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## Is the set of average-position preserving transformations a Lie group

Let $M$ be a compact subset in $\mathbb{R}^n$ and $\mu$ a volume form on $M$. Let $x_i$ denote the function corresponding to the $i$-coordinate. Does the set of diffeomorphisms satisfying $$\int_M{x_i \mu } = \int_M{ \varphi_*(x_i) \mu }, \quad i = 1,\dots,n$$ form a Lie group? In other words, is the set of transformations that preserve the $\mu$-average position a Lie group? Certainly the isotropy group of the $\mu$-average position is a Lie-Group, however this set seems to incude a little more than the isotropy group.

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It's not a Lie group in the traditional sense, in that it's not finite-dimensional. Consider the case $M=[-1,1] \subset \mathbb R$. In this case, any anti-symmetric ($\phi(-x)=-\phi(x))$ diffeomorphism works. – Ryan Budney Aug 1 2011 at 22:18
It's not even a Lie group in Lie's sense, i.e., a (pseudo-)group of (local) diffeomorphisms that is characterizable as the set of solutions of some system of differential equations. Lie didn't regard finite dimensionality as an essential feature; he regularly worked with the (pseudo-)group of contact transformations and other 'infinite groups'. In fact, in the literature of the time, finite dimensional Lie groups were often just called 'finite Lie groups', which can be quite confusing for modern readers. – Robert Bryant Aug 2 2011 at 0:36
Thanks for the quick response guys. I'm not too concerned with the dimensionality concerns, so my question is regarding psuedo-groups. Does this set fail the closure axiom? Is my intuition my enemy? If not, which axiom has failed? – hoj201 Aug 2 2011 at 2:12
If I understand the question right, this set is not even a group. – Tom Goodwillie Aug 2 2011 at 2:29