How to explain why free loop space, or based loop space, is infinite dimensional to non-math people.

Question

I am giving a math talk to non-mathematicians. I was wondering how to explain how the free loop space, or based loop space, of a topological space is infinite dimensional so that a non-mathematician can understand.

Answer 1

If you're giving a talk to a bunch of physicists, you might get mileage from giving intuition for why the infinitesimal deformations from a given point in the loop space form an infinite-dimensional vector space.

Roughly speaking, to deform a map $f:S^1\to M$, to deform $f$, you can "tweak" it by deforming it on a small neighborhood $U\subset S^1$. Such deformations on disjoint subsets will be independent of each other. There are infinitely many such subsets, hence infinitely many different independent directions, hence infinite dimensions.

It may be easier to explain why a vector space of functions (e.g. compactly supported smooth functions on $M$) is infinite dimensional, because in that case the functions themselves are vectors and you don't have the added difficulty of visualizing deformations.

Answer 2

I would start by giving an idea of what some function spaces look like. Perhaps an easy place to start would be the vector space of polynomials from $\mathbb{R}$ to itself. You can pretty easily show that here we need an infinite number of basis elements to generate any given polynomial. From here, you can say that this space lives inside the much larger space of continuous maps from $\mathbb{R}$ to itself and so this space is also infinite dimensional.

With a bit of hand waving you can say "In fact, in general a function space is an infinite dimensional object in all but the most simple of cases". Here you might like to give an example of a finite dimensional function space - perhaps the space of continuous maps from a single point into the reals. Once the audience is suitably convinced that function spaces are normally infinite dimensional you can introduce the loopspace of a space (the problem here is that we don't, in general, have a vector space - so you may have to field questions about what infinite dimensional actually means for an arbitrary topological space).

Answer 3

Naturally, both previous answers are great and expound upon the fact that function spaces are typically infinite dimensional. To elaborate though, another great way can be to show that certain loop spaces admit filtrations into finite dimensional projective varieties of arbitrary dimension.

I do not know if this can be done in general, but if you're working with Lie groups then the trick is effectively the same as showing that $C^\infty(\mathbb R)$ is infinite dimensional by showing that the polynomials are an infinite dimensional subspace.

Details: If your space is a real lie group $K$ there is a well defined notion of polynomial based loops $\Omega_\text{poly}K$. This space is technically defined as $L_{\text{alg}}K_{\mathbb C} \cup \Omega K$ where $\Omega K$ is the based loop space of $K$ and $L_{\text{alg}}K_{\mathbb C}$ are the maps which arise as the restriction of algebraic maps $\mathbb C^\times\to K_{\mathbb C}$ (edit: ie $L_{\text{alg}}K_{\mathbb C} = K_\mathbb{C}(\mathbb C[t,t^{-1}])$).

This space admits a filtration $$\Omega_{\text{poly},1}K\subseteq \Omega_{\text{poly},2}K \subseteq \cdots \subseteq \Omega_{\text{poly}}K$$ where each $\Omega_{\text{poly},n}K$ is isomorphic to $\cup_{k=1}^n Gr(k,2n+1)$ where $Gr_{\mathbb R}(k,2n+1)$ is the Grassmannian of $k$-spaces in $\mathbb R^{2n+1}$. This is certainly finite dimensional, being a union of finite dimensional spaces. It is also clear that you can achieve arbitrarily large dimension, so $\Omega_{\text{poly}}K$ is infinite dimensional.

Now by definition, $\Omega_{\text{poly}}K \subseteq \Omega K \subseteq \Lambda K$, so all of these spaces are also infinite dimensional.