# 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 has some scalar curvature R. Is there a nice formula which relates the lie algebra of the group to the scalar curvature at a point of the manifold?

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Notice that a Lie group does not have a canoncal Riemannian structure, so in a sense, your question is not well-posed. – Mariano Suárez-Alvarez Jul 9 '10 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. – Benoît Kloeckner Jul 9 '10 at 15:01
And not just semisimple, of course: there are Lie groups with bi-invariant metrics whose Lie algebras are not even reductive. – José Figueroa-O'Farrill Jul 9 '10 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.

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I think it should involve the scalar product with $Z$, not the commutator. – Victor Protsak Jul 9 '10 at 21:26
Victor, the result is correct at is stands, except that $X,Y,Z$ are supposed to be left-invariant vector fields. – José Figueroa-O'Farrill Jul 9 '10 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? – Victor Protsak Jul 10 '10 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: $\langle R(X,Y)X, Y \rangle$. – rpotrie Jul 10 '10 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. – rpotrie Jul 10 '10 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.

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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). – Deane Yang Jul 11 '10 at 17:49
It is available here at NumDAM without any subscription: archive.numdam.org/ARCHIVE/AIF/AIF_1966__16_1/… – Giuseppe Apr 12 '11 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!

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