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## The quaternions and Lie algebras II

We now know what a Lie algebra is and we know they are abstractions of infinitesimal symmetries, which are given by derivations. Today we will see what we can say about associating infinitesimal symmetries to continuous symmetries: that is, given a matrix Lie group $G$, we will describe its associated Lie algebra $\mathfrak{g}$ of infinitesimal elements and the exponential map $\mathfrak{g} \to G$ which promotes infinitesimal symmetries to real ones.

As in the other post, I will be ignoring some technical details for the sake of exposition. For example, I am generally not specifying how I’m topologizing various objects, and this is because of the general fact that a finite-dimensional real vector space has a unique Hausdorff topology compatible with addition and scalar multiplication. Whenever I talk about limits in such a vector space, I therefore don’t need to specify how I’m imposing a topology, although it will generally be convenient to induce it via a norm (which I am also not specifying).

The exponential map

Recall that if $G$ is a Lie group, a one-parameter subgroup is a smooth homomorphism $\phi : \mathbb{R} \to G$. Differentiating any such homomorphism at the identity gives a linear map $d \phi_0 : \mathbb{R} \to T_e(G)$, and by evaluating this map we can associate to $\phi$ the tangent vector $d \phi_0(1) \in T_e(G)$, its “infinitesimal generator.” It turns out that in the other direction we can recover $\phi$ completely from its infinitesimal generator, so it is quite sensible to think of the elements of $T_e(G)$ as the infinitesimal symmetries associated to $G$ (and we will make this more precise in the next section). Thus we denote by $T_e(G) = \mathfrak{g}$ the Lie algebra of $G$; we’ll construct its Lie bracket in the next section.

To recover a one-parameter subgroup from its infinitesimal generator, we’ll first look at the most important case $G = \text{GL}_n(\mathbb{R})$. What’s nice about $\text{GL}_n(\mathbb{R})$ is that it comes canonically with a nice embedding into a vector space $\mathcal{M}_n(\mathbb{R})$, which means that its tangent space at any point can be identified with a subspace of $\mathcal{M}_n(\mathbb{R})$ by translation (and is in fact the whole thing). Thus, a tangent vector in $T_e(G) = \mathfrak{gl}_n(\mathbb{R})$ is just a matrix $D \in \mathcal{M}_n(\mathbb{R})$. If a one-parameter subgroup $\phi : \mathbb{R} \to G$ has the property that $d \phi_0(1) = D$, then it must more generally satisfy

$\displaystyle \frac{d \phi}{dt} = D \phi$.

But we know that the unique solution to this differential equation is

$\displaystyle e^{Dt} = \sum_{n \ge 0} \frac{D^n}{n!} t^n$.

(Note that we can’t write down a map like this for an arbitrary Lie group $G$, which doesn’t necessarily come equipped with an embedding into a vector space.) The analytic properties of this function are fairly straightforward to work out: in particular, it’s a smooth map $\mathbb{R} \to \mathcal{M}_n(\mathbb{R})$ with pointwise inverse $e^{-Dt}$, so really does land in $\text{GL}_n(\mathbb{R})$ for all $t$, hence really does define a one-parameter subgroup associated to the tangent vector $D$.

Note: Now is a good time to mention that this is the appropriate general context in which to motivate the definition of the exponential. Indeed, since $\phi$ is a homomorphism, it follows that $\phi(t) = \phi \left( \frac{t}{n} \right)^n$ for all $n \in \mathbb{N}$, hence that

$\displaystyle \phi(t) = \lim_{n \to \infty} \phi \left( \frac{t}{n} \right)^n = \lim_{n \to \infty} \left( 1 + \frac{Dt}{n} + O \left( \frac{1}{n^2} \right) \right)^n = e^{Dt}.$

Now, given an arbitrary matrix Lie group $G \subset \text{GL}_n(\mathbb{R})$, we have an embedding $T_e(G) \subset \mathcal{M}_n(\mathbb{R})$, so we can exponentiate any tangent vector as above, but it’s no longer obvious that the result actually lands in $G$. Rather than prove a general result here (or the more general result about an abstract Lie group $G$), we’ll prove by hand that for all the examples we care about, this is true. The corresponding map $\exp : \mathfrak{g} \to G$ given by sending $D \in \mathfrak{g}$ to $\phi(1) = e^D$ where $\phi : \mathbb{R} \to G$ is the unique one-parameter subgroup with $d \phi_0(1) = D$ is the exponential map.

