SL(2,C) & sl(2,C): Physics Book Reference

[dextercioby]

One of the most important groups in physics is the group of complex unimodular matrices

$$\mbox{SL}\left(2,\mathbb{C}\right)=:\left\{A\in \mathcal{M}_{2}\left(\mathbb{C}\right)|\mbox{det} A=1\right\}$$

This is a Lie group.I haven't seen any decent physics book in which the theory of this group is presented.

So I'm asking you to give me a reference on a book which would contain

1.The irreducible representations of this group.
2.The operators which generate the Lie algebra of this group and the commutation relations which define the Lie algebra.
3.How is the space of this (the group's) representations constructed and what is the action of operators on its basis.
4.Irreducible representations of the Lie algebra of this group.(Operators,base & action of operators on the base).
5.The connection (if any) between irreducible representations of the group & of its algebra.
6.The connections (if any) between the irreducible representations of this group & algebra and the group $\mbox{SU(2)}$ and its Lie algebra $\mbox{su(2)}$...How are all operators connected (if so) ?

Daniel.

 

[matt grime]

The book for Lie Algebra representations is Fulton and Harris "Representation Theory".

But I'll give you a quick precis.

The Lie algebra of any reasonably Lie Group is almost trvial to calculate.

One looks at the set of matrices {X such that 1+eX} is "in the group" in some sense we make precise:

What this means is you put 1+eX into the equations that define the group. 1 is the identity matrix, X is an arbitrary matrix and e is a real variable. Then we formally declare e^2=0, and see what conditions this places on X. This sounds more complicate than it is.

For example SL(n,C) is given by the matrices with determinant 1. So the Lie algebra sl(n,C) is the nxn matrices satisfiying det(1+eX)=1 modulo e^2

det(1+eX)=1+etr(X) + o(e^2), so modulo e^2 we need 1=1+etr(X), ie sl(n,C) is the nxn trace zero matrices.

sl(2,C) is commonly given with basis e,f,h where e is 1 in the top right zero elsewhere, f is 1 in the bottom left zero elsewhere, and h is 1 top left, -1 bottom right, and zero off the diagonal. the bracket is given by matrix commutation:

[ef] = ef -fe

etc.

Its representation theory is very elegant and is the most fundamental in lie algebras, via things called highest weight modules.

All finite dimensional representations are explicitly constructed in Fulton and Harris. They are indexed by the set of natural numbers in a nice way. It is worth buying that book.

If you really want then I can do an intro here for you.


Links to the group: here you must be careful. What kinds of representations do you want to consider? There is a well known 1-1 correspondence between irred reps of the algebra and irred reps of a certain kind of SL(2,C), namely the unitary ones I think.

SU(2) needs care. It is a complex group, but its tangent space is not a complex Lie algebra, only a real lie algebra.

However, there is a very clear 1-1 correspondence between reps of SU(2) and reps of sl(2).

They are both parametrized by "weights". Acutally, the nicest description of either SU(2) or sl(2)'s reps is via polynomials.

The irred reps of sl(2) are given by the space of homogeneous polynomials in two variables.

Let P(n) be the "vector space of homogeneous polys of degree n in two variables x and y". That is the abstract vector space with basis x^ry^{n-r}

Then e, as above, acts via xd_y, f via yd_x (partial differential operators) and h is the commutator [ef] = xd_x - yd_y. This is the unique irred rep of dimension n+1.

Notice that h acts via multiplication by a scalar (n), this n is called the weight.


SU(2) also acts on the space of homogeneous polys of degree n.


If the matrix element T of SU(2) has coeffs a,b top row and c,d bottom row, then if f(x,y) is a polynomial in x and y define T(f) = f(ax+cy,bx+dy).

 

[dextercioby]

Thank you,Matt.I hope to find the book...This group is essential for any relativistic theory as it is the universal covering group of the homogenous Lorentz group $\mbox{SO(3,1)}$ and so all relativistic physics is based on this $\mbox{SL}\left(2,\mathbb{C}\right)$ and its irreducible representations.

Daniel.

 

[matt grime]

sl(2) is the building block of all nice Lie algebras: essentially they are copies of sl(2) glued together in some way that we can describe using geometry of these things that are "weights" which are approximatley just eigenvalues. The Fulton and Harris one is good, and cheap and ubiquitous. It even explains how to get between reps of lie groups and algebras: the group homomorphism into GL(n), if it is differentiable gives a rep of the lie algebra, and vice versa. There are some representations that aren't given this way but I don't think they are of much interest to physicists. At least not looking at the presentation of the material.

