Timeline for Does a finite-dimensional Lie algebra always exponentiate into a universal covering group
Current License: CC BY-SA 2.5
20 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 12, 2010 at 13:49 | comment | added | Greg Kuperberg | The exponential map from $\mathfrak{su}(2)$ to $\text{SU}(2)$ is the entire group in the sense that it is smooth and surjective, and has certain coordinate singularities. It is also possible that the intrinsic radius of convergence of BCH in this case is $\pi/2$ in the basis of Pauli matrices. If it is, then you could obtain the entire group law of $\text{SU}(2)$ by sawing the 3-sphere into two hemispheres, one centered at $I$ and the other at $-I$. (StackOverflow lets you revise comments for a few minutes, but in MO you have to delete them.) | |
| Oct 12, 2010 at 13:39 | comment | added | Victor Galitski | Greg, Re: Hopf fibration and such. I am still resiting to accept that EXP of 2D representation of su(2) would not give me the entire group, just because it is the defining representation. I understand that there is no global coordinate map to cover the sphere, but I don't have to view ${\bf R}$ as global coordinates, but rather as a means to express the constraints, which would give me the sphere. After all, if I go back to basics and define $SU(2)$ group as a set of $2 \times 2$ unitary matrices with $det=1$ I surely get $S^3$. P.S. Can I revise comments w/o deleting and resubmitting them? | |
| Oct 12, 2010 at 13:33 | comment | added | Greg Kuperberg | Well, yes, nilpotent Lie algebras and Lie groups, and even solvable Lie algebras and Lie groups, are "sort of" trivial. They don't have to be as banal as the harmonic oscillator algebra though; you can make an $n$-step nilpotent or solvable Lie algebra for every $n$. And, degenerate though they are, in some ways they still have a rich structure. Arguably, the global stability of BCH and the exponential map are only allowed by some type of degeneracy. | |
| Oct 12, 2010 at 13:21 | comment | added | Victor Galitski | The nilpotent algebras are sort of trivial in my (physicist's) view. Also, all nilpotent/solvable algebras I know descend from simple algebras + an Abelian component via contraction. I.e., you get harmonic oscillator $\mathfrak{h}_4 = span \{ 1, a^+a^-, a^+, a^-\}$ with $[a^+,a^-, a^{\pm}] = \pm a^\pm$, $[a^-,a^+]=1$ and $[1,A}=0$ follows from $u(2)=su(2) + u(1)$ by performing a parameter ($\epsilon$)-dependent linear transform and then take the singular limit $\epsilon \to 0$ (contraction). Geometrically it should correspond to cutting the 3D sphere and unrolling it into the Euclidean space. | |
| Oct 12, 2010 at 13:11 | comment | added | Greg Kuperberg | @Victor The Hopf fibration is a clever point of comparison, but it's not the same thing. The Hopf fibration is a twisted $S^1$ bundle over $S^2$, whereas your periodic coordinate system is an untwisted $S^1 \times S^2$. | |
| Oct 12, 2010 at 13:07 | comment | added | Victor Galitski | The above looks much like the Hopf fibration to me: I.e., represent $g({\bf R}) = \exp \left[ - i {\bf R} \cdot {\sigma \over 2} \right]$ and use "spherical coordinates:" ${\bf R}= R {\bf n}$ with the radius $R \ge 0$ and ${\bf n} in S^2$. We know that $SU(2)=S^3=S^2/S^1$ and know that it is a non-trivial Hopf fibration. We also know that 2D representation is basically a defining representation of $SU(2)$, which is the universal covering group to the underlying $su(2)$ algebra. So it seems to me that we are bound to get the three dimensional sphere via this representation. No? | |
| Oct 12, 2010 at 13:05 | comment | added | Greg Kuperberg | @Victor On the other hand, there are non-trivial Lie groups that don't have any topology and where everything works as you envision. For instance, if $A$ is nilpotent, then the simply connected $G$ is diffeomorphic to $A$, the exponential map is a diffeomorphism, and the BCH formula converges everywhere because it terminates outright. Maybe if $A$ is solvable, then BCH series still has infinite radius of convergence and it all works; I'd have to check. But in semisimple cases, there is an essential topological problem. | |
| Oct 12, 2010 at 13:02 | comment | added | Victor Galitski | Greg: There were typos in my previous comment: What I've been trying to get at is to define the covering group (at least as a topological space first) as Lie algebra $A/?$ modulo something that involves the LIE ALGEBRAIC STRUCTURE ALONE. As to the "loophole," it's a bit more complicated than that. Take "my" representation of SU(2) above. Think of ${\bf R}$ as coordinates in A, with $\sigma_{x,y,z}/2$ generators. Clearly if $R_1−R_2=4πk$ and ${\bf n}_1 = {\bf n}_2$ OR $R_1+R_2=4πk$ and ${\bf n}_1 = -{\bf n}_2$ ($k \in \mathbb{Z}$), then $R_1 \sim R_2$. Don't we get $su(2)/\sim =S^3$ this way? | |
| Oct 12, 2010 at 13:00 | comment | added | Greg Kuperberg | @Victor What I mean by patching and analytic continuation is exactly what you say: First restrict to ball $B(r)$ in $A$ on which BCH converges; the answers then lie in $B(2r)$. Then you can build the entire Lie group $G$ by gluing together copies of $B(r)$, using the rest of $B(2r)$ as an extra lip for the gluings. If you want to be free of coordinate singularities, then this is the best you can do, because even a simply connected $G$ often has a different topology from $A$ (which is always a vector space). | |
