Consider the braid group with $n$ strands $B_n$. Each braid can be drawn (say) from bottom to top as $n$-intertwining strictly monotonic strands. Moreover, the group $B_n$ is generated by $n-1$ generators $\sigma_i$. Geometrically, the strands of $\sigma_i$ are trivial except for the $i^{\text{th}}$ strand which passes over the ${i+1}^{\text{th}}$ strand. Similarly, the pure braid group is generated by $\binom{n}{2}$ elements. Explicitly these are $A_{ij}=\sigma_{j-1}...\sigma_{i+1}\sigma_i^2\sigma_{i+1}^{-1}...\sigma_{j-1}^{-1}$, for $1 \leq i <j \leq n$.

Any element of the braid group $B_n$ can be represented as (the isotopy class of) a diffeomorphism of the $n$-punctured $2$-disc (the diffeomorphim permutes the punctures and fixes the boundary of the disc). If I am not mistaken, $\sigma_i $ is represented by the half Dehn twist around the loop encircling the $i^{\text{th}}$ and ${i+1}^{\text{th}}$ punctures.

Is it correct that the $A_{ij}$ are represented by full Dehn twists around the curves encircling the $i^{\text{th}}$ and ${j}^{\text{th}}$ punctures while avoiding the ones in between?


Your description of $A_{ij}$ isn't quite correct in that it does not uniquely specify the curve. There are infinitely many non-isotopic curves such that the disc they bound in the plane contains only the $i$-th and $j$-th punctures. To specify the curves uniquely, one convention would be to put all the puncture points on a common line, and to demand your curves stay below the line except in a small neighbourhood of the puncture points.

That would be one way to be precise about which curves you're talking about.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.