I'm putting the finishing touches on a post about projective representations of Lie groups and Lie algebras — an essential topic for understanding how symmetries act on quantum systems — and discovered a gap in my own knowledge. See, in discussing projective representations of Lie algebras on a Hilbert space $\mathcal{H}$, it is very convenient to be able to talk about the Lie algebra of the unitary group $\operatorname{U}(\mathcal{H}),$ and even moreso about the Lie algebra of the projective unitary group
$$\operatorname{PU}(\mathcal{H}) = \operatorname{U}(\mathcal{H})/\{\alpha I | \alpha \in \operatorname{U}(1)\}.$$
The trouble is that when $\mathcal{H}$ is infinite-dimensional, as is often the case in quantum systems of interest, $\operatorname{U}(\mathcal{H})$ is infinite-dimensional and thus not a manifold, at least not in the usual sense of a manifold being a space locally diffeomorphic to $\mathbb{R}^n.$ Everything I've ever learned about Lie groups and algebras has been in the finite-dimensional context; clearly a little more knowledge is needed to understand these ideas properly.
So I went and learned a bit about Banach Lie groups, which is one way of defining infinite-dimensional Lie groups, and how this idea can be used to formalize the notion of a Lie algebra for $\operatorname{U}(\mathcal{H})$ or $\operatorname{PU}(\mathcal{H}).$ This material would be too much of a diversion to include in my post about projective representations, but I found it interesting so I wrote up some notes about it that I'm posting here.
In the first section, I'll define the notion of a Banach manifold, which is one way of extending the notion of a manifold to infinite dimensions. In the second section, I'll give some basic definitions and theorems about Banach Lie groups and algebras. In the third section, I'll show that $\operatorname{U}(\mathcal{H})$ and $\operatorname{PU}(\mathcal{H})$ are Banach Lie groups and compute their Lie algebras. I'll give references for various lemmas as they come up, but I cobbled most of this knowledge together from many different sources and don't have a single good pedagogical source to cite.
Prerequisites: Familiarity with the fundamentals of finite-dimensional Lie groups and algebras, and with the basic definition of a smooth manifold and its tangent spaces.
Table of Contents
- Banach Manifolds
- Banach Lie Groups and Algebras
- Banach Lie Theory of the Unitary and Projective Unitary Groups
1. Banach Manifolds
A smooth manifold is a space that is locally diffeomorphic to $\mathbb{R}^n$ for some finite $n.$ So to define infinite-dimensional manifolds, we need an infinite-dimensional generalization of $\mathbb{R}^n.$ Some essential properties of $\mathbb{R}^n$ are: (1) it is a real vector space; (2) it has an inner product; (3) it is complete with respect to the norm induced by the inner product. A space satisfying all three of these properties is by definition a real Hilbert space. If we relax property (2) by forgetting about the inner product, and assume instead only that our vector space has a norm with respect to which it is complete, then the resulting object is called a Banach space.
We will generalize the notion of a manifold to infinite dimensions by replacing $\mathbb{R}^n$ with a general Banach space $X$. Note that there are other possible ways of generalizing $\mathbb{R}^n$ to infinite dimensions, leading to other possible definitions of infinite-dimensional manifolds; for example, the Wikipedia page on Lie groups (accessed Dec 8, 2021) mentions a type of infinite-dimensional manifold for which $\mathbb{R}^n$ is replaced by a general locally convex topological vector space. Because I'm only interested in the unitary group $\operatorname{U}(\mathcal{H})$ at present, which turns out to have a local Banach structure, we'll stick to Banach manifolds.
The definition of a Banach manifold is exactly what you would expect. A (smooth) Banach manifold over the Banach space $X$ is a topological space $M$ together with a collection of charts $\{(U_j, \psi_j)\},$ where:
- $\{U_j\}$ is an open cover of $M$.
- Each $\psi_j : U_j \rightarrow X$ is a homeomorphism onto its image.
- The transition maps $\psi_j \circ \psi_k^{-1} : \psi_k(U_j \cap U_k) \rightarrow \psi_j(U_j \cap U_k)$ are smooth.