 

[dextercioby]

Assuming a Minkowski metric $$g_{\mu\nu}=g^{\mu\nu}=\mbox{diag}\left(+,-,-,-\right)$$,one can prove quite easily that the generators of the restricted homogenous Lorentz group satisfy the comutation relations

$$\left[M_{i},M_{j}\right]_{-}=\epsilon_{ijk}M_{k}$$

$$\left[N_{i},N_{j}\right]_{-}=-\epsilon_{ijk}M_{k}$$

$$\left[M_{i},N_{j}\right]_{-}=\epsilon_{ijk}N_{k}$$

So this is the algebra of the $\mbox{lor(1,3)}$ group.


My QM treacher goes then to prove that this group is homomorphic (locally isomorphic) with the group $\mbox{SL(2,\mathbb{C})}$.Okay.

Question:if these 2 Lie groups are homomorphic,does that mean that their Lie algebras are isomorphic...?[/QUOTE]

Baker[1] makes me think this fact is not automatic.Theorem 6.12,page 175.He himself builds the homomorphism between the 2 groups (in another way as my QM teacher does) and shows its derivative generates an algebra isomorphism.Alright.

To sum up,the 2 groups are homomorphic (locally isomorphic) and their algebras are isomorphic.The kernel of the homomorphism is made up of $\pm I_{2}$.That would suggest the double covering.The homomorphism maps 2 distinct matrices from SL(2,C) into one element of Lor(1,3).

Following Delamotte[2],page 31,he says that in general case one can prove 2 theorems:

"*An n-connected group can have n-valued representations (my note:"representations later").
.
**For all multiple connected groups (labeled $$G$$),there is a unique "smallest" simply connected group (labeled $$\bar{G}$$) which has the same Lie algebra (my note:"he's a physicist,the algebras are isomorphic") and no subgroup of $$\bar{G}$$ is homomorphic to $$G$$.

Then he says that $$\bar{G}$$ is called universal covering group of $$G$$.I guess that's a definition. Then he gives the 2 examples with SO(3) as the $$G$$ and SU(2) as the $$\bar{G}$$ and Lor(1,3) as the $$G$$ and SL(2,C) as the $$\bar{G}$$.

Okay.I'm not interested (for now) in the proofs for the 2 mathematical theorems.

However,other questions pop up.

1.SL(2,C) has other subgroups.SU(2) is one of them.Since SU(2) has a dimension over reals of 3 and the Lor (1,3) has dimension 6 over the reals (it is a group over the reals),i guess that proves the fact that the groups are not homomorphic.That doesn't rule out the possibility that a direct product of 2 SU(2) groups be homomorphic to Lor(1,3).I haven't seen this anywhere,so i'll have to do it myself.Here's the question:does SL(2,C) have 6-dimensional (over the reals) subgroups...?

2.If it does,are they simply connected...?

Daniel.

 

[dextercioby]

Here's the bibliography

[1]A.Baker,"Matrix Groups.An Introduction to Lie Groups",Springer Verlag,2002.
[2]B.Délamotte,"Théorie des Groups de Lie,Poincaré et Lorentz",lecture notes,1997.

Daniel.

 

[matt grime]

Ah, the rule for these things is that the Lie group must be connected and simply connected.

They will have isomorphic Lie algebras because this is a local statement, about things "at the origin", provided the map at 1 is a local isomorphism: its derivative must have the correct rank: L_n and SL_n have the same lie algebras, as do O_n and SO_n. It only matters about the (simply) connected component containing the identity.


This is a false in postive characteristic (just like galois theory, indeed it is a "separability" thing) - there are abstract isomorphisms that are not "algebraic automorphisms", by which we actually properly mean "geometric".

 

[dextercioby]

This whole nebula cracks me up.Before going to representations,i'll have to ask what is the connection (if any) between the # of group generators (6 for Lor(1,3) and only 3 for SL(2,C)) and the rank of the groups (2 for each of them) and what implication dos the fact that the # of generators differs has on the irreducible finite representations.I mean,will i need 2 SL(2,C) algebras (6 basis elements in all,3 from each algebra) to build the irreducible representations of Lor(1,3)...?

Daniel.

Daniel.

