According to Wikipedia, a Convex cone is some subset of a vector space:
A subset C of a vector space V is a cone (or sometimes called a linear cone) if for each x in C and positive scalars α, the product αx is in C. Note that some authors define cone with the scalar α ranging over all non-negative reals (rather than all positive reals, which does not include 0).
A cone C is a convex cone if αx + βy belongs to C, for any positive scalars α, β, and any x, y in C.
But, eventually, forgetting the vector space, convex cone, is an algebraic structure in its own right. It is a set endowed with the addition operation between its elements, and with the multiplication by nonnegative real numbers. The addition is assosiative, commutative and has a unique identity element, and multiplication is distributive over the addition. My questions.
- Can we define convex cone in this way (not mentioning any vector space in the definition)?
- Is it true that any such convex cone can be canonically extended to a vector space, so, that Wikipedia's definition using this vector space is equivalent to the cone's original definition?
- Is there any source that uses this "independent definition"?
My real question is that how can a cone be defined without mentioning a vector space so, that 2. is true.
I think that 2. is not true with my definition above. E.g. take the addition as $x+y=0$ for all $x, y\in C$ and the multiplication as $\lambda x=0$). Then this addition is associative, commutative and the multiplicatio is distributive, so this structure is a cone according my definition, but evidently cannot be a subset of a vector space if $C$ has at least 2 elements. So this definition is wrong. If we want 2. to be true then we must require some invertible-like property of the addition. For example, we can require the following condition:
For all $x,y,z\in C\setminus \{0\}$, if $x+y=x+z$, then $y=z$.
In this case, the map $$\hat y:C\to C: x\mapsto x+y$$ is injective for all $y\in C\setminus \{0\}$ (but not necessarily surjective).
Now let's identify the elemets of $C$ with these maps, and let's call them positive vectors. Then let's denote the map $\hat y(C)\to C: x\mapsto \widehat y^{-1}(x)$ with $-\hat y$, and let's define the addition of these maps by their composition, the multiplication by non-negative real numbers by $\lambda \hat x:=\widehat{\lambda x}$ and by negative real numbers by $-\lambda \hat x:= -(\lambda \hat x)$. Then all these maps and their compositions together would form a vector space if the commutativity were also true for a negative and a positive vector: $$\hat x+ (-\hat y) = (-\hat y) + \hat x$$ But I don't know, what should we require from the original $C$ so that this becomes automatically true.
You may be looking for the notion of a convex space or barycentric algebra (https://ncatlab.org/nlab/show/convex+space).
They provide an abstract definition of the notion of "convex combination", without mentoining vector spaces directly.
Within this framework convex spaces, which arise as convex subsets of vector space are called geometric. Not all convex spaces are geometric however. For example, combinatorial convex spaces are those, for which the convex combination $\lambda x + (1-\lambda)y$ does not dependent on $\lambda$. They are essentially the same thing as $\wedge$-semilattices and (almost) never geometric.
Taking the product of combinatorial and a geometric convex space yields a convex space which is neither geometric nor combinatorial. Another example of a non-geometric, non-combinatorial space is interestingly the set of all convex subsets of a vector space (cf. https://arxiv.org/abs/0903.5522).