Does a finite-dimensional Lie algebra always exponentiate into a universal covering group

Question

Hi,

I am a theoretical physicist with no formal "pure math" education, so please calibrate my questions accordingly.

Consider a finite-dimensional Lie algebra, A, spanned by its d generators, X_1,...,X_d. If it matters, I do not assume that A is simple or semisimple, but for my (practical) purposes, one may assume the following structure constants: [X_k, X_p] = i f_{kp}^q X_q, where all structure constants f's are real. Here and below summation over repeating indices is assumed.

My questions are:

1. Does it make sense to define a stand-alone exponential g=exp(i B^k X_k) WITHOUT a reference to a particular representation and if so, how? [here and below one may assume that all B^k's are arbitrary REAL numbers (i.e., B=B^k X_k is a real form), but does not have to do so].

2. Either way, take two arbitrary elements in the algebra B = B^k X_k and C = C^k X_k and define P= P^k X_k ==BCH(B,C) as exp(iP)=exp(iB)exp(iC) via the Baker-Campbell-Hausdorff (BCH) relation. Is this a mathematically sensible definition of a map A*A->A? Is convergence of the BCH series ever an issue for a finite dimensional Lie algebra?

3. My main question: Is it correct to say that the universal covering group, G[A] (i.e., a simply-connected Lie group to which A is the Lie algebra] is generated by the exponential: G[A]=exp(iA)? By "generated," I mean that every element g in G[A] is represented by at least one element in the algebra via the exponential.

4. More specifically, take the minimal-dimensional faithful matrix representation of the algebra, T[A]. Is it correct that exp(i T[A]) maps the algebra GLOBALLY into the ENTIRE covering group, G[A], via standard matrix multiplication?

Any (professional) advice would be much appreciated. Thank you, Victor

Answer 1

As I read the question, much of it amounts to, what is an abstract Lie group and what is exponentiation in a Lie group. To review, an abstract Lie group is a smooth manifold with a smooth group law, and its Lie algebra is the tangent space at the identity. The Lie bracket comes from the second derivative of the group commutator $ABA^{-1}B^{-1}$ and associativity of the group implies the Jacobi identity for its Lie algebra. For the rest I will only take finite-dimensional Lie algebras and finite-dimensional representations.

Now, you can have real or complex Lie groups and real or complex Lie algebras. Every real Lie algebra has a complexification with identical structure constants. For that reason, you may not see any distinction between a Lie algebra and its complexification in a physics treatment. But the distinction is very important, because the topology of complex Lie groups is better behaved. Note that you can also realify a Lie group or a Lie algebra, which does not change it as a set; but its real dimension is then twice its complex dimension.

There is a converse theorem that every Lie algebra integrates to an abstract Lie group. If you like, you can take it to be the universal cover, the one which is simply connected. Ado's theorem says that every Lie algebra has a faithful matrix representation. (Per the other MathOverflow question, you can make the Lie group a closed set in that faithful Lie algebra representation.) Every compact, real Lie group has a faithful, closed representation, and I suppose that every complex, simply connected Lie group has a faithful, closed, complex representation. But not every real, simply connected Lie group has a faithful representation. An important example is $\text{SL}(2,\mathbb{R})$. It has the homotopy type of a circle and therefore an infinite cyclic universal cover, but none of its cover have a faithful matrix representation. If you complexify to $\text{SL}(2,\mathbb{C})$, then it becomes simply connected.

Another important fact is that the fundamental group of a Lie group is always abelian and lifts to a subgroup of the center of its universal cover. If the center of a Lie group is a discrete subgroup, then you can divide by it to obtain a miminal Lie group model of that Lie algebra, the model that is the most rolled up.

Among the faithful representations of a real Lie algebra, I think that there is always one in which the Lie group has been unrolled as much as possible. Even without knowing that one representation, you can simply say that two points in a Lie group are equivalent if they are equal in all representations, and quotient by this equivalence. I don't know if this intermediate cover has a special name.

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Okay, so that's what Lie groups are, now what about exponentiation. The most important formula for exponentiation is the "limit of compounded interest": $$\exp(A) = \lim_{n \to \infty} \left(1+\frac{A}n\right)^{n}.$$ This formula can be approximated in an abstract Lie group (by approximating the base of the exponential to second order), so that exponentiation is well-defined in any abstract Lie group. (Or you can use the differential equation as algori says.) You can prove that this exponential satisfies the Baker-Campbell-Hausdorff formula with a positive radius of convergence. If you wanted an infinite radius of convergence, then that does not happen and there certainly are issues about radius of convergence. However, (as Theo explains) a non-zero radius of convergence is good enough to build the Lie group in patches; this is one of the ways to prove that a Lie algebra always has a Lie group.

Another formula for the exponential is the Taylor series: $$\exp(A) = 1 + A + \frac{A^2}2 + \frac{A^3}6\cdots.$$ This formula only makes sense in a matrix representation. But, if you have a matrix representation, it agrees with the other formula.

