# nLab SU(2)

Contents

group theory

### Cohomology and Extensions

#### Lie theory

Background

Smooth structure

Higher groupoids

Lie theory

∞-Lie groupoids

∞-Lie algebroids

Formal Lie groupoids

Cohomology

Homotopy

Related topics

Examples

$\infty$-Lie groupoids

$\infty$-Lie groups

$\infty$-Lie algebroids

$\infty$-Lie algebras

# Contents

## Definition

The special unitary group $SU(n)$ for $n = 2$.

###### Proposition

As a matrix group $SU(2)$ is equivalent to the subgroup of the general linear group $GL(2, \mathbb{C})$ on those of the form

$\left( \array{ u & v \\ - \overline{v} & \overline{u} } \right) \;\;\; with \;\; {\vert u\vert}^2 + {\vert v\vert}^2 = 1 \,,$

where $u,v \in \mathbb{C}$ are complex numbers and $\overline{(-)}$ denotes complex conjugation.

## Properties

### General

###### Proposition

The underlying manifold of $SU(2)$ is diffeomorphic to the 3-sphere $S^3$.

###### Proposition

There are isomorphisms of Lie groups

1. of $SU(2)$ with the spin group in dimension 3 and with the quaternionic unitary group in one dimension

$SU(2) \;\simeq\; Spin(3) \;\simeq\; Sp(1)$
2. of the direct product group of $SU(2)$ with itself, to Spin(4)

$SU(2) \times SU(2) \;\simeq\; Spin(3) \times Spin(3) \;\simeq\; Spin(4)$

with respect to which the canonical inclusion $Spin(3) \hookrightarrow Spin(4)$ is given by the diagonal map.

###### Proposition

The maximal torus of $SU(2)$ is the circle group $U(1)$. In the above matrix group presentation this is naturally identified with the subgroup of matrices of the form

$\left( \array{ t & 0 \\ 0 & t^{-1} } \right) \;\; with t \in U(1) \hookrightarrow \mathbb{C} \,.$

### Lie algebra

###### Proposition

The Lie algebra $\mathfrak{su}(2)$ (see also at su(2)) as a complex matrix Lie algebra is the sub Lie algebra on those matrices of the form

$\left( \array{ i z & x + i y \\ - x + i y & - i z } \right) \;\;\; with \;\; x,y,z \in \mathbb{R} \,.$
###### Definition

The standard basis elements of $\mathfrak{su}(2)$ given by the above presentation are

$\sigma_1 \coloneqq \frac{1}{2} \left( \array{ 0 & 1 \\ -1 & 0 } \right)$
$\sigma_2 \coloneqq \frac{1}{2} \left( \array{ 0 & i \\ i & 0 } \right)$
$\sigma_3 \coloneqq \frac{1}{2} \left( \array{ i & 0 \\ 0 & -i } \right) \,.$

These are called the Pauli matrices.

###### Proposition

The Pauli matrices satisfy the commutator relations

$[\sigma_1, \sigma_2] = \sigma_3$
$[\sigma_2, \sigma_3] = \sigma_1$
$[\sigma_3, \sigma_1] = \sigma_2 \,.$

###### Proposition

The coadjoint orbits of the coadjoint action of $SU(2)$ on $\mathfrak{su}(2)$ are equivalent to the subset of the above matrices with $x^2 + y^2 + z^2 = r^2$ for some $r \geq 0$.

These are regular coadjoint orbits for $r \gt 0$.

### Finite subgroups

The finite subgroup of SU(2) have an ADE classification. See this theorem.

### $G$-Structure and exceptional geometry

Spin(8)-subgroups and reductions to exceptional geometry

reductionfrom spin groupto maximal subgroup
Spin(7)-structureSpin(8)Spin(7)
G2-structureSpin(7)G2
CY3-structureSpin(6)SU(3)
SU(2)-structureSpin(5)SU(2)
generalized reductionfrom Narain groupto direct product group
generalized Spin(7)-structure$Spin(8,8)$$Spin(7) \times Spin(7)$
generalized G2-structure$Spin(7,7)$$G_2 \times G_2$
generalized CY3$Spin(6,6)$$SU(3) \times SU(3)$

### Representation theory

To record some aspects of the linear representation theory of SU(2).

