Talk:Heisenberg group

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Please put questions such as those at the end of the article into this talk page.--CSTAR 21:48, 1 May 2005 (UTC)

Contents

[edit] Open questions

This article should address the following topics:

"Are there even dimensional Heisenberg groups? If not, why not?"

[edit] Reply to User:Linas

For the latter question, I think you may be interested in the theory of nilmanifolds. Also look at work of Stephen Semmes. But in any case please refrain from putting these questions in the body of the article.--CSTAR 04:15, 3 May 2005 (UTC)

Thanks for reference; but why refrain from putting questions in articles? linas 04:27, 3 May 2005 (UTC)

[edit] only Lie group?

The article says "the Lie group of any nilpotent Lie algebra is simply connected". Now, I don't know about that statement, but I do know that there is more than one Lie group whose Lie algebra is this Heisenberg algebra. And only one of them is simply connected. In fact, can you ever say "the Lie group of a Lie algebra"? Doesn't any Lie algebra generally have more than one possible Lie group? -Lethe | Talk 20:11, Jun 4, 2005 (UTC)

Yes, obviously any (proper) quotient of a simply-connected Lie group G by a (normal) discrete subgroup N will have the same Lie algebra as that of G. Moreover
\pi_1(G/N) \cong N
so G/N is not simply connected.
By the phrase Lie group associated to a Lie algebra L I suppose is meant (or maybe I meant if it turns out I wrote that part) the simply-connected Lie group whose Lie algebra is L. This wil be the universal covering group of any Lie group with Lie algebra L. --CSTAR 22:14, 4 Jun 2005 (UTC)
Yeah! That's what I had in mind. Fulton & Harris make a point of showing that one of these Lie groups cannot be a matrix group. -Lethe | Talk 08:04, Jun 5, 2005 (UTC)
OK, so I want to repair this statement. As far as I know, nilpotency has nothing to do with whether the Lie group is simply connected; every Lie algebra has a simply connected Lie group, whether nilpotent or not. As for whether the Lie algebra is diffeomorphic to the nilpotent Lie group, well I don't know about that. Is it true? Is that the point of the statement? Maybe we should get rid of the bit about simple connectivity, and just keep that bit? -Lethe | Talk 10:28, Jun 5, 2005 (UTC)
By the way CSTAR, it wasn't you who added the stuff in question. Just in case you thought you were losing your memory. -Lethe | Talk 10:51, Jun 5, 2005 (UTC)
As for whether the Lie algebra is diffeomorphic to the nilpotent Lie group, well I don't know about that. Is it true?
Yep, that's Theorem 2.1 of Chapter XII, Hochschild Structure of Lie Groups, which is my standard reference for Lie Groups.--CSTAR 22:46, 5 Jun 2005 (UTC)
I replaced it with this: "The exponential map of a nilpotent Lie algebra is a diffeomorphism between the Lie algebra and the unique associated connected space simply-connected Lie group." Is it better? -Lethe | Talk 00:31, Jun 6, 2005 (UTC)

Heh. Sorry for the confusion. I'm tired, but that sounds right. It's possible I was thinking 'exponential map' when I wrote this and then inexplicably inverted the sense. It'll hold until the day someone writes a good article on nilpotent lie algebras. linas 06:09, 6 Jun 2005 (UTC)

[edit] suggestions for improvement

This is a very, very limited approach to the concept ,Heisenberg group'. One actually should start with a (finite-dimensional) symplectic vector space as it was done in H. Tilgner, Ann.Inst.H.Poincare A13 (1970) 103-127 ,A Class of Solvable Lie Groups and Their Relation to the canonical Formalism'. This allows for a basisfree description of all canonical concepts in terms of the symplectic form. Then there is the Weyl algebra as a factorization of the universal enveloping algebra of the Heisenberg Liealgebra, in which this Liealgebra together with the symplectic Lie algebra is embedded as a Lie algebra. The simply connected Lie group, with the Heisenber algebra as its Lie algebra should be called Weyl group. Certainly this Weyl group is generated by the exponentials of this embedded Lie algebra. And note that there is a counterpart, if the symplectic form is subsituted by a pseutoorthogonal bilinear form, Jordan algebras playing the role of Lie algebras, the Clifford algebras playing the role of the Weyl algebras. There is no correspondance to the concept of (Lie) group here, but symmetric spaces in the sense of Otmar Loos may overtake this role. Hans Tilgner 13:50, October 19, 2005

