# 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 $\Delta_n$.

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 the above result would greatly help me.)

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$\Delta_n$ is standard $n$-simplex? – Paul Nov 16 '11 at 8:27
Yes, it is the standard n-simplex as a submanifold of R^n. – Piotr Pstrągowski Nov 16 '11 at 19:08
I would remark that this question has been posted even on MathOverflow. There, until now it has received three interesting answers. Cf. mathoverflow.net/questions/81105/… – Giuseppe Nov 18 '11 at 19:11

The answer given by Sergei Ivanov on MathOverflow was to use

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 $x\mapsto \sum m_i\operatorname{dist}(x,x_i)^2$ 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.

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