Let $(M,g)$ be a Riemannian manifold. Is it true that for any symmetric 2tensor $\alpha$ we have: $Trace_g(\alpha)=1/\omega_n\int_{S^{n1}}\alpha(V,V)dvol(V)$ where $w_n$ is the volume of $S^{n1}$? Thank you.

2$\begingroup$ You are off by a factor of $n$, assuming that you are on an $n$manifold. However, this has nothing to do with Riemannian geometry because you are doing your calculations at a point. This is a linear algebra fact. To make it right, you'd need to evaluate the left hand side at a point $x\in M$ (and divide it by $n$), and, on the right hand side, integrate over the unit $V\in S_x\subset T_xM$. $\endgroup$– Robert BryantApr 16, 2013 at 11:38

$\begingroup$ See meta.mathoverflow.net/questions/1367/… $\endgroup$– Todd Trimble ♦Jan 6, 2014 at 12:28

$\begingroup$ An unsuccessful attempt to migrate to MSE was made. $\endgroup$– Todd Trimble ♦Jan 6, 2014 at 12:30
1 Answer
A followup to Robert Bryant's comments. Let $S^{n1}$ be the unit sphere in some tangent space with inner product $g$. Let $\{e_i\}_{i=1}^n$ be an orthonormal frame and let $V_i=\langle V,e_i\rangle$. For $i\neq j$, $\int_{S^{n1}}V_{i}V_{j}\operatorname{dvol}\left( V\right) =0$ since the integrand is odd with respect to reflection about the coordinate hyperplane $\left\{ V_{i}=0\right\} $. Taking $i=j$, we get $\int_{S^{n1}}V_{i}^{2}\operatorname{dvol}\left( V\right) =\frac{\omega_{n}}{n}$ since this expression is independent of $i$ and since $\sum_{i=1}^{n}\int_{S^{n1}}V_{i}^{2}\operatorname{dvol}\left( V\right) =\int_{S^{n1} }\operatorname{dvol}\left( V\right) =\omega_{n}$, using $V^2=1$. We conclude with $\alpha_{ij}=\alpha (e_i,e_j)$ that $$ \int_{S^{n1}}\alpha\left( V,V\right) \operatorname{dvol}\left( V\right) =\sum _{i,j=1}^{n}\int_{S^{n1}}\alpha_{ij}V_{i}V_{j}\operatorname{dvol}\left( V\right) =\sum_{i=1}^{n}\int_{S^{n1}}\alpha_{ii}V_{i}^{2}\operatorname{dvol}\left( V\right) =\frac{\omega_{n}}{n}\operatorname{Trace}{}_{g}(\alpha). $$ I once gave this formula as a homework problem in a Riemannian geometry class at UCSD and learned this short proof from one of the students.
A reference for this formula, in relation to differential geometry, is the book "Riemannian geometry" by Gallot, Hulin, and Lafontaine.
Happy Thanksgiving, November 28, 2013.
In response to the question being put on hold as offtopic by Ricardo Andrade, Andrey Rekalo, Olivier Benoist, Stefan Kohl, David White:
A few elementary examples of tracing in Riemannian geometry are as follows.
Acting on functions and tensors, the trace of the Hessian operator $\nabla ^{2}$ is the (rough) Laplacian $\Delta$. A trace of the covariant derivative of a tensor is its divergence.
A trace of the Riemann curvature $4$tensor $\operatorname{Rm}$ is the Ricci $2$tensor $\operatorname{Ric}$. The trace of the Ricci tensor is the scalar curvature.
For the second Bianchi identity $\nabla_{\lbrack V}\operatorname{Rm}% (W,X],Y,Z)=0$ (square brackets denotes cyclically permuting and summing), tracing twice yields the contracted second Bianchi identity $\operatorname{div}(\operatorname{Ric})=\frac{1}{2}dR$.
The trace is important in the irreducible decomposition of tensors, including both the curvature and its covariant derivatives. It is useful to consider the tracefree part of a tensor. The Weyl tensor is totally tracefree.
The mean curvature is the trace of the second fundamental form.
Under a variation of a metric, its volume form evolves by onehalf of the trace of the variation.
In the study of the partial differential equations arising from equations involving curvature, covariantly differentiating and tracing is useful.

2$\begingroup$ Re: your "response", undoubtedly the closers know that the notion of "trace" is useful in mathematics. But as you indicated yourself in your answer, this is essentially a "homework question", and certainly does not fit the MO of MO (pun intended). (The operative word in the closure reason is "research".) $\endgroup$ Nov 29, 2013 at 11:57

1$\begingroup$ To Willie Wong: I respectfully disagree with you. $\endgroup$– user41263Dec 4, 2013 at 1:08