SU(2) and elementary properites of Lie Algebras

[topsquark]

I've been having fun with my new Lie Algebra text and it occurred to me that working out a couple of basic examples of my own would be a good idea. I got rather large surprise.

The example I'm working with is SU(2) and I'm going through some basic properties it has. For all its uses in Physics SU(2) seems to be a rather boring group Mathematically speaking. But an excellent test-bed for my fledgling Algebra skills.

(1)
The first question is something I'd never noted before. I had thought that the spinor representation of SU(2) using the Pauli matrices
\(\displaystyle x = \left [ \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right ] \)

\(\displaystyle y = \left [ \begin{matrix} 0 & -i \\ i & 0 \end{matrix} \right ] \)

\(\displaystyle z = \left [ \begin{matrix} 1 & 0 \\ 0 & -1 \end{matrix} \right ] \)

were a good "basis" for this. But one crucial aspect of SU(2) is missing: they all have det() = -1, not 1. Should I be using {xy, yz, zx} or something similar?

(2)
A Lie Algebra is based on a vector space. In this case the most general member of the underlying vector space should be (assuming the Pauli matrices) \(\displaystyle a_0 I + a_1 x + a_2 y + a_3 z\), where the a's are complex. Now, what are the elements of g = SU(2)? Would they be the elements of the vector space or would they simply be x, y, and z?

An example of the above comes in when I'm trying to find ideals for SU(2). We need a subalgebra, h, of the g = SU(2) Lie Algebra, and it must have the property \(\displaystyle [g, h] \subseteq h\). Are we dealing with g and h as elements of the vector space? Or do we need to satisfy the condition using only x, y, and z elements? ie. Do I have to calculate \(\displaystyle [ a_0 I + a_1 x + a_2 y + a_3 z, h ] \) or just a couple of [x, z] brackets? (In this case both ways of stating it work out to be the same thing, but I'm assuming this is not a general property.)

As a check, using either way of looking at [g,h] above, I am saying that there are no non-trivial ideals in SU(2) and thus SU(2) is simple. Is this correct?

(3)
Finally, a question about the Lie bracket. For SU(2) we have the "multiplication table"
\(\displaystyle \begin{array}{c|c|c|c|} [,] & x & y & z \\ \hline x & 0 & 2iz & -2iy \\ \hline y & -2iz & 0 & 2ix \\ \hline z & 2iy & -2ix & 0 \\ \hline \end{array}\)

We have that [x, y] = 2iz. But 2iz does not belong to the Algebra, does it? And for that matter, neither does 0. But we need [x, y] to be an element of SU(2). Is this example telling me that we are to use \(\displaystyle a_0 I + a_1 x + a_2 y + a_3 z\) as opposed to the individual [x, y] brackets?

Thank you for not applying "tldr." (Kiss)

-Dan

 

[Euge]

Hi topsquark,

An $\Bbb R$-basis for $SU(2)$ is $\{ix/2,iy/2,iz/2\}$. I only put the $1/2$ there to match the usual convention in physics. In particular, $SU(2)$ is a $3$-dimensional Lie algebra over $\Bbb R$.

There are so many things I can say about $SU(2)$ that are mathematically interesting, but maybe if you delved into the algebraic/geometric topology of $SU(2)$, you may not find it boring? :D

 

[topsquark]

Okay, that takes care of (1). Why has this not come up in my classes? (D*mn all Physicists anyway!). (3) now becomes:
\(\displaystyle \begin{array}{c|c|c|c|} [,] & ix & iy & iz \\ \hline ix & 0 & -2iz & 2iy \\ \hline iy & 2iz & 0 & -2ix \\ \hline iz & -2iy & 2ix & 0 \\ \hline \end{array}\)
I'm not going to quibble about the 2's so this also works for me.

