# How can I prove this relation involving gamma matrices?

Let
$$l^{\mu} = l^{\mu}_{\parallel} + l^{\mu}_{\bot}$$ be a D-dimensional vector living in a Minkowskian space; the only non-zero components of $l^{\mu}_{\parallel}$ are the first four, while the only non-zero components of $l^{\mu}_{\bot}$ are the last $D - 4$.
Let moreover be $\gamma^{\mu}$ an element of a generic D-dimensional Clifford algebra, and define $$\gamma^5 = i\gamma^0\gamma^1\gamma^2\gamma^3;$$ the element $\gamma^5$ therefore commutes with all the elements with index different from $\mu = 0,1,2,3$.
Finally, using Feynman slash notation let's consider the trace $$tr[-2\gamma^5\not{l_{\bot}}(\not{l_{\parallel}} - \not{k})\gamma^{\lambda}\not{l_{\parallel}}\gamma^{\nu}(\not{l_{\parallel}} + \not{p})] ,$$ where $k^{\mu}$ and $p^{\mu}$ are assumed to be D-dimensional vectors behaving like $l^{\mu}_{\parallel}$, i.e. having only the first four component different from zero. According to Peskin & Schroesder's "An introduction to QFT", page 663,

To evaluate this contribution [...] we must retain one factor each of $\gamma^{\nu},\gamma^{\lambda},\not{p}$, and $\not{k}$ to give a nonzero trace with $\gamma^5$.

This is equivalent to say that $$tr[-2\gamma^5\not{l_{\bot}}(\not{l_{\parallel}} - \not{k})\gamma^{\lambda}\not{l_{\parallel}}\gamma^{\nu}(\not{l_{\parallel}} + \not{p})] = tr[2\gamma^5\not{l_{\bot}}\not{k}\gamma^{\lambda}\not{l_{\parallel}}\gamma^{\nu}\not{p}].$$ I don't get why this should be true. I tried to solve the problem assuming specifical values for the indexes $\nu$ and $\lambda$ and working on the fact that $\gamma^5$ commutes with $\gamma^{\alpha}$ if $\alpha > 3$ and anticommutes otherwise, but got nowhere; either each term of the trace becomes zero, or everyone survives.
So the question is: how can I prove the last relation I wrote?

The proposed final equality is in fact not correct - the correct interpretation of this statement given by P&S is in the text, c.f equation (19.59) where the end resolution of the trace is presented. To see this, one needs to be up on their Dirac algebra in $d$ dimensions, particularly with regards to those relations involving $\gamma^5$.
The trace as initialised consists of 18 terms. The relevant non vanishing contributions come from those terms proportional to $\ell_{\perp} \cdot \ell_{\perp}$. One can systematically reduce the number of terms through employment of the following relations $$\text{Tr}(\gamma^5 \gamma^{\mu} \gamma^{\nu}) = 0 \,\,\,,\,\,\text{Tr}(\gamma^5 \gamma^{\mu} \gamma^{\nu} \gamma^{\rho} \gamma^{\sigma}) \sim \epsilon^{\mu \nu \rho \sigma}$$ and $$\text{Tr}(\gamma^5 \gamma^{\mu} \gamma^{\nu} \gamma^{\rho} \gamma^{\sigma} \gamma^{\alpha} \gamma^{\beta}) \sim 4 \delta^{\mu \nu} \epsilon^{\rho \sigma \alpha \beta} + \text{perms.}$$ with $\mu,\nu,\rho,\sigma,\alpha, \beta \in \left\{0,\dots,3\right\}.$
A sacrificing of Lorentz invariance but with the ability to utilise the familiar definition of $\gamma^5$ in $d$ dimensions (as proposed by ‘tHooft and Veltmann circa 1972) means that the following further relations hold $$\left\{\gamma^5, \gamma^{\mu} \right\} = 0\,\,\,\,\text{and}\,\,\,\,[\gamma^5, \gamma^{\mu}] = 0$$ where the former holds if $\mu \in \left\{0,\dots, 3 \right\}$ and the latter if $\mu$ is evaluated in any of the other $d-4$ dimensions.
The important realisation is that in the trace of $\gamma^5$ with four or six other gamma’s in $d=4$, if we make one or more of these gamma’s carry an index outwith $\left\{0,1,2,3\right\}$, the trace vanishes by virtue of their proportionality to the epsilon symbol, as given above, non zero for e.g $\epsilon^{0123}$ and its permutations. This is why the non vanishing contributions come with $\ell_{\perp} \cdot \ell_{\perp}$, we have removed the dependence of remnant $d$ dimensional $\not{\ell_{\perp}}$ through explicit $\not{\ell_{\perp}} \not{\ell_{\perp}}$ terms to produce the scalar product.
As you will see, this leaves us with five terms, four containing a trace involving $\gamma^5, \gamma^{\lambda}, \gamma^{\nu}$ and either $\not{k}$ or $\not{p}$ but under anticommutation of these objects they all cancel in a pairwise fashion. The remaining term is that within (19.59) and the resulting loop integration reduces to (19.56), with a sole dependence on $\ell_{\perp} \cdot \ell_{\perp}$, as mentioned above.