Uniformization Theorem Curvature on a Riemann sphere Given a Riemann surface $X$ and a conformal metric $\lambda(z)^{2} dz d\bar{z}, \lambda(z) > 0$ on it. The curvature of $X$ is defined to be $K = -\Delta \text{log}\lambda$ according to https://www.math.wichita.edu/~ryan/teaching/M829F/syllabus/Jost-book/JJ_ch2.pdf. This seems to be exactly the definition of scalar curvature for a Kahler manifold. Then is the uniformization thereom which states that every Riemann surface admits a conformally related metric of constant curvature referring to the scalar curvature? Isn't the uniformization then a direct consequence of the Yamabe problem?
 A: $\DeclareMathOperator{\ric}{Ric}
\DeclareMathOperator{\trace}{trace}
\DeclareMathOperator{\scal}{Scal}
\DeclareMathOperator{\rm}{Rm}$
In real dimension $2$, the scalar curvature is twice the sectional curvature.
Indeed, if $\{e_1,e_2\}$ is an orthonormal frame, then
\begin{align}
\ric(e_1,e_1) &= \rm(e_1,e_1,e_1,e_1) + \rm(e_1,e_2,e_1,e_2) = 0+ \sec(e_1,e_2) = \sec(e_1,e_2),\\
\ric(e_1,e_2) &= \rm(e_1,e_1,e_2,e_1) + \rm(e_1,e_2,e_2,e_2) = 0 + 0 = 0,\\
\ric(e_2,e_2) &= \rm(e_2,e_1,e_2,e_1) + \rm(e_2,e_2,e_2,e_2) = \sec(e_2,e_1) + 0= \sec(e_1,e_2).
\end{align}
(Note that I use the convention $\sec(X,Y) = \rm(X,Y,X,Y)=g(R(X,Y)X,Y)$ for orthonormal vector fields and $\ric(X,Y) =\trace\left(Z\mapsto R(X,Z)Y\right)$.)
Hence, we have $\ric = \sec \cdot g$.
Tracing this equality gives
$$
\scal = 2 \sec.
$$
It follows that the sectional curvature is constant if and only if the scalar curvature is constant.
Therefore, in dimension $2$, the Yamabe problem for $(M,g)$ is equivalent to finding a metric of constant sectional curvature in the conformal class $[g]$, which solves the uniformization problem.
However, as it is mentioned by @GunnarÞórMagnússon in the comment section, the uniformization problem is much easier to solve than the Yamabe problem. The fact that the uniformization Theorem was true is a main motivation that lead people to think that the Yamabe problem would be solvable.
