Lie Group Curvature

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Section 3.9 Lie Group Curvature

Given the formula (3.8.6) for the natural connection in a Lie group, it is straightforward to compute the curvature. We have

\begin{align} R(X,Y)Z \amp= \nabla_X\nabla_Y Z - \nabla_Y\nabla_X Z - \nabla_{[X,Y]} Z\notag\\ \amp= \frac14 \bigl[X,[Y,Z]\bigr] -\frac14 \bigl[Y,[X,Z]\bigr] - \frac12 \bigl[[X,Y],Z\bigr]\notag\\ \amp= \frac14\bigl[[X,Y],Z\bigr] - \frac12\bigl[[X,Y],Z\bigr] = - \frac14 \bigl[[X,Y],Z\bigr]\tag{3.9.1} \end{align}

where we have used the Jacobi identity in the penultimate equality.

As an example, consider the Lie group \(\SO(3)\text{.}\) Introducing the component notation

\begin{equation} R(u,v)w = R^a{}_{bcd}w^bu^cv^d e_a\tag{3.9.2} \end{equation}

where \(\{e_a\}=\{r_x,r_y,r_z\}\) is a basis of \(\so(3)\) and repeated dummy indices are summed over (the Einstein summation convention), we can compute \(R(r_x,r_y)r_z = 0\text{,}\) so that \(R^a{}_{zxy}=0\text{,}\) and

\begin{equation} R(r_x,r_y)r_x=-\frac14 r_y\tag{3.9.3} \end{equation}

so that \(R^y{}_{xxy}=-\frac14\text{.}\) Since our basis is orthonormal (and Euclidean), we can freely raise and lower indices, so

\begin{equation} R^{xy}{}_{xy} = \frac14\tag{3.9.4} \end{equation}

(using the antisymmetry of the Riemann tensor components, with no sum over \(x\) or \(y\)), leading to

\begin{equation} R^x{}_x = \frac12 = R^y{}_y = R^z{}_z\tag{3.9.5} \end{equation}

for the Ricci tensor (\(R^a{}_b=R^{ma}{}_{mb}\)) and finally \(R=\frac32\) for the Ricci scalar (\(R=R^m{}_m\)). ¹ Not surprisingly, the curvature is constant; \(\SO(3)\) has (locally) the geometric structure of \(\RP^3\) (which is that of \(\SS^3\)), not merely its topology.

As this argument suggests, Lie groups are manifolds with constant curvature; they are symmetric spaces. ²

The numerical value of \(R\) depends on the conventional scale chosen for the Killing form and carries no geometric information.

More precisely, these descriptions are equivalent in the compact, connected case.