Questions about representation theory of Lie algebra

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phoenix95
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Summary:

SL(2, C), SU(2), SU(3)

Main Question or Discussion Point

I have confusions about representation theory. In the following questions, I will try to express it as best as possible.
For this thread say representation is given as
ρ: L → GL(V)
where L is the Lie group(or symmetry group for a physicist)
GL(V) is the general linear transformations of V(of some dimension)

Questions:
1. Often the irreducible representations are labelled along the columns of Reps, dimensions; see page 20 in this. What do they exactly mean? As I see it reps are the representation space V of the given symmetry group(SU(2) in this case) and dimension is referring to the dimension of V. Am I correct?

2. Each entry in the columns is a new irreducible representation(is this correct?), how does the increase in weight relates to the increase in dimensions? What happens here?

3. Take a look at page 35 of this book(attached): Introduction to quarks and Partons by F. E. Close(1979). There he labels the multiplet by different quark names. By doing this he is constructing a representation space V namely (u, d, s) and the SU(3) group acts on this space, correct? So given a higher dimensional representation space(say the octuplet (1, 1)) how does the SU(3) act on it?
 

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  • #2
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Maybe you should have posted this in a physics forum, since at least your reference on slide 9 seems to be physics.
Summary:: SL(2, C), SU(2), SU(3)

I have confusions about representation theory. In the following questions, I will try to express it as best as possible.
For this thread say representation is given as
ρ: L → GL(V)
where L is the Lie group(or symmetry group for a physicist)
GL(V) is the general linear transformations of V(of some dimension)

Questions:
1. Often the irreducible representations are labelled along the columns of Reps, dimensions; see page 20 in this. What do they exactly mean?
I can only assume that this refers to the weight spaces of a representation: how many ladder up and ladder down operators we have.
As I see it reps are the representation space V of the given symmetry group(SU(2) in this case) and dimension is referring to the dimension of V. Am I correct?
Looks like. In general you must be careful whether the dimension of the Lie group, or Lie algebra, or the dimension of the representation is meant.
2. Each entry in the columns is a new irreducible representation(is this correct?), how does the increase in weight relates to the increase in dimensions? What happens here?
See the theorem under section 13 in here:
https://www.physicsforums.com/insights/lie-algebras-a-walkthrough-the-representations/

It is the other way around. The increase of ##\dim V## allows, or better enforces more weight spaces. ##V## is written as a direct sum of invariant subspaces, and the invariance is that of the operators given by the Cartan subalgebra. Ladder up and ladder down is impossible at the end of the ladder, and its length is given by ##\dim V##.
3. Take a look at page 35 of this book(attached): Introduction to quarks and Partons by F. E. Close(1979). There he labels the multiplet by different quark names. By doing this he is constructing a representation space V namely (u, d, s) and the SU(3) group acts on this space, correct? So given a higher dimensional representation space(say the octuplet (1, 1)) how does the SU(3) act on it?
##V_0## is a the two dimensional subspace, which is invariant under the two operators of the Cartan subalgebra. If ##\dim V=8##, then there is one way up and one way down: ##V=V_0\oplus V_{-1} \oplus V_{1}##. This should correspond to ##\mathfrak{su}(3)\cong \operatorname{ad} \mathfrak{su}(3)## itself: diagonal, upper triangular and lower triangular matrices.
 
  • #3
phoenix95
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I understand now. The confusion was that I thinking that the representation space V always has the same dimensions as n in SU(n). I didn't think that there can be higher-dimensional V or even lower ones.

I have one more question: The dimension of V, it has to be either equal or more than the n in SU(n), is this correct? In other words, is it possible to construct a doublet for SU(3)? Or a triplet for SU(4)?

In general you must be careful whether the dimension of the Lie group, or Lie algebra, or the dimension of the representation is meant.
Yes, I see that now.
 