To make sense of condition 3 we need to define the derivative of a function on a Banach space. It is the Fréchet derivative; a map $f : X \rightarrow X$ is called differentiable at $x \in X$ if there exists a bounded operator $\text{D}f : X \rightarrow X$ satisfying
$$f(x + h) = f(x) + (\text{D}f) h + O(h^2)$$
for any $h \in X.$ More formally, $\text{D}f$ should satisfy
$$\lim_{h \rightarrow 0} \frac{\Vert f(x + h) - f(x) - (\text{D}f)h \Vert_X}{\Vert h \Vert_X} = 0.$$
The only thing missing is the definition of what it means for $\text{D}f$ to be bounded; this means that it has finite operator norm
$$\Vert \text{D}f \Vert_{\infty} = \sup_{x \neq 0} \frac{\Vert (\text{D}f) x \Vert_{X}}{\Vert x \Vert_X}.$$
Note: You'll
occasionally see people — including me, later in this post — say that a Banach manifold is a manifold
"(locally) modeled on a Banach space." The phrase "(locally) modeled on a
space S" just means that, in the usual definition of a manifold, you
replace $\mathbb{R}^n$ by $S$.
We are left only with the small problem of defining the tangent space to a point $p \in M.$ For finite-dimensional manifolds, there are two definitions usually employed:
Definition 1: A tangent vector $v_p$ at a point $p \in M$ is a derivation, i.e. a linear map $v : C^{\infty}(M, \mathbb{R}) \rightarrow \mathbb{R}$ satisfying the product rule
$$v_p(f g) = v_p(f) g(p) + f(p) v_p(g).$$
Definition 2: A tangent vector $v_p$ at a point $p \in M$ is an equivalence class of chart-vector pairings $(U_j, v)$ where $U_j$ is a chart that maps $p$ to zero and $v$ is a vector in $X.$ Two chart-vector pairings $(U_j, v), (U_k, w)$ are said to be equivalent if $\psi_j \circ \psi_k^{-1}$, considered as a diffeomorphism on a neighborhood of $0$ in $X$, pushes $w$ forward to $v$:
$$\text{D}(\psi_j \circ \psi_k^{-1})(w) = v.$$ Addition of equivalence classes is defined by pushing $(U_j, v)$ and $(U_k, w)$ to the same chart and adding the push-forwards of $v$ and $w$ in that chart; the choice of chart does not change the resulting equivalence class.
I have been led to understand by these two MathOverflow posts (1), (2) that for Banach manifolds, these two notions of tangent vector are not equivalent. Definition 1 is for "operational tangent vectors," while definition 2 is for "kinematic tangent vectors;" see chapter 6 of Kriegl & Michor for more details. We will adopt definition 2, which is standard (see e.g. chapter 2.2 of Lang's book on differential geometry).
Every smooth map $\phi : M \rightarrow N$ between Banach manifolds induces a push-forward of tangent vectors at $p \in M$ to tangent vectors at $\phi(p) \in N$ in the usual way. Let $X$ and $Y$ be the local Banach spaces of $M$ and $N$, respectively. If $(U, \psi)$ is a chart in $M$ sending $p$ to zero and $(V, \chi)$ is a chart in $N$ sending $\phi(p)$ to zero, then $\chi \circ \phi \circ \psi^{-1}$ is a smooth map from a neighborhood of zero in $X$ to a neighborhood of zero in $Y$; it has a differential $\text{D}(\chi \circ \phi \circ \psi^{-1})_0$ at $0 \in X$, which is a linear operator from $X$ to $Y$. The push-forward of the equivalence class $[(U, v)]$ by $\phi$ is defined by
$$\text{D}\phi_p[(U, v)] = [(V, \text{D}(\chi \circ \phi \circ \psi^{-1})_0(v)].$$
This map can be checked to be well-defined by standard methods. (One applies the chain rule for smooth maps on Banach spaces to show that the derivative $\text{D}(\chi \circ \phi \circ \psi^{-1})_0$ changes the right way under a change of the chart $U$ or $V$.)
Interestingly, by using the push-forward we can interpret every kinematic tangent vector (defined as an equivalence class) as an operational tangent vector (defined as a linear map on smooth functions satisfying the product rule). For any smooth function $\phi : M \rightarrow \mathbb{R},$ and any tangent vector $v_p$ at $p$, we define
$$v_p(\phi) = (\text{D}\phi_p)(v_p).$$
Because $\phi$ has its image in the real numbers, and tangent vectors of $\mathbb{R}$ are canonically identified with $\mathbb{R}$, the quantity $(\text{D} \phi_p)(v_p)$ is a real number. Linearity and the product rule of $v_p(\phi)$ follow from the same properties of $\text{D} \phi.$ So we are free to think of (kinematic) tangent vectors as derivations, but it may not be the case that every derivation is a (kinematic) tangent vector.