Even in the complex group $\text{SL}(2,\mathbb{C})$, the exponential map is not surjective. You cannot reach the element $$\begin{pmatrix} -1 & 1 \\ 0 & -1 \end{pmatrix},$$ although you can reach it and everything else in $\text{GL}(2,\mathbb{C})$. The exponential map is always dense in a complex, connected Lie group and always surjective in a compact, connected Lie group. In a real non-compact Lie group such as $\text{SL}(2,\mathbb{R})$, it isn't even dense (exercise). As Theo emphasizes, the important positive fact is that it's diffeomorphism in a neighborhood of the identity. You can walk to every point in any Lie group by taking products of elements in such a patch, and in this weaker sense every Lie group is generated by its Lie algebra.

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One more remark about exponentials. Not every real Lie group has a faithful finite-dimensional representation, and it is non-trivial that every complex, simply connected Lie group does. However, a faithful infinite-dimensional continuous representation of a real Lie group is easy. If $G$ is a real Lie group, then it acts on a Hilbert space $L^2(G)$ (using its left-invariant Haar measure), which in physics-speak is the vector space of normalizable wave functions on $G$. This is a unitary representation. The Taylor series for the exponential then converges with an infinite radius of convergence. However, if you unearth the actual calculation of the exponential in this big representation, it isn't so different from the geometric exponential that algori described. Because, traveling wave functions in a manifold are only more complicated than classical trajectories in the same manifold.

Note that the unitary representations of a complex Lie group $G$ are important, but what is always meant is a representation of the realification of $G$. A representation which is both complex (in the sense that the matrix entries are complex analytic) and unitary is nonsense. In the latter case, every Lie algebra element has to have imaginary spectrum, whereas in former case, if $A$ has this property then the spectrum of $iA$ is real.

Answer 2

The answers to your first three questions are essentially "yes", although I'm not sure what you mean in the last one.

Let $\mathfrak g$ be any real Lie algebra. Then there exists a Lie group $G$ with $\operatorname{Lie}(G) = \mathfrak g$ (this is Lie's Third Theorem). You can take the connected component of the identity of this group and the simply-connected cover of that, and you get a connected simply-connected group. In fact, the functor $\operatorname{Lie}$ from Lie groups to Lie algebras restricts to an equivalence of categories on the connected simply-connected groups; the inverse functor exactly assigns to each Lie algebra its (unique up to isomorphism) connected simply-connected Lie group.

Now let $G$ be a Lie group and $\mathfrak g = \operatorname{Lie}(G)$. Then there exists a neighborhood of $0\in \mathfrak g$ and a neighborhood of the identity in $G$, and maps $\exp,\log$ between them that are diffeomorphisms. The map $\exp: \mathfrak g \to G$ is actually an entire map; one definition is that each $x\in \mathfrak g$ defines a left-invariant vector field on $G$ (by right translation), and that the corresponding ODE has arbitrary-time solutions, and then $\exp(x)$ is defined to be the image after flowing by $1$ second along the vector field defined by $x$ starting at the identity in $G$. So if you canonically take $G$ to be the connected, simply-connected Lie group integrating $\mathfrak g$, then you can define $\exp$. For any Lie group homomorphism $\phi: G \to H$ with corresponding Lie algebra homomorphism $\varphi : \operatorname{Lie}(G) \to \operatorname{Lie}(H)$, one has $\exp\circ\varphi = \phi \circ \exp$. In particular, if $V$ is any finite-dimensional $\mathfrak g$-module, then the connected, simply-connected $G$ acts on $V$, and $\exp: \mathfrak g \to G$ agrees with the matrix exponential in $\operatorname{End}(V)$.

The function $\exp: \mathfrak g\to G$ is globally defined, but it is only a diffeomorphism in some neighborhood of the identity, and $\log$ is its inverse. Then the Baker-Campbell-Hausdorff function $\operatorname{BCH}: (x,y) \mapsto \log(\exp x \cdot \exp y)$ is defined in some neighborhood of the origin in $\mathfrak g$. It is an analytic function, and the coefficients have been computed explicitly; they are all "Lie series". The BCH function defined on its domain the structure of a "partial group", and on an infinitesimal neighborhood of the origin the structure of a "formal group". One proof of the Lie III theorem takes the partial group structure and uses it to translate the domain around, and then you have to prove that the resulting object formed by pasting closes. (Another proof requires detailed structure theory for Lie algebras, but does not require the analysis and differential equations of the BCH approach.)

Finally, any connected topological group is generated by any nonempty open neighborhood therein: the subgroup generated by an open set is necessarily open, because it is a union of translations of that set, but then all its cosets are open, so the subgroup is also closed, and hence everything if the big group is connected. In particular, $\exp(\mathfrak g) \subseteq G$ generates $G$. It need not be all of $G$: for example, the exponential map $\exp: \mathfrak{sl}(2,\mathbb R) \to {\rm SL}(2,\mathbb R)$ is not onto (it does not hit $\left( \begin{smallmatrix} -1 & 1 \\ 0 & -1 \end{smallmatrix}\right)$, for example).

