# Calculating OPE of Graviton Vertex Operator

Consider Exercise 2.8 in Polchinski's String Theory book. We are asked to compute the weight of $$f_{\mu \nu}:\partial X^{\mu} \bar{\partial}X^{\nu}e^{ik\cdot X}:$$ I have carried out the usual operator product expansion with the energy-momentum tensor defined by $$T(z) = -\frac{1}{\alpha'}:\partial X^{\mu} \partial X_{\mu}:\tag{2.4.4}$$ After carrying out the operator product expansion and collecting the singular terms, I compared with the solution given by http://isites.harvard.edu/fs/docs/icb.topic1136888.files/ps4/ps4-sol.pdf

My solution, for the most part, agrees with the answer given, however I differ in the contraction where they have the extra term $$-4\eta_{\mu \nu}f_{\alpha \beta}:\partial X^{\alpha}(w)\bar{\partial}X^{\beta}(\bar{w})\partial X^{\mu}(z){e^{ik\cdot X(w,\bar{w})}}\partial X^{\nu}(z) e^{ik\cdot X(w,\bar{w})}:\tag{8}$$ where the exponential at the end is contracted with $\partial X^{\nu}$ and the exponential before that is contracted with $\partial X^{\mu}(z)$.

Now, I don't see how the above follows since our given combination of operators only contains one copy of the exponential. How does this follow?

In particular, it is stated that in order for $G_{f,k}(w,\bar{w})$ to be a tensor, we must have the condition $$k^{\alpha}f_{\alpha \beta} = k^{\beta}f_{\alpha \beta}.$$ In order for $G_{f,k}$ to be a tensor, wouldn't we need the condition that $$k^{\alpha}f_{\alpha,\beta} = 0,$$ since we need the more singular terms $$O(\frac{1}{z^{3}})$$ to vanish?