What Are Some Alternative Sources for Help with Lie Algebra Definitions?

[topsquark]

First off: I am not complaining about any of the members here or elsewhere. I do not post questions expecting help. When I post a question I would like an answer but I do not require it. Getting an answer depends on who is around and who is willing to help when I post the question.

Still...

I have posted a number of questions about basic Lie Algebra lately and the threads have gotten lots of views but no responses. (This is true of the other two Math forums I lurk around on.) If I can't get an answer here does anyone have any suggestions of where else I might be able to get the help? I have skimmed the rest of the chapter and it pretty much depends on the material that I have questions on so I would have difficulty pressing on without the answers. I mean, I'm not looking for someone to sit down with me and tutor the whole course, just someone that can walk me through some of the basic definitions and how to apply them for one or two special cases.

Any thoughts to share?

-Dan

 

[Euge]

Hi Dan,

I appreciate that you've expressed your frustration, but don't give up yet! :D I've taken a look at all your recent Lie algebra posts. They are all missing important pieces of information. You haven't defined any of the symbols $T_0, T^{\pm}, H, E^{\pm}$, and while you haven't made clear exactly which model you're using for $A1$. Like we've discussed with $E8$, $A1$ is a type, or classification.

Having said all that, it appears you're using $A_1^{(1)}$, i.e., $\mathfrak{sl}(2,\Bbb C)$, which consists of all trace zero complex $2 \times 2$ matrices. Further,

$$T^0 = \begin{bmatrix}1&0\\0&-1\end{bmatrix},\quad T^{+} = \begin{bmatrix}0&1\\0&0\end{bmatrix},\quad T^{-} = \begin{bmatrix}0&0\\1&0\end{bmatrix}$$

and the Lie bracket you used is the commutator, i.e., $[A,B] := AB - BA$. Are these correct assertions so far?

 

[topsquark]

Hi Dan,

I appreciate that you've expressed your frustration, but don't give up yet! :D I've taken a look at all your recent Lie algebra posts. They are all missing important pieces of information. You haven't defined any of the symbols $T_0, T^{\pm}, H, E^{\pm}$, and while you haven't made clear exactly which model you're using for $A1$. Like we've discussed with $E8$, $A1$ is a type, or classification.

Having said all that, it appears you're using $A_1^{(1)}$, i.e., $\mathfrak{sl}(2,\Bbb C)$, which consists of all trace zero complex $2 \times 2$ matrices. Further,

$$T^0 = \begin{bmatrix}1&0\\0&-1\end{bmatrix},\quad T^{+} = \begin{bmatrix}0&1\\0&0\end{bmatrix},\quad T^{-} = \begin{bmatrix}0&0\\1&0\end{bmatrix}$$

and the Lie bracket you used is the commutator, i.e., $[A,B] := AB - BA$. Are these correct assertions so far?

Thanks for the reply and looking things over. Yes, you are correct on all counts.

I have been using the T's recently as a symbol for the generators because the text insists on using H and the E's to point out that the basis they are using is the same as for the Cartan-Weyl basis (in all the examples they've given... I know that isn't a general property) and it's difficult for me to generalize to other algebras when they use the H and E's for all of them. I didn't think it was necessary to be more clear on that...I was taking for granted that if I gave the Lie brackets you wouldn't need it. Sorry about that! Anyway, if I have to use a particular representation I've been leaning toward the adjoint representation. But no matter. What I've been calling \(\displaystyle A_1\) (isomorphic to \(\displaystyle SL(2, \mathbb{C} )\)) uses the 2x2 matrix rep. as you've defined it above and the Lie brackets I've been using are the "usual" commutators.

I've been using \(\displaystyle SL(2, \mathbb{C} )\) to represent \(\displaystyle A_1\). That's part of the problem with the text I'm using...it occasionally gloss over things (strange for a grad level text). I don't know the difference between \(\displaystyle A_1\) and \(\displaystyle A_1^{(1)}\), unless you are talking about a derived algebra of \(\displaystyle A_1\). That would make sense with the text's terminology: \(\displaystyle g^{(1)} \equiv [ g, g ]\) where g is a Lie algebra and [,] is the Lie bracket defined on g, but it sounds like you are referring to something else. (In the case of \(\displaystyle SL(2, \mathbb{C} )\) the algebra and derived algebra are both the same anyway.) The text hasn't used the words "type" or "classification."

Hmmm. I had thought my posts were relatively clear except for the stuff I was asking about. Either the conventions in my text aren't all that general or maybe I do need that tutor after all.

-Dan

 

[Euge]

What I've been calling \(\displaystyle A_1\) (isomorphic to \(\displaystyle SL(2, \mathbb{C} )\)) uses the 2x2 matrix rep. as you've defined it above and the Lie brackets I've been using are the "usual" commutators.

