Topological space representing anti-periodic boundary conditions It is well-known that we often use the circle $\mathbb{S}^1$ to represent boundary conditions, i.e., if $f$ is $2\pi$-periodic, then we can define it on the circle. In analogy, what if $f$ is $2\pi$-anti-periodic, i.e., $f(x+2\pi) = -f(x)$, is there a natural topological space to define $f$ on?
 A: A function with "anti-period" $2\pi$ is also periodic with period $4\pi$, since we have $$f(x+4\pi)=f((x+2\pi)+2\pi)=-f(x+2\pi)=-(-f(x))=f(x).$$ So we can still view $f$ as defined on the unit circle, just "scaled" appropriately.
A: It sounds like your background is physics so I don't know how much of the relevant mathematical background you have here. There are these things called vector bundles over a topological space $X$: you can think of them as being like a "continuous family" $V_x, x \in X$ of vector spaces; the $V_x$ are called the fibers of the bundle. Given a vector bundle you can also consider "sections" of it, which are like a continuous choice $f_x \in V_x, x \in X$ of vector in each of these vector spaces; these generalize continuous functions on $X$. For example, any smooth manifold has a tangent bundle whose sections are vector fields. 
Vector bundles are formally defined by defining their total space $E$, which looks like the disjoint union $\bigsqcup_{x \in X} V_x$ of all the fibers but with a particular topology, and then a projection map $\pi : E \to X$ sending $V_x$ to $x$. The fibers of the vector bundle are then given by the fibers $\pi^{-1}(x) \cong V_x$ of the projection map $\pi$, and sections of the bundle can be defined as right inverses to $\pi$: that is, as maps $f : X \to E$ such that $\pi \circ f = \text{id}_X$. You can check that this condition is equivalent to requiring $f(x) \in V_x$ as expected. 

All of that was background for the following construction. $S^1$ admits a nontrivial real line bundle (a vector bundle whose fibers are $1$-dimensional real vector spaces) which "twists by $-1$" as you go around the circle. You can visualize its  as the (open) Mobius strip; somewhat more explicitly, you can define it as the quotient of the space $\mathbb{R} \times [0, 2\pi]$ by the equivalence relation
$$(x, 0) \sim (-x, 2\pi).$$
The projection map to $S^1$ is given by projecting to the second coordinate, thinking of $S^1$ as $[0, 2\pi]$ with the endpoints identified. Now you can check that a (continuous) section of this line bundle is basically the same thing as a (continuous) antiperiodic function. Similarly, if we had taken the trivial line bundle $\mathbb{R} \times S^1$, its (continuous) sections would just be (continuous) functions on $S^1$, or $2\pi$-periodic (continuous) functions. 
