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.


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.

  • $\begingroup$ 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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