2. Banach Lie Groups and Algebras
A Lie group, in brief, is a space $G$ that is both a group and a manifold, and for which those two structures agree. I.e., the multiplication map $(g, h) \mapsto gh$ and the inverse map $g \mapsto g^{-1}$ are both smooth. A Banach Lie group is a space $G$ that is both a group and a Banach manifold, for which the multiplication and inverse maps are smooth in the sense that they are smooth in every chart.
The definition of a Banach Lie algebra is intuitive — it is a Banach space $\mathfrak{g}$ with a bilinear, antisymmetric bracket $[\cdot, \cdot] : \mathfrak{g} \times \mathfrak{g} \rightarrow \mathfrak{g}$ such that (1) the bracket satisfies the Jacobi identity, and (2) the bracket is bounded:
$$\Vert [\cdot, \cdot] \Vert_{\infty} = \sup_{x, y \neq 0} \frac{\Vert [x, y] \Vert_{\mathfrak{g}}}{\Vert x \Vert_{\mathfrak{g}} \Vert y \Vert_{\mathfrak{g}}} < \infty.$$
In order for these two notions to be compatible, it ought to be the case that the tangent space of a Banach group $G$ at the identity has a natural bracket that makes it a Banach Lie algebra $\mathfrak{g}.$ This is not hard to show using the usual tools for finite-dimensional Lie groups. Let $(U, \psi)$ be any chart sending the identity $e \in G$ to $0 \in X$, and denote by $L_g : G \rightarrow G$ the diffeomorphism on $G$ corresponding to left-multiplication by $g$. For any point $x \in \psi(U),$ we can define a smooth map $L_x = \psi \circ L_{\psi^{-1}(x)} \circ \psi^{-1}$ that sends $0$ to $x$. The differential of this map at the origin, which is a linear map on $X$, should be thought of as a push-forward from tangent vectors at $0$ to tangent vectors at $x$. For any vector $v \in X$, the family of push-forwards $(\text{D}L_{x})_0(v)$ for varying $x$ defines a smooth, "left-invariant" vector field on $\psi(U).$ Since we are working in charts, we can make this very explicit; associated to any $v \in X$ there is a smooth function $\xi_v : X \rightarrow X$ given by
$$\xi_v(x) = \text{D}(\psi \circ L_{\psi^{-1}(x)} \circ \psi^{-1})_0(v). \tag{1}$$
This map is smooth in $x$ because $L_g$ depends smoothly on $g$.
Since $\xi_v$ is a smooth map from $X$ into $X$, its differential $(\text{D} \xi_v)_{x}$, which generally may depend on $x \in X$, is a linear operator on $X$. We can now define the bracket of two vector fields $\xi_v$ and $\xi_w$ in the usual way:
$$[\xi_v, \xi_w](x) = (\text{D} \xi_w)_x (\xi_v(x)) - (\text{D} \xi_v)_x (\xi_w(x)).$$
Restricting to the tangent space of the origin gives a bracket on $X$:
$$[v, w] = [\xi_v, \xi_w](0).$$
This bracket is clearly continuous with respect to the norm on $X$, since it was constructed from smooth functions, and in Banach spaces continuity and boundedness are equivalent. So we have produced a bounded Lie bracket on a Banach space $X$, which means that $X$ is a Banach Lie algebra. (The proofs that the bracket of smooth vector fields satisfies linearity, antisymmetry, and the Jacobi identity are the same as in finite dimensions.) With a bit of bookkeeping, it is not hard to show that the Lie bracket that this construction induces on equivalence classes $[(U, v)]$ makes a Banach Lie algebra out of the tangent space $T_e G.$
3. Banach Lie Theory of the Unitary and Projective Unitary Groups
We can now move on to the main point of this note, which is giving a Banach Lie group structure to the groups $\operatorname{U}(\mathcal{H})$ and $\operatorname{PU}(\mathcal{H})$ for any Hilbert space $\mathcal{H},$ and computing their Lie algebras.
