Curvature of a Lie group Since a lie group is a manifold with the structure of a continuous group, then each point of the manifold [Edit: provided we fix a metric, for example an invariant or bi-invariant one] has some scalar curvature R.  

Question [Edited] Is there a nice formula which expresses the scalar curvature at a point of the manifold in terms of the lie algebra of the group?

 A: One result which I think will be what you are interested in is this,
(corrected and clarified in response to Jose's pointers) 


*

*For a Lie Group with a bi-invariant Riemannian metric the Riemann-Christoffel connection is half the Lie Algebra, i.e $\nabla _ X Y = \frac{1}{2}[X,Y]$. This follows from a combination of Koszul's identity and the fact that bi-invariant metrics on Lie Groups are Ad-invariant

*For a compact semi-simple Lie Group the negative of the Killing form gives a natural candidate for such a bi-invariant Riemannian metric. 
This mapping of the connection in terms of the Lie Algebra can be fruitfully used to achieve simpler expressions for various other quantities, like most beautifully the statement that the scalar curvature becomes one-fourth of the dimension of the Lie Group! 
A: See Exercice 1 in Chapter 4 of Do Carmo's "Riemannian Geometry". 
The formula is $R(X,Y)Z = \frac 1 4 [[X,Y], Z]$.
In particular, if $X$ and $Y$ are orthonormal, the sectional curvature of the generated plane is 
$K(\sigma)= \frac 1 4 \|[X,Y]\|^2$ 
Which is always $\geq 0$. 
EDIT: In view of the comments, it is important to add that this is for a bi-invariant metric. 
A: For left-invariant (or right-invariant) metrics, this paper of Arnold gives a formula for  the sectional and Riemannian curvatures, in terms of the adjoint of the Lie bracket operation in the metric.
