2
$\begingroup$

I've recently took interest in morphism and category theory and I'm amazed how it offers a very general notion. However, I'm struggling to apply this for the affine vector spaces.

I've seen that a linear application could be thought as a morphism between two vector spaces: Let ($V_1, \oplus_1, \odot_1$) and ($V_2, \oplus_2, \odot_2$) be two vector spaces of dimension $n\in \mathbb{N}$ over a field $\mathbb{F}$, where $\odot_{\bullet}$ and $\oplus_{\bullet}$ denotes the scalar multiplication and the vector addition of the vector space $V_\bullet$ respectively. From what I understand, a morphism $\varphi$ from $V_1$ to $V_2$ is a map that preserves the structure of these spaces, i.e. that is compatible whith their respective operations: $$ \begin{cases} \forall(u,v)\in V_1,\quad \varphi(u \oplus_1 v) = \varphi(u) \oplus_2 \varphi(v)\\ \forall\lambda\in\mathbb{F},\quad\forall u\in V_1,\quad\varphi(\lambda \odot_1 u) = \lambda \odot_2 \varphi(u) \end{cases}, $$ We can see that this definition coincides with the definition of a linear map. Let ($A_1, V_1, \boxplus_1$) and et ($A_2, V_2, \boxplus_2$), two affine spaces with respective vector space direction $V_1$ and $V_2$. Note that $\boxplus_\bullet$ denotes the transitive and free action of the additive group of $V_\bullet$ on the set $A_\bullet$ (https://en.wikipedia.org/wiki/Affine_space#Definition). Now this is where I struggle, let $\varphi$ be a morphism from $A_1$ to $A_2$. So I assume that it means somehow something like: $$ \forall \mathrm{P}\in A_1,\quad \forall u \in V_1, \quad\varphi(\mathrm{P}\,\boxplus_1\,u) = \varphi(P)\,\boxplus_2 \,? $$ What should I put instead of the interrogation point in the right handside of the equation above? Indeed, $\varphi$ only acts on points of $A_1$, not vectors of $V_1$. I know that it will eventually coincide with the definition of an affine map, i.e. that the "$?$" will be eventually replaced by $\vec{\varphi}(u)$, with $\vec\varphi$ a linear map from $V_1$ to $V_2$, but how can I demonstrate it directly from the definition of a morphism of affine spaces?


$\endgroup$
3
  • 1
    $\begingroup$ The problem with that is having a "definition of a morphism of affine spaces" in the first place. Category theory does not provide a way to deduce what it means to be a morphism from what it means to be an object, even though for categories occurring in algebra it often seems there is only one reasonable definition. Maybe universal algebra can make that notion precise, but I know nothing about it. $\endgroup$ Commented Jun 19, 2024 at 15:41
  • $\begingroup$ It's not what you asked, but affine maps are precisely the maps preserving affine combinations, just like linear maps preserve linear combinations. I think (but can't find a reference right now) that you can define an affine space to be a set equipped with an "affine combination operation" without mentioning a vector space. These alternative definitions for objects and morphisms would fit together in the way you want. $\endgroup$ Commented Jun 19, 2024 at 15:53
  • $\begingroup$ By the way, I prefer to say "affine space" instead of "affine vector space", since it is not a vector space. $\endgroup$ Commented Jun 20, 2024 at 15:15

2 Answers 2

1
$\begingroup$

ORIGINAL ANSWER REPLACED BY: $\newcommand\vcal{\mathcal{V}} \newcommand\acal{\mathcal{A}} \newcommand\F{\mathbb{F}} $

Let $\vcal$ be the category of vector spaces over the field $\F$ of real or complex scalars. The morphisms are linear maps between vector spaces. Observe that using the standard topology of a real or complex vector space, a map $L: V \rightarrow W$ is linear if it is additive, i.e., for any $v_1,v_2 \in V$, \begin{align*} L(v_1+v_2) = L(v_1)+L(v_2). \end{align*}

An object in the category $\acal$ of affine spaces over $\F$ consists of a pair $(A,V)$, where $V$ is a vector space over $\F$, with an operation \begin{align*} A \times V &\rightarrow A\\ (a,v) &\mapsto a+v \end{align*} that satisfies the following properties:

  1. For each $a \in A$, the map \begin{align*} V &\rightarrow A\\ v &\mapsto a+v \end{align*} is bijective. The inverse map is denoted by subtraction: \begin{align*} A &\rightarrow V\\ b &\mapsto b-a. \end{align*}
  2. For each $v \in V$, the map \begin{align*} A &\rightarrow A\\ a &\mapsto a+v \end{align*} is bijective.
  3. For any $a \in A$, $v, w \in V$, \begin{align*} a+(v+w) &= (a+v)+w.\\ \end{align*}