Let $X$ be a Banach space. The set of all linear maps on $X$ bounded in the operator norm, denoted $B(X)$, is itself a Banach space with respect to the operator norm. In fact, it is a Banach algebra; for $P, Q$ in $B(X)$, we have
$$\Vert P Q \rVert_{\infty} \leq \Vert P \Vert_{\infty} \Vert Q \Vert_{\infty}. \tag{2}$$
The proof is standard; it appears e.g. as exercise 2.5 in Douglas' book on Banach algebra techniques in operator theory. One of the basic results on Banach algebras is that the set of invertible elements $\operatorname{GL}(X) \subset B(X)$ is open with respect to the topology induced by the operator norm. This can be found as proposition 2.7 in Douglas. Because $\operatorname{GL}(X)$ is an open subset of $B(X)$, and $B(X)$ is obviously a Banach manifold over itself, we conclude that $\operatorname{GL}(X)$ has a natural Banach manifold structure over the Banach space $B(X)$. (The familiar fact that open subsets of manifolds are manifolds works for Banach manifolds as well; the proof is identical.)
Furthermore, one can see that $\operatorname{GL}(X)$ is a Banach Lie group. Any bounded linear map on a Banach space is smooth, since it is its own derivative. Multiplication is a linear map on $B(X) \times B(X)$ that is bounded by equation (2), so it is smooth, and multiplication on $\operatorname{GL}(X)$ is smooth by virtue of being a restriction of a smooth map. Inversion is a little more subtle; one has to use the Neumann series, an algebraic generalization of the geometric series, which for any operator $P \in B(X)$ with $\Vert P \Vert_{\infty} < 1$ gives the convergent series
$$\sum_{n=0}^{\infty} P^n = (e - P)^{-1}.$$
So for any fixed $T \in \operatorname{GL}(X)$, so long as we have $\lVert T^{-1} P \rVert_{\infty} < 1$ there is a convergent series
$$(T - P)^{-1} = \sum_{n=0}^{\infty} (T^{-1} P)^n T^{-1}.$$
Since there is a convergent series for the inverse function in a neighborhood of any point in $\operatorname{GL}(X),$ it is analytic and therefore smooth.
It shouldn't be surprising to hear that the Banach Lie algebra $\mathfrak{gl}(X)$ is just the full Banach algebra $B(X).$ Indeed, the Lie algebra of a Banach Lie group depends only on a neighborhood of the identity, and any sufficiently small neighborhood of the identity in $B(X)$ lies entirely within $\operatorname{GL}(X).$ (In fact, the Neumann series implies that the open unit ball of $e \in B(X)$ lies in $\operatorname{GL}(X).$) The bracket is just the commutator $[P, Q] = P Q - Q P.$
So far we have only considered $X$ as a Banach space; let us relabel it $\mathcal{H}$ and let it be a Hilbert space as well. We now have a notion of the adjoint of a bounded operator, and can define the unitary group $\operatorname{U}(\mathcal{H})$ as the set of operators $U$ in $\operatorname{GL}(\mathcal{H})$ satisfying $U^{\dagger} = U^{-1}.$ We'd like to show that this is a Banach Lie group, which we can do using a version of the closed subgroup theorem. First, in the unit ball of the identity in $\operatorname{GL}(\mathcal{H})$, every operator has a logarithm defined by the usual power series
$$\log(e+P) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1} P^n}{n}. \tag{3}$$
Generally speaking, for any fact about finite-dimensional matrix algebras that is expressed in terms of absolutely convergent series, the proofs go through exactly the same way in Banach algebras. So with the exponential defined as
$$e^{P} = \sum_{n=0}^{\infty} \frac{P^n}{n!},$$
one can show $e^{\log(e+P)} = e+P.$ Similarly, it is straightforward to show that the operator $e^P$ is unitary if and only if $P$ is antihermitian — you start with the equation
$$e^{P} (e^{P})^{\dagger} = e,$$
write out two power series, manipulate the series freely since they are absolutely convergent, and conclude that this equation is equivalent to $P^{\dagger} = -P.$
With these basic facts established, it is easy to construct a Banach manifold structure on $\operatorname{U}(\mathcal{H}).$ At each point $P \in \operatorname{GL}(\mathcal{H}),$ we consider the set
$$S = \{Q \in \operatorname{GL}(\mathcal{H})\, | \, \lVert Q P^{-1} - e \rVert_{\infty} < 1\}.$$
This set is the preimage of the unit ball in $\mathbb{R}$ under a continuous map, so it is open. On this subset, the map
$$\psi : Q \mapsto \log(Q P^{-1})$$
is a diffeomorphism from $S$ into $B(\mathcal{H}).$ Furthermore, the unitary subset of $S$ is in exact bijection with the antihermitian operators in $\psi(S).$ By restricting these charts to $\operatorname{U}(\mathcal{H}),$ we obtain a sub-Banach manifold modeled over the Banach space of antihermitian operators in $B(\mathcal{H}).$ Because the charts on $\operatorname{U}(\mathcal{H})$ are just restrictions of charts on $\operatorname{GL}(\mathcal{H}),$ smoothness of multiplication and inversion are immediate.