Hello Hans. Thank you for making these suggestions for improvements to the article. I agree with them. I would like to see the article treat the general Heisenberg group on any symplectic vector space. I'm reading a related paper by Babbit at the moment, and hope to update the article shortly. I have also looked up your paper and hopefully will be able to learn some more things to help this article from that. In the mean time, you are welcome to revise the article. -lethe talk + 17:12, 3 July 2006 (UTC)

[edit] vectors in your matrix

The article states:

By choosing a basis e1, ..., en of Rn, and letting

p_i = \begin{bmatrix} 0 & \operatorname{e}_i & 0 \\ 0 & 0_n & 0 \\ 0 & 0 & 0 \end{bmatrix}

The idea of putting your basis vectors in the matrix disturbs me deeply. Surely components of the vector are meant? -lethe talk + 01:25, 4 July 2006 (UTC)

Why? The top row consists of a scalar a row vector and a scalar.
The righthand column consists of a scalar, a column vector a and a scalar.
Is this too "matrixy" for your taste? I used this when I pput this in because that's the refernce I had available--CSTAR 02:45, 4 July 2006 (UTC)
It is too "matrix-y" for my tastes, and I've got the non-matrixy version almost finished. But I'm now trying to get the two approaches to sync up, and I'm having some problems. Anyway, I still don't like putting basis vectors into a matrix. Once you choose a basis, vectors can be identified with n-tuples, and those n-tuples can be put into matrices, but never the basis vectors themselves, right? -lethe talk + 03:18, 4 July 2006 (UTC)
Well the top row for instance is the basis vector with a scalar "prepended" and a scalar "postpended". It's certainly abuse of notation, but isn't it obvious what it means?
I've had an epiphany which has made it perfectly clear what the notation means, and I know the perfect minor modification of wording which would have avoided the confusion for me. The problem is, in general vector spaces, vectors, including basis vectors, are not n-tuples. But in Rn, of course, all vectors are tuples. I was being stupid and not remembering we are in Rn. See how you like this diff. That's fixed it for me. -lethe talk + 04:42, 4 July 2006 (UTC)


[edit] canonical form of the group

Also, the following part bothers me

\begin{bmatrix} 1 & a & c \\ 0 & I_n & b \\ 0 & 0 & 1 \end{bmatrix} \times \begin{bmatrix}1 & a' & c' \\ 0 & I_n & b' \\ 0 & 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & a+ a' & c+c' +a b' \\ 0 & I_n & b+b' \\ 0 & 0 & 1 \end{bmatrix}

OK, that's just matrix multiplication, but the product in the corner there isn't showing any of the antisymmetry that I expect. -lethe talk + 01:35, 4 July 2006 (UTC)

It may seem strange for th ereason you mention, but this is indeed a (non-canonical) form for the Heisenberg group; non-canonical in the sens that the relation to a symplectic form is not entirely obvious, But it comes straight from a good source:
* A Kirillov, Éléments de la théorie des représentations, Éditions MIR, 1974 p 320.
--CSTAR 02:41, 4 July 2006 (UTC)
Well, I'm going to add the canonical version, and then maybe you can help me figure out how to see that they're the same thing. -lethe talk + 03:18, 4 July 2006 (UTC)

Alright, I've added it. The first part of my addition I got from a paper, the group law in terms of the symplectic form. The second bit, the isomorphism between the coordinate free approach and the matrix approach, where I broke the vectors up into components and turned them into matrices, well that was just my guess as to how it should look, but according to the group law, there should be an antisymmetric product in the corner, so something's wrong. I'll think about it some more, and I won't leave the article in this uncoherent state for too long. Your input is welcome. -lethe talk + 03:36, 4 July 2006 (UTC)

