# SU(2) and differential forms [closed]

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.

• This is not a curvature, as there's no metric a priori. In fact, this is the Lie bracket of invariant vector fields. – Alex Degtyarev Feb 1 '15 at 0:06

$SU(2)$ 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

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.

@Alex Degtyarev's comment gives the right idea. The tangent bundle of $SU(2)$ can be given a left-trivialization, which will identify it with $SU(2)\times\mathfrak{su}_2$. In this way, you can think of $\mathfrak{su}(2)$ as the Lie algebra of all left-invariant vector fields on $SU(2)$. The Lie bracket of two such vector fields is the left-invariant vector field coming from the bracket of the corresponding vectors in $\mathfrak{su}(2)$. However, this Lie algebra is non-abelian, meaning that the bracket of two vectors in $\mathfrak{su}_2$ may not vanish.

Note that a connected Lie group is abelian if and only if it's Lie algebra is abelian (as the image of the exponential map generates the identity component). The fact that $SU(2)$ is non-abelian is then reflected in the fact that $\mathfrak{su}_2$ is a non-abelian Lie algebra.