Example. Let $G = \text{SL}_n(\mathbb{R}) \subset \text{GL}_n(\mathbb{R})$ be the special linear group. If $\phi : \mathbb{R} \to G$ is a one-parameter subgroup, it must satisfy $\det \phi(t) = 1$ for all $t$. If $\phi(t) = e^{Dt}$, then differentiating gives $\text{tr } D = 0$. Conversely, using the identity

$\det \exp(M) = \exp \text{tr } M$

we conclude that if $\text{tr } D = 0$ then $e^{Dt} \in \text{SL}_n(\mathbb{R})$ for all $t$. Hence the Lie algebra $\mathfrak{sl}_n(\mathbb{R})$ of the special linear group consists of precisely the $n \times n$ matrices with zero trace.

Note that even for $n = 2$ the exponential map $\exp : \mathfrak{sl}_2(\mathbb{R}) \to \text{SL}_2(\mathbb{R})$ is not surjective: the exponential of any diagonalizable matrix is diagonalizable, and the only elements of $\text{sl}_2(\mathbb{R})$ which are not diagonalizable are nilpotent, so exponentiate to unipotent matrices, hence the exponential map misses matrices like $\left[ \begin{array}{cc} -1 & 1 \\ 0 & -1 \end{array} \right]$.

Example. Let $G = \text{O}(n) \subset \text{GL}_n(\mathbb{R})$ be the orthogonal group. If $\phi : \mathbb{R} \to G$ is a one-parameter subgroup, it must satisfy $\phi(t)^T \phi(t) = I$ for all $t$. If $\phi(t) = e^{Dt}$, differentiating this condition gives

$\displaystyle e^{D^T t} D e^{Dt} + D^T e^{D^T t} e^{Dt} = 0 \Rightarrow D + D^T = 0$.

Conversely, if $D = -D^T$ then $(e^{Dt})^T = e^{-Dt}$. Hence the Lie algebra of the orthogonal group consists of precisely the $n \times n$ skew-symmetric matrices. Note that any such matrix automatically has zero trace, hence the image of the exponential map already lies in the special orthogonal group $\text{SO}(n)$, the connected component of the identity of the orthogonal group. For this reason we denote their Lie algebras by the same symbol $\mathfrak{so}(n)$ since in fact they are identical.

Example. Let $G = \text{U}(n) \subset \text{GL}_{n}(\mathbb{C})$ be the unitary group. The discussion above about the exponential map for $\text{GL}_n(\mathbb{R})$ carries over word-for-word for $\text{GL}_n(\mathbb{C})$. In addition, if $\phi : \mathbb{R} \to G$ is a one-parameter subgroup, then it must satisfy $\phi(t)^{\dagger} \phi(t) = I$ for all $t$ (where $M^{\dagger}$ denotes the conjugate transpose). If $\phi(t) = e^{Dt}, D \in \mathcal{M}_n(\mathbb{C})$ then differentiating this condition gives

$e^{D^{\dagger} t} D e^{Dt} + D^{\dagger} e^{D^{\dagger} t} e^{Dt} = 0 \Rightarrow D + D^{\dagger} = 0$.

Conversely, if $D^{\dagger} = -D$ then $(e^{Dt})^{\dagger} = e^{-Dt}$. Hence the Lie algebra $\mathfrak{u}(n)$ consists precisely of the skew-Hermitian matrices. Note that a skew-Hermitian matrix need not have zero trace: it need only have a trace with zero real part. The skew-Hermitian matrices with zero trace naturally give the Lie algebra $\mathfrak{su}(n)$ of the special unitary group. Note also that both of these Lie algebras are real, not complex.

