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## SU(2) and the quaternions

The simplest compact Lie group is the circle $S^1 \cong \text{SO}(2)$. Part of the reason it is so simple to understand is that Euler’s formula gives an extremely nice parameterization $e^{ix} = \cos x + i \sin x$ of its elements, showing that it can be understood either in terms of the group of elements of norm $1$ in $\mathbb{C}$ (that is, the unitary group $\text{U}(1)$) or the imaginary subspace of $\mathbb{C}$.

The compact Lie group we are currently interested in is the $3$-sphere $S^3 \cong \text{SU}(2)$. It turns out that there is a picture completely analogous to the picture above, but with $\mathbb{C}$ replaced by the quaternions $\mathbb{H}$: that is, $\text{SU}(2)$ is isomorphic to the group of elements of norm $1$ in $\mathbb{H}$ (that is, the symplectic group $\text{Sp}(1)$), and there is an exponential map from the imaginary subspace of $\mathbb{H}$ to this group. Composing with the double cover $\text{SU}(2) \to \text{SO}(3)$ lets us handle elements of $\text{SO}(3)$ almost as easily as we handle elements of $\text{SO}(2)$.

The quaternions

Recall that $\text{SU}(2)$ is the subgroup of $\text{Aut}(\mathbb{C}^2)$ preserving an inner product on $\mathbb{C}^2$ with determinant $1$. Equivalently, the inner product induces an adjoint (conjugate transpose) operation $M \mapsto M^{\dagger}$ on $\text{End}(\mathbb{C}^2)$, and $\text{SU}(2)$ is the subgroup satisfying $M^{\dagger} M = I$ and $\det(M) = 1$.

$\text{SU}(2)$ naturally embeds into the real subalgebra of linear transformations $M$ satisfying $M^{\dagger} M \in \mathbb{R}$ (that is, is a real multiple of the identity) and $\det(M) \in \mathbb{R}_{\ge 0}$, and this will be our definition of the quaternions $\mathbb{H}$. (It’s not the best definition, but I am not comfortable with the ones that are better-motivated, and it’s a big improvement over the definition by generators and relations.) This is analogous to the definition of the complex numbers as the subalgebra of $\text{End}(\mathbb{R}^2)$ satisfying $M^T M \in \mathbb{R}$ and $\det(M) \in \mathbb{R}_{\ge 0}$, in which $\text{SO}(2)$ naturally embeds. The rest of the analogies to the construction of the complex numbers are left as an exercise.

In the standard basis, and relative to the standard inner product, the elements of $\mathbb{H}$ are given by $2 \times 2$ complex matrices of the form

$\left[ \begin{array}{cc} \alpha & - \bar{\beta} \\ \beta & \bar{\alpha} \end{array} \right]$.

As a real vector space, the quaternions are spanned by the four matrices $1, \mathbf{i}, \mathbf{j}, \mathbf{k}$, where $1$ is the identity matrix and

$\mathbf{i} = \left[ \begin{array}{cc} i & 0 \\ 0 & -i \end{array} \right], \mathbf{j} = \left[ \begin{array}{cc} 0 & 1 \\ -1 & 0 \end{array} \right], \mathbf{k} = \left[ \begin{array}{cc} 0 & i \\ i & 0 \end{array} \right]$.

Abstractly, the quaternions can be defined by the relations $\mathbf{i}^2 = \mathbf{j}^2 = \mathbf{k}^2 = \mathbf{ijk} = -1$. In particular, $\mathbf{ij} = \mathbf{k}, \mathbf{jk} = \mathbf{i}, \mathbf{ki} = \mathbf{j}$, and the generators anticommute.

The quaternions are closed under adjoint: in particular, we see that $\mathbf{i}^{\dagger} = - \mathbf{i}, \mathbf{j}^{\dagger} = - \mathbf{j}, \mathbf{k}^{\dagger} = - \mathbf{k}$, so the generators are all skew-adjoint. The adjoint defines a norm on quaternions as follows: if $q \in \mathbb{H}$, we define $|q| = \sqrt{q^{\dagger} q}$. If we write $q = a + b \mathbf{i} + c \mathbf{j} + d \mathbf{k}$, then $q^{\dagger} = a - b \mathbf{i} - c \mathbf{j} - d \mathbf{k}$ and

$|q| = \sqrt{a^2 + b^2 + c^2 + d^2}$.