That leaves (2). So when we do the check for an ideal, we have as the most general element of SU(2) to be \(\displaystyle g = a_1(ix) + a_2(iy) + a_3(iz)\) and given a subalgebra h of g we need to verify that
\(\displaystyle [ a_1(ix) + a_2(iy) + a_3(iz), h] \subseteq h\) for all \(\displaystyle a_1, ~a_2, ~a_3\)

As a negative example for clarity: Let \(\displaystyle h = p (ix) + q (iy)\). Then to test h we have

\(\displaystyle [a_1(ix) + a_2(iy) + a_3(iz), p(ix) + q(iy)] = a_1q~ (ix)(iy) + a_2p ~(iy)(ix) + a_3p~ (iz)(ix) + a_3 q ~(iz)(iy)\)

\(\displaystyle = - a_3q(ix) - a_3p(iy) + (-a_1q + a_2p) (iz) \nsubseteq b(ix) + c(iy)\)

Thus h is not an ideal of SU(2). Did I do this right?

-Dan

 

[Math Amateur]

Okay, that takes care of (1). Why has this not come up in my classes? (D*mn all Physicists anyway!). (3) now becomes:
\(\displaystyle \begin{array}{c|c|c|c|} [,] & ix & iy & iz \\ \hline ix & 0 & -2iz & 2iy \\ \hline iy & 2iz & 0 & -2ix \\ \hline iz & -2iy & 2ix & 0 \\ \hline \end{array}\)
I'm not going to quibble about the 2's so this also works for me.

That leaves (2). So when we do the check for an ideal, we have as the most general element of SU(2) to be \(\displaystyle g = a_1(ix) + a_2(iy) + a_3(iz)\) and given a subalgebra h of g we need to verify that
\(\displaystyle [ a_1(ix) + a_2(iy) + a_3(iz), h] \subseteq h\) for all \(\displaystyle a_1, ~a_2, ~a_3\)

As a negative example for clarity: Let \(\displaystyle h = p (ix) + q (iy)\). Then to test h we have

\(\displaystyle [a_1(ix) + a_2(iy) + a_3(iz), p(ix) + q(iy)] = a_1q~ (ix)(iy) + a_2p ~(iy)(ix) + a_3p~ (iz)(ix) + a_3 q ~(iz)(iy)\)

\(\displaystyle = - a_3q(ix) - a_3p(iy) + (-a_1q + a_2p) (iz) \nsubseteq b(ix) + c(iy)\)

Thus h is not an ideal of SU(2). Did I do this right?

-Dan

Just BTW and out of interest, Dan, what is your new Lie Algebra text?

Peter

 

[topsquark]

Just BTW and out of interest, Dan, what is your new Lie Algebra text?

Peter

I actually have two of them now. For the moment I'm concentrating on "Affine Lie Algebras and Quantum Groups" by Fuchs. It's geared toward Physics applications rather than just the Mathematical ones. I've been doing my usual skim though and the text sometimes boggs me down with definitions but every other page contains Math that I've seen in Quantum Mechanics... it's just that my QM classes never bothered to present the material in more general terms. I like it. :)

-Dan

 

[topsquark]

There are so many things I can say about $SU(2)$ that are mathematically interesting, but maybe if you delved into the algebraic/geometric topology of $SU(2)$, you may not find it boring? :D

Okay. That "topology" stuff. Didn't that go out of style when the Indian rubber sheet guys went out of business?

I'm just working through the Algebra stuff. My Topology is seriously weak so I thought I'd run through the Lie Algebra stuff before trying to tackle it.

(Tychonoff. Who needs him anyway? I like how compact I am already.)

-Dan

 

[topsquark]

I'm posting this just to round out the conversation. If anyone has any idle moments and wants to double check me feel free, but I'm reasonably confident that I've got a handle on this.