| Oct 12, 2010 at 12:52 | comment | added | Greg Kuperberg | @Victor It is true that there is a periodicity: If $\vec{v}$ is a unit vector, then the function $\exp(it\vec{v} \cdot \vec{\sigma})$ is periodic with period $2\pi$. You can quotient $\mathbb{R}^3$ by this periodicity, and you almost get the right answer. You only miss the fact that the $t = \pi$ sphere crunches to a single point. Moreover the quotient as stated is a little bit violent because $t=0$ is also only a single point, and the entire sphere $t=2\pi$ is crunched onto it. The topology here is that $S^3$ is already simply connected and does not unroll to $\mathbb{R}^3$. | |
| Oct 12, 2010 at 12:47 | comment | added | Victor Galitski | RE: Greg's example above. Actually, it's not the Pauli matrices $\sigma_z= X$ and $\sigma_y=-iY$ that are 2D generators of $su(2)$, but those DIVIDED BY TWO, i.e., $\sigma_{x,y,z}/2$ (you get extra factors then). Anyway, it seems self-contradictory to me to attempt extension of the BCH relation to the whole algebra in a complicated way via analytical continuation or patching, while we now in advance that the object that we are supposed to get is a group, which is "smaller" than the algebra. Would not it make more sense to narrow down the range of elements in A first and then do BCH? | |
| Oct 12, 2010 at 7:28 | comment | added | Greg Kuperberg | @Victor Re radius of convergence of BCH. This is a good question and not an easy one because the expansion is not unique. (Actually, $\text{BCH}(X,Y)$ is unique in the free Lie algebra generated by $X$ and $Y$, but the literal formula in brackets is not unique.) If you Google for ' "Baker-Campbell-Hausdorff" "radius of convergence" ', you can find various papers, many of them in the math physics literature. | |
| Oct 12, 2010 at 7:21 | comment | added | Greg Kuperberg | @Victor Re global validity. I was careful to keep my parameters $s$ and $t$ less than $\pi$ to avoid the loophole that $\exp(isX) = \exp(i(s - 2\pi)X)$. You can't cover $\text{SU}(2)$ with just one chart within the radius of BCH. Your suggestion "define a map...via rotation" is promising, but it already resembles the standard approach of defining a Lie group in patches. | |
| Oct 12, 2010 at 7:12 | comment | added | Greg Kuperberg | @Victor Re "lost in translation". The main incompatibility in notation is a factor of $i$: You request Hermitian generators $X_k$; mathematicians would instead want anti-Hermitian generators $Y_k = iX_k$. (After all, this eliminates the factor of $i$ in your Lie bracket formula.) A Lie algebra of anti-Hermitian matrices generates a subgroup of $U(n)$. If it is a closed subgroup, then it is compact. In mathematics one often just says "compact", because every compact Lie group has a faithful unitary representation. So when you said "real", you meant "compact". | |
| Oct 12, 2010 at 5:50 | comment | added | Dick Palais | Victor, there is a paper by Masuo Suzuki in the 1977 Comm. in Math. Physics called "On the Convergence of Exponential Operators---the Zassenhaus Formula, BCH Formula and Systematic Approximants" that has some information on this. I have sent it to you as an email attachment | |
| Oct 12, 2010 at 5:17 | comment | added | Victor Galitski | Your very illuminating explanation (thanks!) brings up the question about the radius of convergence of BCH series. What is known about it and what is the best reference? Thanks again! I am really impressed with the level of expertise I am getting here on this web-site. Unfortunately, we do not have anything close to it in physics. | |
| Oct 12, 2010 at 5:14 | comment | added | Victor Galitski | Greg: I checked your example and now see that there is no way to construct a globally valid BCH identity as a series. Although, if we were to restrict ourselves to equivalence classes that actually matter for SU(2) (which is a 3D sphere), we perhaps would have been fine? Eg, $\exp(i\pi X)$ and $\exp(5i \pi X)$ are really the same thing. Perhaps a way to go is to define a map $A \times A \to A$ via rotation: $R(X_1,X_2) = e^{i ad\, X_1} X_2$ and they say that $X \sim Y$ if $\forall Z \in A$ $R(X,Z) = R(Y,Z)$. I wonder if the radius of convergence of BCH is related to the quotient $A/\sim$? | |
| Oct 12, 2010 at 4:27 | comment | added | Victor Galitski | I think something was "lost in translation" from physics to math and vice versa. Basically, what I want (or most other reasonable physicists for this matter) is a representation of a Lie algebra, $A= {\rm span} \left\{X_1, X_2,\cdot,X_d \right\}$ such that in this representation $X$'s are Hermitian operators/matrices, which ensure that for any linear combination (think, ``Hamiltonian''): $$ H = \sum\limits_{i=1}^d c_i X_i $$ its spectrum is real. For any such representation, $\exp(i A)$ is likely a group. I have wanted to know if $e^{iA}$ understood in abstract sense is the covering group. | |
| Oct 12, 2010 at 4:13 | comment | added | Victor Galitski | Greg: Actually, we call $X=\sigma_z$ and $−iY=\sigma_y$ Pauli matrices, which together with $\sigma_x$ form 3 generators of 2D representation of su(2). Now, consider the exponentials $$ g({\bf R}) = \exp\left[ {i \over 2} {\bf R} \cdot {\bf S} \right] = I \cos{R \over 2} + i \left({\bf n} \cdot {\bf S} \right) \sin{R \over 2}, $$ where ${\bf S}=\sigma_x {\bf e}_x+ \sigma_y {\bf e}_y +\sigma_z {\bf e}_z$ and ${ \bf R} =R {\bf n}$ is a 3D vector. They seem to give rise to full SU(2) (but higher-d representations don't always do that). I don't know if this generalizes to other groups. | |
| Oct 11, 2010 at 14:51 | history | answered | Greg Kuperberg | CC BY-SA 2.5 |