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Let $N$ and $M$ be $n$-dimensional and $n+1$-dimensional Riemannian manifolds and let $F : N \to M$ be an isometric immersion. What is the standard definition of the Laplace-Beltrami operator? $$ \Delta F = ? $$ I am interested in this particular case where the co-dimension is 1, but is there a more general definition?

Thanks!

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This is known as the tension field or harmonic map Laplacian. As you'd expect, it's just $\mathrm{tr_g} (\nabla^2F)$, but we need to work out how to rigorously define this covariant second derivative of a map between manifolds.

The derivative $DF$ can be viewed as a section of the bundle $E = T^*N \otimes F^*TM$. Pulling back the Levi-Civita connection of $M$ to a connection on $F^* TM$ and then using the usual extension of connections to tensor bundles, we get a natural connection on $E$ which allows us to define $\nabla^2 F = \nabla D F$ as a section of $T^* N \otimes E = T^* N \otimes T^* N \otimes F^* TM.$ Taking the trace with respect to the metric of $N$ yields $\Delta F \in \Gamma(F^* TM)$; i.e. a vector field along $F$.

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    Thank you very much for your answer! I am still a bit confused. According to your definition I have that for every $V, W \in \Gamma(TN)$: \begin{align}(\nabla^E DF)_V(W) &= \nabla^{F^*TM}_V( DF (W)) - F^*(DF (\nabla^N_V DF(W))) \\ &= F^*(\nabla^M_{DF(V)}DF(W)) - F^*(DF (\nabla^N_V DF(W)))\end{align} right? Then taking the trace w.r.t. the metric of N means that if $E_1, ..., E_n$ is a orthonormal frame of $N$, then $$ \Delta F = \sum_{i=1}^n (\nabla^E DF)_{E_i}E_i. $$ Is this correct? – 2017-02-24
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    Yeah, that looks right (though all the explicit pullbacks makes my brain hurt). In coordinates we get $\Delta F^\alpha = g_N^{ij} \left( \partial_i \partial_j F^\alpha +\Gamma(M)^\alpha_{\beta\gamma} \partial_i F^\beta \partial_j F^\gamma - \Gamma(N)_{ij}^k \partial_k F^\alpha\right)$ - note the first two terms come out of your first term $F^*(\nabla_{DF(V)} DF(W))$. – 2017-02-24
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    Yes, it took me a while to understand all the formalities. :D Thank you very much! I need this because I just started studying the mean curvature flow. :) – 2017-02-24
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    Actually there is a mistake in what I wrote.. There should be $F^*(DF(\nabla^N_VW))$ instead of $F^*(DF(\nabla^N_V DF(W)))$.. – 2017-02-24