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(Work over a fixed field $k$.)

The nLab offers a list of definitions of the concept "affine space". Here's two of them:

  1. An affine space is a set $A$ together with a vector space $V$ and an action of (the additive group or translation group of) $V$ on $A$ that makes $A$ into a $V$-torsor (over the point).

    nLab goes on to say that an affine linear map is a $V$-equivariant map; for our purposes, however, we want all $k$-affine spaces to form into a single category. So let us stipulate that a morphism $(\varphi,f) : (V_0,A_0) \rightarrow (V_1,A_1)$ is a linear transform $\varphi : V_0 \rightarrow V_1$ together with a function $f : A_0 \rightarrow A_1$ such that $f(ax) = \varphi(a)f(x)$.

  2. An affine space over the field $k$ is an inhabited set $A$ together with, for every natural number $n≥0$ and every $(n+1)$-tuple $(r_0,…,r_n)$ of elements of $k$ such that $r_0+…+r_n=1$, a function $γ(r_0,…,r_n):A^n \rightarrow A$ (thought of as $γ(r_0,…,r_n)(x_0…,x_n) = r_0 x_0 + \cdots + r_n x_n$), satisfying some equations; an affine linear map is a function that preserves these operations.

Okay, viewing affine spaces (in the manner of (2)) as sets equipped with affine combination functions, we get a forgetful functor $U_\mathbf{Aff} : \mathbf{Vect} \rightarrow \mathbf{Aff}$ that forgets the non-affine combinations, which is faithful but not full. And viewing affine spaces (in the manner of (1)) as sets equipped with the regular action of a vector space, we get a forgetful functor $U_\mathbf{Vect} : \mathbf{Aff} \rightarrow \mathbf{Vect}$ that forgets everything but the vector space, which is full but not faithful.

There's something bugging me about this situation. I think that $U_\mathrm{Vect} \circ U_\mathrm{Aff}$ is isomorphic to $\mathrm{id}_\mathbf{Vect}$. However, this would imply that $U_\mathrm{Aff} \circ U_\mathrm{Vect}$ is not isomorphic to $\mathrm{id}_\mathbf{Aff}.$ Okay, so what on earth does $U_\mathrm{Aff} \circ U_\mathrm{Vect}$ look like? Clearly both $U_\mathrm{Aff}$ and $U_\mathrm{Vect}$ are essentially bijective on objects. So $U_\mathrm{Aff} \circ U_\mathrm{Vect} : \mathbf{Aff} \rightarrow \mathbf{Aff}$ is either non-full, or non-faithful.

I think that it is neither full nor faithful. Is this correct? If so, I find the non-fullness difficult to stomach and/or comprehend. It means that for no apparent reason, certain affine transforms arbitrarily fail to be obtainable from linear transforms under this functor. Well, which ones are they?

Guys, help me out here. What's up with the endofunctor $U_\mathrm{Aff} \circ U_\mathrm{Vect} : \mathbf{Aff} \rightarrow \mathbf{Aff}$?

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  • $\begingroup$ I don't know what you mean by «set equipped with affine combination functions». $\endgroup$ – Mariano Suárez-Álvarez Apr 9 '15 at 5:19
  • $\begingroup$ @MarianoSuárez-Alvarez, briefly and imprecisely: if $a \in \mathbb{R}^n$ and $V$ is a vector space, then $V$ is equipped with a linear combination function $f_a : V^n \rightarrow V$ given by $f_a(x) = \sum_{i=1}^n a_i x_i$. Now throw away all such linear combination functions except the $f_a$'s such that $\sum_{i = 1}^n a_i = 1.$ The result is a set $V$ equipped with only some of the linear combination functions that we started with, namely the affine-combination ones. $\endgroup$ – goblin Apr 9 '15 at 5:24
  • $\begingroup$ @MarianoSuárez-Alvarez, see the final definition here where they speak of the "unbiased" definition. $\endgroup$ – goblin Apr 9 '15 at 5:26
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    $\begingroup$ Can you please add the relevant information to the body of the question itself? These are not exactly well-known definitions... $\endgroup$ – Mariano Suárez-Álvarez Apr 9 '15 at 5:30
  • $\begingroup$ @MarianoSuárez-Alvarez, sure. Is that better? $\endgroup$ – goblin Apr 9 '15 at 5:39
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I work with definition (1) only. The "forgetful" functor $\mathbf{Vect} \to \mathbf{Aff}$ is the one sending a vector space $V$ to $(V, V)$, where $V$ acts on itself by translation. Thus the composite $\mathbf{Vect} \to \mathbf{Aff} \to \mathbf{Vect}$ is indeed the identity.

Since the "forgetful" functor $\mathbf{Vect} \to \mathbf{Aff}$ is essentially surjective on objects, to understand $\mathbf{Aff} \to \mathbf{Vect} \to \mathbf{Aff}$, it is enough to see what it does to affine spaces of the form $(V, V)$. But that is clear: a morphism $(V, V) \to (W, W)$ is simply a map of the form $v \mapsto \phi (v) + w_0$ where $\phi : V \to W$ is a linear map and $w_0 \in W$, and the endofunctor $\mathbf{Aff} \to \mathbf{Aff}$ sends this to the map $v \mapsto \phi (v)$.

Essentially, what we are seeing here is a "global" version of the semidirect product decomposition $\mathrm{Aff} (V) \cong V \rtimes \mathrm{GL} (V)$.

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    $\begingroup$ I like the idea of focusing just on definition (1). But Its not making a lot of sense. Which morphisms are in the image of $\mathbf{Aff} \rightarrow \mathbf{Aff}$? Which parallel morphisms are identified by $\mathbf{Aff} \rightarrow \mathbf{Aff}$? The answer to both questions seems to be completely evil. Its as if the Dark One himself is reaching through the formalism to touch the world once more... $\endgroup$ – goblin Apr 9 '15 at 12:51
  • $\begingroup$ I explained that in my answer already. Read carefully and think about what I wrote. $\endgroup$ – Zhen Lin Apr 9 '15 at 13:10
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    $\begingroup$ @goblin: One reason things may seem "evil" is that $\mathbf{Aff}\to\mathbf{Aff}$ is essentially surjective on objects and in fact objectwise isomorphic to the identity, but it is not naturally isomorphic to the identity. So even though every affine space is isomorphic to one in the image of $\mathbf{Aff}\to\mathbf{Aff}$, it is not very helpful to think of every object as literally being in the image (doing so requires choosing unnnatural isomorphisms). $\endgroup$ – Eric Wofsey Apr 9 '15 at 18:31

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