Actually, $SL(2,\Bbb C)$ consists of those $2 \times 2$ complex matrices with determinant $1$. This is a (Lie) group, but it is not an algebra over $\Bbb C$ with the usual operations. What you mean is $\mathfrak{sl}(2,\Bbb C)$ -- this might be an annoying notational distinction, but the difference in meaning is great enough to make this point. :D

I don't know the difference between \(\displaystyle A_1\) and \(\displaystyle A_1^{(1)}\), unless you are talking about a derived algebra of \(\displaystyle A_1\). That would make sense with the text's terminology: \(\displaystyle g^{(1)} \equiv [ g, g ]\) where g is a Lie algebra and [,] is the Lie bracket defined on g, but it sounds like you are referring to something else.

Sorry, you can disregard my use of the symbol $A_1^{(1)}$. I was being a bit lax in it's use, but there's more structure to it as an algebra than $\mathfrak{sl}(2,\Bbb C)$.

Now, regarding your Abelian Cartan subalgebra post, keep this in mind. While a simple Lie algebra may be decomposed into subalgebras, it cannot be decomposed into nontrivial ideals. This is where the difference lies -- a subalgebra of an algebra need not be an ideal (however, every ideal is a subalgebra). The (root space) decomposition you wrote is incorrect, but I'd say you're on the right track. I can explain it further but it'll have to be a bit later.

 

[topsquark]

Actually, $SL(2,\Bbb C)$ consists of those $2 \times 2$ complex matrices with determinant $1$. This is a (Lie) group, but it is not an algebra over $\Bbb C$ with the usual operations. What you mean is $\mathfrak{sl}(2,\Bbb C)$ -- this might be an annoying notational distinction, but the difference in meaning is great enough to make this point. :DSorry, you can disregard my use of the symbol $A_1^{(1)}$. I was being a bit lax in it's use, but there's more structure to it as an algebra than $\mathfrak{sl}(2,\Bbb C)$.

Now, regarding your Abelian Cartan subalgebra post, keep this in mind. While a simple Lie algebra may be decomposed into subalgebras, it cannot be decomposed into nontrivial ideals. This is where the difference lies -- a subalgebra of an algebra need not be an ideal (however, every ideal is a subalgebra). The (root space) decomposition you wrote is incorrect, but I'd say you're on the right track. I can explain it further but it'll have to be a bit later.

This might have to be moved to the Algebra forum. As long as I have you captive anyway. (Wink)

Evidently my text isn't using the same fonts as you. \(\displaystyle \mathfrak{sl}(2,\mathbb{C})\) is news to me, but I get what you are talking about. Is this a general kind of thing? That we use an "ordinary" font to represent a group and that "torturous" font for a Lie algebra? The only fonts that look complicated like that in Physics are for tensors, and even then you only find them in the older books. I do understand there's a big difference between Lie algebras and Lie groups, which actually is why I'm not currently studying Lie groups: Lie algebras look like they're easier to work with than Lie groups, at least at an introductory level. Though I could easily be wrong...I'm still working through chapter 1. (And learning just how much the Physics people aren't telling me. Every other page or so I see something in the text that generalizes something my Quantum instructors never saw fit to mention.)

You have actually mentioned one of my questions. The text specifically mentions theorems and definitions for semi-simple Lie algebras (root systems for example), but uses the simple algebras \(\displaystyle \mathfrak{sl}(2,\mathbb{C})\) and \(\displaystyle \mathfrak{sl}(3,\mathbb{C})\) as examples. I'm lost on why the text is doing this.

-Dan

 

[Euge]

Evidently my text isn't using the same fonts as you. \(\displaystyle \mathfrak{sl}(2,\mathbb{C})\) is news to me, but I get what you are talking about. Is this a general kind of thing? That we use an "ordinary" font to represent a group and that "torturous" font for a Lie algebra?

Generally, a Lie group is represented by a capital letter in Roman font, and its associated Lie algebra is represented by the corresponding lower case letter in Fraktur font. For instance, if I let $G$ be a Lie group, then it's lie algebra is $\mathfrak{g}$. I would understand if you wrote $sl(2,\Bbb C)$ instead of $\mathfrak{sl}(2,\Bbb C)$, for the Lie algebra associated with the matrix group $SL(2,\Bbb C)$, but it's confusing to use $SL(2,\Bbb C)$ for both the group and its associated lie algebra.

You have actually mentioned one of my questions. The text specifically mentions theorems and definitions for semi-simple Lie algebras (root systems for example), but uses the simple algebras \(\displaystyle \mathfrak{sl}(2,\mathbb{C})\) and \(\displaystyle \mathfrak{sl}(3,\mathbb{C})\) as examples. I'm lost on why the text is doing this.

