Lie Groups – Volume of SU(2) and Haar Integral

Question

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.

Answer 1

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.