Volume of SU(2) and Haar Integral The Pauli matrices are given by
$$\sigma_1 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}, \quad \sigma_2 = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}, \quad \sigma_3 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}.$$
Then the following matrices form a basis for $\mathfrak{su}(2)$ over $\mathbb{R}$.
$$u_1 = i\sigma_1 = \begin{pmatrix} 0 & i \\ i & 0 \end{pmatrix}, \quad u_2 = - i\sigma_2 = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}, \quad u_3 = + i\sigma_3 = \begin{pmatrix} i & 0 \\ 0 & -i \end{pmatrix}.$$
Then their duals $u_1^*$, $u_2^*$, and $u_3^*$ are left-invariant $1$-forms on $\mathrm{SU}(2)$. I want to find the value of the following integral.
$$\int_{\mathrm{SU}(2)} u_1^* \wedge u_2^* \wedge u_3^*.$$
Since $u_1^* \wedge u_2^* \wedge u_3^*$ is left-invariant, this is a Haar integral (right?). So, if I know one pair (a left-invariant volume form, its volume), maybe I can evaluate the integral.
Could you help me? Thank you.
 A: By the well known isomorphism of $SU(2)$ with the unit quaternions the volume of $SU(2)$ is the surface area of the $3$-sphere $S_3=2\pi^2\,.$
What follows is the approach along your lines:
Following A. Gorodnik's Lecture Notes
let's write the unit quaternions as
$$
(a,\boldsymbol{b})\,,~~\text{ where }~~a\in\mathbb R,~\boldsymbol{b}\in\mathbb R^3,~b=|\boldsymbol{b}|<1\,,~a=(1-b^2)^{1/2}
$$
so that $a^2+|\boldsymbol{b}|^2=1\,.$
The isomporphism between unit quaternions and $SU(2)$ is given by
$$
\left(\begin{matrix}a+ib_3&b_2+ib_1\\-b_2+ib_1&a-ib_3\end{matrix}\right)\,.
$$
Note that $\boldsymbol{b}$ characterizes $a$ only up to a sign but it is sufficient to consider the Haar measure on the "upper half" $SU(2)\cap\{a>0\}$ so that $\Phi(\boldsymbol{b})$ characterizes the matrix uniquely.
Multiplication in $SU(2)$ is quaternion multiplication and is given by
$\Phi(\boldsymbol{b}_1)\Phi(\boldsymbol{b}_2)=\Phi(\boldsymbol{c})$ where
$$
\boldsymbol{c}=a_1\boldsymbol{b}_2+a_2\boldsymbol{b}_1-\boldsymbol{b}_1\times\boldsymbol{b}_2\,.
$$
When restricted to $SU(2)\cap\{a>0\}$ a left invariant vector field can be seen as a map
$$
X:\quad SU(2)\cap\{a>0\}\simeq\mathbb R^3\to\mathbb R^3\simeq\mathfrak{so}(2)
$$
which must satisfy
\begin{align}\tag{1}
X(\boldsymbol{b})=J(\boldsymbol{b})\,X(\boldsymbol{0})
\end{align}
where $J(\boldsymbol{b})$ is the matrix
$$
J(\boldsymbol{b})=\left(\begin{matrix}a & -b_3&b_2\\b_3&a&-b_1\\-b_2&b_1&a\end{matrix}\right)\,.
$$
Clearly (1) can be written as
\begin{align}\tag{2}
X(\boldsymbol{b})=a\,X(\boldsymbol{0})+\boldsymbol{b}\times X(\boldsymbol{0})\,.
\end{align}
The proof of (1) is not hard. It just requires the calculation of a Jacobian matrix. The unit quaternion $(1,\boldsymbol{0})$ corresponds to the identity matrix so that $X(\boldsymbol{0})$ is the vector field in the unit element of $SU(2)$ which characterises the left invariant field $X$ on $SU(2)$ completely.
Let's now pick three basis vectors in $\mathbb R^3\:$
$$
X_1(\boldsymbol{0}):=\left(\begin{matrix}1 \\ 0\\0\end{matrix}\right)\,,~~X_2(\boldsymbol{0}):=\left(\begin{matrix}0 \\ 1\\0\end{matrix}\right)\,,~~X_3(\boldsymbol{0}):=\left(\begin{matrix}0 \\ 0\\1\end{matrix}\right)\,.
$$
By (1) they define three vector fields $X_1(\boldsymbol{b})\,,X_2(\boldsymbol{b})\,,X_3(\boldsymbol{b})$ on $SU(2)\cap\{a>0\}\,.$
Lets denote by
$$
X_1^*(\boldsymbol{b})\,,~~X_2^*(\boldsymbol{b})\,,~~X_3^*(\boldsymbol{b})\,
$$
their duals. The relation
$$\tag{3}
\langle X_i(\boldsymbol{b}),X_j^*(\boldsymbol{b})\rangle=\delta_{ij}
$$
clearly holds for $\boldsymbol{b}=\boldsymbol{0}\,.$ In order that
(3) holds for all $\boldsymbol{b}$ it is easy to see that
\begin{align}
X_i^*(\boldsymbol{b})=[J(\boldsymbol{b})^\top]^{-1}X_i^*(\boldsymbol{0})
=[J(-\boldsymbol{b})]^{-1}X_i^*(\boldsymbol{0})\tag{4}
\end{align}
must hold. Vectors and one-forms are now identified as usual
$$
\frac{\partial}{\partial b_i}\leftrightarrow X_i(\boldsymbol{0})\,,\quad
db_i\leftrightarrow X_i^*(\boldsymbol{0})\,.
$$
In particular, the basis one-forms $db_i$ extend to $SU(2)\cap\{a>0\}$ by  (4).
Standard exterior algebra tell us that
the volume form on $SU(2)\cap\{a>0\}$ is then
\begin{align}
X_1^*(\boldsymbol{b})\wedge X_2^*(\boldsymbol{b})\wedge X_3^*(\boldsymbol{b})&=\frac{1}{{\rm det}J(-\boldsymbol{b})}\,db_1\,db_2\,db_3
=\frac{1}{a}\,db_1\,db_2\,db_3\\
&=\frac{1}{\sqrt{1-b_1^2-b_2^2-b_3^2}}\,db_1\,db_2\,db_3\,.
\end{align}
This form is left invariant. It is also right invariant because it does not depend on the signs of the $b_i$. The volume of $SU(2)$ in this Haar measure  is now
$$\tag{5}
2\int\limits_{b_1^2+b_2^2+b_3^2\le 1}\frac{1}{\sqrt{1-b_1^2-b_2^2-b_3^2}}\,db_1\,db_2\,db_3
$$
which is best calculated using polar coordinates
$$
\left(\begin{matrix}b_1\\b_2\\b_3\end{matrix}\right)=\left(\begin{matrix}r\cos\theta\\r\sin\theta\cos\phi\\r\sin\theta\sin\phi\end{matrix}\right)\,,~~\theta\in[0,\pi]\,,~~\phi\in[0,2\pi)\,.
$$
The integral (5) becomes
$$
2\int_0^1\int_0^{2\pi}\int_0^\pi\frac{1}{\sqrt{1-r^2}}\,r^2\,\sin\theta\,d\theta\,d\phi\,dr=2\pi^2
$$
which is the standard volume of $S_3$ as expected.
