I found cited in a physics paper that $H^*(B\mathbb{Z}_2) = \mathbb{Z}_2[a]$ and I wanted to know what geometric shapes this corresponded to. Notice the answer is a polynomial ring. Here $\mathbb{Z}2 = \mathbb{Z}/2\mathbb{Z}$.
Wikipedia's article on classifying space refers to homotopy theory and gives the example that the circle $S^1$ is the classifying space for the infinite cycle group $\mathbb{Z}$. This is written as $E\mathbb{Z} = \mathbb{R}$ and they are discussing the "universal bundle".
I'd like a more down-to-earth explication. What kind of objects are being represented by the classifying space of $\mathbb{Z}_2$ and what do the equivalences classes in the cohomology ring look like?
What could mod-2 cohomology be measuring about the shape of the space? Here it says $M_3$ could we could have $M = S^3 / i(S^1)$ with some reasonable map $i : S^1 \to S^3 = \mathbb{R}^3 \cup \{ pt\}$ (a "knot"). This is taken from a Physics paper from arXiv which I am still looking at [1].
Just to collect from the comments:
- $B\mathbb{Z}_2$ is the classifying space of principal $\mathbb{Z}_2$-bundles over paracompact spaces. What I am seeing is an perfectly good abstract definition:
For a discrete group $G = \mathbb{Z}/2\mathbb{Z}$ is ... a path-connected topological space $X$ such that the fundamental group of $X$ is isomorphic to $G$ and the higher homotopy groups are trivial.
Since I have only read through Munkres and Hatcher the last parts of that sentence are challenging.
It sounds like we could just say "two-sheet cover".
The cohomology ring of real projective space $H^*(\mathbb{R}P^\infty, \mathbb{Z}_2) = \mathbb{Z}_2[a]$ which is thes same as the ring in question. Here they do mention that for the configuration space of three points on a sphere $H^*(S^{m-1}) \simeq \mathbb{Z}[a]/(a^2) $ so this is already a much smaller ring than the answer that we have.
Reading through the definition of paracompact space we have, at least that metric spaces are paracompact and CW complexes are paracompact.
Here "geometric" just means "shape". Here's another possible question... why are 2-sheeted covers of CW complexes classified by a polynoial ring?
It's unclear how to respond to your question without knowing more about your background in topology. In any case, this answer will mostly repeat things that have already been said in the comments.
$B \mathbb{Z}_2$ can be thought of as either the classifying space of principal $\mathbb{Z}_2$-bundles (which correspond to double covers) or as the classifying space $BO(1)$ of real line bundles. Maybe the most "geometric" description of it is that it is an infinite real projective space $\mathbb{RP}^{\infty}$, so you can get some intuition for it by studying the finite real projective spaces.
If $M$ is path-connected and maybe satisfies some other mild hypotheses then homotopy classes of maps $M \to B \mathbb{Z}_2$ are computed by
$$H^1(M, \mathbb{Z}_2) \cong \text{Hom}(\pi_1(M), \mathbb{Z}_2) \cong \text{Hom}(H_1(M), \mathbb{Z}_2).$$
As far as the cohomology calculation goes, the generating class $a$ corresponds to the first Stiefel-Whitney class $w_1 \in H^1(-, \mathbb{Z}_2)$ of a real line bundle. Loosely speaking it measures how the line bundle "twists" (more formally, it is measuring monodromy, which the identification with $\text{Hom}(\pi_1(M), \mathbb{Z}_2)$ also suggests), and is zero iff the line bundle is trivial(izable). The fact that the cohomology (over $\mathbb{Z}_2$) is the polynomial algebra on $w_1$ says that every characteristic class of real line bundles is a polynomial in $w_1$ and that $w_1$ is free in the sense that it satisfies no polynomial relations. This reflects the fact that, by the computation above, it can take on any value in $H^1(M, \mathbb{Z}_2)$ for any $M$.
The fact that $BO(2) \cong B \mathbb{Z}_2$ says that $w_1$ is a complete invariant of real line bundles: two real line bundles $L, L'$ are isomorphic iff $w_1(L) = w_1(L')$.
It's unclear whether any of this constitutes an answer to your question; it would help if you were more specific about which of this most sounds like what you're looking for.