# Talk:Scalar curvature

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## Notation -- R vs S

I originally switched from S to R because every other article on WP that talks about scalar curvature uses R not S (since S is already taken, used to denote the "action").

Also: I have yet to see any book where the scalar curvature is denoted by S. Neither of the two mathbooks I have on Riemannian geometry call it that. So in what sense is "S" the "usual" notation? linas 22:18, 7 October 2005 (UTC)

It is usual in Riemannian geometry, and this is an article in Riemanninan geometry. Almost any modern book use S or Sc. Ricci flow guys use and inex oriented intos use R. See Curvature of Riemannian manifolds and Weyl curvature. Also if you do not use index notation it has no sense to use R. Tosha 22:50, 9 October 2005 (UTC)

## Pseudo-Riemannian case

It ought to be mentioned that scalar curvature is defined for pseudo-Riemannian metrics as well. There is at least one sentence in the article which implies scalar curvature is specific to Riemannian metrics only. Jjauregui (talk) 20:01, 7 April 2008 (UTC)

## Section on 2-dimensional case needs fine tuning

The section titled "2 dimensions" starts as follows:

"In 2 dimensions, scalar curvature is exactly twice the Gauss curvature:

S = 2/(ρ1 ρ2)


where ρ1, ρ2 are principal radii of the surface."

This isn't quite describing the case of "2 dimensions" but rather the case where a 2-dimensional manifold is assumed to have an embedding in 3-space. So -- either the equation should not assume the surface is embedded in 3-space, or the title of this section should be changed to reflect this assumption. (Or else both the embedded and intrinsic cases could be addressed.)Daqu (talk) 19:50, 12 March 2009 (UTC)

## Definition via triangles?

I don't know differential geometry. I've heard of a scalar-valued quantity related to curvature defined something like this: Take the sum of the angles in an equilateral triangle. Divide by the length of the side. (Or do we need the square of the length?) Take the limit as the triangle shrinks.

How is that related to the scalar curvature defined in this article? Michael Hardy (talk) 03:01, 30 August 2009 (UTC)
Up to a constant, that would be the sectional curvature. It's not really a scalar because it depends on the directions of the legs of the triangle, even if they are tending to zero in magnitude. There are, of course, relationships between the two things, but it is not quite so simple. Sławomir Biały (talk) 03:05, 30 August 2009 (UTC)
Actually, I take the last part back. Up to a factor of the dimension, the scalar curvature is the average value of the sectional curvature. So, perhaps, if one selected a triangle "at random", then the scalar curvature represents the expectation value of the construction you mention. Sławomir Biały (talk) 03:22, 30 August 2009 (UTC)
In a space with constant curvature Κ, the sum of the interior angles of a triangle is equal to π+ΚA where A is the area of the triangle. See spherical excess. From this one can show that the sum of the interior angles of an n-gon is (n-2)π+ΚA. This can be generalized to arbitrary closed curves in a space with variable curvature. See Gauss–Bonnet theorem. JRSpriggs (talk) 14:17, 1 September 2009 (UTC)
Just to clarify the above post, as I think JRSpriggs may have misunderstood Michael Hardy's original misapprehension: What he said holds for constant sectional curvature, which is not the same thing as the scalar curvature (the subject of this article). The purpose of my post was to indicate how the sectional curvature (that the angular excess measures) is related to the scalar curvature. Sławomir Biały (talk) 15:25, 1 September 2009 (UTC)

I think I should have said the amount by which the sum exceeds π, rather than just the sum. Michael Hardy (talk) 14:24, 1 September 2009 (UTC)