Technically, the one thing we haven't shown is that these "logarithm charts" are actual charts on $\operatorname{GL}(\mathcal{H}),$ i.e., that they are compatible with the standard charts on $\operatorname{GL}(\mathcal{H}),$ which are open balls contained in $\operatorname{GL}(\mathcal{H})$ with the chart map being inclusion into $B(\mathcal{H})$. But this follows immediately from the fact that the logarithm is diffeomorphic (in fact, analytic) on our charts.
So $\operatorname{U}(\mathcal{H})$ is a Banach Lie group modeled over the space of bounded antihermitian operators on $\mathcal{H}$ — let's call that Banach space $\operatorname{ah}(\mathcal{H})$ — and as a vector space its Banach Lie algebra is just $\operatorname{ah}(\mathcal{H}).$ Because the Banach Lie algebra $\mathfrak{u}(\mathcal{H})$ is a subalgebra of $\mathfrak{gl}(\mathcal{H})$, and we know that $\mathfrak{gl}(\mathcal{H})$ is $B(\mathcal{H})$ equipped with the standard commutator, we know that $\mathfrak{u}(\mathcal{H})$ must be $\operatorname{ah}(\mathcal{H})$ equipped with the standard commutator. But there's a nice little calculation we can do to see this explicitly, which I think is informative enough to include.
We will work in the logarithmic chart around the identity in $\operatorname{U}(\mathcal{H}),$ which is the set $S = \{ P\, |\, \Vert P - e \rVert_{\infty} < 1\},$ with the chart $\psi : S \rightarrow \operatorname{ah}(\mathcal{H})$ given by $\psi(P) = \log(P).$ Recall from equation (1), back in section 2, that the vector field $\xi_A$ associated with any $A \in \operatorname{ah}(\mathcal{H})$ is defined by
$$\xi_A : \psi(S) \rightarrow \operatorname{ah}(\mathcal{H}),$$
$$\xi_A(x) = D(\psi \circ L_{\psi^{-1}(x)} \circ \psi^{-1})_0(A).$$
Here $x$ is a point in $\psi(S)$; $\psi$ is the logarithm, and $\psi^{-1}$ is the exponential map. For antihermitian operators $A, B$, the Lie bracket is defined by
$$[A, B] = [\xi_A, \xi_B](0) = (\text{D} \xi_B)_0 (\xi_A(0)) - (\text{D} \xi_A)_0 (\xi_B(0)).$$
This can be computed using a trick, which is that when $f$ is a map between vector spaces we can compute $\text{D}f$ by a limit:
$$(\text{D}f)_{p}(q) = \lim_{\epsilon \rightarrow 0} \frac{f(p + \epsilon q) - f(p)}{\epsilon}.$$
This gives
$$\xi_A(0) = A,$$as expected, and
$$(\text{D}\xi_B)_0(\xi_A(0)) = \lim_{\epsilon \rightarrow 0} \frac{\xi_B(\epsilon \xi_A(0)) - \xi_B(0)}{\epsilon} = \lim_{\epsilon \rightarrow 0} \frac{\xi_B(\epsilon A) - B}{\epsilon}.$$
As for $\xi_B(\epsilon A),$ we have
$$\xi_B(\epsilon A) = D(\psi \circ L_{\psi^{-1}(\epsilon A)} \circ \psi^{-1})_0(B) = \lim_{\delta \rightarrow 0} \frac{\log(e^{\epsilon A} e^{\delta B}) - \epsilon A}{\delta}.$$
Combining these expressions gives
$$(\text{D}\xi_B)_0(\xi_A(0)) = \lim_{\epsilon \rightarrow 0} \lim_{\delta \rightarrow 0} \frac{\log(e^{\epsilon A} e^{\delta B}) - \epsilon A - \delta B}{\epsilon \delta}.$$
The leading order Taylor expansion of $\log(e^{\epsilon A} e^{\delta B})$ is
$$\log(e^{\epsilon A} e^{\delta B}) = \epsilon A + \delta B + \frac{1}{2} \epsilon \delta (A B - B A) + O(\epsilon^2, \delta^2).$$
Plugging this into the previous expression gives
$$(\text{D}\xi_B)_0(\xi_A(0)) = \frac{1}{2} (A B - B A),$$
from which we may immediately conclude
$$[A, B] = (\text{D} \xi_B)_0 (\xi_A(0)) - (\text{D} \xi_A)_0 (\xi_B(0)) = A B - B A,$$
as desired.
