What is left of a Vector Space if we remove the sum?

Question

Consider a mathematical model for physical space, let's say a (static) 2-dimensional plane. Such a model may start from a given set (of points) with some algebraic structure over it. Usually, as a minimum, this structure is assumed to be an Affine Space; then, once a point is chosen as the "origin" (which amounts to placing an "observer" somewhere in the physical plane), the points of the affine space are mapped to vectors in the associated Vector Space.

This works if it is assumed that the observer may deduce a "relationship" (in position) between any two points. But let's assume that it is only possible to know the position of a point relative to the origin - think of the observer as a "radar". At this point it doesn't make sense to add vectors. Basically we have a vector space structure stripped off of the sum operation, and having $\mathbb R^+$ as the "field" of scalars.

To clarify, I'm assuming that the observer cannot move to a different location to locate points from there. If it is assumed that relative positions add as vectors, the relative position between two points out of the origin may be inferred from their positions with respect to the origin. But actually this is something that I want to prove by adding homogeneity to the space.

I know that if we remove the product a vector space becomes a group, but what does it become if we remove the sum? I was thinking about a Cone, but such a concept is defined as a subset of a vector space, so with a concept of sum actually.

Answer 1

I'm inclined to say that 'cone' is indeed the best answer; however we need to be clear about that we are only talking about the surface of the cone here. The convex-cone you used as a tag would refer to the massive filled up cone.

Cones can be defined without reference to a bigger space to embed them into, see Wikipedia: https://en.wikipedia.org/wiki/Cone_(topology)

Following wikipedia we could make 'the cone of a circle' as the object found by making the cylinder $S^1 \times [0, 1]$ and then collapsing the boundary $S^1 \times \{0\}$ to a single point. This yields a closed cone.

Your cone is topologically equivalent to the interior of this thing, i.e. the open cone you are left with when removing the other boundary $S^1 \times \{1\}$. Of course we could construct this directly by first making the half-open cylinder $S^1 \times [0, 1)$ and then collapsing the boundary $S^1 \times \{0\}$ to a point.

This is dealing with your $\mathbb{R}^2$-example, for general $\mathbb{R}^n$ we would use the same construction with $S^{n-1}$ in place of $S^1$.

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Now one could argue that in doing this we are throwing away too much information, being left only with the topology and forgetting all the geometry, but this could be remedied by 'remembering' the distance to $0$ for each point.

By construction each point has already coordinates of the form $(\sigma, x)$ with $\sigma \in S^{n-1}$ and $x \in [0, 1)$. We understand that $x$ is the only info we need to 'measure' the distance to $0$.

The question remains what distance function feels natural. Any function $f$ that maps $[0, 1)$ into $[0, \infty)$ monotonously will do, so I guess something with $\log$ or $\tan$ could be used to make this work. The fact that we have many options to choose from here is already an indication that we did not really lose any information when passing to the topological picture, the idea being that the fact that the distance only depends on the second coordinate, which is 'still there' indicates that nothing was lost. But someone else might perhaps argue that the fact that we can choose many different distance functions here in fact indicates that we did lose information since originally there was only one choice. Both standpoints make sense to me, which indicates that 'forgetting structure' is a subtle topic.

Answer 2

This is a tentative partial answer to my own question, containing also open points which require further investigation. If anybody has any suggestion, you are welcome to comment or post a different answer.

The construction of a cone is indeed eye-catching, but the problem there is defining the "base" set. If we start with a metric, such a set may simply be the unit circle; however, as I also explained in my edits to my original question, I want to deal with the scenario before introducing a metric.

I have been elaborating on the first comment by @psl2Z (does the mention work here?):

You would have an action of the monoid $(,\cdot)$ on a set $$, where $(,+,\cdot)$ is the field.

Let us indeed start from a bare set $V$, whose elements are thought to represent (relative) positions (with respect to a chosen spatial point, i.e. to an "observer"). Instead of a generic field, I want to consider $\mathbb R$ from the outset; more precisely, let us assume that we can multiply any element of $V$ by any positive real number and obtain an element of $V$. (The heuristic idea is that space extends indefinitely in each direction.)