The norm implies that any nonzero quaternion has inverse $\frac{q^{\dagger}}{|q|^2}$, so the quaternions form a normed division algebra over $\mathbb{R}$ with an involution $q \mapsto q^{\dagger}$ related to its norm by the B*-identity.

This is a lot of structure, and there are several theorems asserting that it is quite rare to find all this structure together: if one insists on associativity, $\mathbb{R}, \mathbb{C}, \mathbb{H}$ are the only finite-dimensional real division algebras as well as the only normed real division algebras. This has several important consequences:

1. Over any field, Schur’s lemma asserts that the endomorphisms of a finite-dimensional irreducible representation of an algebra form a division algebra. It follows that over $\mathbb{R}$ the possible endomorphism rings are $\mathbb{R}, \mathbb{C}$ and $\mathbb{H}$. This leads to a natural classification of, say, real representations of compact groups as real, complex, or quaternionic depending on their endomorphism ring.
2. One can define Hilbert spaces over all of these algebras. John Baez has been writing a series about this.
3. The automorphisms of the above Hilbert spaces, in the finite-dimensional case, are a natural source of Lie groups. Finite-dimensional real Hilbert spaces give the Lie groups $\text{SO}(n)$, finite-dimensional complex Hilbert spaces give the Lie groups $\text{SU}(n)$, and finite-dimensional quaternionic Hilbert spaces give the Lie groups $\text{Sp}(n)$, the symplectic groups. The Lie algebras of these groups are precisely the infinite series of simple Lie algebras, and the exceptional Lie algebras can still be constructed from real division algebras if one also brings in the octonions.

The multiplicativity of the quaternion norm (since it comes from determinants) implies Euler’s four-square identity, which allows us to reduce Lagrange’s four-square theorem to the prime case. One can even approach this theorem using the Hurwitz quaternions, which satisfy an analogue of unique factorization.

A standard reference for this and many other properties of the quaternions is Conway and Smith’s On quaternions and octonions.

The exponential map

$\text{SU}(2)$ naturally embeds into $\mathbb{H}$ as the group of elements of norm $1$, and this allows us to construct a nice representation of $\text{SU}(2)$. Under the adjoint $q \mapsto q^{\dagger}$, which is involutive, the quaternions decompose into a direct sum $\text{span}(1) \oplus \text{span}(\mathbf{i}, \mathbf{j}, \mathbf{k})$ where the first subspace is elements fixed under adjoint and the second is elements negated under adjoint. It follows from here that if $q \in \text{SU}(2) \subset \mathbb{H}$, then the conjugation action

$\displaystyle \text{Ad}(q) : v \mapsto qvq^{-1}, q \in \text{SU}(2), v \in \mathbb{H}$

preserves this direct sum decomposition, hence acts on the imaginary subspace $\text{span}(\mathbf{i}, \mathbf{j}, \mathbf{k})$. Moreover, by the multiplicativity of the norm, this representation acts on the unit sphere in this subspace, so lands in $\text{SO}(3)$. This is a special representation of $\text{SU}(2)$ called its adjoint representation.

The adjoint representation is behind the use of quaternions to describe rotations which is prevalent in many applied fields. It’s not hard to guess that it is just the double cover $\text{SU}(2) \to \text{SO}(3)$ we’ve already encountered, but in a more manageable form (one where explicit computations are easier to do).

Let’s check that this is the case. Since $v \mapsto qvq^{-1}$ fixes $q$ and $q^{\dagger}$, it also fixes the imaginary part $\text{Im}(q) = \frac{q - q^{\dagger}}{2}$ of $q$, which lies in the imaginary subspace. The imaginary part therefore determines the axis of rotation of the rotation $v \mapsto qvq^{-1}$. Moreover, if the imaginary part is zero then the rotation is the identity, and so it is reasonable to expect that the relative size of the real and imaginary parts of $q$ determines the angle of the rotation.