To conclude, we will turn our attention to the projective unitary group $\operatorname{PU}(\mathcal{H})$, defined as the quotient of $\operatorname{U}(\mathcal{H})$ by the $\operatorname{U}(1)$ subgroup $\{\alpha e | \alpha \in \operatorname{U}(1)\}.$ Points in $\operatorname{PU}(\mathcal{H})$ are equivalence classes $[U]$ of unitary operators that differ by a phase. Around each point $U$ in $\operatorname{U}(\mathcal{H})$ we defined the chart
$$S_U = \{V \in \operatorname{U}(\mathcal{H})\, | \, \Vert V U^{\dagger} - e \Vert_{\infty} < 1\}$$
with chart map $\psi : S_U \rightarrow \operatorname{ah}(\mathcal{H})$ given by
$$\psi_U(V) = \log(V U^{\dagger}).$$
Around each point $[U]$ in $\operatorname{U}(\mathcal{H})$ we will define the chart
$$S_{[U]} = \{[V] \in \operatorname{U}(\mathcal{H})\, | \, \exists \widetilde{V} \in [V] \text{ with } \Vert \widetilde{V} U^{\dagger} - e \Vert_{\infty} < 1\}.$$
The chart map $\psi_{[U]}$ will map from $S_{[U]}$ into $\operatorname{ah}(\mathcal{H})/i \mathbb{R}$, the Banach space of equivalence classes of antihermitian operators whose difference is an imaginary multiple of the identity. Basically, when we quotient out by a phase on $\operatorname{U}(\mathcal{H})$, we have to quotient out by a "trace term" in $\operatorname{ah}(\mathcal{H}).$ The chart map will be defined by
$$\psi_{[U]}([V]) = [\log(\widetilde{V} U^{\dagger})].$$
So, for an equivalence class $[V]$, we find a representative $\tilde{V}$ with $\widetilde{V} U^{\dagger}$ in the unit ball of the identity, take the logarithm of $\widetilde{V} U^{\dagger},$ and pass to the equivalence class in $\operatorname{ah}(\mathcal{H})/i \mathbb{R}.$ The choice of representative $\widetilde{V}$ doesn't matter; if we rephase $\tilde{V} \mapsto e^{i \phi} \widetilde{V},$ so long as $e^{i \phi} \widetilde{V} U^{\dagger}$ is still in the unit ball of the identity, the logarithm as defined by our power series in equation (3) only adds a multiple of the identity $i \phi e$, keeping us within the same equivalence class. Similarly, this map is independent of the choice of representative $U \in [U].$
The transition maps of $\operatorname{PU}(\mathcal{H})$ on $\operatorname{ah}(\mathcal{H})/i \mathbb{R}$ are just quotients of transition maps of $\operatorname{U}(\mathcal{H})$ on $\operatorname{ah}(\mathcal{H})$; from this, one can show smoothness. Similarly, smoothness of multiplication and inversion on $\operatorname{PU}(\mathcal{H})$ follow from the same properties on $\operatorname{U}(\mathcal{H}),$ so $\operatorname{PU}(\mathcal{H})$ is a Banach Lie group modeled over $\operatorname{ah}(\mathcal{H})/i\mathbb{R}$. Its Banach Lie algebra therefore has as its vector space $\operatorname{ah}(\mathcal{H})/i\mathbb{R}$, and it shouldn't be hard to convince yourself that the Lie bracket is given by the commutator,
$$[[A], [B]] = [A B - B A],$$
which is well defined on equivalence classes.
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