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I got pretty confused during my attempt to use the set mentioned in the title, namely the set of all vector spaces that use the same set of scalars $S$, as the set of 'vectors' to make a new vector space, $V$. I defined the 'sum' of two 'vectors' (vector spaces), $V_i$ and $V_j$ as their tensor product space $V_i\otimes V_j$, and used the set $S$ as my set of scalars, defining 'multiplication by a scalar' by $cV_p$, $c\in S$ meaning the vector space $V_q$ containing all the vectors in $V_p$ except multiplied by $c$, which should be OK I think since all of the vector spaces in $V$ have a clear meaning to multiplying by an element of $S$ since that's what 'multiplication by a scalar' in those vector spaces means (although given that those vector spaces are closed under this operation, it seems that my scalar multiplication basically means 'do nothing' (which is fine by me)). In the book I'm reading it says the set $V_i\otimes V_j$ is a vector space, and it seems that $\otimes$ does all the things you'd want vector addition to do, and so it seems $V$ satisfies all the axioms needed for it to be a vector space, apart from two: I don't see how it can contain a zero element, nor an additive inverse for each element in $V$.

The reason I'd want to do such a crazy thing is because if I can make this set $V$, the set of all vector spaces using the $S$ for their concept of scalar multiplication, into a vector space, in particular one that uses $S$ as I've described, then surely $V\in V$ !

So is there some way to fix this, either by coming up with some clever vector spaces to be additive inverses and the zero element or perhaps by redefining vector addition? Or is this set $V$ doomed to never be a vector space?

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I can't see any natural way to turn the set of vector spaces over a fixed field $k$ into a vector space itself. In particular as you notice, $cV=V$, so the scalar action is not invertible.

But I don't see what vector spaces have to do with your goal. If you want to talk about a set of all widgets, and notice that this set is itself a widget, and should therefore contain itself, you can already have that. The set of all sets would contain itself (if it existed). The set of all ordinals would itself be an ordinal. The category of all categories is itself a category and one may wonder whether it should therefore also contain itself, by this reasoning.

Because such objects can lead to contradictions, they are barred from existence in ZFC set theory, by using the axiom of restricted comprehension, one is only allowed to construct subsets of known existing sets, not all sets. But I think it is still useful to think about them.

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  • $\begingroup$ So is the 'set of all vector spaces using the same set of scalars' not allowed to be a thing? $\endgroup$ – Kaonicping Nov 14 '17 at 20:11
  • $\begingroup$ @Kaonicping Well the scalar action you've proposed certainly doesn't do it, since it's not a ring homomorphism. But I'm not saying there's no way to do it. There may a way to endow the set of vector spaces with a vector space structure (and if I were looking for one, I would start with direct sum instead of tensor product). But the point is, there are lots of places where the set of all widgets would seem to contain itself, even if vector space is not such a widget. $\endgroup$ – ziggurism Nov 14 '17 at 22:09
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You might be looking for the tensor algebra

which consists of $$T(V):=\bigoplus_{i=0}^{\infty} T^i(V)$$

with $T^i(V)=V \otimes V \dots \otimes V$, and $T^0:=k$, the base field.

This is not the "set of all vector spaces," but it it basically looks at how one can get a "vector space" out of all "possible tensor products." The salient feature however is the multiplication given by tensor product.


Anyhow, if you ignore the previous comment we can always make a purely formal vector space, it is just not interesting. In particular, let $S$ be the collection of $\mathbb R$-vector spaces up to isomorphism. Then $S:=\{\mathbb R, \mathbb R^2, \dots\}$. Now, just let these be objects and define scalar multiplication formally, and addition in the same way. I want to reiterate that this is not an interesting construction in my mind.


The problem you're identifying is basically the equivalent of Russells Paradox.

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