In one of the lecture notes on conformal field theory, the author derives an infinitesimal version of conformal transformation for dimension $d \geq 3$ and its as follows, $x'^{\mu}= x^{\mu}+a^{\mu}+ b^{\mu}_{\nu}x^{\nu}+c^{\mu}_{\rho\sigma}x^{\rho}x^{\sigma}$. Then after some further calculations, we can find $b_{\mu \nu}=\alpha\eta_{\mu\nu}+m_{\mu\nu}$. Where $\eta$ is a flat metric with signature $ \{-1,1,1,1\}$ and all terms other than $x$ are constants.
Now, here is my question, if we focus on the infinitesimal conformal transform by symmetric part ($m_{\mu\nu}$) it has the form, $x'^{\mu}= x^{\mu}+m^{\mu}_{\nu}x^{\nu}$.
Now the author claims "The generator corresponding to this infinitesimal conformal transformation is angular momentum $L_{\mu\nu}=i(x_{\mu}{\partial}_{\nu} -x_{\nu}\partial_{\mu})$".
Since this is a vector field I believe it should be a lie algebra generator. I know how to get the lie algebra generators from the lie group but here we are somehow getting the lie algebra generator using the infinitesimal conformal transforms on the base manifold. Could someone explain how the lie algebra has been derived in this case? (The author states this should be familiar and doesn't explain how it's derived) (I have some familiarity with principal fiber bundles and associated fiber bundles, I truly appreciate it if someone can give a detailed explanation).
Thank you.