ of $S^3$ with unit normal vector field $eta$ is, at a point $p in Sigma$, defined by
$$ A_p(v) = - nabla_v eta, quad v in T_p Sigma.$$
differential-geometry manifolds riemannian-geometry smooth-manifolds tensors
asked Aug 11 at 23:31
Eduardo Longa
1,5162518
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I found this notion on this part of a survey by S. Brendle:
What exactly is the definition (without using basis) of the “trace-free part of the second fundamental formâ€�
For me, the second fundamental form of a hypersurface $Sigma$ of $S^3$ with unit normal vector field $eta$ is, at a point $p in Sigma$, defined by
$$ A_p(v) = - nabla_v eta, quad v in T_p Sigma.$$
differential-geometry manifolds riemannian-geometry smooth-manifolds tensors
asked Aug 11 at 23:31
Eduardo Longa
1,5162518
4
I guess Brendle is using a nonstandard definition of mean curvature, i.e., the sum of the principal curvatures rather than half that. At any rate, given a linear map $T$ on a $2$-dimensional vector space, its trace-free part is $T-frac12texttr(T) I$. You can check this identity by taking the second fundamental form to be diagonal with entries $lambda_1,lambda_2$. Then both sides will be $1+frac14(lambda_1+lambda_2)^2$.
– Ted Shifrin
Aug 12 at 0:06
@TedShifrin You are right. Simon Brendle uses that $H=lambda_1+lambda_2.$ See page 2 arxiv.org/pdf/1307.6938.pdf
– mfl
Aug 12 at 0:13
add a comment |Â
up vote
0
down vote
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up vote
0
down vote
favorite
I found this notion on this part of a survey by S. Brendle:
What exactly is the definition (without using basis) of the “trace-free part of the second fundamental formâ€�
For me, the second fundamental form of a hypersurface $Sigma$ of $S^3$ with unit normal vector field $eta$ is, at a point $p in Sigma$, defined by
$$ A_p(v) = - nabla_v eta, quad v in T_p Sigma.$$
differential-geometry manifolds riemannian-geometry smooth-manifolds tensors
asked Aug 11 at 23:31
Eduardo Longa
1,5162518
I found this notion on this part of a survey by S. Brendle:
What exactly is the definition (without using basis) of the “trace-free part of the second fundamental formâ€�
For me, the second fundamental form of a hypersurface $Sigma$ of $S^3$ with unit normal vector field $eta$ is, at a point $p in Sigma$, defined by
$$ A_p(v) = - nabla_v eta, quad v in T_p Sigma.$$
differential-geometry manifolds riemannian-geometry smooth-manifolds tensors
asked Aug 11 at 23:31
Eduardo Longa
1,5162518
edited Aug 11 at 23:40
asked Aug 11 at 23:31
Eduardo Longa
1,5162518
asked Aug 11 at 23:31
Eduardo Longa
1,5162518
asked Aug 11 at 23:31
Eduardo Longa
1,5162518
1,5162518
4
I guess Brendle is using a nonstandard definition of mean curvature, i.e., the sum of the principal curvatures rather than half that. At any rate, given a linear map $T$ on a $2$-dimensional vector space, its trace-free part is $T-frac12texttr(T) I$. You can check this identity by taking the second fundamental form to be diagonal with entries $lambda_1,lambda_2$. Then both sides will be $1+frac14(lambda_1+lambda_2)^2$.
– Ted Shifrin
Aug 12 at 0:06
@TedShifrin You are right. Simon Brendle uses that $H=lambda_1+lambda_2.$ See page 2 arxiv.org/pdf/1307.6938.pdf
– mfl
Aug 12 at 0:13
add a comment |Â
4
I guess Brendle is using a nonstandard definition of mean curvature, i.e., the sum of the principal curvatures rather than half that. At any rate, given a linear map $T$ on a $2$-dimensional vector space, its trace-free part is $T-frac12texttr(T) I$. You can check this identity by taking the second fundamental form to be diagonal with entries $lambda_1,lambda_2$. Then both sides will be $1+frac14(lambda_1+lambda_2)^2$.
– Ted Shifrin
Aug 12 at 0:06
@TedShifrin You are right. Simon Brendle uses that $H=lambda_1+lambda_2.$ See page 2 arxiv.org/pdf/1307.6938.pdf
– mfl
Aug 12 at 0:13
4
4
I guess Brendle is using a nonstandard definition of mean curvature, i.e., the sum of the principal curvatures rather than half that. At any rate, given a linear map $T$ on a $2$-dimensional vector space, its trace-free part is $T-frac12texttr(T) I$. You can check this identity by taking the second fundamental form to be diagonal with entries $lambda_1,lambda_2$. Then both sides will be $1+frac14(lambda_1+lambda_2)^2$.
– Ted Shifrin
Aug 12 at 0:06
I guess Brendle is using a nonstandard definition of mean curvature, i.e., the sum of the principal curvatures rather than half that. At any rate, given a linear map $T$ on a $2$-dimensional vector space, its trace-free part is $T-frac12texttr(T) I$. You can check this identity by taking the second fundamental form to be diagonal with entries $lambda_1,lambda_2$. Then both sides will be $1+frac14(lambda_1+lambda_2)^2$.
– Ted Shifrin
Aug 12 at 0:06
@TedShifrin You are right. Simon Brendle uses that $H=lambda_1+lambda_2.$ See page 2 arxiv.org/pdf/1307.6938.pdf
– mfl
Aug 12 at 0:13
@TedShifrin You are right. Simon Brendle uses that $H=lambda_1+lambda_2.$ See page 2 arxiv.org/pdf/1307.6938.pdf
– mfl
Aug 12 at 0:13
add a comment |Â
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4
I guess Brendle is using a nonstandard definition of mean curvature, i.e., the sum of the principal curvatures rather than half that. At any rate, given a linear map $T$ on a $2$-dimensional vector space, its trace-free part is $T-frac12texttr(T) I$. You can check this identity by taking the second fundamental form to be diagonal with entries $lambda_1,lambda_2$. Then both sides will be $1+frac14(lambda_1+lambda_2)^2$.
– Ted Shifrin
Aug 12 at 0:06
@TedShifrin You are right. Simon Brendle uses that $H=lambda_1+lambda_2.$ See page 2 arxiv.org/pdf/1307.6938.pdf
– mfl
Aug 12 at 0:13