How does vectorialisation work?

Question

My textbook says:

Let $X$ be an affine space over $V$ with related free, transitive action $\alpha: V \to S(X)$. For $v \in V$, we write $\alpha_v(x) = x + v$. Let $x,y$ be given. The unique $v \in V$, such that $x+v = \alpha_v(x) = y$, is written as $y - x$.

We obtain a function $\theta= X \times X \to V: (x,y) \mapsto y -x$ and when we fix $x$, we obtain the function $\theta_x: X \to V: y \mapsto y -x$

This is a bijection ($\rho_x: V \to X: v \mapsto x +v$ is the inverse function).

By structure transport, we can place the unique $\mathbb{K}$-vectorspace structure on the set $X$ such that $\theta_x$ (and also $\rho_x)$ become isomorphisms. We then call the set $X$ with this vector space structure $X_x$ and we call $X_x$ the vectorialisation of $X$ in $x$.

I understand everything except the last paragraph (starting from by structure transport). Can someone explain this in detail please?

Thanks in advance.

Answer 1

The function $\theta_x : X \to V$ is a bijection, and $V$ is a vector space. You can consider the problem:

What vector space $X_x$ has the property that $X$ is its set of vectors and $\theta_x$ becomes an isomorphism $X_x \to V$?

In this problem, the unknowns are the vector space structure in $X_x$; e.g. what is addition, and what is scalar multiplication?

You can solve for the unknowns by simply looking at the definition of isomorphism and "solving" them for the relevant information.

For example, part of the definition of isomorphism is

$$ \theta_x( a +_{X_x} b) = \theta_x(a) +_V \theta_x(b) $$

which we can solve to get

$$ a +_{X_x} b = \theta_x^{-1}(\theta_x(a) +_V \theta_x(b)) $$

and now we have a formula that defines the addition operation on ${X_x}$.

Answer 2

This simply means that the vector structure on X defined by the bijections \begin{align}\theta_x \colon X&\longrightarrow V, &\rho_x\colon V&\longrightarrow X,\\ y&\longmapsto y-x&v&\longmapsto x+v\end{align} is given by the following: \begin{align} y+y'&\stackrel{\text{def}}{=}\rho_x\bigl(\theta_x(y)+\theta_x(y')\bigr)=x+\bigl((y-x)+(y'-x)\bigr)\\ \lambda\mkern1mu y&\stackrel{\text{def}}{=}\rho_x\bigl(\lambda\theta_x(y)\bigr)=x+\bigl(\alpha(y-x)\bigr) \end{align}