Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

(This question was originally asked on StackExchange: http://math.stackexchange.com/questions/82437/can-one-average-close-smooth-functions )

Suppose $M$ is a connected, smooth, second-countable manifold.

Let $U \subset M^n$ be some neighbourhood of the diagonal. We will call a function $a: U \times \Delta_n \rightarrow M$ an "averaging operator of order n" if $a|_{U \times v_i} = \pi_i$, where $v_i$ denotes the i-th vertex of the standard $n$-simplex $\Delta_n$.

Intuitively, $a$ should tell us how to take a weighted average of "close" functions.

Does every $M$ admit a smooth averaging operator of order $n$, for some $U$ and all $n$?

(I am trying to prove that smooth functions are dense in $Hom(M,N)$ without embedding $N$ into any $\mathbb{R}^n$ and I think it would easily follow from the above.)

share|improve this question
add comment

3 Answers 3

up vote 5 down vote accepted

There is a couple of standard methods.

First, one can embed $M$ into some $\mathbb R^N$ and fix a smooth neighborhood retraction onto the image. Then let $a$ be the composition of the weighted average in $\mathbb R^N$ and the retraction.

Second (this is an extended version of Nikite Kalinin's answer), you can take the barycenter with respect to a Riemannian metric on $M$. For $x_1,\dots,x_n\in M$ and $(m_1,\dots,m_n)\in\Delta_n$, define the barycenter as a point where the function $$ x\mapsto \sum m_i\cdot dist(x,x_i)^2 $$ attains its minimum. Since the square of the Riemannian distance is smooth near the diagonal, and in normal coordinates it has the same first and second derivatives at the diagonal as the Euclidean counterpart, an easy application of the implicit function theorem shows that the barycenter is well-defined and depends smoothly on its arguments in a neighborhood of the diagonal. Moreover the size of the neighborhood can be estimated in terms of the injectivity radius and sectional curvatures of the Riemannian metric.

share|improve this answer
add comment

You have some Riemannian metric on $M$, so, for each point there is it small neighbourhood where any two points can be connected by unique geodesic. So, if you have $n$ points in this neighbourhood you can canonically map simplex to it.

share|improve this answer
    
Could you elaborate a little bit? I agree, that since there is a unique geodesic, I can connect canonically two close points using a path. This gives seems to a priori only give me a map from the edges of n-simplex..? Thank you! –  Piotr Pstrągowski Nov 16 '11 at 20:06
    
then you connect (by geodesic) vertex and point on edge, so, these lines sweep all 2-faces, etc. –  Nikita Kalinin Nov 16 '11 at 20:33
add comment

Equip $M$ with some Riemannian metric and define $a$ the following way: $x=(x_0,x_1,\dots,x_n)\in U$, and point $\lambda=(\lambda_0,\lambda_1,\dots,\lambda_n)\in\Delta^n$ written in barycentric coordinates set $a(x,\lambda)$ to be the minimum point of the function $$f(z)=\sum_i \lambda_i |x_i-z|^2,$$ where $| x - y |$ denotes the distance in $M$.

The function $a$ is uniquelly defined in a neighborhood of the diagonal. For the details, see K. Grove, H. Karcher, How to conjugate $C^1$-closed group actions, Math. Z. 132 (1973), 11–20.

share|improve this answer
    
P.S. Nikita's answer also works, but mine is better, it is symmetric with respect to permutations of the factors. –  ε-δ Nov 16 '11 at 20:25
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.