One place to learn more are the lecture notes on my website: http://math.berkeley.edu/~theojf/LieQuantumGroups.pdf

Answer 3

Re 1: yes, the exponential map can be defined for any Lie group as follows: we take an element $X$ of the algebra to the $g(1)$, where $t\mapsto g(t)$ is the 1-parametric subgroup such that $g'(0)=X$. This makes sense for any Lie group and depends only on the Lie group structure.

Re 3: any finite dimensional Lie algebra is the Lie algebra of a Lie group. This is a classical theorem on Lie groups, but unfortunately, the only proof I know of proceeds via the classification of semi-simple Lie groups and algebras. The exponential is a local isomorphism, so a sufficiently small neighborhood of the unit is in the image of the exponential map, and so every element of the group is a product of exponentials. Moreover, the Lie group can be assumed simply-connected (or else take the universal cover). Then a faithful representation of the algebra integrates to a locally faithful representation of the simply-connected group.

So the only thing that can prevent the answer to 4. from being "yes" (if I understand the question correctly) is that the image of a map of Lie groups may not be a Lie group. Off hand I'm not sure that a minimal-dimensional representation would do the trick but some representation would, as follows from Greg Kuperberg's answer here Is every finite-dimensional Lie algebra the Lie algebra of a closed linear Lie group?.

Answer 4

The short answer to 3. is "no". The simplest example is the circle group, $e^{it}$ of complex numbers of absolute value 1, (thought of as a $1 \times 1$ matrices), its Lie algebra $A$ consists of the line $it$, and when you exponentiate it you get the circle, NOT its universal covering group, the real line. More generally, if you take any Lie subgroup $G$ of a matrix group and exponentiate its Lie algebra you get back $G$, not the universal covering group. Moreover, there are examples where the universal covering group does not have a faithful linear representation, so you cannot get it by exponentiating a matrix Lie algebra.

Answer 5

Here is a separate and more technical answer meant to address question 2, as I think that I now understand it, and some of the issues touched upon by 3 and 4. I think that question 1 is more or less settled; given any $A$ in the Lie algebra of a Lie group $G$, there is an exponential ray in $G$ in the direction of $A$. (But note that mathematicians like to absorb the factor of $i$ into the real form of the Lie algebra. E.g., the Lie algebra of $U(n)$ consists of anti-Hermitian, not Hermitian, matrices.)

The Lie group $\text{SL}(2,\mathbb{R})$ is certainly a real form of $\text{SL}(2,\mathbb{C})$. You can make the structure constants real. As mentioned, an accepted way to build any Lie group is to first define it in a coordinate patch using the exponential map, define multiplication in a possibly smaller patch using the BCH formula, and then stitch together different patches to make the entire Lie group. So, you could ask, is this really necessary; could you use the exponential map and the BCH formula once to define everything?

In this case, no. If $A \in \text{SL}(2,\mathbb{R})$ satisfies $\text{Tr}(A) < -2$, then it is not in the exponential map. On the other hand, the Lie group is connected, so there must be two elements in the exponential map whose BCH product does not converge, because their actual product is not in the exponential map. For example, if $$X = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} \qquad Y = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix},$$ then you can check that $$\text{Tr}(\exp(\frac{\pi}{2} X)\exp(\frac{2\pi}{3}Y)) = -\cosh \frac{\pi}2 < -2.$$ There actually many different power series forms of BCH, but none of them can work. Even for these specific matrices $X$ and $Y$ and scalars $s$ and $t$, any power series for $$f(s,t) = \text{BCH}(sX,tY)$$ has to be outside of the radius of convergence at $(s,t) = (\frac{\pi}{2},\frac{2\pi}{3})$, because the actual answer is not reachable.

You might hope that such an example is less relevant to physics because $\text{SL}(2,\mathbb{R})$ is not a compact group and has no finite-dimensional unitary representations. However, $\text{SU}(2)$ is another real form of the same complex Lie group $\text{SL}(2,\mathbb{C})$. It is compact and simply connected, and as relevant to physics as any Lie group ever is. Its Lie algebra contains $iX$ and $Y$. Since it is compact, the exponential map is surjective, which seems encouraging. However, you have to accept the exponential map out to a radius of $\pi$ for the Lie generator $iX$ or $Y$ (or any other unit-length vector combination of Pauli matrices), in order to reach all of $\text{SU}(2)$. But then, you cannot have a convergent power series for $$g(s,t) = \text{BCH}(isX,tY)$$ out to $(s,t) = (\frac{\pi}{2},\frac{2\pi}{3})$, because we already know that it cannot converge out to the same radius after an imaginary rotation of this $s$. You could look for some sort of resummation method --- but it is clear that analytic continuation does work and that is already similar in spirit to resummation.