(…)

###### Lemma

We have

$\wedge^2 \mathbf{4} \;\simeq\; 3 \cdot \mathbf{1} \;\oplus\; 1 \cdot \mathbf{3} \;\;\;\in\; \mathrm{RO}(\mathrm{Sp}(1))$
###### Proof

Consider the canonical linear basis

(1)$\mathbf{4} \;\simeq\; \big\langle \{q_0, q_1, q_2, q_3\}\big\rangle_{\mathbb{R}}$
$q_0, q_1, q_2, q_3 \;\in\; \mathbb{H}$
\begin{aligned} & q_0 = 1\,, \\ & q_i q_i = -1 = - q_0 \;\; \text{for} i \in \{1,2,3\} \,, \\ & q_{\sigma(1)} q_{\sigma(3)} = \mathrm{sgn}(\sigma) q_{\sigma(3)} \,, \text{for} \sigma \in \mathrm{Sym}(3) \end{aligned}

On this, the action by Sp(1)

$\mathrm{Sp}(1) \;\simeq\; \{ q \in \mathbb{H} \;\vert\; q \bar q = 1 \}$

is by left quaternion multiplication.

Consider then the following linear basis of $\mathbf{4}\wedge \mathbf{4}$:

$\mathbf{4}\wedge \mathbf{4} \;\simeq\; \left\langle \left\{ \array{ a_1^+ := q_0 \wedge q_1 + q_2 \wedge q_3, \\ a_2^+ := q_0 \wedge q_2 + q_3 \wedge q_1, \\ a_3^+ := q_0 \wedge q_3 + q_1 \wedge q_2, } \array{ a_1^- := q_0 \wedge q_1 - q_2 \wedge q_3, \\ a_2^- := q_0 \wedge q_2 - q_3 \wedge q_1, \\ a_3^- := q_0 \wedge q_3 - q_1 \wedge q_2 } \right\} \right\rangle$

with induced $\mathfrak{sp}(1)$ Lie algebra action given by

$q_i \cdot ( q_j \wedge q_k ) \;=\; (q_i q_j) \wedge q_k + q_j \wedge (q_i q_k) \,.$

From this we find

\begin{aligned} q_1 \cdot a_1^{\pm} & = \; q_1 \cdot \big( q_0 \wedge q_1 \pm q_2 \wedge q_3 \big) \\ & =\; \big( \underset{ = 0 }{ \underbrace{ q_1 \wedge q_1 } } + \underset{ = 0 }{ \underbrace{ q_0 \wedge (-q_0) } } \big) \pm \big( \underset{ = 0 }{ \underbrace{ q_3 \wedge q_3 } } + \underset{ = 0 }{ \underbrace{ q_2 \wedge (- q_2) } } \big) \\ & =\; 0 \end{aligned}

and

\begin{aligned} q_1 \cdot a_2^{\pm} & =\; q_1 \cdot \big( q_0 \wedge q_2 \pm q_3 \wedge q_1 \big) \\ & = \; \big( \underset{ = q_1 \wedge q_2 }{ \underbrace{ q_1 \wedge q_2 } } + \underset{ = q_0 \wedge q_3 }{ \underbrace{ q_0 \wedge q_3 } } \big) \pm \big( \underset{ = q_1 \wedge q_2 }{ \underbrace{ (- q_2) \wedge q_1 } } + \underset{ = q_0 \wedge q_3 }{ \underbrace{ q_3 \wedge (- q_0) } } \big) \\ & =\; \left\{ \begin{array}{l} 2 a_3^+ \\ 0 \end{array} \right. \end{aligned}

and

\begin{aligned} q_1 \cdot a_3^{\pm} & =\; q_1 \cdot \big( q_0 \wedge q_3 \pm q_1 \wedge q_2 \big) \\ & = \; \big( \underset{ = - q_3 \wedge q_1 }{ \underbrace{ q_1 \wedge q_3 } } + \underset{ = - q_0 \wedge q_2 }{ \underbrace{ q_0 \wedge (-q_2) } } \big) \pm \big( \underset{ = - q_0 \wedge q_2 }{ \underbrace{ (- q_0) \wedge q_2 } } + \underset{ = - q_3 \wedge q_1 }{ \underbrace{ q_1 \wedge q_3 } } \big) \\ & =\; \left\{ \begin{array}{l} -2 a_2^+ \\ 0 \end{array} \right. \end{aligned} \,.

