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Let $X$ be an affine space over a vector space $V$ such that $\emptyset \neq Y \subset X$ and consider the following 2 statements:

(1) $Y \leq_a X$

(2) There exists a subspace $W \leq V$ such that $Y$ is an affine space over $W$

It is not hard to show that (2) follows from (1), but my teacher told me the converse is not true.

However, I argued in the following way to prove it and I can't see where my proof fails. If someone can point me out the flaw in my proof, and if possible can provide a counterexample to the converse, I would be very glad.

So, here is my attempt:

Let $y_1, \dots, y_n \in Y, a_1, \dots a_n \in F$ such that $\sum_i a_i = 1$.

By definition of affine space, we then have an action $\lambda: W \times Y \to Y: (w,y) \mapsto y + w$.

Now, let $y \in Y$ and consider the affine combination $\sum_i a_i y_i = y + \sum_i a_i(y_i - y)$, which can be written as $\lambda\left(\sum_i a_i(y_i -y),y\right) \in Y$, such that indeed $Y \leq_a X$

Interesting to know:

One can define an affine space $X$ over a vector space $V$ if there exists an action $\lambda: V \times X \to X$ such that:

1) $\lambda(0,x) = x$

2) $\lambda(v, \lambda(w,x)) = \lambda(v+w,x)$

3) $\forall x,y \in X: \exists ! v \in V: \lambda(v,x) = y$

The following are then certainly equivalent:

(1) $Y \leq_a X$

(2) There exists a subspace $W \leq V$ such that $Y$ is an affine space over $W$ under the left action $\lambda \vert_{W \times Y}^Y: W \times Y \to Y: (w,y) \mapsto \lambda(w,y)$

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The mistake is in the claim:

$y+\sum_ia_i(y_i-y)=\lambda\left(\sum_i a_i(y_i -y),y\right)$

The first parameter of $\lambda$ should be an element of $W$, whereas the term $\sum_i a_i(y_i -y)$ is in $Y$.

For a counterexample, consider a curved line in the plane, viewed as an affine space through its parametrization. For example, the graph of a parabola $$f:\mathbb{R}\to\mathbb{R}^2,\quad f(t)=(t,t^2)$$ equipped with the action $$\lambda:\mathbb{R}\times f(\mathbb{R})\to f(\mathbb{R}),\quad \lambda(s,f(t)) = f(t+s)$$ satisfies your requirements 1)-3).

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  • $\begingroup$ This looks like a promising answer. I will study it when I have time. $\endgroup$ – user370967 Oct 17 '17 at 14:16

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