The Lie bracket

Suppose a Lie group $G$ acts on a suitably nice real algebra $A$ via a suitably nice action $\rho : G \to \text{Aut}(A)$. Given any one-parameter subgroup $\phi : \mathbb{R} \to G$ we get a one-parameter subgroup $\rho \circ \phi : \mathbb{R} \to \text{Aut}(A)$ which we can differentiate to get a derivation $d(\rho \circ \phi)_0$. This assignment gives a linear map $\mathfrak{g} \to \text{Der}(A)$ given by differentiating the one-parameter subgroup generated by the corresponding tangent vector, and in fact we can canonically place a Lie algebra structure on $\mathfrak{g}$ such that the above map is a homomorphism of Lie algebras.

This is done as follows. The action of $G$ on itself by conjugation descends to a canonical representation $\text{Ad}_g : \mathfrak{g} \to \mathfrak{g}$ on the tangent space at the identity, the adjoint representation of $G$, and differentiating the corresponding map $G \to \text{Aut}(\mathfrak{g})$ at the identity gives a linear map $\mathfrak{g} \to \text{End}(\mathfrak{g})$, or equivalently a bilinear map

$\displaystyle [ \cdot, \cdot ] : \mathfrak{g} \times \mathfrak{g} \to \mathfrak{g}$.

Note that differentiating the conjugation action is exactly how we defined the Lie bracket on $\text{Der}(A)$. Thus, for the same reasons as before, the bracket satisfies $[x, x] = 0$, hence it is skew-symmetric, and it gives an action of $\mathfrak{g}$ on itself by derivations, so it satisfies the Jacobi identity. Finally, since both the Lie bracket on $\mathfrak{g}$ and the Lie bracket on $\text{Der}(A)$ are defined by differentiating conjugation, it follows that they are compatible, so the natural map $\mathfrak{g} \to \text{Der}(A)$ is a Lie algebra homomorphism as desired.

It is straightforward to check that, like the Lie bracket of derivations, the Lie bracket on $\mathfrak{gl}_n(\mathbb{R})$ is also given by the commutator $[a, b] = ab - ba$. Thus for all of the matrix Lie groups we described above, their Lie algebras also have bracket given by the commutator in the corresponding subalgebras of $\mathfrak{gl}_n(\mathbb{R})$.

An extremely important property of the Lie bracket, which we will not prove, is the Baker-Campbell-Hausdorff formula, which shows that the Lie bracket is enough to recover the multiplication in the Lie group in a neighborhood of the identity.

Functoriality

Let $f : G \to H$ be a smooth homomorphism of Lie groups. Differentiating induces a map $df_e : \mathfrak{g} \to \mathfrak{h}$. Since $f$ is a homomorphism, it respects group multiplication, hence conjugation, hence the adjoint action on $\mathfrak{g}$ and $\mathfrak{h}$. Since $f$ is smooth, it respects taking the derivative of the adjoint action, hence it is a morphism of Lie algebras. All of the natural compatibility conditions are met, giving a functor

$\text{LieGrp} \to \text{LieAlg}_{\mathbb{R}}$

from the category of Lie groups to the category of real Lie algebras. The derivative of the exponential map $d \exp_0 : \mathfrak{g} \to \mathfrak{g}$ at the origin is the identity, so by the inverse function theorem it is a local diffeomorphism. Hence the image of a small open ball in the Lie algebra gives a small open neighborhood of the identity in the Lie group $G$. If $G$ is connected, we know that open neighborhoods of the identity generate $G$, hence $f : G \to H$ is completely determined by the induced map $\mathfrak{g} \to \mathfrak{h}$. In other words, for connected $G$ the above functor is faithful.

It is appropriate to pause for a moment and think about how lucky this is. The category of Lie groups is a priori a messy category whose objects are infinitary and full of differential-geometric information, yet the category of Lie algebras is completely algebraic and the ones relevant to the above discussion are all finite-dimensional: they can be completely specified by the structure constants of the Lie bracket, hence by a finite list of numbers. And from the above discussion it seems that many properties of connected Lie groups can be determined just by looking at Lie algebras.

