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This question is a continuation of this one, where I did not receive a complete answer so am moving to MO. I am trying to understand a paper by JP Szaro that was referred to there, and in particular, what is meant by the action of a Lie group on an affine manifold by affine transformations.

I'm assuming Szaro's definition of $n$-dimensional affine manifold $M$ coincides with the Wikipedia definition, meaning that $M$ can be covered by an atlas of coordinate charts whose transition functions are in $Aff(\Bbb R^n)$. The general statement I'm having trouble with is "assume that a Lie group $G$ acts on $M$ (complete in sense of Wikipedia) by affine transformations", e.g., see Corollary 3.7 of Szaro's paper.

Obviously $Aff(\Bbb R^n)$ acts on the universal cover of any complete affine manifold (just $\Bbb R^n$) and so $G$ may act by affine transformations after factoring through this one. However, I have no feel whatsoever for the conditions under which $G$ descends to an action on $M$—nor precisely what is meant by "acts by affine transformations". Is the statement a local one saying that in each affine coordinate chart $\{x_i\}$ the components of the generating vector field are given by $V= (a_{ij}x_i+b_j) \partial/\partial x_j$, or is it simply saying that the $G$ action factors through an action of the affine group on $M$ (to what extent are these the same thing)?

If the latter is correct, then does this necessarily imply $M$ admits an action of the affine group and we therefore must restrict our attention to a select set of complete affine manifolds?

An explanation with explicit examples of affine manifolds other than $\Bbb R^n$ would be very useful. Hopefully the lack of answers to my previous question warrants the transfer to MO. Thanks.

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If the definition of Szaro corresponds to the definition of Wikipedia, then the affine structure of $M$ is defined by a connection $\nabla$ whose curvature and torsion forms vanish identically. An affine transformation of $(M,\nabla)$ is a diffeomorphism which preserves the connection $\nabla$. When $M$ is complete, it is the quotient of $\mathbb{R}^n$ by a subgroup $H$ of affine transformation which acts properly and freely on $R^n$. In this case the group of affine transformations of $(M,\nabla)$ is the quotient of the normalizer of $H$ in $\operatorname{Aff}(\mathbb{R}^n)$ by $H$. When $(M,\nabla)$ is compact and complete, I have shown in my thesis that the connected component of the group of affine transformations of $(M,\nabla)$ is nilpotent. This implies that the connected component of $G$ is nilpotent in this case.

To show that the connected component of $\operatorname{Aff}(M,\nabla)$ (the group of affine transformations of $(M,\nabla)$) is nilpotent when $(M,\nabla)$ is compact and complete, one remarks firstly that if $M$ is the quotient of $\mathbb{R}^n$ by the group of affine transformations $H$, then the action of $H$ is irreducible, that is, does not preserve a proper affine subset. To see this, suppose that $H$ preserves an affine subset $V$, then $V/H$ and $M$ are $K(\pi,1)$-Eilenberg McLane space. Thus the cohomology of $M$ and $V/H$ is the cohomology of $H$, since $M$ is compact. (We can assume $M$ is oriented up to a 2-cover.) Thus $H^n(M,\mathbb{R})=H^n(\pi,\mathbb{R})\neq 0$. This implies that $H^n(V/H,\mathbb{R})\neq 0$, thus the dimension of $V/H$ is at least $n$.

Next, you remark that $N(H)_0$ the connected component of the normalizer of $H$ in $\operatorname{Aff}(\mathbb{R}^n)$ commutes with $H$. You deduce that this implies that it acts freely on $\mathbb{R}^n$ since the action of $H$ on $\mathbb{R}^n$ is irreducible. Now, let $\operatorname{aff}(M,\nabla)$ be the Lie algebra of $\operatorname{Aff}(M,\nabla)$. For every vector $X,Y\in \operatorname{aff}(M,\nabla)$, $\nabla_XY$ is again in $\operatorname{aff}(M,\nabla)$. Since $\operatorname{Aff}(M,\nabla)$ is the quotient of $N(H)$ by the discrete group $H$, their Lie algebras are isomorphic. Remark that the product induced by $\nabla$ on $n(H)$ the Lie algebra of $N(H)$ is the associative product of $\operatorname{aff}(\mathbb{R}^n)$ the Lie algebra of $\operatorname{Aff}(R^n)$ defined by $(A,a).(B,b)=(AB,A(b))$. The associative algebra $n(H)$ does not have an idempotent $(C,c)$ because if $(C,c)$ is such an idempotent, $(C,c).(C,c)=(C^2,C(c))=(C,c)$. This implies that the 1-parameter group generated by $(C,c)$ fixes $-c$. This is in contradiction with the fact that the action of $N(H)_0$ on $\mathbb{R}^n$ is free. An associative algebra which doesn't have a nilpotent element is nilpotent so $n(H)$ is nilpotent, hence $\operatorname{aff}(M,\nabla)$ and $\operatorname{Aff}(M,\nabla)_0$ are nilpotent.

When $(M,\nabla)$ is compact and complete, I have also shown that the natural action of $\operatorname{Aff}(M,\nabla)_0$ on $M$ is locally free, thus its orbits define on $M$ a foliation.

The typical example is the torus $T^n$ which is the quotient of $\mathbb{R}^n$ by the group generated by $n$-translations $t_{e_1},...,t_{e_n}$ such that $(e_1,...,e_n)$ is a basis of $\mathbb{R}^n$. The connected component of $\operatorname{Aff}(T^n)$ is $T^n$. Remark that $\operatorname{Aff}(T^n)$ is not nilpotent although its connected component is nilpotent since the action of $\operatorname{GL}(n,\mathbb{Z})$ on $\mathbb{R}^n$ induces an action of $\operatorname{GL}(n,\mathbb{Z})$ on $T^n$.

Tsemo Aristide

Dynamique des variétés affines. Journal of the London Mathematical Society 63.2 (2001): 469-486.

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  • $\begingroup$ Hi Tsemo, thank you for your answer and the reference to your paper. Unfortunately I don't read any French! Perhaps you could provide a reference to your statement about constructing the group of affine transformations of $(M,\nabla)$? Or possibly expand your answer a bit to indicate why this is true. An example would be very useful. $\endgroup$ – hhu89 Apr 7 '16 at 16:53

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