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Suppose I have a 2D surface $S=(x,y,f(x,y))$ embedded in a Riemannian manifold $(M,g)=(\mathbb{R}^3,g)$, i.e. in a curved 3-space.

Then the induced metric tensor $h$ on $f$ is $ h_{ij} = g_{ab}\, \partial_i S^a \partial_j S^b $ in local coordinates.

Now I want the surface normals. Normally one takes the cross-product of the tangent vectors. What is the Riemannian analogue here?

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    $\begingroup$ It's a bad idea to use the letter $h$, as in this context it often refers to the second fundamental form of the submanifold. You want a vector that is $g$-orthogonal to both vectors $\partial_i S^a$, $i=1,2$, so you want the kernel of the $2\times 3$ matrix $\sum_a g_{ab}\partial_i S^a$ ($i=1,2$, $b=1,2,3$), so a cross-product of $\sum_a g_{ab}\partial_1S^a$ and $\sum_a g_{ab}\partial_2 S^a$ will do it. $\endgroup$ – Ted Shifrin Jul 17 '17 at 17:19
  • $\begingroup$ @TedShifrin Thank you Prof. Shifrin, that absolutely makes sense. If you don't mind me asking, I'm interested in the flux of a vector field $v$ through $S$; should I use $\iint v\cdot n dA=\iint g_{ij} v^i n^j dx dy$? I'm just not sure if $n=(g_{ab}\partial_1 S^a)\times(g_{ab}\partial_2 S^a)$ is contravariant. It seems to usually be written as $\iint \sqrt{|\tilde{g}|} v^i n_i dx dy $, where $\tilde{g}$ is the induced metric. $\endgroup$ – user3658307 Jul 17 '17 at 19:23
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    $\begingroup$ Right. This is the usual issue that the cross-product of vectors is a pseudovector (because it is hiding under the guise of the wedge product of two covectors). The induced metric here is giving the area 2-form on the surface as $\sqrt{|\tilde g|}dx\wedge dy$. Note that $n_i = \sum g_{ij}n^j$, so everything is consistent. $\endgroup$ – Ted Shifrin Jul 17 '17 at 19:33
  • $\begingroup$ @TedShifrin Ah, I see! I think that $n$ should be a unit vector in the latter expression I wrote above. This gives $\iint \sqrt{ | \tilde{g} | } g_{ik} (n^k / ||n||) v^i dx\,dy $, where $ ||n|| = \sqrt{g_{ab} n^a n^b} $. (So in $\mathbb{E}^3$, $g=I$ and $||n||=\sqrt{ | \tilde{g} | }$). Does that seem correct? Thanks again! $\endgroup$ – user3658307 Jul 17 '17 at 23:25
  • $\begingroup$ Yes, of course, in the flux computation, $n$ needs to be a unit vector. $\endgroup$ – Ted Shifrin Jul 18 '17 at 0:09

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