I am a physicist with some background in differential geometry and I apologize for any possible unprecise terminology. 


Consider the Lie group $SU(2)$ and its tangent space $su(2)$ forming a tangent bundle.


It is well known that SU(2) is noncommutative, thus $ a \circ b \circ
a^{-1} \circ b^{-1} \neq 0$ in general ($a,b \in SU(2)$). 


This operation may be associated with surrounding a surface in the tangent space, resulting in a nonzero curvature. In particluar, there should be a curvature 2-form that reflects the above nonlinearity. How is this expressed precisely?


Can anyone point me to such a specific treatment of SU(2) in differential forms language? Hints to the literature are appreciated, but please no general textbooks about lie groups or DG.


I try to specify. There should be a common metric in SU(2), such as the angle in the Euler axis- angle representation when projecting to SO(3). Then, a metric is not even necessary, one just needs a connection. There should be a canonical connection.



SU(2) and differential forms
$SU(2)$ is a compact Lie group, so it has a bi-invariant Riemannian metric, whose Levi-Civita connection and Riemann curvature can be expressed using the Lie bracket on its Lie algebra $su(2)$. See, for example,


[Lie groups with bi-invariant Riemannian metric](https://cuhkmath.wordpress.com/2011/06/30/lie-groups-with-bi-invariant-riemannian-metric/)


The book *Comparison Theorems in Riemannian Geometry* by Cheeger and Ebin also explains all of this. The whole story is particularly elegant using the dual description with differential forms and the Maurer-Cartan equations, but I don't know a reference for this.


As it happens, $SU(2)$ is the 3-dimensional sphere and its bi-invariant metric is, up to a constant scale factor, the standard one. All of this can be worked out nicely using quaternions, as explained in *Naive Lie Theory* by Stillwell.


Actually, the differential form version is not so hard to explain. First, the Riemannian geometry part: If you have an orthonormal frame of tangent vector fields and let $\omega^1, \dots, \omega^n$ be the dual $1$-forms, then there is a unique set of $1$-forms, $\omega^i\_j = -\omega^j\_i$, satisfying
$$
d\omega^i + \omega^i\_j\wedge\omega^j = 0.
$$
These $1$-forms represent the Levi-Civita connection. The Riemannian curvature tensor $R$ is then given by $2$-forms $\Omega^i\_j = -\Omega^j\_i$, where
$$
\frac{1}{2}R^i\_{jkl}\omega^k\wedge\omega^l = \Omega^i\_j = d\omega^i\_j + \omega^i\_k\wedge\omega^k\_j.
$$


As for $SU(2)$, note that if $A$ denotes the map from an element in the group to the the element written as matrix, then the differential form $\Theta = A^{-1}\,dA$ is a trace-free skew-hermitian matrix of $1$-forms. In other words,
$$
\Theta = \begin{bmatrix} i\omega^1 & \omega^2 - i\omega^3\\
\omega^2 + i\omega^2 & -i\omega^1 \end{bmatrix}
$$
 It is easy to verify that $\Theta$ is invariant under left translations, and the Riemannian metric where $\omega^1, \omega^2, \omega^3$ are orthonormal is bi-invariant. $\Theta$ is called the Maurer-Cartan form and satisfies the Maurer-Cartan equation:
$$
d\Theta = d(A^{-1}\,dA) = -A^{-1}\,dA\wedge A^{-1}\,dA = -\Theta \wedge\Theta.
$$
Using this and the formula for $\Theta$ above, you can figure out what the connection $1$-forms $\omega^i\_j$ associated with $\omega^1, \omega^2, \omega^3$ are and compute the curvature. You can use the definition of the exterior derivative of a $1$-form $\theta$
$$
\langle d\theta, X\otimes Y\rangle = X\langle\theta,Y\rangle
 - Y\langle\theta,X\rangle - \langle\theta,[X,Y]\rangle
$$
with $\theta = \omega^1, \omega^2, \omega^3$ and vector fields $X$ and $Y$ equal to two of the left invariant vector fields $e\_1, e\_2, e\_3$, which are dual to $\omega^1, \omega^2, \omega^3$, to figure out how to express the Riemann curvature in terms of the Lie bracket.


Or you could represent each element of $SU(2)$ by a unit imaginary quaternion $u$. Then you would write $\Theta = \bar{u}\,du$, which is an imaginary quaternion-valued $1$-form and therefore of the form $\Theta = i\omega^1 + j\omega^2 + k\omega^3$. The rest is similar to the calculations described above.


