Examples of Computations of Curvature Operator I seem to be struggling with this. My understanding is that the curvature operator $\mathfrak{R}(x \wedge y)$ returns the bivector which makes the equation $g(\mathfrak{R}(x \wedge y), (v \wedge w)) = g(R(x,y)w,v)$ true for every bivector $v \wedge w \in \Lambda^2 M$. 
What it seems to me is that given an orthonormal basis, the way to actually compute this is to compute the left hand side on a basis of $\Lambda^2 M$, and solve the resulting linear system to figure out which bivector $\mathfrak{R}(x \wedge y)$ is. 
If I understand this correctly, it should not be terribly hard. But while working through Petersen, I am trying to reproduce the computations he omits on a Berger sphere (a warped metric on the 3-sphere), I am unable to make the calculations work out. 
I would really appreciate a worked example where the particular bivector can be found. I'd also like to be able to recover the curvature tensor from this operator, which seems like it should be only a little more work from there, by the definition. The example doesn't have to be from a Berger sphere, but in case that example is a particularly easy example, the Berger sphere is defined by declaring $\{ \epsilon^{-1} X_1, X_2, X_3 \}$ to be an orthonormal frame on $SU(2)$, with $[X_i, X_{i+1}] = 2X_{i+2}$, indices taken mod $3$.
 A: I figured out what was going wrong here with some help from faculty at my university. I'll post one example.
As said, the curvature operator takes a bivector $x \wedge y$ to the bivector $\mathfrak{R}(x \wedge y)$ which is defined to satisfy the equation $g(\mathfrak{R}(x \wedge y), (v \wedge w)) = g(R(x,y)w, v) = R(x,y,w,v)$ for every pair of vectors $w,v$
The thing that I was confused about is the abuse of notation. The elements on the left live in $\Lambda_p^2 M$, so the inner product on the left is not the same as the inner product on the right. Obvious thing when you say it, but necessary. The inner product on the left can be evaluated in terms of the standard inner product on $T_pM$ as $g(x,v)g(y,w) - g(x,w)g(y,v)$.
Now since $\{\epsilon^{-1}X_1, X_2, X_3 \}$ are declared to be orthonormal, they give a basis for the tangent space at identity and hence a basis for the tangent space everywhere since we're dealing with a Lie group. Now this basis induces a basis on the vector space(s) $\Lambda_p^2 M$ for every $p$. The basis is given by $\epsilon^{-1}X_1 \wedge X_2$, $\epsilon^{-1}X_1 \wedge X_3$ and $X_2 \wedge X_3$. It is enough to describe the curvature operator on these guys. Let us find $\mathfrak{R}(\epsilon^{-1} X_1 \wedge X_2)$.
We may write $$\mathfrak{R}(\epsilon^{-1}X_1 \wedge X_2) = a\epsilon^{-1}X_1 \wedge X_2 + b\epsilon^{-1}X_1 \wedge X_3 + cX_2 \wedge X_3$$, and we must determine $a,b,c$.  Now the key fact is that the orthonormal basis for $T_pM$ gives an orthonormal basis for $\Lambda_p^2 M$. To check this, observe that $g(\epsilon^{-1}X_1 \wedge X_2, \epsilon^{-1}X_1 \wedge X_3) = (1)(0)-(0)(0) = 0$. Similarly, $g(\epsilon^{-1}X_1 \wedge X_2, \epsilon^{-1}X_1 \wedge X_2) = (1)(1) - (0)(0) = 1$. The last pair is obviously $0$.
Now then we can take inner products of the big equation above with each of these three bases elements, and then orthonormality will kill off each term in turn.
$$ g(\mathfrak{R}(\epsilon^{-1}X_1 \wedge X_2), \epsilon^{-1}X_1 \wedge X_2) = g(a\epsilon^{-1}X_1 \wedge X_2 + b\epsilon^{-1}X_1 \wedge X_3 + cX_2 \wedge X_3, \epsilon^{-1}X_1 \wedge X_2)$$
Now by the bilinearity of $g$ and the orthonormality condition we just saw above, everything vanishes except $a$. On the other hand, this must be equal to $R(\epsilon^{-1}X_1, X_2)X_2, \epsilon^{-1}X_1)$ by the definition.and this quantity has already been computed in Petersen. We have $R(\epsilon^{-1}X_1, X_2)X_2, \epsilon^{-1}X_1) =\epsilon^{-2}g(R(X_1,X_2)X_2,X_1) =\epsilon^{-2}g(\epsilon^2 X_1, X_1) = g(X_1, X_1) = \epsilon^2$
Thus $a = \epsilon^2$. 
It is much less laborious to show $b = c = 0$, since everything else is $0$.
