# Prove that $x + V$ is an affine subspace of $X$, where $V\subset T$ is a subspace (Dodson & Poston)

Here is the definition of an affine space in Tensor Geometry: The Geometric Viewpoint and its Uses by Dodson, Christopher T. J., Poston, Timothy:

An affine space with vector space $$T$$ is a non-empty set $$X$$ of points and a map $$\newcommand{\inv}{^{-1}} d: X\times X\to T$$ called a difference function, such that for any $$x,y,z\in X$$:

1. $$d(x,y) + d(y,z) = d(x,z)$$
2. The restricted map $$d_x: \{x\} \times X\to T$$, $$(x,y)\mapsto d(x,y)$$ is bijective.

Given $$x\in X$$ and $$t\in T$$, there is a unique point $$z\in X$$ satisfying $$d_x(z) = t$$. We denote this point by $$x + t$$. If $$V\subset T$$, we write $$x + V := \{x+t: t\in V\}$$.

Here's an affine subspace:

$$X'\subset X$$ is an affine subspace or flat of $$X$$ if

1. $$d(X'\times X')$$ is a vector subspace of the vector space $$T$$ for $$X$$, and
2. $$X'$$ is an affine space with vector space $$d(X'\times X')$$ and difference function $$d: X'\times X' \to d(X'\times X')$$, $$(x,y) \mapsto d(x,y)$$.

I want to prove that $$x + V$$ is an affine subspace of $$X$$, where $$V\subset T$$ is a subspace.

For the first condition:

1. Given $$d(x+v_1, x+v_2)\in T$$ and $$d(x+v_3,x+v_4)\in T$$, we want to show $$d(x+v_1,x+v_2) + d(x+v_3, x+v_4) = d(x+v_5,x+v_6)$$ for some $$v_5,v_6\in V$$.
2. Similarly for scalar multiplication, given $$d(x+v_1, x+v_2)\in T$$ and $$a\in \mathbb R$$, we want $$ad(x+v_1, x+v_2) = d(x+v_1', x+v_2')$$ for some $$v_1',v_2'\in V$$.

For the second, I think only the bijectivity of $$d_x$$ needs to be checked?

• There are several (equivalent) rather abstract definitions of affine spaces. If one works on linear spaces, here is a natural way to look at them Commented Sep 25, 2021 at 16:19
• Thanks - I understand the intuition. How do I prove the above result though? @OliverDiaz Commented Sep 25, 2021 at 16:20

The assumption on the difference map can be re-written as follows. For all $$x,y,z \in X$$

$$d(y,z) = d(x,z) - d(x,y).$$

Using this, we have that for all $$\hat{x}$$

$$d(x+v, x+u) = d(\hat{x},x+u)-d(\hat{x},x+v).$$

In particular, we can pick $$\hat{x}= x$$ and get that

$$d(x+v,x+u) = d(x,x+u) - d(x,x+v) = u-v \in V.$$

(Note that if we let $$v = 0$$, we see that for all $$u \in V$$ we have that $$u \in d(X' \times X')$$. Thus, $$V \subset d(X' \times X')$$ and we also have the other inclusion since $$u-v \in V$$.)

From this, we can easily that $$x+V$$ is a vector space. Specifically, let $$x+v_1, x+v_2, x+v_3, x+v_4 \in x+V$$ for $$v_1, v_2, v_3,v_4 \in V$$. Then we have that since $$V$$ is a vector space it contains $$v_2+v_4$$, $$v_1+v_3$$ and $$v_2+v_4-(v_1+v_3)$$.

Thus, we have that

$$d(x+v_1,x+v_2)+d(x+v_3,x+v_4) = v_2-v_1+v_4-v_3 = v_2+v_4 - (v_1+v_3) = d(x+v_1+v_3, x+v_2+v_4).$$

Similarly

$$ad(x+v_1, x+v_2) = d(x+av_1, x+av_2).$$

Thus, not only is it a vector space, but addition and multiplication function as you would expect them to.

As you've already noted for the second condition, we only need to check bijectivity. Let us fix $$x+v \in X'$$.

For injectivity, assume that there exists $$x+v_1, x+v_2 \in X'$$ such that

$$d(x+v, x+v_1) = d(x+v, x+v_2).$$

Then, we have that this implies that $$v_1 - v = v_2 - v$$ which implies that $$v_1 = v_2$$. Or we just note that $$x+v \in X$$ and get infectivity from the difference function on $$X$$.

For surjectivity, let $$u \in d(X' \times X') = V$$. Let $$w = u + v \in V$$. Thus, $$x + w \in x+V$$. Finally $$d(x+v, x+w) = w-v = u+v-v = u$$. Thus, the map is surjective.