TL;DR
Is there a well-defined closed formula from the braid group $B_n$ to $\{-1,1\} \left(\text{ or }\{0,1\}\right)$ which represents whether all the strings have returned to their initial position?
First, let me set up the problem.
Recall the presentation for $B_n$: $$\langle\,\sigma_1,\ldots,\sigma_{n-1}\;|\;\sigma_{i}\sigma_{i+1}\sigma_i=\sigma_{i+1}\sigma_i\sigma_{i+1}\text{ for all } i,\;\sigma_i\sigma_j=\sigma_j\sigma_i\text{ for }|i-j|\geq 2\,\rangle$$
I use the 'over-hand rule'. That is, $\sigma_i$ represents the $i$th string passing over the $(i+1)$th string.
Given a braid $g\in B_n$, let's call a product of elements $g_1^{e_1}\ldots g_m^{e_m}$ with $e_i\in\Bbb Z$ a basic representation for $g$ if $g=g_1^{e_1}\ldots g_m^{e_m}$ and each $g_i=\sigma_j$ for some $j$.
Now I can ask a more narrow question:
Is there a closed form indicator function from $B_n$ which, given a $g\in B_n$, can operate on a basic representation of $g$ (in a well-defined manner) and tell us whether the strings are back in their original positions? By closed form, I mean that if $$\sigma_{i_1}^{e_1}\cdots\sigma_{i_m}^{e^m}$$ is a basic representation of $g$, then our formula only operates on the terms $n$ and $i_1,\ldots, i_m$ and $e_1, \ldots, e_m$ and tells us whether the strings have returned to starting position.
As an example, let's work with the simplest braid pattern that some of you probably learned in elementary school. This braid is an element of $B_3$ and goes 'right over middle then left over middle'. We can represent this braid by $\sigma_1\sigma_2^{-1}$ if we are labeling our strings right-to-left. A more visual approach is here.
Our closed form equation $f$ would tell us that $f(\sigma_1\sigma_2^{-1})=-1$ and $f(\sigma_1\sigma_2^{-1}\sigma_1\sigma_2^{-1})=-1$ but will tell us that $f(\sigma_1\sigma_2^{-1}\sigma_1\sigma_2^{-1}\sigma_1\sigma_2^{-1})=1$.
We thus see that this function cannot be a homomorphism.
As a more complicated example from $B_4$ exemplified by this pattern, our formula would tell us that $f(\sigma_1\sigma_2^2\sigma_3^{-1}\sigma_2^{-2})=-1$ but $f(\sigma_1\sigma_2^2\sigma_3^{-1}\sigma_2^{-2}\sigma_1\sigma_2^2\sigma_3^{-1}\sigma_2^{-2})=1$ (in right-to-left numbering).
Why do I care?
I recently began studying braid groups on my own. I was never much of a crafty kid and never got the hang of braiding, but this self-study surprisingly gave me a language I understood that enabled me to do it (who says mathematics always abstracts thoughts and actions?).
I examined the common homomorphisms related to $B_n$ and was surprised to see that the canonical morphism onto $S_n$ post-composed with the sign homomorphism didn't represent whether the strings were back where they started (in fact, this function isn't a homomorphism at all).
Also, I couldn't find any formula which represented this function. However, I know that there are crazier formulas in mathematics concerning much more abstruse notions, and I was wondering whether someone talented here could figure it out and satiate my curiosity.
The pure braids $P_n$ (those for which each strand terminates in its original position) comprise the kernel of the projection map $B_n\to S_n$. To see if a braid $\sigma_{i_1}^{e_1}\cdots\sigma_{i_m}^{e_m}$ is pure, simply compute its residue in the symmetric group, $\tilde{\sigma}_{i_1}^{e_1}\cdots\tilde{\sigma}_{i_m}^{e_m}$ where $\tilde{\sigma}_k$ is the transposition swapping $k\leftrightarrow k+1$: if this is the identity permutation then our braid was pure, else it wasn't.