This suggests the following idea. Let $\mathbf{h} = \frac{ \text{Im}(q) }{| \text{Im}(q) |}$. Then $\mathbf{h}^2 = -1$ (exercise), so the subalgebra generated by $\mathbf{h}$ is isomorphic to $\mathbb{C}$. Then we can write

$q = \cos \theta + \mathbf{h} \sin \theta$

for some real $\theta$. The quaternion norm descends to the subalgebra generated by $\mathbf{h}$, giving the usual complex norm, and this implies that analytic arguments are available and we can safely say that Euler’s formula

$\displaystyle \cos \theta + \mathbf{h} \sin \theta = \exp \left( \mathbf{h} \theta \right)$

must hold! This gives a smooth homomorphism from the copy of $S^1$ inside $\text{SU}(2)$ above to the copy of $S^1$ inside $\text{SO}(3)$ corresponding to rotation about $\text{Im}(q)$, and the only smooth homomorphisms from a circle to itself are the multiplication maps $z \mapsto z^n$ (where $z \in \mathbb{C}, |z| = 1$). It follows that $q$ is a rotation by $n \theta$ for some integer $n$. To compute $n$ it suffices to find the smallest $\theta$ for which the map $v \mapsto qvq^{-1}$ is trivial, i.e. $q$ lies in the center of $\mathbb{H}$. We leave as an exercise the verification that the center of $\mathbb{H}$ is the subalgebra of scalar multiples of the identity, so this first occurs when $q = -1$, hence $n = 2$ and we get a double cover $\text{SU}(2) \to \text{SO}(3)$ inducing a double cover of maximal tori as before.

The exponential $e^q$ is generally defined for every $q \in \mathbb{H}$ by the usual power series. Since $\left( e^q \right)^{\dagger} = e^{q^{\dagger}}$, in order for its image to land in $\text{SU}(2)$ it is necessary and sufficient that $e^q e^{q^{\dagger}} = 1$, and since $q$ and $q^{\dagger}$ commute it is necessary and sufficient that $e^{q + q^{\dagger}} = 1$; that is, $q$ must be purely imaginary. (It is not generally true that $e^{a+b} = e^a e^b$, since the quaternions are noncommutative. In general this only holds if $a$ and $b$ commute.) It follows that the restriction of the exponential $e^q$ to the imaginary subspace defines an exponential map $\mathbb{R}^3 \to \text{SU}(2)$ neatly generalizing the exponential map $\mathbb{R} \to \text{SO}(2)$, and also giving an exponential map $\mathbb{R}^3 \to \text{SO}(3)$.

This map is really about as simple as possible: to describe a rotation about some axis by an angle $\theta$, write down the corresponding unit vector $\mathbf{h}$ in the imaginary subspace, and then the desired rotation is just

$\displaystyle v \mapsto qvq^{-1}, q = e^{ \mathbf{h} \frac{\theta}{2} }$.

If we don’t want to use the exponential, we can just directly write $q = \cos \frac{\theta}{2} + \mathbf{h} \sin \frac{\theta}{2}$; I think this is often done in computer graphics. Note that it only requires $4$ (slightly redundant) real parameters to store a rotation this way, as opposed to a $3 \times 3$ matrix which requires $9$ (highly redundant) real parameters, and moreover the quaternion parameterization is a covering map, so its local behavior is maximally nice.

By contrast, the naive parameterization of $\text{SO}(3)$ via Euler angles (which corresponds to thinking about $e^{b \mathbf{i} } e^{c \mathbf{j}} e^{ d \mathbf{k}}$ instead of $e^{b \mathbf{i} + c \mathbf{j} + d \mathbf{k} }$) does not come from a covering map. At certain points it has problematic local behavior which is responsible for gimbal lock.

The fundamental group, redux

The quaternion description of the double cover $\text{SU}(2) \to \text{SO}(3)$ leads to a complete visual proof that $\pi_1(\text{SO}(3)) \simeq \mathbb{Z}/2\mathbb{Z}$ (instead of the result that it is either this or trivial). First, an element $q \in \text{SU}(2)$ can almost be identified with its imaginary part $\text{Im}(q)$, which determines its real part up to sign. It follows that $\text{SU}(2) \cong S^3$ can be identified with a pair of solid $3$-balls $B^{-}, B^{+}$ with their boundaries identified, one for each possible sign of the real part. This is just the familiar choice for a pair of charts on $S^3 \subset \mathbb{R}^4$, and it might be easier to visualize the corresponding identification of $S^2$ with a pair of solid circles with their boundaries identified.