To define a morphism of $\acal$, first observe that given affine spaces $(A,V)$ and $(B,W)$ and a map $F: A \rightarrow B$, there is for each $a \in A$, an induced map \begin{align*} dF_a: V &\rightarrow W\\ v &\mapsto F(a+v)-F(a). \end{align*} Call $dF_a$ the differential of $F$ at $a$. A map $F: A \rightarrow B$ is a morphism of $\acal$ if for any $a \in A$ and $v \in V$, \begin{align*} F(a+v) &= F(a) + dF_a(v). \end{align*} From this it follows that if $v_1,v_2 \in V$, \begin{align*} dF_a(v_1+v_2) &= F(a+v_1)-F(a)+F(a+v_2)-F(a)\\ &= dF(v_1)+dF(v_2). \end{align*} It follows that $dF$ is linear. Moreover, if $a_1, a_2 \in A$, then \begin{align*} dF_{a_2}(v) &= F(a_2+v)-F(a_2)\\ &= F(a_1+(a_2-a_1)+v) - F(a_1+(a_2-a_1))\\ &= F(a_1) + dF_{a_1}((a_2-a_1)+v) - F(a_{1}) - dF_{a_1}(a_2-a_1)\\ &= dF_{a_1}(v). \end{align*} Therefore, the differential of $F$ does not depend on $a\in A$ and can be denoted simply by $dF$.

It follows that a map $F: A \rightarrow B$ is a morphism if and only if there is a linear map $L: V \rightarrow W$ such that for any $a \in A$, \begin{align*} F(a+v) = F(a)+L(v). \end{align*}

$\endgroup$
4
  • $\begingroup$ Thank you for your answer, however how would you define a "morphism of the affine category" then ? I'm sorry if it is trivial, but I'm quite new to the subject $\endgroup$ Commented Jun 20, 2024 at 8:55
  • $\begingroup$ Sorry, my answer was too vague. I've replaced it by a more carefully written one. $\endgroup$ Commented Jun 20, 2024 at 15:11
  • $\begingroup$ Your claim Observe that using the standard topology of a real or complex vector space, a map $L:V\to W$ is linear if it is additive does not make sense to me. There is no standard topology on an infinite dimensional vector space and anyway, additivity and linearity have nothing to do with topology. There are additive but non-linear maps even $\mathbb R\to\mathbb R$. For the definition of morphisms you should require that the differentials are linear and independent of the point $a$. $\endgroup$ Commented Mar 13 at 11:35
  • $\begingroup$ @Jochen, good point. I was assuming that the vector spaces are finite dimensional. But I also agree that it’s better to assume $dF$ is linear and independent of the point. $\endgroup$ Commented Mar 13 at 13:52
1
$\begingroup$

There are a couple ways to do this. Personally I find it unsatisfying to think of affine spaces in terms of their associated vector spaces: it is actually possible to give an axiomatization of affine spaces that does not require this. In the same way that a vector space over a field $K$ is a set equipped with operations that allow us to define arbitrary $K$-linear combinations

$$(v_1, \dots v_n) \mapsto \sum c_i v_i,$$

an affine space over a field $K$ can be defined as a set equipped with operations that allow us to define only affine $K$-linear combinations

$$(p_1, \dots p_n) \mapsto \sum c_i p_i, \sum c_i = 1.$$

In the vector space case, instead of talking about all linear combinations at once we conventionally reduce to talking only about addition and scalar multiplication, which can be used to build all linear combinations. In the affine case we can no longer do this. We can restrict our attention to binary affine combinations

$$(p_1, p_2) \mapsto c p_1 + (1 - c) p_2, c \in K$$

since arbitrary affine combinations can be built out of compositions of these. These parameterize the line containing $p_1$ and $p_2$ (whereas general affine combinations parameterize the hyperplane containing some set of points). But I don't think the resulting axioms look all that nice. It seems cleaner to me to use all affine combinations at once.

Formally, the collection of all affine combination operations organize into a gadget called a Lawvere theory, and we can define an affine space as what is called a model of this Lawvere theory, which means a set equipped with affine combination operations which satisfy exactly the relations that affine combinations do. (Vector spaces can be defined in the same style, using linear combinations.) Then a morphism of affine spaces is a function which respects all these affine combination operations. This may look like a lot of structure to carry around but in practice it just means we can do completely ordinary-looking calculations with affine combinations. For example it is straightforward to verify that any affine subspace of a vector space is an affine space in this sense.

Then you can prove as a theorem that every affine space is obtained from a vector space, rather than making it part of the definition, and also prove that morphisms of affine spaces correspond to affine maps in the usual sense. The details seem a bit tedious to spell out though.

A very similar idea can be used to define the notion of a convex space, where you only allow convex linear combinations.

$\endgroup$
3
  • $\begingroup$ Nice. I always forget about the affine combination approach. $\endgroup$ Commented Jun 20, 2024 at 18:23
  • $\begingroup$ Oh! I didn't know that this abstract definition of affine spaces existed! Thank you! $\endgroup$ Commented Jun 20, 2024 at 20:38
  • $\begingroup$ @Plop: no problem. I'm pretty sure I learned about this from John Baez (math.ucr.edu/home/baez/TWF.html) but I don't remember exactly where in twf it is. $\endgroup$ Commented Jun 20, 2024 at 20:47

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.