Someplace I've got this isomorphism written down (The statement on p 320 in Kirillov is an exercise, which I did at some point in my life, and I've got the notes for its solution somewhere).--CSTAR 04:12, 4 July 2006 (UTC)
Could maybe be that the last term in the corner should be aJb′, where J is the standard symplectic matrix? Edit: Well, that would make the product agree with what I think it should be, but then our group product wouldn't be matrix multiplication, so nevermind. I don't know the book by Kirillov (nor can I read French, even if I had the book), but tomorrow I'll look around and see if I can find it in another book. -lethe talk + 04:31, 4 July 2006 (UTC)
There's an English version of Kirillov's book (I'm not sure, however, if it's the English translation of the same book) published by Springer. I'm sure it's in your University library.--CSTAR 04:44, 4 July 2006 (UTC)
Yeah, apparently they do, though they claim it's translated from Russian rather than French. Go figger. OK, I'll be back tomorrow. -lethe talk + 04:52, 4 July 2006 (UTC)
Well yeah, what I meant was, I wasn't sure whether both were translated from the same source (ie russian).--CSTAR 04:54, 4 July 2006 (UTC)


So check it out.

\left[e^{ik_1\cdot Q}e^{ia_1\cdot P}e^{i\theta_1E}\cdot e^{ik_2\cdot Q}e^{ia_2\cdot P}e^{i\theta_2E}f\right](x)=e^{i(\theta_1+\theta_2+a_1\cdot k_2)}e^{i(k_1+k_2)\cdot x}f(x+a_1+a_2) =\left[e^{i(k_1+k_2)\cdot Q}e^{i(a_1+a_2)\cdot P}e^{i(\theta_1+\theta_2+a_1\cdot k_2)E}f\right](x)

while on the other hand

\left[e^{i(k_1\cdot Q+a_1\cdot P+\theta_1E)}\cdot e^{i(k_2\cdot Q+a_2\cdot P+\theta_2E)}f\right](x) = \left[e^{i(k_1+k_2)\cdot Q+i(a_1+a_2)\cdot P+i(\theta_1+\theta_2+(a_1\cdot k_2-a_2\cdot k_2)/2)}f\right](x)

and the latter equality is calculated by means of the commutation relations and the Baker-Hausdorff formula (or just use the Weyl form of the commutation relations).

We can get the matrix representation of the canonical form with the association

(p,q,t)\mapsto \begin{pmatrix} 1 & p & t+\frac{1}{2}q\cdot p\\ 0 & 1 & q\\ 0 & 0 & 1\end{pmatrix}

which satisfies the same group law as above. With this correction, the article will probably be free of mistakes. I guess we should go ahead and make the correction, but there's still something confusing me about this business which I can't quite put my finger on. I haven't gone to the library yet. Hopefully get to that shortly. But I guess the upshot of this calculation is that the two different group laws are simply due to different choices of parametrization. Does that sound plausible? The non-canonical group law parametrization seems analogous to a multiplication of rotation matrices written in terms of Euler angles, while the canonical group law would be analogous to a group law for rotations in terms of their axes and rotation angles about their axes. -lethe talk + 08:14, 6 July 2006 (UTC)

I've had a look at Kirillov. He defines everything in terms of the Weyl algebra. The homework problem is to show that the (noncanonical form) matrix group is isomorphic to the Lie group obtained from the subalgebra of polynomials of degree ≤ 1. The map shown above does accomplish this. I also like Kirillov's definition of the Weyl algebra better than the one we have. He defines it as the tensor algebra over v mod ω(u,v) = [u,v]. I might update Weyl algebra. -lethe talk + 14:02, 10 July 2006 (UTC)

I second that, the right way to think about Heisenberg groups is as some vector space V and a copy of the Real numbers twisted by the symplectic pairing. If the canonical example hides this behaviour then it just seems confusing. I work with Higher Dimensional Heisenberg groups a a fairly regualr basis and I was looking at the canonical example and saying to myself "where is the symplectic pairing?" So I don't think this is obvious. —Preceding unsigned comment added by 86.27.127.97 (talk) 14:56, 18 December 2007 (UTC)

Oh, by the way, if you want to see the symplectic structure in the matrix representation then you need to add a minus ;), otherwise you don't get antisymmetry.

