A braid element $b$ in the braid Group $B_n$ is an essential braid if it does not have at least one of the generators in it. For example, $b=\sigma_1\sigma_2\sigma_4 \in B_5$ is essential since $\sigma_3$ is not in $b$. I want to prove the following claim:

Claim: Let $b$ be an essential braid in $B_n$. Let $K(b) \in GL_m (\mathbb{Z}[q^{\pm 1},t^{\pm 1}]$ be its Krammer representation. Then, $\det(K(b)-I)=0$.

Recall that the Krammer representation $K(t,q)$ is a representation of the braid group $B_n$ in $GL_m (\mathbb{Z}[q^{\pm 1},t^{\pm 1}])=Aut(F_m)$ where $m={\frac{n(n-1)}{2}}$ and $F_m$ is the free module of rank $m$ over $\mathbb{Z}[q^{\pm 1},t^{\pm 1}]$. The representation can formulated as follows:

$K(\sigma_k)(e_{i,j})= \begin{cases} tq^2e_{k,k+1} & i=k, j=k+1; \\ (1-q)e_{i,k} + qe_{i,k+1} & j=k, i<k; \\ e_{i,k} + tq^{k-i+1}(q-1)e_{k,k+1} & j=k+1, i<k; \\ tq(q-1)e_{k,k+1} + qe_{k+1,j}, & i=k, k+1<j; \\ e_{k,j} + (1-q)e_{k+1,j} & i=k+1, k+1<j; \\ e_{i,j}, & i<j<k \textit{ or } k+1<i<j;\\ e_{i,j} + tq^{k-i}(q-1)^2e_{k,k+1} & i<k<k+1<j \end{cases} $

where $\{e_{i,j}\}_{1 \leq i < j \leq <n}$ is the free basis of $F_m$.

Thanks in advance.

  • $\begingroup$ So $m$ is both a braid and an integer? $\endgroup$ – Luc Guyot Oct 12 '16 at 18:47
  • $\begingroup$ Sorry for the confusion. I edited it. Now, $b $ is a braid and $m $ is an integer. Thank you for pointing. $\endgroup$ – Mehmet Aktas Oct 12 '16 at 21:55

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