I am trying to follow the derivation of the Jone's Polynomial from a braid representation presented in chapter 2 of Ohtsuki's Quantum Invariants.
The representation of the braid $b$ with $n$ strands is a linear operator (which we treat as a matrix after choosing some basis) on $(\mathbb{C}^2)^{\otimes n}$ is $\psi_n(b)$. Let $\sigma_i$ denote the crossing of the $i$-th strand under the $(i+1)$-th strand for $i \in \{1, \cdots , n-1 \}$. Then, $$\psi_n(\sigma_i)= \pmatrix {1 & 0\\ 0 &1}^{\bigotimes (i-1)}\otimes R \otimes \pmatrix {1 & 0\\ 0 &1}^{\bigotimes (n-i-1)},$$ where, after many calculations (which I omit) having to do with the requirements on $R$ in order for it to properly represent the braid generator, we find that $$R= \pmatrix {t^{1/2} & 0 & 0 & 0\\ 0 & 0 & t & 0\\ 0 & t & t^{1/2}-t^{3/2} & 0\\ 0 & 0 & 0 & t^{1/2}}.$$
We want the trace of a braid representation to be invariant under Markov moves, but this is not the case. So we introduce an auxiliary matrix $$h= \pmatrix {t^{-1/2} & 0\\ 0 & t^{1/2}},$$ which allows for $\mathrm{trace}(h^{\otimes n}\cdot\psi_n(b))$ invariant under both Markov moves.
This is what I don't understand: I can figure that $$\mathrm{trace}_2((\pmatrix {1 & 0\\ 0 &1} \otimes h)\cdot R) = \pmatrix {1 & 0\\ 0 &1}$$ by direct calculation. I do not understand the last two of the following lines which justify the invariance of the trace. $$\begin{align} \mathrm{trace}( h^{\otimes (n+1)}\cdot\psi_{n+1}(\sigma_n b)) & = \mathrm{trace}( h^{\otimes (n +1)}\cdot (\pmatrix {1 & 0\\ 0 &1}^{\bigotimes (n-1)} \otimes R ) \cdot \psi_{n+1}(b))\tag{1}\\ & = \mathrm{trace}(h^{\otimes n}\cdot\psi_n(b))\tag{2}. \end{align} $$ Going from (1) to (2) should use the fact above about the partial trace, but I don't see how. A more explicit derivation of this would be much appreciated. Thank you for your time.
Use the fact that $(A\otimes B)(C\otimes D) = (AC)\otimes (BD)$. Now $$h^{\otimes(n+1)}\cdot( I_2 ^{\otimes (n-1)}\otimes R) = (h\otimes h\otimes \cdots\otimes h )\cdot(I_2\otimes I_2\otimes \cdots \otimes I_2 \otimes R) = (h\cdot I_2)^{\otimes (n-1)}\otimes (h^{\otimes 2}\cdot R) = h^{\otimes (n-1)}\otimes (h^{\otimes 2}\cdot R)$$
Now we plug it back to the LHS of equation $(1)$. $$\mathrm{trace}( h^{\otimes (n+1)}\cdot\psi_{n+1}(\sigma_n b)) = \mathrm{trace}( h^{\otimes (n +1)}\cdot (I_2^{\bigotimes (n-1)} \otimes R ) \cdot \psi_{n+1}(b)) =\mathrm{trace}(( h^{\otimes (n-1)}\otimes (h^{\otimes 2}\cdot R))\cdot \psi_{n+1}(b)) $$
