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Dec 16, 2019 at 21:47 comment added mr_e_man I think you could extend $\phi$ and $N$ to a neighbourhood of $f(M)$; take the ordinary 3D gradient: $v=\nabla\phi$; project it onto the surface: $w=v-(v\cdot N)N$; take the 3D divergence: $\Delta\phi=\nabla\cdot w$; and restrict this to $f(M)$.
Oct 25, 2012 at 20:07 vote accept TerronaBell
Oct 22, 2012 at 21:44 comment added TerronaBell Interestingly enough, (df(X) \cdot dN(Y)) is already symmetric -- i.e., you don't even need to symmetrize.
Oct 22, 2012 at 1:19 comment added Robert Bryant @fuzzytron: Actually, the second fundamental form is easy in these terms: For any pair of tangent vectors $X,Y$, one has $$ II(X,Y) = =-\tfrac12\bigl(df(X)\cdot dN(Y) + df(Y)\cdot dN(X)\bigr) $$ @Kofi: How you are going to use $II$ to get the Levi-Civita connection? While I know how to do it, I don't think it's straightforward.
Oct 21, 2012 at 19:44 answer added Rbega timeline score: 4
Oct 21, 2012 at 18:37 answer added Robert Bryant timeline score: 4
Oct 21, 2012 at 18:03 comment added TerronaBell Thanks Kofi. The tricky part there is that you then need an expression for the second fundamental form that satisfies the criteria above. In other words, imagine that you start with just $f$, $N$, and the differential. How can you build up all the objects you need to define $\Delta$ without appealing to coordinates or exterior calculus? (Also note that something like "$\Delta = \mathrm{div} \circ \mathrm{grad}$" doesn't satisfy the first criterion above.)
Oct 21, 2012 at 17:57 comment added Matthias Ludewig You could use the formula $\Delta = \nabla^* \nabla$, where $\nabla$ is the Levi-Civita connection. The Levi-Civita connection can in turn be represented via the second fundamental form.
Oct 21, 2012 at 17:42 history edited TerronaBell CC BY-SA 3.0
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Oct 21, 2012 at 17:36 history asked TerronaBell CC BY-SA 3.0