 

[matt grime]

What do you mean by "generators" for SL(2,C)? It is an uncountable group, so it can't have a finite set of generators, at least not in the sense I'd think of.

 

[dextercioby]

Don't the infinitesimal generators of a Lie group form a Lie algebra...?The SL(2,C) has 3 infinitesimal generators,E,F,G or H,X,Y (this is the notation in Fulton & Harris [1]).



Daniel.


-------------------------------------------------------------
[1]W.Fulton,J.Harris,"Representation Theory.A first course",Springer Verlag,1991.

 

[dextercioby]

So basically,i don't need an irreducible representation of the algebra sl(2,C) (and that chapter from Fulton & Harris (note [1],post #10),but i need the irreducible finite dimensional representations of the group SL(2,C),because I'm interested in finding the finite dimensional irreducible representations of Lor(1,3).

So the Sl(2,C) has 3 generators,while the Lor(1,3) has 6.I haven't figured out the connection between this part & the representations...I can only guess (and hope that someone could prove me right/wrong) that the double valuedness of the universal covering homomorphism between the 2 groups implies that one irreducible representation of Lor(1,3) corresponds to 2 irreducible representations of SL(2,C).It looks an incorrect conclusion,because both SO(3) and SU(2) have 3 infinitesimal generators and the SU(2) realizes a similar double valued universal covering of SO(3) as is the case with SL(2,C) & Lor (1,3).

Help...?

Daniel.


Daniel.

 

[matt grime]

SL(2,C) does not have 3 generators, sl(2,C) does, is that what you mean? There is a 1-1 corresponedence with reps of SL and sl, as Fulton and Harris give, and these all give reps of *any* group G with a hom into SL. Perhaps with a little fiddling you can show that there is some Ind functor that is inverse to the Res functor and that classifying irred reps of SL classifies those of Lor too, or perhaps it turns out that the Lia alg of Lor is semisimple and equal to, sayl sl direct product sl, whence the reps are given by slot wise actions: (x,y).(m,n)=(x.m,y.n)

One thing to consider, since I know f.a. about the lorentz group is in what sense do you mean "double cover", for instance as a formal analogy, consider the groups C_2xC_2 and C_4, these both are a "double cover" of C_2 via (x,y) maps to x and z maps to z^2 resp. But they are different extensions of C_2. I don't know what kind of extension Lor is of SL(2,C)

Oh, and we recover the reps of G be exponetiating reps of g where g is G's lie alg.

 

[dextercioby]

SL(2,C) does not have 3 generators, sl(2,C) does, is that what you mean?

So those E,F,G (H,X,Y) basis in sl(2,C) are not infinitesimal generators of SL(2,C)...?

There is a 1-1 corresponedence with reps of SL and sl, as Fulton and Harris give, and these all give reps of *any* group G with a hom into SL.Perhaps with a little fiddling you can show that there is some Ind functor that is inverse to the Res functor and that classifying irred reps of SL classifies those of Lor too, or perhaps it turns out that the Lia alg of Lor is semisimple and equal to, sayl sl direct product sl, whence the reps are given by slot wise actions: (x,y).(m,n)=(x.m,y.n)

The Lor(1,3) group is semisimple,therefore all his reducible finite dimensional representations are completely reducible,so it suffices to use only the irreducible ones.I'm not interested in those infinite dimensional ones.

As for his algebra,i don't know if it's semisimple...My Qm teach. says that,due to the homomorphism between the 2 groups,the irreducible finite dimensional reps of Lor(1,3) can be found knowing the ones for SL(2,C)...Is that correct (meaning,the motivation)...?

One thing to consider, since I know f.a. about the lorentz group is in what sense do you mean "double cover", for instance as a formal analogy, consider the groups C_2xC_2 and C_4, these both are a "double cover" of C_2 via (x,y) maps to x and z maps to z^2 resp. But they are different extensions of C_2. I don't know what kind of extension Lor is of SL(2,C)

I'll think about that.

Oh, and we recover the reps of G be exponetiating reps of g where g is G's lie alg.

Apparently it's another way around it.I'm given 2dim irreducible reps of SL(2,C) and with them i want to get to 2dim irreducible reps of the Lie algebra lor(1,3).How do i do it...?

Daniel.

 

[matt grime]

We know there is a homomorphism from Lor to SL(2,C), right?

So given a rep of SL(2,C) the composite map

Lor --> SL(2,C)--> End(V)

is a rep of Lor.