(p,q,t)\mapsto \begin{pmatrix} 1 & p & t+\frac{1}{2}q\cdot p\\ 0 & 1 & -q\\ 0 & 0 & 1\end{pmatrix}

now the symplectic paring on R^2 is given by the matrix

\begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix} —Preceding unsigned comment added by 86.27.127.97 (talk) 15:15, 18 December 2007 (UTC)

[edit] moving on

So I've added everything I wanted to add to this article. It remains to make my additions coherent with the existing article. For one thing, we have to choose a single notation. My preferred way to accomplish this would involve among other things deleting a large amount of the existing material. In particular, I think that the verification of the group law and inverse and the exponentiation in terms of matrices is tedious and not appropriate for this article. We could just mention what the group and algebra representations are, and leave homework problems to the reader. But before undertaking anything drastic, I would like to solicit your input, CSTAR. I will also mention that I am going offline shortly for what may turn out to be an extended amount of time, so I may not attend to this matter further in a timely fashion. -lethe talk + 09:43, 11 July 2006 (UTC)

Though the symplectic form of the Heisenbeg group is certainly the right one from a technical point of view, I'm not so sure that it's really appropriate for an encyclopedia article. My view of these things is "what could be simpler than matrices"? I Imagine some individual asking me what the Heisenberg group is. Knowing the technical advantages of the symplectic form, I probably will still respond with the matrix definition, even if it's non-canonical. --CSTAR 17:28, 11 July 2006 (UTC)
I certainly can appreciate the view that the matrix representation is the easiest one to grasp. It also seems to be the standard definition most often used in the literature, at least from what I've seen. But the article already starts off with the 3x3 matrix case in the examples section, and then has a later section which sells itself as having the more "general" Heisenberg group, I think the more abstract formulation can fit nicely there. I don't agree that the canonical formulation is inappropriate for the article, I think it might be relevant base material for some articles about geometric quantization (which has become my new favorite subject recently) that I envision in the distant future. I think with some reorganization of the article we could arrive at a state which would satisfy both of us, and in fact some reorganization is necessary because my additions have left the article in a disorganized state with too much repetition. The one complaint I had about the calculation of the matrix multiplication, inverse, and exponentiation being too remedial for this article would be mitigated if we lost the canonical formulation, or even moved it to a later section, but I might still like to get rid of it or at least tighten it up. Seems like the sort of thing the reader ought to do for herself. But I won't press the issue if you disagree. -lethe talk + 21:44, 11 July 2006 (UTC)
BTW another book which I thoroughly recommend (and kick myself for not buying when it was on a yellow sale) is Varadarajan's book "Geometry of Quantum Theory" vols 1 and 2, republished as 1 volume by (surprise) Springer. It treats beautifully and carefully a vast number of subjects, including Mackey's theory of systems of imprimitivity, Gleason's theorem, Induced representations.--CSTAR 21:54, 11 July 2006 (UTC)
I think I actually picked up the copy you failed to! I bought an old used copy of the pre-Springer two volume set a while ago. I've only gotten around to reading the first couple of chapters from volume 1, which seemed to be about Stone duality of Boolean algebras and quantum logic. I was looking more for stuff about polarizations and the Moyal product on manifolds, so I ended up spending more time with a book by Woodhouse. I guess I will give Varadarajan another look based on your recommendation, but for now all my books have disappeared into storage. -lethe talk + 22:06, 11 July 2006 (UTC)

[edit] A group extension of the Heisenberg group

I'm studying a group with the presentation M=(x,y| yz=zy) -- that is, its like the Heisenberg group, except that the other of the pair, zx does not equal xz (here, z is defined as in the article). Letting <z> denote the centralizer of z, I then get the Heisenberg group as the quotient group H3 = M / < z > where M is as above. Does this group M have a name? linas 17:12, 11 March 2007 (UTC)