  • #4
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I have one more question: The dimension of V, it has to be either equal or more than the n in SU(n), is this correct?
No. The question is, which kind of representation you are looking for. E.g. you can always define ##\mathfrak{g}.V\equiv 0##, and ##V## can have any dimension. Sure, the trivial representation is not very illuminating, but allowed. The dimension must at least be the dimension of the Lie algebra if we require a faithful representation, i.e. an embedding ##\mathfrak{g}\hookrightarrow \operatorname{GL}(V)##. I suppose there are cases in between, too, i.e. with ##0< \dim V < \dim \mathfrak{g}##, although not faithful.
In other words, is it possible to construct a doublet for SU(3)? Or a triplet for SU(4)?
I'm not sure about the representations of ##SU(n)## with ##n>2## in general. The case ##n=2## is rather simple and easy to calculate, see the theorem in the link, or this one here: https://arxiv.org/pdf/math-ph/0005032.pdf, chapter 5. I had to do it first, or search the internet. The plan goes as follows:

  1. Write ##\mathfrak{su}(n)= \mathfrak{h} +\sum_{\alpha\in \phi} \mathfrak{su}(n)_\alpha## with a Cartan subalgebra ##\mathfrak{h}## and an according root system ##\phi##.
  2. In case of ##\mathfrak{g}=\mathfrak{sl}(n)## this decomposition is diagonal, upper and lower triangular matrices. The basis change towards ##\mathfrak{su}(n)## makes this is a bit more inconvenient, as the matrices look different.
  3. Then choose a representation space ##V## and calculate the subspace ##V_0## defined by ##\mathfrak{h}.v=0##.
  4. Look what repeated applications ##Y^k.V_0## do (ladder down), where ##Y## is a basis vector of ##\mathfrak{su}(n)_\alpha## and determine the weights.
  5. Do the same for ##X^k.V_0## (ladder up).
  6. This should give you a decomposition of ##V## along the weights and how the various operators act among them.
So or similar is the task. It's a bit nasty to actually calculate it, so I would probably rather search for the solution on the internet. I'm sure the irreducible, finite dimensional representations for ##\mathfrak{su}(n)## are fully cataloged.

Let's see whether your question has an easy answer. Assume we have ##V=V_0 \oplus V_+ \oplus V_-## all one dimensional, say spanned by ##v_0,v_+,v_-##, and ##\mathfrak{h}.V_0=0## with ##\mathfrak{h}\subseteq \mathfrak{su}(n)## CSA. Let us further assume we have at least three positive roots ##\gamma \in \{\,\alpha, \beta, \alpha+\beta \,\}## and $$[H,X_\gamma]=\gamma(H)X_\gamma\, \; , \; \,[H,Y_\gamma]=-\gamma(H)Y\; , \;[X_\gamma,Y_\gamma]=H_\gamma\; , \;[X_\alpha,X_\beta]=X_{\alpha+\beta}\; , \;[Y_\alpha,Y_\beta]=-Y_{\alpha+\beta} $$
This means we have for ##v_i \in \{\,v_0,v_+,v_-\,\}##
\begin{align*}
H_\gamma(v_0) &=0\, , \,H_\gamma(v_+)=h_1(\gamma)v_+\, , \,H_\gamma(v_-)=h_{-1}(\gamma)v_-\\
X_\gamma(v_0) &=x_0(\gamma)v_+\, , \,X_\gamma(v_+)=0\, , \,X_\gamma(v_-)=x_{-1}(\gamma)v_0\\
Y_\gamma(v_0) &=y_0(\gamma)v_-\, , \,Y_\gamma(v_+)=y_1(\gamma)v_0\, , \,Y_\gamma(v_-)=0
\end{align*}
Now we have to check whether the Jacobi identity holds or if there need to be adjustments on the coefficients. But beside some signs the multiplication table should be ok. Finally we have to check, whether there are values ##h_{\pm 1}(\gamma), x_{0}(\gamma), x_{-1}(\gamma), y_{0}(\gamma), y_{1}(\gamma)## such that the above satisfy the conditions of a representation, i.e. ##[A,B](v)=A(B(v))-B(A(v))##.