To Sławomir Biały: As the article says, "in two dimensions, scalar curvature is exactly twice the Gaussian curvature". And as the article on sectional curvature says, "it is the Gaussian curvature of that section". JRSpriggs (talk) 16:48, 1 September 2009 (UTC)
Of course, in two dimensions no one ever talks about "scalar curvature", as that is completely redundant. To focus on that case in the article would indeed be perverse. It is only in higher dimensions in which scalar curvature becomes an interesting object of study independently of other notions of curvature. The question is: how is the scalar curvature of a Riemannian manifold related to its sectional curvature. See my original post for an explanation. Here is slightly more detail:
$S_p = (const.)\int_{\operatorname{Gr}_2(T_pM)} K_p(\sigma)\,d\mu(\sigma)$
where the integral is taken with respect to the unit normalized rotationally-invariant measure on the Grassmannian of two-planes in the tangent space of M at p. Or, more explicitly, in an orthonormal frame
$S_p = \sum_{i,j} K_p(e_i\wedge e_j).$
--Sławomir Biały (talk) 17:17, 1 September 2009 (UTC)
Yes, that is true. But the question was about triangles, and they live in two dimensions.
Perhaps it would have been better if I had given the formula in terms of exterior angles. In that case, the sum of the exterior angles of a closed polygon (with winding number 1) is 2π-ΚA. JRSpriggs (talk) 17:50, 1 September 2009 (UTC)
Of course, but one can also make sense of triangles in higher dimensions: they simply live on a two dimensional subvariety. When understood in this way, the angle excess does not measure the scalar curvature, rather it measures the sectional curvature. Sławomir Biały (talk) 20:11, 1 September 2009 (UTC)
OK. JRSpriggs (talk) 12:36, 3 September 2009 (UTC)

## Direct geometric interpretation

The formula:

$\frac{\operatorname{Vol} (B_\varepsilon(p) \subset M)}{\operatorname{Vol} (B_\varepsilon(0)\subset {\mathbb R}^n)}= 1- \frac{S}{6(n+2)}\varepsilon^2 + O(\varepsilon^4)$

is somehow not derived comprehensible. Can someone show how this comes out using

$d\mu_g = \Big[ 1 - \frac{1}{6}R_{jk}x^jx^k+ O(|x|^3) \Big] d\mu_{{\rm Euclidean}}$

from the corresponding article of the Ricci-Tensor? —Preceding unsigned comment added by Dbfrosch (talkcontribs)

Hint. If A is a symmetric n x n matrix, then the average value of xTAx for x in the unit ball of Euclidean space is tr A/(n+2). 71.182.247.220 (talk) 21:40, 27 October 2009 (UTC)
Details. Integrate xTAx in spherical coordinates over the unit ball. Denote by T(A) the integral
$T(A) = \int_{SO(n)} R^TAR\, dR$
taken with respect to the invariant probability measure on the special orthogonal group SO(n). Then T(A) commutes with all rotations, and therefore by Schur's lemma,
$T(A) = \lambda I.$
Taking a trace on both sides gives
$\operatorname{tr} A = n\lambda.$
so
$T(A) = \frac{\operatorname{tr} A}{n} I.$
Now, let ωn−1 be the (n−1)-volume of the (n−1)-sphere. The average value of xTAx over the unit ball is given by
$\frac{\omega_{n-1}\int_0^1 t^{n+1} e_1^TT(A)e_1\, dt}{\omega_{n-1}\int_0^1 t^{n-1}\, dt} = \frac{\operatorname{tr} A}{n+2},$
as required. 71.182.247.220 (talk) 16:27, 28 October 2009 (UTC)

## Presentation is a bit confusing

The first equation defines S = tr[g] Ric, but doesn't define Ric. The second equation gives another expression for S, which doesn't use Ric, but where Ric is defined. I'm sure the experts here understand what is going on, but to me, trying to figure out, it is confusing. I also don't understand what it means to take the trace of the tensor with respect to the metric. The reference to the trace article only refers to the trace with respect to a basis. Is the metric a basis? I thought it was a bilinear form. Is their an obvious interpretation of a bilinear form as a set of basis vectors? The columns of the matrix? --Thinkor (talk) 22:36, 9 January 2010 (UTC)

Ric is not defined anywhere. It is the Ricci tensor. There is a link to follow. Sławomir Biały (talk) 15:41, 11 January 2010 (UTC)