We know that the above functor is faithful. Is it full? A simple counterexample shows what’s wrong. It is well known (and we will prove later) that the only continuous homomorphisms $f : S^1 \to S^1$ are given by the maps $z \mapsto z^n, n \in \mathbb{Z}$ (where we think of $z \in \mathbb{C}$ as a unit complex number). The associated Lie algebra is $\mathfrak{so}(2) \cong \mathfrak{u}(1) \cong \mathbb{R}$ with the trivial bracket, however, so among the morphisms $\mathbb{R} \to \mathbb{R}$ of Lie algebras only those of the form $v \mapsto nv, n \in \mathbb{Z}$ come from maps between the corresponding Lie algebras. The problem is that since Lie algebras only give information about what’s happening in a neighborhood of the identity, they can’t distinguish a Lie group from any of its covering spaces, and in particular can’t distinguish $S^1$ from $\mathbb{R}$. However, as it turns out (and we will not prove this), if $G$ is simply connected then every Lie algebra homomorphism $\mathfrak{g} \to \mathfrak{h}$ lifts to a homomorphism $G \to H$ of Lie groups.

In particular, associated to any Lie group $G$ is its category of finite-dimensional complex representations, or smooth maps $G \to \text{GL}_n(\mathbb{C})$ with commutative triangles as morphisms. Applying the above functor gives Lie algebra homomorphisms $\mathfrak{g} \to \mathfrak{gl}_n(\mathbb{C})$. The corresponding category is the category of Lie algebra representations of $\mathfrak{g}$. By the above discussion, differentiating gives a functor $\text{Rep}(G) \to \text{Rep}(\mathfrak{g})$, and if $G$ is simply connected (again, we won’t prove this) we actually get an equivalence of categories. But $\mathfrak{g}$ being a nice linear object, it is much easier to study the category of representations of $\mathfrak{g}$ than the category of representations of $G$.

The quaternions

The special cases of Euler’s formula and the exponential map for $\text{SU}(2)$ can be explained in the above context as follows. We saw above that we can write the exponential map as an actual exponential whenever we can embed a given Lie group $G$ into an algebra in a nice way. Above we did this for $\mathcal{M}_n(\mathbb{R})$. In the case of Euler’s formula and $\text{SU}(2)$:

1. The exponential map for $\text{SO}(2) \cong \text{U}(1)$ is particularly nice because this group embeds nicely into the algebra $\mathbb{C}$.
2. The exponential map for $\text{SU}(2)$ is particularly nice because this group embeds nicely into the algebra $\mathbb{H}$.

This realizes the Lie algebra $\mathfrak{so}(2) \cong \mathfrak{u}(1)$ as the imaginary complex numbers with the trivial bracket, and realizes the Lie algebra $\mathfrak{su}(2) \cong \mathfrak{so}(3)$ as the imaginary quaternions with the bracket

$\displaystyle [i, j] = ij - ji = 2k, [j, k] = jk - kj = 2i, [k, i] = ki - ik = 2j.$

These relations perhaps look nicer when written

$\displaystyle \left[ \frac{i}{2}, \frac{j}{2} \right] = \frac{k}{2}, \left[ \frac{j}{2}, \frac{k}{2} \right] = \frac{i}{2}, \left[ \frac{k}{2}, \frac{i}{2} \right] = \frac{j}{2}.$

Indeed, the one-parameter subgroup $\phi(t)$ generated by $\frac{i}{2}$ acts on imaginary quaternions by $v \mapsto e^{ \frac{it}{2} } v e^{ - \frac{it}{2} }$, so by a rotation of $t$ radians about the $i$-axis, and the analogous statement is true for any purely imaginary quaternion of norm $1$. The operators $\frac{i}{2}, \frac{j}{2}, \frac{k}{2}$ can therefore rightfully be called the infinitesimal generators of rotation about the $i, j, k$-axes, and this is precisely why they appear (up to normalization) as the angular momentum operators in quantum mechanics, which is very important to our story.