Christoffel Symbols of Bi-invariant Metric on $SO(3)$ I'd like to calculate directly the Christoffel symbols for $SO(3)$. Let us suppose I have the metric on $SO(3)$ given by 
$$g(X,Y)=1/2 [\operatorname{tr}(X^tY)].$$
I'm aware of Milnor paper and I know I can get the Christoffel symbol of a bi-invariant metric of a generic Lie Group through Koszul formula, but in this case is there a way to find the Christoffel Symbols o the connection forms directly?
Taking the standard basis $E_1,E_2,E_3$ of $\text{so}(3)$ where $[E_1,E_2]=E_3$, $[E_2,E_3]=E_1$ and $[E_3,E_1]=E_2$, if I'm not mistaking $g_{ij}$ at the identity is the identity matrix... so where do I go from there? I'm getting quite confused... 
 A: There is no Christoffel symbols when you have only a local (or global) basis, as explained in the answer to your other question. Indeed, Christoffel symbols is not invariant under change of coordinates (i.e., it's not a tensor)
If the goal is to calculate the curvature, then one can use directly the (orthonormal) basis $\{E_1, E_2,E_3\}$. The following is true for all Lie group equipped with a bi-invariant metric (which include all compact Lie groups). 
Let $\{ E_1, \cdots, E_n\}$ be an orthonormal basis on $\mathfrak g$ with respect to a metric $h$. We always identified $\mathfrak g$ with the set of left-invariant vector fields and the metric $h$ is identified with the bi-invariant metric on $G$. The structural constant are denoted by $\epsilon_{ij}^k$, that is, 
$$ [E_i, E_j] = \epsilon_{ij}^k E_k.$$
Since $h$ is a bi-invariant metric, the exponential map $\exp : \mathfrak g \to G$ is also the exponential mapping $T_e G \to G$ with respect to the bi-invariant metric $h$ (This is the only place we use that $h$ is bi-invariant, see here for a proof). 
Lemma 1: $\nabla _XX = 0$ for all $X\in \mathfrak g$. 
Proof of lemma 1: By definition of the exponential mapping $\exp(\cdot) = e^{(\cdot)}$, we have
$$ e^{tX} e^{sX} = e^{(s+t)X}.$$
If we write $\gamma (t)  =e^{tX}$, then
$$\gamma'(t) = (e^{tX})_* X.$$
This implies that $\gamma(t)$ is the intgral curve of the vector fields $X$. Thus $\nabla_X X = 0$ since $\gamma$ is a geodesics. 
Lemma 2: We have $\nabla_X Y = \frac 12 [X,Y]$.
Proof of Lemma 2: For all $X, Y\in \mathfrak g$, from Lemma 1, 
$$\begin{split}
0 &= \nabla_{X+Y}(X+Y) \\ 
&= \nabla_X X + \nabla_XY + \nabla_YX + \nabla _YY \\ &= \nabla_XY + \nabla_YX
\end{split}$$
Together with the torsion free condition
$$\nabla_XY - \nabla _YX = [X,Y],$$
the lemma is proved. 
Note that Lemma 2 gives much more refined information about the connection than that given by the Koszul formula. Now we can somehow calculate the "Christoffel symbols" of the connection, note that since $\{E_1, \cdots E_n\}$ forms a basis of $T_gG$ for all $g\in G$, so in general
$$\nabla_{E_i} E_j = T_{ij}^k(g) E_k$$
for some global functions $T_{ij}^k : G\to \mathbb R$. Lemma 2 implies that 
$$ T_{ij}^k =\frac 12 \epsilon_{ij}^k$$
(In particular, $T_{ij}^k$ is a constant function).
We can go on and calcaulate the curvature:
Lemma 3: The Riemann curvature tensor is given by 
$$ R(X, Y, Z, W) =  \frac 14 h([[X,Y],Z],W)$$
Proof of Lemma 3: Using Lemma 2 and Jacobi identity,
$$\begin{split}
R(X, Y, Z, W) &:= h( -\nabla_X \nabla _Y Z + \nabla_Y \nabla _X Z + \nabla_{[X,Y]} Z, W) \\ 
&= \frac 12 h(- \nabla _X [Y,Z] + \nabla_Y [X,Z] + [[X,Y],Z],W) \\
&= \frac 12 h(-\frac 12 [X,[Y,Z]] + \frac 12 [Y,[X,Z]] + [[X,Y],Z], W) \\
&= \frac 14 h([[X,Y],Z],W). 
\end{split}$$
Now we can represent $R$ using the structural constant: 
$$\begin{split} 
R_{ijkl} &= R(E_i, E_j, E_k, E_l)\\
&= \frac 14 h([[E_i, E_j], E_k], E_l) \\
&= \frac 14 h([\epsilon_{ij}^m E_m, E_k], E_l) \\
&= \frac 14 h(\epsilon_{ij}^m \epsilon_{mk}^n E_n , E_l) \\
&=\frac 14 \epsilon_{ij}^m \epsilon_{mk}^l.
\end{split}$$
Going back to your case $G= SO(3)$, we have (by case by case checking) 
$$ R_{1212} =R_{1313} = R_{2323} = \frac 14,$$
and others are zero (we are assuming that $i < j$ and $k <l$). 
