As of recent, I've been studying Differential Geometry per the Dover Publication on the subject, and I've ran into a bit of an issue with tangent vectors to a parametric surface $ \mathbf{x}(u^1,u^2) $ at a point $ \mathbf{P}\ $ in $\mathbb{R}^3$ with the standard right handed Cartesian basis.

The book was asserting that the tangent vectors

$$ \mathbf{x_\alpha } \equiv \frac{\partial \mathbf{x} }{\partial u^\alpha }\ $$

are orthogonal at all regular (non-singular) points of the surface. I was working through an example ( I don't have the particular example, but the it's immaterial ), and I found that for the particular example:

$$ g_{\alpha\beta} \neq 0 \: \: , \: \: \alpha \neq \beta $$

Where $ g_{\alpha\beta} $ is the metric tensor, thus implying that the tangent vectors at P are not orthogonal. The surface in question was nothing special, and was regular at all points on the surface. I double checked my calculations, and everything seemed to be ok, but of course to err is human.

The question that I have then--assuming my calculations are incorrect--( I'll have to check more later ), and that the book is correct in it's assertion is: What forces the tangent vectors at a point on a surface to be orthogonal?

I haven't gotten much into proofing, but can follow proofs for the most part, and my best conjecture is that they're orthogonal because each parameter $ \: u^\alpha $ is really $ u^\alpha(x_1,x_2,x_3) $, and since $$ x_\alpha x^\beta= {\delta_{\alpha}}^{\beta} $$ ie all $ \: x_\alpha $ are orthogonal, the same must be true for the parameters. Seeing as I lack a proof in this, I don't want to just take that as unfounded fact. Please do correct any notational errors if any are found. As stated, I've just began learning the course, and am applying what I've learned thus far in asking this question, otherwise can anyone shed a bit of light on the matter?

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    $\begingroup$ I suspect there's some missing information here. There's no reason for tangent vectors to be orthogonal for a general case of arbitrary coordinates. Given two tangent vector fields, you can construct a third that is orthogonal to one or the other. There should be a coordinate associated with such a constructed field, which I'm guessing can be recovered through integration. $\endgroup$ – Muphrid Apr 28 '13 at 6:46
  • $\begingroup$ Dear Doryan, do not refer to books by publisher but by author and title: Dover has published many books on differential geometry and anyway it would be absurd for you to demand that a user consult their catalogue in order to find your book. $\endgroup$ – Georges Elencwajg Apr 28 '13 at 7:38
  • $\begingroup$ By the way, it does not make sense to say that "the tangent vectors at $P$ are not orthogonal": there are infinitely many such tangent vectors and some pairs are orthogonal and some are not. The $g_{\alpha, \beta}$ correspond to the choice of some local basis for the tangent vectors, which have no reason to be mutually orthogonal. $\endgroup$ – Georges Elencwajg Apr 28 '13 at 7:47
  • $\begingroup$ I agree with Muphrid: the coordinate tangent vectors need not be orthogonal in general. A simple example is given by $\mathbf{x}(u^1, u^2) = (u^1 + u^2, u^2, 0)$. Perhaps you're missing some additional assumptions that the authors make? $\endgroup$ – Jesse Madnick Apr 28 '13 at 7:48
  • $\begingroup$ Sorry about that Georges. Didn't mean for that to imply that someone needs to check their catologue for the particular book. Didn't have the book on me at the time, and had very limited resources as to acquiring the title and the author. I have the book on me now, and It's called Differential Geometry by Erwin Kreyszig. The section in question begins on page 101 ( Section 32 ) if anyone has the book. Further inspection of that particular section leads to believe I may have just misread it. But I was referring to the basis for the tangent space at a point P, which after reading the section $\endgroup$ – Doryan Miller Apr 28 '13 at 20:17

After further study, and since everything is determined by the metric, and thus the geometric surface, said manifolds do not come equipped with the requirement that the tangent bundle needs to set up an orthogonal basis. This would be a further limitation that we would have to create ourselves.


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