# Why these particular numerical factors in the definition of Gaussian curvature?

Wikipedia tells me that:

Gaussian curvature is the limiting difference between the circumference of a geodesic circle and a circle in the plane:

$K = \lim_{r \rightarrow 0} (2 \pi r - \mbox{C}(r)) \cdot \frac{3}{\pi r^3}$

Gaussian curvature is the limiting difference between the area of a geodesic circle and a circle in the plane:

$K = \lim_{r \rightarrow 0} (\pi r^2 - \mbox{A}(r)) \cdot \frac{12}{\pi r^4}$

Can anyone explain to me why we are dividing by the factors $\frac{3}{\pi r^3}$ and $\frac{12}{\pi r^4}$ respectively? I don't understand why we are dividing by these particular factors?

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First, I guess it should say "geodesic disc" rather than "circle". At least to me, a geodesic circle is a closed geodesic loop in your surface, whereas a geodesic disc of radius r is all the points distance r from a fixed point (at least for r smaller than the injectivity radius). Note the boundary of a geodesic disc is not a geodesic.

As for the factors in those formulae, well, there's no absolute scale for Gaussian curvature. People have just agreed on the convention that the curvature of the unit sphere should be 1. (EDIT: As Greg Kuperberg points out in his answer, there are some good reasons for this convention. E.g., Gauss-Bonnet.) That then forces those factors to be what they are. It amounts to the statement that, for a small geodesic disc on the unit sphere of radius r , $$C(r) \sim 2\pi\left( r - \frac{1}{6} r^3\right),$$ and a similar formula for the area. There really is no deeper reason than that.

So, to see if the factors are right (and you should never trust what you read on the internet!) I would suggest doing exactly those calculations for the unit sphere. I've checked the first formula involving the circumference and it looks good to me. If you have problems with the calculation, leave a comment and I'll write my version down for you, but I have a feeling it's best to do these things yourself.

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Cheers. The constants seem to agree with what I have calculated. That also explains why I have seen any definitions being used with different normalising constants. Thanks again. –  alext87 Dec 21 '09 at 10:44
You're welcome! –  Joel Fine Dec 21 '09 at 10:51

Joel is right that it is partly just a convention to scale Gaussian curvature so that the curvature of a unit sphere is $1$. However, there are three natural motivations for this scale besides matching 1 to 1 in the case of a sphere.

First, Gauss defined his curvature as the product of the extrinsic curvatures of a surface in $\mathbb{R}^3$. So there is a coefficient of 1 in this natural formula.

Second, the unit sphere has the property that the deviation from Euclid's parallel postulate has a factor of 1. In other words, the area $A$ of a triangle with angles $\alpha, \beta, \gamma$ is $\alpha + \beta + \gamma - \pi$. In general, if you have a very small triangle with area $A$ at a point of local curvature $K$, its angle deviation is $KA$ to first order. This factor of 1 leads to a factor of $2\pi$ in the Gauss-Bonnet theorem, that the integral of Gaussian curvature is $2\pi \chi$.

Third, Gaussian curvature is the ratio of the Ricci curvature tensor to the metric, and it is also half of the scalar curvature.

In comparing these formulas, the most reasonable scales for Gaussian curvature are the standard choice, the standard choice times 2 to match scalar curvature, and the standard choice divided by $2\pi$ to match the Gauss-Bonnet theorem. The volume and area formulas are some justification for a 1/3 or a 1/12 or similar, but these are taken to be less fundamental scales.

(One irony of the discussion is that $\pi$ itself is half of the most important value in trigonometry.)

Also the volume and surface area ratios are given in Wikipedia in $n$ dimensions. It is also worth looking at the generalized Gauss-Bonnet theorem in $2n$ dimensions.

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There should not be a symbol $\pi$. Instead, there should be a symbol for what we now call $2\pi$. It would make calculations so much easier, e.g. by better illuminating when there really are strange factors of $2$. –  Theo Johnson-Freyd Dec 21 '09 at 21:47
You're right, I concede there is a good choice of scale for curvature. The Gauss-Bonnet theorem is particularly convincing. I guess you can also see it infinitesimally: for a surface in R^3, the curvature is the the limit of the ratio of solid angle prescribed by the normal to surface area (which is just infinitesimal Gauss-Bonnet for a surface in R^3). I'll amend my answer now. –  Joel Fine Dec 22 '09 at 12:45