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I often hear that the coefficients of the 1st FF are the 1st FF. But the I also hear that the dot product is the 1st FF. But make sense to me but...which one is which? If the dot product is the 1st FF, then how come people say that Gaussian curvature can be written in terms of the 1st FF? This isn’t true since it cannot. It can be written in terms of the coefficients of the 1st FF. Is this just a colloquialism then?

Spencer Kraisler
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Given a surface $S \subseteq \mathbb{R}^3$, the first fundamental form is the inner product on the tangent space (at each point) on the surface. The first fundamental form doesn't have "coefficients" by itself. Once you choose some parametrization which covers some part of the surface, you get an induced basis for the tangent space and then you can represent the first fundamental form by a $2 \times 2$ matrix which has coefficients (the "famous" $E,F,G$). Different parametrizations (even if they cover the same part of the surface) will give you different matrices and different coefficients.

If you know the parametrization and the matrix representing the first fundamental form, you can reconstruct the first fundamental form itself on the image of the parametrization so sometimes people identify the first fundamental form (restricted to the image) with its coefficients in some (implicitly understood) parametrization.

Finally, your question about the curvature is a good one. The easiest way to introduce the curvature is to do it locally via the coefficients and not via the inner product itself (which is a global thing defined on the whole surface). Then it is not so clear why the curvature depends on the inner product and not on the coefficients. However, to show that the curvature is a notion which is defined globally on the whole surface, one actually needs to verify that it is given by the "same" formula when you use different parametrizations and doing so implicitly shows that the curvature actually depends on the inner product.

More sophisticated approaches derive the curvature globally directly from the inner product (via the covariant derivative which by itself is defined in terms of the inner product) and so make it clear that the curvature can be constructed from the inner product without even talking about the coefficients.

levap
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  • Fascinating. So indeed curvature can be written in terms of the 1st FF. I have yet to see such an expression in regards to regular surfaces. I have seen things like the Brioschi formula, but as before this is just written in terms of the coefficients, not the 1st FF itself. Do you know how to describe K in terms of the 1st FF? It seems wherever I look on the internet, I can't find such an expression! Just people saying it can be written. If you show me how to write K in terms of the 1st FF, I would greatly appreciate it! – Spencer Kraisler Apr 14 '20 at 23:00
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    @SpencerKraisler: What I have in mind is called the Riemann curvature tensor (https://en.wikipedia.org/wiki/Riemann_curvature_tensor) which is a generalization of the Gaussian curvature from surfaces to Riemannian manifolds. In the particular case of a surface, after performing some identifications it becomes the Gaussian curvature and it is defined using the covariant derivative (the Levi-Civita connection) which itself is defined using the inner product. – levap Apr 14 '20 at 23:08
  • Thank you for your answer. So I have one more question. I guess to me, we aren't exactly writing things in terms of the 1st FF. We're writing things in terms of the 1st FF AND the chosen parameterization. Could you possibly explain to me why its just the 1st FF we're writing things in terms in? – Spencer Kraisler Apr 16 '20 at 21:30
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    @SpencerKraisler: I think this is more of a meta-mathematical question. Let me give you an analogy. Let's say you want to define the determinant of an linear operator $T \colon V \rightarrow V$ (where $V$ is finite dimensional). The "easiest" definition is to choose some basis for $V$, use it both for the domain and range and represent $T$ by a matrix $A$ and then define $\det(T) = \det(A)$. Then one checks that it is "well-defined" so that if you choose a different basis, you get a different matrix $B$ but still $\det(A) = \det(B)$. Great, so that's well-defined but is really the – levap Apr 17 '20 at 18:09
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    "determinant of $T$" depends only on $T$? I convinced you that it doesn't depend on the matrix representing $T$, but we still choose some auxiliary data (a basis), did something and extracted a number. Is this really a "property of $T$" and not of our process? For the determinant of $T$, you can also give a more complicated definition which doesn't depend on choosing a basis and since it doesn't "depend on anything", it's pretty clear that it it is a property of $T$ and not of anything else. The same goes for the curvature or other things you write in terms of $E,F,G$. – levap Apr 17 '20 at 18:16
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    I can write down a very complicated formula giving the curvature of $S$ at a point using a local parametrization, $E,F,G$ and their derivatives. Then I need to convince you that this formula doesn't depend on the parametrization so that if you use different parametrization and $E',F',G'$, the resulting number will be the same. But will this convince you that it actually depends on the first fundamental form and not on "anything else"? I'm not sure it will, unless you see a construction which will not involve $E,F,G$ at all but only the first fundamental form (just like defining $\det T$ – levap Apr 17 '20 at 18:18
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    without ever going through a matrix $A$). – levap Apr 17 '20 at 18:18
  • You clarified a lot of confusion for me, thank you! :) I have one more question, but I may just be falling to analogical fallacy. You defined the determinant in a way that required choosing some arbitrary basis. However, I could define the determinant as being the product of the eigenvalues. This requires no basis (I think). Is there a way to we can define these intrinsic properties without any parameterization? Not just Gaussian curvature, but also, say, length of a curve, angle of two tangent vectors, etc.? – Spencer Kraisler Apr 17 '20 at 18:47
  • You could define the determinant as the product of the eigenvalues but an operator might not have "enough" eigenvalues in the field (i.e a non-trivial rotation in the plane doesn't have any (real) eigenvalues) so you need to move to the algebraic closure but then maybe the definition depends on the algebraic closure? And yes, you can define everything without any parametrizations. You don't need $E,F,G$ to define the angle of two tangent vectors or the length of curves, you can just use the first fundamental form. It it also possible define the curvature without – levap Apr 18 '20 at 19:16
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    referring to $E,F,G$, only the first fundamental form, only it is much more complicated. – levap Apr 18 '20 at 19:18