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Let's start with a denumerable set $S$. The usual process to define a free vector space over $S$ is as follows:

A free vector space over $S$ consists of a tuple $(V,\alpha)$, satisfying the following universal property: Given any other $(W,\beta)$ there exists a unique linear map $\mu: V \to W$ that makes the following diagram commute.

Of course to allow for the extra linear structure, $V$ should be "bigger" than $S$, so $V$ consists of all functions $S\to \mathbb{R}$. These functions are considered as "formal sums", namely a typical element of $V$ can be seen as $\phi=\sum a_i s_i$, $a_i \in \mathbb{R}$ and $s_i$ is a basis, $s_i : S \to \mathbb{R}$ that satisfies $s_i(s_j)=\delta_{ij}$. Aha, so $\sum a_i s_i$ acts on an element of $S$, say $s_k$ and gives $\left(\sum a_i s_i \right)(s_k)=a_k$. So far so good.

What happens if I try to free a vector space over a non-denumerable set $S$, say an interval of $\mathbb{R}$. By naive analogy should I get "formal integrals" as elements of the free vector space? Does this connect by any means to distributions?

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First, the free vector space on $S$ is not the set $F(S, \mathbb{R})$ of functions $S \to \mathbb{R}$. That's too big. Indeed, in the universal property, consider $(W, \beta)$ to be $W = \mathbb{R}$ and $\beta(s) = 1$ for all $s \in S$. Then you cannot define $\mu(v)$ where $v \in V = F(S, \mathbb{R})$ is defined by $v(s) = 1$ for all $S$, because you would be computing $\sum_{s \in S} 1$ which doesn't converge in $\mathbb{R}$, because $S$ is infinite.

Instead, the free vector space $V$ on $S$ is a subset of $F(S, \mathbb{R})$. More precisely, it is the subset of functions $f : S \to \mathbb{R}$ such that $f(s) = 0$ for all but a finite number of members $s \in S$. I'll let you check the universal property, carefully this time.

Now, this doesn't depend at all on $S$ being denumerable. You can make the exact same definition for any set $S$, even not denumerable, and you'll still get the free vector space on $S$.

If you want to have a little bit more intuition about the free vector space, you were almost right with your description as formal sums $\phi = \sum_{s \in S} \phi_s s$, except that the coefficients $\phi_s$ need to be zero for all but a finite number of indices $s \in S$. Indeed, recall that in the definition of a basis, every element can be written as a linear combination of elements of the basis. And a linear combination is always a finite sum.

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  • $\begingroup$ You're absolutely right. What is the missing ingredient to connect the free vector space with "formal integrals"; does that even make sense? My question is basically how do infinite dimensional vector spaces pop up in this categorical language? $\endgroup$ – EEEB Mar 29 '17 at 16:19
  • $\begingroup$ A categorical way of seeing this is any set $S$ is the $S$-fold coproduct of $1$. Since the free vector space functor, $F$, is a left adjoint, it preserves colimits and thus coproducts. So $$F(S)\cong F(\sum_{s\in S} 1)\cong\bigoplus_{s\in S}F(1)$$ This "reduces" the problem to an infinite direct sum. $\endgroup$ – Derek Elkins Mar 29 '17 at 16:20
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    $\begingroup$ There's no need for formal integrals, whatever that would mean, here, because all the sums in the free vector space are finite. You can ask for a free Hilbert space on a denumerable set, and that brings in "formal" denumerable sums (though formality is questionable when the infinite sums are not arbitrary,) but it's hard to get some construction on an uncountable set yielding formal integrals. If you squint, you might be able to see this as describing $L^1[0,1]$ as a free Banach space on $[0,1]$, but that's not literally true.maths.ed.ac.uk/~tl/cambridge_ct14/cambridge_ct14_talk.pdf $\endgroup$ – Kevin Carlson Mar 29 '17 at 16:30
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Najib Idrissi has answered your main question. Let me say a little about why your intuition that you should get "formal integrals" is wide off the mark. You seem to think that integrals are to uncountable sets as sums are to countable sets. However, this is simply not true. Integrals are a certain very special way to sum over infinite sets which are equipped with a lot of extra structure.

For example, if I give you a nice function on $\mathbb{R}$, you can integrate it. But this uses a huge amount of special structure on $\mathbb{R}$, and also a huge amount of special properties of the "nice" function we're integrating. If I handed you some totally random uncountable set $S$ that has no connection to $\mathbb{R}$ or any familiar constructions from analysis, there is no natural way to define integrals of functions on $S$. Or if I hand you a totally random function on $\mathbb{R}$ which has no nice analytic properties (is not continuous, or even remotely close to being continuous (aka "measurable")), it is not possible to define its integral.

(Actually, what I said above is not quite true: there is always a natural way to define "integrals" of functions on $S$, but it is just to define integrals as sums. Infinite sums indexed by any set make sense; this has nothing to do with whether the set is countable. However, it turns out that an infinite sum can never converge unless all but countably many of its terms are zero, which is one reason you don't hear about uncountable sums very often.)

Now when we're defining the free vector space on a set $S$, all we know about $S$ is that it is a set. The definition works for any set, and does not use any structure of $S$ besides the fact that $S$ is a collection of elements. So you should not expect integrals to pop up at all, since they don't make sense for arbitrary sets (unless by "integral" we just mean "sum"). Even if $S$ is a "nice" uncountable set like $\mathbb{R}$, you still shouldn't expect integrals to pop up, because the definition of the free vector space doesn't use any of the nice structure of $\mathbb{R}$ that lets us define integrals. As far as the free vector space is concerned, $\mathbb{R}$ is just a set of unrelated points with no geometric or analytic structure at all.

There is also another reason not to expect integrals to show up, which is that in a vector space, you can only take finite sums. There is no operation on a vector space that allows you to take a sum of infinitely many elements: it's not just that such a sum might not converge, but that the term "converge" is meaningless because the definition of a vector space does not include any such concept. So even in the countable case, you get only finite formal sums, not infinite formal sums. If you're only ever able to take finite sums and have no notion of infinite sums or convergence, you certainly shouldn't expect to be encountering any integrals.

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  • $\begingroup$ Thanks for taking the time to give a comprehensive answer Eric. How do infinite dimensional vector spaces pop up? Do you start with an unstructured set $S$, free a vector space on it, namely $V$, and then endow this $V$ with whatever nice extra structure is necessary, i.e. topology, measure and so on? It is a bit unclear to me how starting from a finite dimensional vector space you get to infinite dimensional ones. $\endgroup$ – EEEB Mar 29 '17 at 16:57
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    $\begingroup$ No nice extra structure is necessary to have an infinite-dimensional vector space. For example, if $S$ is any infinite set, the free vector space on $S$ is infinite-dimensional. An infinite-dimensional vector space is just a vector space that happens to not have a finite basis. This doesn't mean it has to be nice or have a topology or any other structure we often see on specific spaces in analysis. I'm not sure what you mean by "starting from a finite dimensional vector space you get to infinite dimensional ones". $\endgroup$ – Eric Wofsey Mar 29 '17 at 17:05

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