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?

• Notice that a Lie group does not have a canoncal Riemannian structure, so in a sense, your question is not well-posed. Jul 9, 2010 at 15:00
• Your question makes no sense as it is: to get a scalar curvature, you need a Riemannian structure. For a Lie group, a natural choice is to take a left-invariant metric. You could edit your question in this direction. If you are interested in the curvature of pseudo-riemannian metrics, then in the semi-simple case you can also consider the --bi-invariant-- Killing form. Jul 9, 2010 at 15:01
• And not just semisimple, of course: there are Lie groups with bi-invariant metrics whose Lie algebras are not even reductive. Jul 9, 2010 at 21:42

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.

• I think it should involve the scalar product with $Z$, not the commutator. Jul 9, 2010 at 21:26
• Victor, the result is correct at is stands, except that $X,Y,Z$ are supposed to be left-invariant vector fields. Jul 9, 2010 at 21:43
• I don't understand: the formula doesn't involve the metric, but if you rescale the metric, surely the curvature will rescale, too? Jul 10, 2010 at 7:16
• Yes, the metric is involved in the formula for the sectional curvature (notice that is one quarter of the square of the norm of the bracket). The curvature tensor, $R: \mathcal{X}(M) \times \mathcal{X}(M) \times \mathcal{X}(M) \to \mathcal{X}(M)$ does not depend on the metric (it is defined as $R(X,Y)Z=\nabla_Y \nabla_X Z - \nabla_X \nabla_Y Z + \nabla_{[X,Y]} Z$. However, the sectional curvature (naturally) starts to depend on the metric: If $X$ and $Y$ are orthonormal, it is defined as: $\langle R(X,Y)X, Y \rangle$. Jul 10, 2010 at 9:01
• Well, you are right (thanks!), of course it depends on the metric for Levi-Civita conections. The thing about the Lie group is that one can work directly in the Lie algebra and the connection "dissappers" there in the opearator $R$ which can be writen only using brackets. Jul 10, 2010 at 15:27

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.

• It's also given by Proposition 3.18 in the book "Comparison Theorems in Riemannian Geometry" by Cheeger and Ebin (a book I recommend for one of the cleanest expositions of Riemannian geometry). Jul 11, 2010 at 17:49
• It is available here at NumDAM without any subscription: archive.numdam.org/ARCHIVE/AIF/AIF_1966__16_1/…
– agt
Apr 12, 2011 at 15:20

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!