How to tell whether a combinatorial surface is orientable.

Question

I am learning about combinatorial surfaces, and I've encountered this question:

What surface is represented by $a_1a_2\cdots a_na_1^{-1}a_2^{-1}\cdots a_n^{-1}$.

I know the Euler characteristic is $0$ so it's between a Torus and a Klein Bottle, if only I can determine whether it is orientable or not. An answer in this post: What surface is represented by $a_1 a_2 \cdots a_n a_1^{-1} a_2^{-1} \cdots a_n^{-1}$? declares that this is orientable because there are no "twists," but I'm not quite sure how they discern that from the algebraic presentation. Do I only have a twist when I have an edge, say $a_1$, then a bunch of other edges, and then the same edge again in the same direction: $a_1\cdots a_1\cdots$?

Is there an obvious way of telling whether there's a mobius strip in your surface other than having to cut&paste it out explicitly?

Thanks

Answer 1

If there is any letter pair neither of which is an inverse, like $a$ and $a$, or both of which are inverses, like $a^{-1}$ and $a^{-1}$, then the surface is nonorientable. The curve which connects two identified points on those edges is orientation reversing after gluing, i.e. its neighborhood is a Mobius band.

Conversely, if all letter pairs consist of a letter and its inverse --- like $a$ and $a^{-1}$ --- then the surface is orientable.