Now, we can differentiate this map to get a map of lor, which I'll define to be Lor's tangent space at its origin. Edit: we;re both usign that notation, good, saves confusion.

I don't know if this gives all of the irred reps.

I don't know anything about "infinitesimal generators", sorry. Edit: a quick search tells me they are indeed infinitesimal generators, which apparently means they are just a basis of the Lie Alg.)

 

[dextercioby]

It looks like we're now on the same page,at least apparently...

So does that double valued homomorphism require 2 inequivalent irreducible & finite dimensional reps of SL(2,C) to get a similar representation for Lor(1,3)...?Or does that homomorphism require only one rep.of SL(2,C)...?

Daniel.

 

[matt grime]

In the one I've given the composite map you've given from Lor to SL to V is a rep of Lor, which will be irreducible if V is, I think, and differentiating gives a rep of lor, its Lie Algebra. Ie "requires 1 irred rep of SL(2,C)".

If you were to know that Lor = SLxSL, then some of the reps of Lor would be exactly the tensors VxW for reps of SL. Again, in the non-finite group case I don't know if this is all of them. Actually, scrap that, the reps of linear algebraic groups were characterised by Chevally et al weren't they, won't they tell you all you need to know?

There are many ways of getting reps of bigger groups from smaller ones (the two here mentioned, plus induction and tensor induction). Checking what the structure of Lor is would be a good start I guess.

I'm not sure what "double valued... requires..." means, but as you may see this is becuase there are many was of getting reps of Lor here from SL. I doubt thet Lor is SLxSL, so it'll probably just be the composite map method.

 

[matt grime]

Right, I appear to have this the wrong way round.

The Lorentz group is O(3,1), it is not connected. Take Lor to be the connected component of the identity. It is not somply connected, it's cover is SL(C,2), and Lor is just PSL(C,2)

So there is a map from SL(C,2) to Lor, not the other way round; see me learn physics as we go... Many apologies for making you deal with an ignorant mathematician. I really should read all of the posts more closely rather than just some of them and forgetting to check back.

So, from a rep of SL we get one of Lor by quotienting out.

 

[dextercioby]

Alright,before going further,of u know about a book which gives/builds the finite dimensional irreducible reps of SL(2,C),please give me a reference.I want to have an alternative to my lecture notes (which i'll discuss further) and to Weyl [1].

Daniel.

----------------------------------------------------------
[1]H.Weyl,"Group Theory and Quantum Mechanics",Dover,1931.

 

[matt grime]

Again, my apologies for having to learn this as I go. I can't think of any book for SL(2,C) that derives the reps directly (SL(2,R), yes).

Try John Baez This weeks finds 181 and 162 for some hints.

And most things by Gelfand on this should help (Greame Segal says Gelfand, Graev and Vilenkin's book treats these things in a nice way: Generalized Functions, 1966)

this book here

http://www.cambridge.org/uk/catalogue/catalogue.asp?isbn=0521005515

was recommended to "those who didn't appear to know a lot about the reps of the lorentz group" in a newsgroup:

http://www.lns.cornell.edu/spr/2000-02/msg0021686.html


and i think i'd have to be labelled one of those.




Or there's this one

Noncommutative Harmonic Analysis, Math. Surveys and Monogr., No. 22, American Math. Society, Providence, R.I., 1986. which has a chapter on SL(2,C) and the lorentz group.

 

[dextercioby]

I took a week off for further documentation.I found a great treatment on the reps of SL(2,C) in a book written by a physicist,Moshe Carmeli:"Group Theory and General Relativity",McGraw-Hill,1977.

I'll study and resume posting shortly.

Daniel.

 

[Haelfix]

For the Poincare group, the task of getting the irreducible reps is found in the classical paper by Wigner (you can find a reference in Weinberg I). Its quite readable, he actually breaks the paper in two sections, one for mathematicians, one for physicists. Highly amusing!

Anyway for physicists, we restrict our attention to reps that are unitary and continuous. Now in general, most of the time these turn out to be infinite dimensional.

Basically the 'baby problem' is as follows, look at the orbits of SL(2,C) that act on say Minkowski space.. Pick a point on each orbit and find the isotropy group, now list all representations thereof. This is what physicists call the "little group" (for instance see Weinberg) Most of the time this is all you need.

However in general, for finite irreducible representations you need to use Young diagrams and to find a unique highest weight vector to output an exhaustive list.