We are then given an operation from $\mathbb R^+\times V$ into $V$. In other words, if $\alpha\in\mathbb R^+$ and $x\in V$ then $\alpha x\in V$. Of course, we want $1x=x$ and $\beta(\alpha x)=(\beta\alpha)x$. Since $(\mathbb R^+,\cdot)$ is a group (and not only a monoid), this defines a group action. Further, if $\alpha\neq1$ we want $\alpha x\neq x$ for all $x\in V$, so the action is free. It is not transitive (heuristically, positions in different directions are not proportional to each other).

Let us introduce an equivalence relation $\sim$ in $V$ by $y\sim x$ iff $y=\alpha x$ for some $\alpha>0$. The equivalence classes are the (oriented) directions. We may even use $\mathbb R^*$ ($=\mathbb R-\{0\}$) for the group, so that each direction has its opposite-oriented one.

I would like now to introduce a concept of dimension and a concept of isotropy. For the former, the problem is that I don't have the sum in $V$ (to form linear combinations); for the latter, I need to introduce rotations, but I don't have a metric (to define them as isometries).

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EDIT

I have meanwhile elaborated a bit on my ideas, and here's what I came up with. I am hereafter classifying the two actions as "Dilations" and "Rotations". Maybe in a "physical-based" exposition the dilations should be seen as existence of "all" "distances" from the origin, while the rotations as possibility to "look around".

Dilations

Let $V$ be a set (of (relative) positions). Let

$$\cdot: \mathbb R^\times \times V \rightarrow V$$

be a free action of the commutative group $(\mathbb R^\times,\cdot)$ on $V$, i.e.

1. $1\cdot\mathbf x=\mathbf x$;
2. $\beta(\alpha\mathbf x)=(\beta\alpha)\mathbf x$;
3. $\alpha\mathbf x\neq x$ if $\alpha\neq1$.

For $\mathbf x,\mathbf y\in V$, let $mathbf y\sim\mathbf x$ iff $\mathbf y=\alpha\mathbf x$ for some $\alpha\gt0$ (equivalence relation). Let $D:=V/\sim$ (set of directions). For each $d\in D$ and $x\in d$, let $-d:=[(-1)\cdot\mathbf x]$ ($\in D$) (opposite direction with respect to $d$).

Rotations

Let $R:=\mathbb R[\text{mod2}]$ (set of rotation parameters). A complete set of representatives is $]-1,1]$. $(R,+)$ is a commutative group. Let

$$ \times: R \times V \rightarrow V $$

be an action of $(R,+)$ on $V$. In formulae:

1. For $[r],[s]\in R$, $[s]=[r]$ iff $s-r\in2\mathbb{Z}$;
2. For $\mathbf x\in V$, $0\times\mathbf x=\mathbf x$;
3. For $r_1,r_2\in R$ and $\mathbf x\in V$, $r_2\times(r_1\times\mathbf x)=(r_1+r_2)\times\mathbf x$.

Let us assume that $\cdot$ and $\times$ commute, i.e.

$$ r \times (\alpha \mathbf x) = \alpha \cdot (r \times \mathbf x) $$

for $\mathbf x\in V$, $\alpha\in\mathbb R^\times$, $r\in R$.

At this point, $\times$ is $\sim$-preserving, i.e., for $\mathbf x,\mathbf y\in V$ with $\mathbf y\sim\mathbf x$ and $r\in R$, $r\times\mathbf y\sim r\times\mathbf x$. In other words, for $r\in R$, $d\in D$ and $\mathbf x,\mathbf y\in d$ (i.e. $\mathbf x,\mathbf y\in V$ and $\mathbf y\sim\mathbf x$), $[r\times\mathbf y]_\sim=[r\times\mathbf x]_\sim$ (which we call $r\times d$).

Further, let us assume that $\times$ is simply transitive on $D$, i.e., if $d_1,d_2\in D$, there is a unique $r\in R$ s.t. $d_2=r\times d_1$.

As a consequence, if $r\times\mathbf x=\mathbf x$ for some $\mathbf x\in V$, then $d=r\times d$ where $d=[\mathbf x]$ (because $\times$ is $\sim$-preserving), thus $r=0$ (because $\times$ is simply transitive on $D$, and then free on $D$), i.e. $\times$ is free on $V$ (in particular, if given $\mathbf x_1,\mathbf x_2\in V$ there is an $r\in R$ s.t. $\mathbf x_2=r\times\mathbf x_1$, such an $r$ is unique).

Finally, let us note that, for $r\in R^\times$ and $d\in D$, $r\times(r\times d)=d$ iff $(r+r)\times d=d$ iff $(r+r)=0$ ($\times$ is free on $D$) iff $r=1$. Now, there is a unique $r\neq0$ s.t. $r\times d=-d$. Let us choose an $\mathbf x\in d$; then, $r\times\mathbf x=\alpha\mathbf x$ for some $\alpha\lt0$ and $r\times(r\times\mathbf x)=\alpha^2\mathbf x$ ($\cdot$ and $\times$ commute). Since $\alpha^2\gt0$, $\alpha^2\mathbf x\in d$; thus, $r\times(r\times d)=d$ and $r=1$. Shortly,

$$ 1\times d = -d $$

for $d\in D$. Also, $1\times\mathbf x=\alpha\mathbf x$ ($\alpha\lt0$) and $\mathbf x=0\times\mathbf x=(1+1)\times\mathbf x=1\times(1\times\mathbf x)=\alpha^2\mathbf x$, i.e. $\alpha^2=1$, $\alpha=-1$ and in the end $1\times\mathbf x=-\mathbf x$ for $\mathbf x\in V$.

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Let's discuss isotropy. I think that in the setup above this concept is already captured by the fact that I'm using the "normal" sum (modulo 2) over $R$. Let us relax this assumption and just pretend that there is some operation $\boxplus:R\times R\rightarrow R$, forming a group and acting simply transitively on the set $D$ of directions. I suppose that the condition of isotropy is:

If $d_1,d_2,d'_1,d'_2\in D$ and $r,\bar{r}\in R$, with $d_2=r\times d_1$, $d'_1=\bar{r}\times d_1$ and $d'_2=\bar{r}\times d_2$ (we rotate everything by $\bar{r}$), then $d'_2=r\times d'_1$.

From here, it is easy to show that $\boxplus$ is abelian. Question: is this sufficient to get the ordinary sum over the reals (reduced to the sum mod 2 over $R$)?

BTW, the answer is "almost" yes, based on this question. That needs to be extended to account for the mod2.

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One other note is that what above will for sure lead to a 2-dim result, so it has to be extended to (at least) 3-dims.

Answer 3

Maybe this is more of a comment, but I post as an answer because it will be pretty long I think.

I think one way to interpret your question is this: if I have a vector-space $V = \mathbb{R}^n$, forget about addition but describe (some of) the things I still remember to someone, what does the class of all mathematical objects that fit my description look like?

In particular: are all these objects obtained by taking the same $V$ and forgetting addition or could we find truly differen examples in the same class?

Here are two ways of making this concrete.

Approach 1.

Suppose I 'list' all elements of $V$ (easy enough, as they are $n$-tuples) and definine a 'scalar multiplication' with elements of $\mathbb{R}$ on this set simply by stating for each $\alpha \in \mathbb{R}$ and $\mathbf{v}$ which element in the set $V$ equals the product $\alpha \mathbf{v}$.

Now when given this information someone can come up with a addition $+$ on $V$ so that the scalar multiplication distributes over $+$ and so that together the newly invented $+$ and all the info you already have turn $V$ back into the vector space you started with. But...

Question: could you also invent a different $+$ that also distributes over the scalar multiplication (which then necessarily also turns $V$ back into a vectorspace) but such that with this different addition the dimension of the vectorspace is something else than $n$?

