Sorry for the strange question.

If you take a vector space, which has addition and multiplication by scalar (e.g. real or complex) and a certain finite dimension, and you remove the addition operation. What do you end up with? Does it have a name?

I know that this is probably a very useless space but the question is mainly about nomenclature. Perhaps "scalable space".

It looks to me that you end up with set sum of (infinite) one dimensional spaces (rays?) and the elements can be in one of them at time and it has a null element in each, which is the same in all rays.

$1\cdot \mathbf{v} = \mathbf{v} \quad \forall \mathbf{v}$

$0\cdot \mathbf{v} = 0\cdot \mathbf{w} \quad \forall \mathbf{v}, \mathbf{w}$

Not many other interesting properties it seems.

(The reason I ask is because, in a certain computer language I have a "interface" that multiplies values by a scalar-like thing, but there is no summation in general.)

  • 1
    $\begingroup$ Sounds like a field acting on a set. $\endgroup$
    – Hayden
    Nov 3, 2020 at 2:19
  • $\begingroup$ @Hayden, yes, with a little bit of extra though, for example in 3 (original) dimensions, it is a field acting on the projective sphere. $\endgroup$
    – alfC
    Nov 3, 2020 at 2:27
  • $\begingroup$ @Hayden I'd say an action of its underlying monoid. That would also need the associativity. $\endgroup$ Nov 3, 2020 at 2:27
  • $\begingroup$ I think it is some kind product set $\mathbb{R} \times RP^{n-1}$ (and the complex version, not sure about notation.) $\endgroup$
    – alfC
    Nov 3, 2020 at 2:30
  • 1
    $\begingroup$ This kind of set embedded in a vector space is often called a cone. $\endgroup$
    – Joppy
    Nov 3, 2020 at 2:33

1 Answer 1


Have a look at the axioms that are satisfied if you were to remove $+: V \times V \to V$, where $V$ is the original vector space in question. We would still have:

  1. $\forall x \in V: 1 \cdot x = x$

  2. $\forall \alpha, \beta \in \mathbb{F}: \forall x \in V:(\alpha \beta)x = \alpha(\beta x)$

That is all you would be left with if you only had scalar multiplication defined on the set $V$. As far as I can see, this sort of structure doesn't have a specific name (probably because it doesn't seem to be extremely interesting). It's really just a field $\mathbb{F}$ that's providing scalars so that the two statements above hold.

In such a space, it wouldn't even make sense to talk about a basis. Remember, every vector in a vector space can be written as some linear combination of its basis. Since addition is crucial to forming linear combinations, there's no sense in which we can talk about such things in the set above.

In that way, talking about dimensionality would also be pointless. I guess "scalable space" would be a good name for such a structure.

  • $\begingroup$ Yes, there is no concept of basis or dimensionality in this set. $\endgroup$
    – alfC
    Nov 4, 2020 at 3:12

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