Consider, as a motivating example, the multiset $\left(\mathbb{P}_n,2\setminus\mathbf{0}\right)$ consisting of the underlying set of polynomials of order $n$ and lower, all with multiplicity $2$—expect for the polynomial identically equal to $0$, $\mathbf{0}$. (Heuristically, this is intended to be what occurs when you "clone" everything in $\mathbb{P}_n$ except for the zero vector; I apologize for incorrect notation.)

If I look at the axioms that a set—combined with typical addition and scalar multiplication operations—would need to satisfy in order to qualify as a vector space, I would find that this multiset satisfies most if not all of them. For example, addition and/or scalar multiplication of the elements in this multiset will always produce another element in the multiset.

With this in mind, and given that many other such examples can be generated, is there any meaningful extension of the notion of a vector space to multisets?

A comment: to clarify the properties I'd like to see retained in such an extension, I'd attempt to preserve at all costs the notion of closure (that addition and scalar multiplication acting on two elements in the multiset always creates another element in the multiset) while attempting as much as possible to keep "non-unique" versions of the other axioms; for example, that there still exists at least one additive identity element $e$ in the multiset, but that identity element is perhaps not unique.

  • $\begingroup$ How willing are you to give up uniqueness? Gerhard "One Won't Be Loneliest Number" Paseman, 2018.06.18. $\endgroup$ Jun 18 '18 at 23:49
  • $\begingroup$ Willing and able. My expectation is that a host of uniqueness theorems for these vector space analogues would fail, but other theorems would hold. Consider the Hilbert projection theorem for some Hilbert space $H$; for some multiset corresponding to a "cloned Hilbert space", the norm-minimizing element uniqueness portion of the theorem fails but the orthogonality portion holds. (If $\color{red}x$ and $\color{blue}x$ are both minimizing elements of $||x-y||$ where $y$ is in some sub"space" $C$, I expect both $\color{red}x - y$ and $\color{blue}x - y$ to be orthogonal to the elements in $C$.) $\endgroup$ Jun 19 '18 at 0:08
  • $\begingroup$ I'd need much more complete definitions. $\endgroup$
    – Wlod AA
    Jun 19 '18 at 0:56

It is hard to posit more than one additive identity by the usual proof:

$e+f=f$ if $e$ is an additive identity and $e+f=e$ if $f$ is an additive identity hence $e=f$ if both are.

If the underlying field is $F=\{0,1\}$ then scalar multiplication is no issue and the vector space $W=F^{n+1}$ is essentially the double of $V=F^n.$ I'll start with a description which preserves some of the mystery: For each vector $x \in V$ we have two copies in $W,$ say $\color{red}x$ and $\color{blue}x.$ The rules are that for $x+y=z$ in $V$ we have

  • $\color{red}x+\color{red}y=\color{blue}x+\color{blue}y=\color{red}z$
  • $\color{red}x+\color{blue}y=\color{blue}x+\color{red}y=\color{blue}z$

The axioms are easy to verify. The way this diverges slightly from what you describe is that also additive identity $\mathbf{0} \in V$ yields two vectors: $\color{red}{\mathbf{0}}$ is the additive identity while $\color{blue}{\mathbf{0}}$ is the color flipper.

Of course this is basically saying that if $u \in W=F^{n+1}$ is any fixed non-zero vector then $D=\{0,u\}$ is a subspace , $W/D $ is isomorphic to $F^n$ and the map which sends $x$ to $\{x,x+u\}$ is $2$-to-$1$

A similar thing can be done over any finite field and can be extended to infinite dimensional vector spaces over finite fields. The scalar multiplication (not an issue when the only non-zero scalar is $1$) should be easy to figure out.


The way I see it, a "multiset" is essentially a set $\tilde X$ together with a surjective map $\tilde X \buildrel\varphi\over\to X$, where $X$ is the "underlying set" of the multiset, such that the map $\varphi$ has finite fibers (the cardinal of each fiber $\varphi^{-1}(x)$ being the "multiplicity" of the element $x$ in the multiset).

So if I were to try to speculate and define a "multigroup", it would be a surjective map $\tilde G \to G$, which we certainly want to be a morphism of groups, and which has finite kernel. So, a short exact sequence $1 \to M \to \tilde G \to G \to 1$, where the kernel $M$ is finite and determines the multiplicity of every element of the multigroup (which has to be the same). And I'm tempted (although perhaps not with good reason) to demand that $M$ be central, in which case we have the classical notion of a central extension.

In the case of vector spaces, however, all short exact sequences split, so I would say that the only "vector multispaces" are the $M \oplus V$ with $V$ the underlying vector space and $M$ a "multiplicity space" that must be finite (which means it is either trivial or the base field is also finite). Not very interesting.

Of course, this is essentially demanding that scalar distributivity and commutativity holds exactly, and maybe that is too much: you might ask yourself, for example, whether you consider the cyclic group of order $4$, seen as an extension $1 \to Z_2 \to Z_4 \to Z_2 \to 1$, to be a "vector multispace" (where each element has multiplicity $2$) over $\mathbb{F}_2$ whose underlying vector space is $\mathbb{F}_2$ (note that $x + x = 2x = 0$ holds only "at the base"), and similarly, whether the quaternion group $1\to Z_2 \to Q \to (Z_2)^2\to 1$ should be a "vector multispace" (where, again, each element has multiplicity $2$) over $\mathbb{F}_2$ whose underlying vector space is $(\mathbb{F}_2)^2$ (here, commutativity holds only "at the base"). Maybe even associativity can be weakened by considering short exact sequences of loops or some such thing.

In the end, I think it really depends what you are willing to admit and, even more fundamentally, what you want to do with such objects.

  • $\begingroup$ Instead of $\mathbb{Z}_4$ and $Q$ why not actual vector spaces with $4$ or $8$ elements? $\endgroup$ Jun 20 '18 at 4:15
  • $\begingroup$ @AaronMeyerowitz I'm not sure I understand your comment, but the vector spaces with $4$ or $8$ elements are unproblematic and fall under the purview of the split (direct sum) case I mentioned in my previous paragraph (and which is also what you mention in your own answer); I don't think there's much to say about them. What I was trying to give a sample of, with $Z_4$ and $Q$, were various problematic cases where I think one needs to think about what one wants to admit or reject. (Should $Z_4$ be considered a "vector multispace"? I don't know. But it makes sense to think about it.) $\endgroup$
    – Gro-Tsen
    Jun 20 '18 at 12:35

For any vector space $V$ and any function $m:V\to\mathbb N$ you could say you have a vector space $V$ whose underlying set is a multiset with multiplicities given by $m$. This doesn't mean that the vector space structure and multiplicity function interact at all, though.

On the other hand, you could turn your multiset into a set by labeling the "identical" elements, for instance calling the two instances of $x^2+3x$ by names like $\color{blue}{x^2+3x}$ and $\color{red}{x^2+3x}$. But then you get into some issues like:

Which is true: $\color{blue}x+\color{red}x=\color{blue}{2x}$ or $\color{blue}x+\color{red}x=\color{red}{2x}$?

  • $\begingroup$ Is the first suggestion to try and create the vector space $V$ out of the underlying set of whatever multiset one is working with? If so, that should undoubtedly work if the underlying set can form a vector space but potentially tricky to work with if, for example, the multiplicity function is unknown. For the second, I could potentially say both are true and discard uniqueness; the key is that such an addition of elements is "closed" in the sense that you always produce another element of the multiset. $\endgroup$ Jun 19 '18 at 0:30

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