0
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ 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 $\endgroup$
    – Mittens
    Commented Sep 25, 2021 at 16:19
  • $\begingroup$ Thanks - I understand the intuition. How do I prove the above result though? @OliverDiaz $\endgroup$ Commented Sep 25, 2021 at 16:20

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .