Lie groups as manifolds

Question

In Weinberg's Classical Solutions in Quantum Field Theory, he states Lie groups, such as $SU(2)$ or $SO(3)$, may be viewed as a manifold. My questions are,

1. If we can interpret, e.g. $SU(2)$ as a manifold, how does one determine the metric?
2. From a differential geometry perspective, the Riemann tensor encodes the curvature of a particular manifold. If our manifold is a Lie group, is there a group theory interpretation of the curvature of that manifold, i.e. a different way of viewing it?
3. If a Lie group as a manifold is Ricci flat, what conclusions can we draw from that, regarding the group in question, if any?

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Edit: For future math S.E. users reading this question in the future, see http://www.rmki.kfki.hu/~tsbiro/gratis/LieGroups/LieGroups.html for an explicit example of a particular metric for $SU(2)$.

Answer 1

Manifolds don't have metrics on them by default. A metric is extra structure making a manifold a Riemannian manifold. It's interesting to study metrics on Lie groups but they don't need to be there.

In particular it's interesting to study bi-invariant metrics (metrics invariant under both left and right multiplication). For compact semisimple Lie groups there is a particularly nice choice of such a metric coming from the Killing form and one can express various things about this metric in terms of the Lie algebra; see, for example, this MO question. In particular the curvature tensor can be written in terms of the Lie bracket.