In a sense this is a reformulation of the question at the end of your answer if dimension can be reconstructed even after forgetting addition: the answer to that is yes if the answer to my question here is no.

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Approach 2.

Suppose I define the class of things I am looking for as follows:

objects in this class consist of

• a set $V$,
• a special point $\mathbf{0} \in V$
• a group-action of the group $\mathbb{R}^\times$ on $V$ satisfying:
  • The orbit of $\mathbf{0}$ under this orbit is the singleton $\{\mathbf{0}\}$
  • The action of $\mathbb{R}^\times$ on the orbit of $\mathbf{x} \in V$ for any $\mathbf{x} \neq \mathbf{0}$ is free

Question: does this class contain objects that are not obtained by taking $\mathbb{R}^n$ for some $n$ and forgetting the addition?

Remark: the group-action of $\mathbb{R}^\times$ on $V$ can be extended to a scalar multiplication of $\mathbb{R}$ on $V$ as in approach 1 by simply defining $0\mathbf{x} = \mathbf{0}$. I didn't add this as a fourth bullet in the definition because the other bullets already make it so that you can always do this.

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How are 1 and 2 different?

I don't have a full answer, but for one thing I think that question 2 is a lot easier to answer. Here is an example: just let $V$ be the union of 3 lines through the origin in $\mathbb{R}^n$ for some $n \geq 2$. Also in this example it is clear that we cannot reconstruct the dimension of the ambient space, since this ambient space is not part of the object and neither are the angles between the lines.

Note that the 3-line example that answers question 2 can never be created in the way described in approach 1. So apparently we rememeber more there. But how much more? Is it just the number of lines, of even more than that? In a sense that is what question 1 is about.

But the whole 'how are approaches 1 and 2 different' is also a way of asking you: which of the questions is closer to what you are asking?

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Note on topology

I didn't do it now, but I think you should always remember the topology. So in approach 1 the topology on $V$ is one of the things you remember and in approach 2 I should have said that $V$ is a topological space (rather than set) and that the groupaction is continuous (rather than just a random action).

The reason that I say that you should remember the topology is that you are working over $\mathbb{R}$. You can't define $\mathbb{R}$ without introducing its topology. This is particular clear if you define $\mathbb{R}$ by taking $\mathbb{Q}$, imposing the topology it (in our view) inherits from $\mathbb{R}$ and creating $\mathbb{R}$ by adding all limits (the standard Cauchy definition) but it is also true if you define $\mathbb{R}$ be somehow specifying it as a set and telling how addition and multiplication work. The cute thing about $\mathbb{R}$ is that a number is non-negative if and only if it is a square, and that $a \geq b$ if and only if $a - b$ is non-negative. So we can define the order purely in terms of the algebraic properties of real numbers and it turns out that the so called 'order topology' (which can be defined on any ordered set) turns out (in the specific case of $\mathbb{R}$) to be equivalent to the standard metric topology.

Long story short:the topology is so intimately connected to $\mathbb{R}$ and hence $\mathbb{R}^n$ that you cannot assume to have forgotten it if you 'only' forget addition. I think. Maybe this is up for debate.

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Back to question 1

You maybe noticed that I could immediately answer my own question 2 but didn't even attempt to answer question 1. I do not know what the answer is. My intuition is torn here. Based on what I wrote in the third paragraph I would think that 'the number of lines' is the only information that approach 1 remembers and approach 2 does not; in that case it should be possible to give the set $V$ any dimension. On the other hand, given what I wrote in the 4th paragraph I would expect that you can only return to the dimension $n$ you started with - this is because you can read off the dimension of any open subset of $\mathbb{R}^n$ remembering only its topology, which means forgetting a lot more than we are talking about here. But on the other hand it might also mean that more of the topology is encoded in the addition than I thought.

I would be very interested if you could solve question 1, but, as I wrote before, in a sense this whole long post is just a long way of asking you: 'what is your question? Is it more like question 1 or more like question 2?' in the latter case, maybe what I wrote there is the answer and this class of objects does, I think, not have its own name.