Since everything here is invariant under cyclic permutation of the three non-zero indices it follows generally that

$(\tfrac{1}{2} q_i) \cdot a_j^+ \;=\; \underset{k}{\sum} \epsilon_{i j k} a_k^+ \,, \;\; (\tfrac{1}{2} q_i) \cdot a_j^- \;=\; 0 \;\; \text{for all} i,j \in \{1,2,3\}$

But this means that

$\big\langle \{a^+_1, a^+_2, a^+_3\} \big\rangle \;\simeq\; \mathbf{3} \,, \phantom{aa} \big\langle \{a^-_i\} \big\rangle \;\simeq\; \mathbf{1} \;\;\;\in \mathrm{RO}(\mathrm{Sp}(1)) \,.$

###### Lemma

We have

$\wedge^3 \mathbf{4} \;\simeq\; \mathbf{4} \;\; \in RO(Sp(1)) \,.$
###### Proof

Consider the following linear basis

$\wedge^3 \mathbf{4} \;\simeq_{\mathbb{R}}\; \left\langle \left\{ \begin{array}{l} b_0 := + q_1 \wedge q_2 \wedge q_3 \\ b_1 := - q_0 \wedge q_2 \wedge q_3 \\ b_2 := + q_0 \wedge q_1 \wedge q_3 \\ b_3 := - q_0 \wedge q_1 \wedge q_2 \end{array} \right\} \right\rangle \,.$

We claim that in terms of these basis elements and those of (1) the isomorphism is given by $b_0 \mapsto q_0$, $b_i \mapsto q_i$. This follows by direct inspection. For instance for the Lie algebra action of $q_1$ we find:

\begin{aligned} q_1 \cdot b_0 & \; = \; q_1 \cdot ( q_1 \wedge q_2 \wedge q_3 ) \\ & \; = \; \underset{ = b_1 }{ \underbrace{ (- q_0) \wedge q_2 \wedge q_3 } } \;+\; \underset{ = 0 }{ \underbrace{ q_1 \wedge q_3 \wedge q_3 } } \;+\; \underset{ = 0 }{ \underbrace{ q_1 \wedge q_2 \wedge (- q_2) } } \\ & \;=\; b_1 \end{aligned}
\begin{aligned} q_1 \cdot b_1 & \;=\; q_1 \cdot ( - q_0 \wedge q_2 \wedge q_3 ) \\ & \;=\; - \underset{ b_0 }{ \underbrace{ q_1 \wedge q_2 \wedge q_3 } } - \underset{ = 0 }{ \underbrace{ q_0 \wedge q_3 \wedge q_3 } } - \underset{ = 0 }{ \underbrace{ q_0 \wedge q_2 \wedge (-q_2) } } \\ & \;=\; - b_0 \end{aligned}
\begin{aligned} q_1 \cdot b_2 & \;=\; q_1 \cdot (q_0 \wedge q_1 \wedge q_3) \\ & \;=\; \underset{ = 0 }{ \underbrace{ q_1 \wedge q_1 \wedge (- q_2) } } + \underset{ = 0 }{ \underbrace{ q_0 \wedge (- q_0) \wedge q_3 } } + \underset{ = b_3 }{ \underbrace{ q_0 \wedge q_1 \wedge (- q_2) } } \\ & \;=\; b_3 \end{aligned}
\begin{aligned} q_1 \cdot b_3 & \;=\; q_1 \cdot (- q_0 \wedge q_1 \wedge q_2) \\ & \;=\; - \underset{ = 0 }{ \underbrace{ q_1 \wedge q_1 \wedge q_2 } } - \underset{ = 0 }{ \underbrace{ q_0 \wedge (- q_0) \wedge q_2 } } - \underset{ = b_2 }{ \underbrace{ q_0 \wedge q_1 \wedge q_3 } } \\ & \;=\; - b_2 \end{aligned}
Dynkin labelsp. orth. groupspin grouppin groupsemi-spin group
SO(2)Spin(2)Pin(2)
B1SO(3)Spin(3)Pin(3)
D2SO(4)Spin(4)Pin(4)
B2SO(5)Spin(5)Pin(5)
D3SO(6)Spin(6)
B3SO(7)Spin(7)
D4SO(8)Spin(8)SO(8)
B4SO(9)Spin(9)
D5SO(10)Spin(10)
B5SO(11)Spin(11)
D6SO(12)Spin(12)
$\vdots$$\vdots$
D8SO(16)Spin(16)SemiSpin(16)
$\vdots$$\vdots$
D16SO(32)Spin(32)SemiSpin(32)