So. Let's define a 2-D spinor representation of the vector space of SU(2) to be given by the basis:
\(\displaystyle x = \left ( \begin{matrix} 0 & i \\ i & 0 \end{matrix} \right ) \text{, } y = \left ( \begin{matrix} 0 & 1 \\ -1 & 0 \end{matrix} \right ) \text{, } z = \left ( \begin{matrix} i & 0 \\ 0 & -i \end{matrix} \right ) \)

The most general element of this vector space is \(\displaystyle g = a_0 I + a_1 x + a_2 y + a_3 z\) where I is the identity matrix, the \(\displaystyle a_i\) are real, and \(\displaystyle a_0 ^2 + a_1 ^2 + a_2 ^2 + a_3 ^2 = 1\). I am defining the Lie bracket to be [a, b] = (1/2) (ab - ba). The structure factors are \(\displaystyle f^{ab}_c = - \epsilon ^{ab}_c\) where \(\displaystyle \epsilon ^{ab}_c \) is the usual Levi-Cevita symbol.

SU(2) contains no proper ideals, is simple and non-Abelian. The derived algebra g' = [g, g] = g and the lower central series \(\displaystyle g_1 = g' \text{, } g_2 = [g, g_1] = g \) (etc). SU(2) is simple, not solvable, and not nilpotent. The center, centralizer, and normalizer are all trivial. It has three sub-Algebras which are simple, not solvable, not nilpotent, etc.

The only confusion I have left is that the Physics representation used is Hermitian, but not unitary. I know that a set of Hermitian matrices are easier to work with in Physics for a measurable property, but Physics still calls the Algebra to be unitary. I'm presuming that as long as the Hermitian representation can be transformed into a unitary representation simply by a change of basis that we can roughly say that the Hermitian representation still belongs to SU(2).

-Dan

 

[Euge]

Sorry for not responding sooner. I had some internet connection issues, but now they're resolved. :D

It looks like you have the gist of things, but there isn't much more I can add to assist you, since you haven't shown your work e.g. in computing the derived series, the centralizer, normalizer, etc.. What I can say is that the computation you made in the negative example looks incorrect (it looks like you've taken the algebra product, not the commutator), disregarding the fact that your $h$ as written is not an algebra (although I know what you mean by it).

 

[topsquark]

Sorry for not responding sooner. I had some internet connection issues, but now they're resolved. :D

It looks like you have the gist of things, but there isn't much more I can add to assist you, since you haven't shown your work e.g. in computing the derived series, the centralizer, normalizer, etc.. What I can say is that the computation you made in the negative example looks incorrect (it looks like you've taken the algebra product, not the commutator), disregarding the fact that your $h$ as written is not an algebra (although I know what you mean by it).

Thanks for the post. My earlier posts had some mistakes in it, which I have since resolved. I revamped how I did the stuff in order to write my last post. I could post my work for all of that, but as it stretches four pages I decided to post just the results. As I said I am just posting my last comment to round it out.

I'm planning to work out one of the dihedral groups this afternoon. If I remember correctly D6 has at least one non-trivial ideal so it'll give me some more "meat to chew on." If nothing else it will be interesting to see what kind of Lie bracket to use. :)

-Dan

 

[Euge]

What do you mean by an ideal of $ D6$?

 

[topsquark]

What do you mean by an ideal of $ D6$?

I could be wrong about the existence of the ideal.

For a definition I have:

And ideal h of a Lie algebra g is a sub-algebra with the property \(\displaystyle [g,h] \subseteq h\).

I'm trying to work out a Lie algebra with an ideal to see what happens to the structure. But I may just go ahead and do D6 anyway, just for the practice.

-Dan

 

[Euge]

I was asking because one usually speaks of normal subgroups of groups, not ideals of groups. Since $ D6$ isn't naturally an algebra, how did you make the group $ D6$ into a Lie algebra?

 

[topsquark]

I was asking because one usually speaks of normal subgroups of groups, not ideals of groups. Since $ D6$ isn't naturally an algebra, how did you make the group $ D6$ into a Lie algebra?

As it happens I didn't. I used one from my text.

Still, we can easily make a vector space out of D6 elements. The problem, of course, is to find a way to define a usable Lie bracket. I thought it would be simpler than it apparently is. I wasn't able to find an easy one so I decided to try another example. I'll post that in a different thread.

-Dan