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I'm interested in the derivation of the Gauss equation (Gauss-Codazzi). Usually we consider the definition of the Riemann tensor on the hypersurface.


where $D$ is the connexion associated to the induced metric ($h$) on the hypersurface.

We have

\begin{align} D_aD_bw_c=D_a\Bigl[h_b^{~d} h_c^{~e}\nabla_d w_e\Bigr]=h_a^{~f}h_b^{~g}h_c^{~e}\nabla_f\Bigl[h_g^{~d}h_k^{~e}\nabla_dw_e\Bigr] \end{align}

where $\nabla$ is the connexion of the metric ($g$) in the full space.

Hence after some algebra we can arrive at the desired result. What I don't understand is why can we not do:

$$ D_aD_bw_c=D_a\Bigl[h_b^{~d} h_c^{~e}\nabla_d w_e\Bigr]=h_b^{~d} h_c^{~e}D_a\Bigl[\nabla_d w_e\Bigr] $$

because $D_a h^b_c=0$.

But in that case I would have

$$ D_aD_bw_c=h_b^{~d} h_c^{~e}D_a\Bigl[\nabla_d w_e\Bigr]=h_b^{~d} h_c^{~e}h_a^{~\mu}h_d^{~\nu}h_e^{~\sigma}\nabla_\mu\Bigl[\nabla_\nu w_\sigma\Bigr]=h_a^{~\mu}h_b^{~\nu}h_c^{~\sigma} \nabla_\mu\nabla_\nu w_\sigma $$

Therefore we would have

$$^{(n-1)}R_{abc}^{~~~~~~~d}~w_d=[D_a,D_b]w_c=h_a^{~\mu}h_b^{~\nu}h_c^{~\sigma} ~^{(n)}R_{\mu\nu\sigma}^{~~~~~~~~d}w_d$$

which is not the desired result.

So I understand the standard derivation, but I don't understand why in the way that I wrote, it doesn't work ?

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You need to distinguish between the manifold indices and the submanifold indices. The submanifold tangent space is different from the manifold tangent space, and the way it sits in the manifold tangent space is changing from place to place. The first formula is finding this change, and it is different in character from taking the intrinsic covariant derivative of the induced metric in the induced manifold--- that's not what the first formula means. The first formula isn't in the induced manifold. You can work it out explicitly for a 2 sphere in 3 space, where the second derivation gives R=0. – Ron Maimon Apr 14 '12 at 8:40
Ohhhhhh I see. When in my second equation I wrote $h_b^d$, I should write it as $h_b^\delta$ where $b$ runs on the hypersurface and $\delta$ on the manifold. So it's not a 3-tensor or a tensor, I guess that I had some confusion between the induced metric and the projection tensor. – anubis Apr 14 '12 at 10:05
Are you satisfied? I was going to write a whole answer, but if this answers your confusion, then I won't bother. – Ron Maimon Apr 14 '12 at 15:14
I'd suggest just writing down $\mathbb{R}^{3}$'s metric in spherical coordinates, and calculating both tensors where the induced metric is that of a sphere with constant $r$. – Jerry Schirmer Apr 14 '12 at 21:53
I understood thanks. – anubis Apr 16 '12 at 2:07

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