Then use the fact that $\psi_{n+1}(b) = \psi_{n}(b)\otimes I_2$ because $b$ is a braid with $n$ strands only, so the $(n+1)^{st}$ term in the tensor product is just $I_2$: $$\mathrm{trace}( h^{\otimes (n+1)}\cdot\psi_{n+1}(\sigma_n b)) =\mathrm{trace}( (h^{\otimes (n-1)}\otimes (h^{\otimes 2}\cdot R))\cdot (\psi_n(b)\otimes I_2)) $$
Now this middle term can be calculated directly: $$h^{\otimes 2} = \pmatrix{t^{-1}\\ & 1 \\ & & 1 \\ & & & t}$$ When multiplied by $R$ this gives $$B = \pmatrix {t^{-1/2}&0&0&0 \\ 0& 0 &t & 0\\ 0&t&t^{1/2}-t^{3/2}&0 \\ 0 &0& 0& t^{3/2}}$$
Now we break this matrix $B$ into the matrix $h\otimes \pmatrix{1&0\\0&0}$ and a traceless part $$B = h\otimes \pmatrix{1&0\\0&0} + \pmatrix {0&0&0&0 \\ 0& 0 &t & 0\\ 0&t&-t^{3/2}&0 \\ 0 &0& 0& t^{3/2}}$$
Plugging back in gives $$\mathrm{trace}( h^{\otimes (n+1)}\cdot\psi_{n+1}(\sigma_n b)) =\mathrm{trace}( (h^{\otimes (n-1)}\otimes ( h\otimes \pmatrix{1&0\\0&0} + \pmatrix {0&0&0&0 \\ 0& 0 &t & 0\\ 0&t&-t^{3/2}&0 \\ 0 &0& 0& t^{3/2}})\cdot (\psi_n(b)\otimes I_2)) = \mathrm{trace}( (h^{\otimes n}\cdot \psi_n(b))\otimes \pmatrix{1&0\\0&0}) + \mathrm{trace}( (h^{\otimes (n-1)}\otimes \pmatrix {0&0&0&0 \\ 0& 0 &t & 0\\ 0&t&-t^{3/2}&0 \\ 0 &0& 0& t^{3/2}})\cdot (\psi_n(b)\otimes I_2)) $$
Using $\mathrm{trace}(A\otimes B) = \mathrm{trace}(A)\mathrm{trace}(B)$, the first term gives $$\mathrm{trace}( h^{\otimes n}\cdot \psi_n(b))\mathrm{trace}( \pmatrix{1&0\\0&0}) = \mathrm{trace}( h^{\otimes n}\cdot \psi_n(b)) $$
For the second term we have no way out besides writing the matrices out. However, note that the matrix $(h^{\otimes (n-1)}\otimes \pmatrix {0&0&0&0 \\ 0& 0 &t & 0\\ 0&t&-t^{3/2}&0 \\ 0 &0& 0& t^{3/2}})$ is block-diagonal, and since we are only concerned with traces, we can focus only on the $(i,i)^{th}$ block entry. Thus we only need consider the product $$x_{ii}\pmatrix {0&0&0&0 \\ 0& 0 &t & 0\\ 0&t&-t^{3/2}&0 \\ 0 &0& 0& t^{3/2}}\pmatrix {y_{2i-1,2i-1}&0&y_{2i-1,2i}&0 \\ 0& y_{2i-1,2i-1}&0 & y_{2i-1,2i}\\ y_{2i,2i-1}&0& y_{2i,2i}&0 \\ 0 & y_{2i,2i-1}& 0& y_{2i,2i}}$$ where $h^{\otimes (n-1)} = (x_{ij})$ and $\psi_n(b)=(y_{kl})$
Multiplying out gives $$x_{ii}\pmatrix{0&*&*&*\\*&0&*&*\\*&*&-t^{3/2}y_{2i,2i}&* \\ *&*&*&t^{3/2}y_{2i,2i}}$$ which is traceless. Thus each block on the main diagonal of the matrix $M = (h^{\otimes (n-1)}\otimes \pmatrix {0&0&0&0 \\ 0& 0 &t & 0\\ 0&t&-t^{3/2}&0 \\ 0 &0& 0& t^{3/2}})\cdot (\psi_n(b)\otimes I_2)$ is traceless. The matrix $M$ is thus traceless.