Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I had referred to this structure earlier in a previous question which went unanswered.

If $u$ and $v$ are in $T_pM$ and $n \in (T_pM)^\perp$ then the extrinsic curvature form $K$ be defined as,

$K(u,v) = g(n,\nabla _u V)$

(where $V$ is a local extension of the vector $v$)

Similarly use say $U$ as a local extension of $u$ for defining $K(v,u)$

Then one claims that the following equalities hold,

$K(u,v) = K(v,u) = -g(v,\nabla _u N)$

(where $N$ is a local extension of $n$)

The second equality can hold if one has metric compatible connection and $U(g(N,V)) = V(g(N,U)) = 0$.

  • I am wondering if $n$ being orthogonal to $u$ and $v$ at $p$ is enough to guarantee the above.

  • I am unable to understand how the symmetric nature of $K$ can be proven. It doesn't seem to follow just by assuming that the connection is Riemann-Christoffel.

share|improve this question
add comment

1 Answer 1

This works only if $\nabla$ is the Levi-Civita connection of the Riemannian structure $g$. The Levi-Civita connection has the property that it is torsion-free, in addition to being compatible with the metric. Torsion-freeness means $\nabla_U V-\nabla_V U=[U,V]$, and $[U,V]$ is tangential to $M$ and thus orthogonal to $n$, which proves the symmetry of $K$.

Edit: If $\nabla$ is not the Levi-Civita connection, your $K$ will not even be a tensor in general (that is, $g(n,\nabla_u V)$ will depend on the choice of the "local extension" $V$ of $v$).

share|improve this answer
    
Thanks for your reply. I see your point but I don't understand how to argue that the local extensions will retain the point wise orthogonality given that it held at the point where the initial vectors was defined. (similar to the first part of my question) –  Anirbit Jan 18 '11 at 8:59
    
I don't understand what you mean by "the local extensions will retain the point wise orthogonality given that it held at the point where the initial vectors was defined". Orthogonality between which vectors and why? $u$ and $v$ are nowhere required to be orthogonal, and $n$ is orthogonal to both by design. –  Florian Jan 18 '11 at 13:24
    
I mean that how do I argue that if $n$ is orthogonal to $u$ and $v$ then $N$ is point wise orthogonal to $U$ and $V$? Also can you elaborate on why if the connection is not Levi-Civita then $K$ will not be a tensor. I don't have much experience with arguments about vector field extensions of vectors. –  Anirbit Jan 18 '11 at 15:03
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.