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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.

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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 $v$).

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    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.2011-01-18