I don't understand the notation. What are x y z? --CSTAR 17:51, 11 March 2007 (UTC)
Perhaps I was too brief. x and y are two abstract group generators, and z = xyx − 1y − 1. So
M=\langle x,y,z\vert yz=zy, z=xyx^{-1}y^{-1} \rangle
Group presentations are abstract. I can give a concrete example:
y=\begin{bmatrix}1&0&0&0\\ 0&1&0&0 \\ 0 &0&1&1 \\ 0&0&0&1 \end{bmatrix}
x=\begin{bmatrix}1&1&1&0\\ 0&1&2&0 \\ 0 &0&1&0 \\ 0&0&0&1 \end{bmatrix}
This then has that zy = yz but zx \ne xz
One can get higher dimensional reps by extending the "121" structure in x by the binomial coefficients tipped on their side in this upper-triangular fashion. (so e.g. the next column would be 1331) (y stays the same, with just more 1's added to the diagonal). (and you can see how the Heisenberg group is the 3D analog of this). I've even got an inf-dimensional rep. Concretely, this is the monodromy group of the polylogarithm. When the published lit on the monodromy of the polylog talks about it, it is invariably in some super-highly-abstract fashion, appealing to K-theory and what-not. I was hunting for more info on the more concrete case, above. linas 18:46, 11 March 2007 (UTC)
I'm also interested in the "abelian varietalization" of the above, i.e. taking the quotient space of the above, with,
y=\begin{bmatrix}1&0&0&0\\ 0&1&0&0 \\ 0 &0&1&b \\ 0&0&0&1 \end{bmatrix}
x=\begin{bmatrix}1&a&a^2&0\\ 0&1&2a&0 \\ 0 &0&1&0 \\ 0&0&0&1 \end{bmatrix}
for real or complex numbers a and b (so clearly, no longer an abelian variety or complex torus, but some kind of variety or homogeneous space or something...) linas 18:59, 11 March 2007 (UTC)
Maybe I'm being a bit thick here, but I still don't understand. Are you describing a nilpotent Lie Group or what?--CSTAR 19:44, 11 March 2007 (UTC)
Yes, for the continuous version, it would be some nilpotent Lie group. I was hoping it might be distinct enough to have some unique name that can be googled. Put it another way: the group M is a group extension of the Heisenberg group. What other group extensions of the Heisenberg group are there? Are they all isomorphic to this one? linas 20:12, 11 March 2007 (UTC)
There are lots. That's what group cohomology is for.--CSTAR 20:20, 11 March 2007 (UTC)
I should say I really don't know, although I'm sure the nswer is known, and it is true, as I mentioned earlier that this can be reduced to a question in the cohomology of Lie Algebras.--CSTAR 20:30, 11 March 2007 (UTC)
This page is to discuss how to improve the article. A good place to ask about the contents of the article are at the reference desk.
In case anyone is still interested: The group <x,y,z|yx=xyz,zy=yz> is not nilpotent, but the quotients you gave as matrix groups happened to be. You can learn a bit about the group by looking at its free nilpotent quotients. If you take the nilpotentcy class 1 quotient, the abelianization, you get Z x Z = <x,y,z|yx=xy,z=1>. If you take the nilpotency class 2 quotient, you get the Heisenberg group, <x,y,z|yx=xyz,zy=yz,zx=xz>. If you take the nilpotency class 3 quotient, you get < x,y,z,a | yx=xyz, zx=xza, ax=xa, zy=yz, ay=ya, az=za >, which has Hirsch length 4 and the generators x,y,z,a form a basis of the Lie ring, which is a free abelian group. The class 4 quotient is < x,y,z,a,b | yx=xyz, zx=xza, ax=xab, bx,=xb, zy=yz, ay=ya, by=yb, az=za, bz=zb, ba=ab >, which has Hirsch length 5 and the generators x,y,z,a,b form a basis of the Lie ring which is a free abelian group. However, at the class 5 quotient and beyond the group is no longer metabelian, so the generators get messier. If you agree to take the presentation in the variety of metabelian groups (so you include the identical relation [[g,h],[k,l]]=1 for all g,h,k,l in the group), then your group has Hirsch length infinity and the Lie ring is free abelian with basis x,y,[y,x],[y,x,x],[y,x,x,x],.... At any rate, the group <x,y,z|yx=xyz,zy=yz> has this large non-nilpotent group as a quotient, so it is not nilpotent either. JackSchmidt (talk) 16:49, 18 December 2007 (UTC)