Explicit computation of a 3-form on SU(2) I am reading Dan Freed's lectures on Quantum Groups on Path Integrals. I am picking up the required math as I go along and I am finding certain calculations hard to follow. 
On page 7, he considers the Maurer-Cartan form $\theta = g^{-1} dg$ on $G=SU(2)$. Then he defines a 3-form $$\omega = c \text{Tr}(\theta^3).$$ The constant $c$ is chosen so that $\int_{G}\omega = 1$.
In an exercise (page 7, Ex 1.19), he asks to compute $c$ for $SU(2)$. I am not able to do this exercise.
How do we calculate the value of $c$?  

My attempt:
An element $g$ in SU(2) can be represented by
$$g=\begin{pmatrix}
\alpha & \beta\\
-\overline{\beta} & \overline{\alpha}
\end{pmatrix}$$ with the condition $|\alpha|^2 + |\beta|^2=1$. Then 
$$\theta =g^{-1} dg = 
\begin{pmatrix}
\overline{\alpha} d\alpha - \beta d\overline{\beta} & \overline{\alpha}d\beta + \beta d\overline{\alpha}\\
-\overline{\beta}d\alpha-\alpha d\overline{\beta} & -\overline{\beta}d\beta + \alpha d\overline{\alpha}
\end{pmatrix}.$$
Calculating with all the variables was getting incredibly messy so I decided to do the following:
If we set 
$$\theta = \begin{pmatrix}
a & b\\
c & d 
\end{pmatrix},$$
then we can cube the matrix and use the rules of exterior algebra to obtain
$$\theta^3 = \begin{pmatrix}
2abc-bcd & abd\\
-acd & abc-2bcd 
\end{pmatrix}.$$
Thus the trace of $\theta^3$ is $3(abc-bcd)$. I am not sure if this approach is right, but I finally get the trace as the following expression after using the constraint:
$$\text{Tr}(\theta ^3) = -6(\overline{\alpha}|\alpha|^2d\beta d\overline{\beta}d\alpha - \overline{\beta}|\beta|^2d\alpha d\overline{\alpha}d\beta + \alpha\beta (\overline{\alpha}d\alpha + \overline{\beta}d\beta)d\overline{\alpha}d\overline{\beta})$$
I have two questions now:
1) Is the expression correct? 
2) How do we integrate this 3 form on SU(2)?
 A: It is somewhat easier to get to this by writing the Maurer-Cartan form $\theta$ as 
$$\theta = \begin{bmatrix} \theta_{11} & \theta_{12} \\ \theta_{21} & \theta_{22}\end{bmatrix},$$
where $\theta_{22} = -\theta_{11}$ and $\theta_{12} = -\bar\theta_{21}$. Now,
$$\text{tr}(\theta^3) = \sum_{i,j,k} \theta_{ij}\wedge\theta_{jk}\wedge\theta_{ki},$$
and, using the symmetries I listed, you easily check that this comes out
$$\text{tr}(\theta^3) = 6\theta_{11}\wedge\theta_{12}\wedge\theta_{21} 
 = 6\theta_{11}\wedge\theta_{21}\wedge\bar\theta_{21}.$$
Now, there are various ways to compute the integral. One is to use your coordinates $\alpha,\beta$, write this $3$-form out, and then use Stokes's Theorem to integrate its exterior derivative over the unit ball in $\Bbb C^2$. When I use $\alpha,\beta$ as the coordinates in the first column of $g$, I get 
\begin{align*}
\theta_{11} &= \bar\alpha\,d\alpha + \bar\beta\,d\beta \\
\theta_{21} &= -\beta\,d\alpha + \alpha\,d\beta \\
\theta_{12} &= \bar\beta\,d\bar\alpha - \bar\alpha\,d\bar\beta.
\end{align*}
Then (after a bit of simplification) the $3$-form $\text{tr}(\theta^3) = -6\,d\alpha\wedge d\beta\wedge(-\bar\alpha d\bar\beta + \bar\beta d\bar\alpha)$, whose exterior derivative is 
$$\psi=-12 d\alpha\wedge d\bar\alpha\wedge d\beta\wedge d\bar\beta.$$
Since $d\alpha\wedge d\bar\alpha = -2\sqrt{-1}dx\wedge dy$, I conclude that $\psi$ is $48$ times the volume element of $\Bbb C^2$, and so the integral over the unit ball is $48$ times the volume of the unit ball in $\Bbb R^4$. Thus, the answer is $48\pi^2/2! = 24\pi^2$.
Alternatively, we can approach it in terms of the forms $\theta_{ij}$. When I think of the columns $e_1,e_2$ of $g$ as giving a unitary frame at the point $[e_1]\in\Bbb CP^1$, I can integrate (removing a set of measure $0$ in $SU(2)$) by applying Fubini's Theorem on $S^1\times\Bbb CP^1$.  $\theta_{11}$ is $\sqrt{-1}\,d\varphi$ for $\varphi$ the usual coordinate on $S^1$; $\frac{\sqrt{-1}}2 \theta_{21}\wedge\bar\theta_{21}$ is the usual Kähler form on $\Bbb CP^1$, whose integral is $\pi$. [Here it's important that we're working with (holomorphic) column vectors of $g$, not row vectors.] Thus, I end up with $6(2\pi\sqrt{-1})(-2\pi\sqrt{-1}) = 24\pi^2$. 
Side comment: It follows from what I just said that the $3$-form $\theta_{11}\wedge\theta_{21}\wedge\bar\theta_{21}$ is twice the invariant volume form on $SU(2)$. And the volume of $SU(2) = S^3$ is, in fact, $2\pi^2$.