As this is basically the same as in the case of ##\mathfrak{su}(2)##, it will be sufficient to concentrate on what is new, and that are the composite roots ##\gamma=\pm(\alpha+\beta)##.

Now you can either try yourself whether there is a solution for the coefficients ##h_{\pm 1}(\gamma), x_{0}(\gamma), x_{-1}(\gamma), y_{0}(\gamma), y_{1}(\gamma)## which are linear in ##\gamma##, or search for "representations + SU(3)". Playing a bit with ##[X_{\alpha+\beta},Y_{\alpha+\beta}]=H_{\alpha+\beta}## on ##\{\,v_0,v_+,v_-\,\}## are only three equations, nine if we add ##[X_\alpha,X_\beta]=X_{\alpha+\beta},[Y_\alpha,Y_\beta]=Y_{\alpha+\beta}##, i.e. doable.
 
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  • #5
phoenix95
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No. The question is, which kind of representation you are looking for. E.g. you can always define g.V≡0g.V≡0\mathfrak{g}.V\equiv 0, and VVV can have any dimension. Sure, the trivial representation is not very illuminating, but allowed. The dimension must at least be the dimension of the Lie algebra if we require a faithful representation, i.e. an embedding g↪GL(V)g↪GL⁡(V)\mathfrak{g}\hookrightarrow \operatorname{GL}(V). I suppose there are cases in between, too, i.e. with 0<dimV<dimg0<dim⁡V<dim⁡g0< \dim V < \dim \mathfrak{g}, although not faithful.
Actually this was sufficient(and I understood the rest as well), thank you very much for your time.
 
  • #6
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Actually this was sufficient(and I understood the rest as well), thank you very much for your time.
I may have to correct my statement a bit. If ##\dim V = \dim \mathfrak{g}## then we always have ##\operatorname{ad}\, : \,\mathfrak{g} \hookrightarrow \mathfrak{gl(g)}## in case of semisimple Lie algebras. I wrote ##\operatorname{GL}(V)## which is a mistake. That would have been a group representation, but we are in the world of Lie algebras here. The condition on the dimensions was the lazy one, will say if ##\mathfrak{g}## and ##V## are of equal dimension, then we can get a faithful representation.

But(!), if e.g. ##\mathfrak{g}=\mathfrak{su}(3)## then we have an eight dimensional Lie algebra. If we choose ##\dim V= 3## then ##\dim \mathfrak{gl}(V)=9## and we can have a faithful representation at least in principle, i.e. ##\mathfrak{su}(3)## fits in ##\mathfrak{gl}(V)##. And since ##su(3)## are ##3\times 3## matrices, they act on ##\mathbb{C}^3## naturally, and the commutator delivers a Lie algebra representation. Hence there is a three dimensional and faithful representation. So your question had indeed an easy answer:

Yes there is a three dimensional, faithful and irreducible (triplet) representation for ##\mathfrak{su}(3)##, defined by: ##A.v=A(v)## and ##[A,B]=AB-BA## where ##A,B \in \mathfrak{su}(3)## and ##v\in \mathbb{C}^3##.

Sorry, that I haven't seen it earlier.

Here is another important fact: A representation is a Lie algebra hiomomorphism ##\varphi\, : \,\mathfrak{g}\longrightarrow \mathfrak{gl}(V)##. This implies that its kernel is an ideal of ##\mathfrak{g}##. The Lie algebras ##\mathfrak{su}(n)##, however, are simple, i.e. they have no ideals. Thus ##\varphi## is either trivial or faithful in this case. Hence there are no non-trivial representations of ##\mathfrak{su}(n)## with a too small representation space. Say ##\dim V=m## and ##\dim \mathfrak{su}(n)=n^2-1##. Then we must have ##n^2-1 \leq m^2 = \dim \mathfrak{gl}(V)##. This means ##m \geq n## for non trivial representations.
 
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