Characterizing "almost-vector spaces" that don't assume the axiom $1\cdot v=v$ for scalar multiplication

Question

Consider a structure satisfying all the axioms of a vector space except for $1\cdot v=v$. I don't know if there might might be a name for these, but lest call them "almost-vector spaces". Now, following exercise 1 of §43 in Halmos' Finite-Dimensional Vetor Spaces, 2. edition, let $V$ be a vector space with a projection $P$. Using juxtaposition to denote the scalar multiplication in $V$, define another scalar product $\cdot$ by $$ \alpha\cdot v := \alpha (Pv). $$ Then $V$ with this new scalar multiplication is an almost-vector space. The problem then goes on to ask (slightly paraphrased):

To what extent is it true that this method is the only way to construct an almost-vector space?

This is a kind of open-ended question, but I at least find that the answer is "not always". I'll make an answer myself with the observations I have, and I'll be grateful for any input, review, or further insights.

Answer 1

An almost-vector space can be viewed as a pair $(V, G)$, where $V$ is a vector space and $G$ is an Abelian group. More precisely, let $AVec$ be the the category of almost-vector spaces, $Vec$ be the category of vector spaces, and $AbGrp$ the category of Abelian groups. Then $AVec \cong Vec \times AbGrp$.

As Milten has noted in their answer, given an almost-vector space $W$, we can define a vector space $V_W = 1 \cdot W = \{1 \cdot w \mid w \in W\} = \{w \in W \mid w = 1 \cdot w\}$. And we can define an Abelian group $G_W = \ker(1 \cdot) = \{w \in W \mid 1 \cdot w = 0\}$. Then $W = V_W \oplus G_W$.

Intriguingly, consider two almost-vector spaces $W, X$ and an almost-linear map $f : W \to X$ (an almost-linear map is one that preserves addition and scalar multiplication, but is not necessarily between vector spaces proper). It turns out that for all $w \in V_W$, we have $f(w) \in V_X$, since $1 \cdot f(w) = f(1 \cdot w) = f(w)$. So $f$ restricts to a linear map $f_V : V_W \to V_X$. Similarly, for all $w \in G_W$, we have $1 \cdot f(w) = f(1 \cdot w) = f(0) = 0$. So $f$ restricts to a group homomorphism $f_G : G_W \to G_X$.

Thus, we have a functor $AVec \to Vec \times AbGrp$, which takes $W$ to $(V_W, G_W)$, and which takes $f$ to $(f_V, f_G)$.

To go the other way, note that every vector space is an almost-vector space, and that every linear map is an almost-linear map. Furthermore, note that every group $G$ can be made into an almost-vector space $A(G)$ by equipping it with the scalar multiplication law $r \cdot x = 0$; group homomorphisms become almost-linear maps. This gives us functors $Vec \to AVec$ and $AbGrp \to AVec$, which combine to give us the a functor $Vec \times AbGrp \to AVec$ which sends $(V, G)$ to $V \times A(G)$.

These functors are inverse equivalences. We have $W = V_W \oplus G_W = V_W \oplus A(G_W) \cong V_w \times A(G_W)$; it's easy to see that (binary) direct sums are direct products in the category of almost-vector spaces, just like in the category of vector spaces. It's also fairly straightforward to verify that $V \cong V_{V \times A(G)}$, and that $G \cong G_{V \times A(G)}$. And both these isomorphisms are natural.

Answer 2

Here is a collection of my results - and as I said, comments and further insights are very welcome! I won't really prove anything unless requested, since the results are more interesting than the quite elementary individual proofs.

Prop 1. Let $W$ be an almost-vector space. Then the map $P:w\mapsto 1\cdot w$ is a projection, in the sense that $P$ is linear and $P^2=P$.

Prop 2. With things defined as in Prop 1, let $$U=im(P)=\{w\mid 1\cdot w=w\}.$$ Then $U$ is a vector subspace of $W$. Similarly, let $$V = ker(P) = im(I-P) = \{w\mid 1\cdot w=0\}.$$ Then $V$ is an almost-vector subspace of $W$ (and $I-P$ is a projection). Furthermore, $$W=U\oplus V,$$ meaning that each $w\in W$ can be written uniquely as $u+v$.

Prop 3. Forgetting previous definitions now, let $U$ and $V$ be vector spaces over the same field. Then let $W=U\oplus V$ as a group, and define scalar multiplication component-wise, i.e. $$ \alpha(u+v) = \alpha u + \alpha v. $$ Then $W$ is a vector space.

Corollary to Prop 3. In particular, in the original setting again, if the $V$ from Prop 2 can be endowed with some scalar multiplication that makes it a vector space, then $W$ can be made a vector space (let's call it $W'$ for explicitness) using the component-wise scalar multiplication. Then the almost-vector space $W$ is constructed from $W'$ via the projection $P:u+v\mapsto u$.

I'd say the main results are Prop 2 and the corollary to Prop 3. To summarise, Prop 2 tells us that any almost-vector space decomposes in a canonical way into the direct sum of a vector subspace (in fact the unique maximal vector subspace) and an almost-vector subspace.

The corollary gives a necessary and sufficient condition (which may, however, not be easy to check) for when an almost-vector space is induced by a vector space with a projection. That the condition is necessary really just follows from the fact that kernels of linear maps (in particular $V=ker(P)$) are subspaces.

Finally, let's give an example of an almost-vector space not induced by a vector space. Let $W = \Bbb R\times \Bbb Z_2$ as a group, and define real scalar multiplication by $$\alpha(r, x) = (\alpha r, 0).$$ Then $W$ is an almost-vector space with $U=\Bbb R\times\{0\}$ and $V=\{0\}\times \Bbb Z_2$ as in Prop 2. However, $V$ cannot be endowed with a scalar product making it a vector space (since $(0,1)$ has additive order $2$), so $W$ is not induced by a vector space. Indeed, no scalar multiplication can make $W$ a real vector field for the same reasons.

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Let me add a slightly more interesting and illustrative example, and then I think I'll leave this be.

Consider $W=\Bbb C^2$ as a complex almost-vector space with scalar multiplication defined by $\alpha(z_1,z_2) = (\alpha(z_1+\Re z_2),0)$. Using the notation of Mark Saving's answer, we have $V_W = \{(z,0)\}$ and $G_W=\{(-\Re z,z)\}$. In this case, the standard scalar multiplication on $\Bbb C^2$ (i.e. component-wise multiplication) agrees with our definition on $V_W$, but not on $G_W$ (or anywhere else). Therefore, $W$ is not induced (in the sense of Halmos' projection construction) by the standard $\Bbb C^2$, even though $V_W$ is a subspace.

However, we can endow $G_W$ with a scalar multiplication (induced by the group isomorphism $(-\Re z,z)\mapsto z$) making it a complex vector space, which leads via our corollary to an alternative scalar multiplication on $\Bbb C^2$ given by $$ \alpha(z_1,z_2) = (\alpha z_1+\alpha\Re z_2 - \Re(\alpha z_2), \alpha z_2). $$ And indeed, $W$ is induced by this exotic version of $\Bbb C^2$ via the projection $P:(z_1,z_2)\mapsto(z_1+\Re z_2,0)$.