Just as there is a stereographic projection from $S^2$ to $\mathbb{R}^2$ that misses a single point, there is a stereographic projection from $S^3$ to $\mathbb{R}^3$ that misses a point, identifying $S^3$ with $\mathbb{R}^3$ together with a point at infinity. This stereographic projection can be chosen to send the positive unit ball $B^{+}$ to the standard unit ball in $\mathbb{R}^3$ and to send the negative unit ball $B^{-}$ to its complement in $\mathbb{R}^3$ (up to their intersection). There is a natural way to do this which uniquely specifies the projection: if a quaternion

$\displaystyle q = a + b \mathbf{i} + c \mathbf{j} + d \mathbf{k}$

has positive real part then it is sent to $\text{Im}(q)$, and we will send its negative counterpart $- q^{\dagger} = -a + b \mathbf{i} + c \mathbf{j} + d \mathbf{k}$ to the point

$\displaystyle - \frac{1}{\text{Im}(q)} = \frac{b \mathbf{i} + c \mathbf{j} + d \mathbf{k}}{b^2 + c^2 + d^2}$.

Then $q$ is always sent inside the unit ball $B^{+}$, while $- q^{\dagger}$ is always sent outside it into the complement $B^{-}$ (up to their intersection), and the two agree if and only if $q$ has zero real part, on the intersection of the positive and negative balls, the unit $2$-sphere.

Thus we have identified $\text{SU}(2)$ with $\mathbb{R}^3 \cup \{ \infty \}$ in such a way that the quaternions with positive real part are sent inside the unit $2$-sphere and the quaternions with negative real part are sent outside the unit sphere. We have also chosen this particular identification because it sends negation to sphere inversion. This is important because the covering map $\text{SU}(2) \to \text{SO}(3)$ is precisely the quotient by negation!

So the quotient $\text{SU}(2) \to \text{SO}(3)$ identifies the positive and negative balls with each other; in other words, $\text{SO}(3)$ the quotient of $\mathbb{R}^3 \cup \{ \infty \}$ by sphere inversion, which is naturally homeomorphic to projective space $\mathbb{P}^3(\mathbb{R})$ (which is, after all, the quotient of $S^3$ by negation). In this picture, the identity $I \in \text{SO}(3)$ is the origin (identified with the point at infinity), and it is now obvious that there are two homotopy classes of paths from the identity to itself since homotopy lifting allows us to lift them to paths in $\mathbb{R}^3 \cup \{ \infty \}$: one of them is from the origin to itself, and the other is from the origin to the point at infinity.

This visualization was explained to me by Ryan Budney on math.SE, and it is somewhat easier to work with since it allows a natural visualization of the universal cover at the same time. We can identify elements of $\text{SO}(3)$ with points in the positive unit ball $B^{+}$ up to identification of antipodes on the unit $2$-sphere. The position of the point relative to the origin is the axis of rotation, and the distance from the point to the origin is the angle of rotation.

### 9 Responses

1. Thank you for the explanation, it helps me a lot.

2. I don’t quite see why we get smooth homomorphism $S^1\subset SU_2\to S^1\subset SO_3$. The only thing I agree it’s clear is that it is a well defined map, how do you argue that it is homomorphism and smooth?

• A composition of smooth maps is smooth and a composition of homomorphisms is a homomorphism.

3. Hi Qiaochu: There is an error of omission in both of the analogous descriptions of the complex numbers and quaternions as matrices.

In both cases, It’s necessary to require $det(M)$ to be a nonnegative real number. Otherwise you allow matrices like $\begin{bmatrix}-1&0\&1\end{bmatrix}$, which add too many real dimensions to the set.

• Whoops. Thanks for the correction!

4. hey, I’m a math student and I am very interested this topic! I wanted to ask but only if you have material about the complex structure J (j as an element of the quaternions of Hamilton). In the sense that if you have material on the endomorphism between the complex and real when I identify V with R ^ 2 and the group of linear transformations of the plan in accordance with decisive places! thanks

5. Could you please tell me which books you are reading from for lie algebra? Is it from “representation theory: a first course” ? Are there any other resources you would recommend for this kind of material?

• Fulton and Harris is good, but I really think it needs to be supplemented with other books. I’ve also been reading Kirillov’s book and Stillwell’s Naive Lie Theory, which is a really nice introduction to the basics.

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