I can't comment much on the text since I don't have it (in fact, I don't even know which one you're using), but a semi-simple Lie algebra is one that can be decomposed into a direct sum of simple Lie algebras. The Dynkin diagrams for simple Lie algebras are connected, but for general semisimple Lie algebras, they are not necessarily connected: each component corresponds to a direct summand of the decomposition of the algebra.

 

[topsquark]

I can't comment much on the text since I don't have it (in fact, I don't even know which one you're using), but a semi-simple Lie algebra is one that can be decomposed into a direct sum of simple Lie algebras. The Dynkin diagrams for simple Lie algebras are connected, but for general semisimple Lie algebras, they are not necessarily connected: each component corresponds to a direct summand of the decomposition of the algebra.

I have two texts on Lie Algebras. The one I am using right now is "Affine Lie Algebras and Quantum Groups", Fuchs, Cambridge University Press, 1992. It's a Math text, but geared a bit toward Physics. I occasionally refer to the other text, a Math book, but the order of topics is much different than in the other so it's hard to compare.

Yes, the direct sum bit is confusing in the examples I'm working with. So far as I have been able to figure both \(\displaystyle \mathfrak{sl} (2, \mathbb{C})\) isn't a direct sum of anything and \(\displaystyle \mathfrak{sl} (3, \mathbb{C})\) may be: It obviously contains an \(\displaystyle \mathfrak{sl} (2, \mathbb{C})\) subalgebra, but I don't yet know if that splits apart as a direct sum with anything. My preliminary work says it doesn't but I need to check that more thoroughly. (Please don't give me an answer to that one...it's my topic for the day.)

Anyway the text doesn't have questions at the end of the chapter so my "homework" is to come up with the Dynkin diagram for the "A" algebras, take a look at the D's to see how they differ, and then finally (and hopefully) get a better idea of what \(\displaystyle E_6\) is about. It's kind of intense for me as algebras like \(\displaystyle \mathfrak{sl} (3, \mathbb{C})\) are rather unwieldy, but I figure it's worth the extra effort since a lot of this has a bearing on QFT. And hey, I just upgraded my TI-92. Let's see what the new one can do. :)

-Dan

 

[Euge]

Some of the homework questions you're assigning yourself are, in my opinion, difficult for the absolute beginner, but I'd be happy to hear that you've solved them. :D

 

[topsquark]

Well, when I'm self-teaching and I have material which is very relevant to my interests, such as how often Lie Algebras crop up in QFT, I get a bit masochistic. (Nerd)

For the record I just spent 3 hours tearing \(\displaystyle \mathfrak{sl}(3, \mathbb{C})\) apart and putting it back together. (Thanks to my new calculator!) I can't see any way to write it as a direct sum. Pretty much what I already knew, but it never hurts to check.

Here's a question. What makes a Lie algebra distinct from other Lie algebras? For example, I would have thought that any representation of the algebra would always have the same structure factors. But a change of basis changes the structure factors...We can write a semi-simple algebra \(\displaystyle \mathfrak{g}\) in the Cartan-Weyl basis and still call it \(\displaystyle \mathfrak{g}\). Perhaps it will come up later in the text but are there properties of an algebra that define the algebra (up to isomorphism, of course)? (Yes, the definition of \(\displaystyle \mathfrak{sl}(3, \mathbb{C} )\) has a defining representation of traceless 3x3 matrices, but I'm trying to look at properties of the algebra in general.)

Please bear in mind that I'm still slogging through Chapter 1 so my question might be a bit premature. Let me know if this is something I can just push under the rug until later.

-Dan

 

[topsquark]

Okay, you told me that I might have a misunderstanding about root systems, which I will take up with you later. For now I think the only problem I have left with the notation deals with the killing form. I am looking at \(\displaystyle \mathfrak{sl}(3, \mathbb{C})\). I will give you the specifics if you need it, but I think that I can be general about most of it.

\(\displaystyle \mathfrak{sl}(3, \mathbb{C}) = \{ H^1, H^2, E^{\pm \alpha}, E^{\pm \beta}, E^{\pm \theta} \}\) (in the Cartan-Weyl basis) has a maximal Abelian subalgebra \(\displaystyle \mathfrak{g}_0 = \{H^1,H^2 \}\). We can then define the root system as two vectors \(\displaystyle \gamma\) \(\displaystyle [H^1, x] = \gamma(H^1) \cdot x\) (where x runs over all the generators E) and similar for \(\displaystyle H^2\). The vectors \(\displaystyle \gamma (H^1)\) and \(\displaystyle \gamma (H^2)\) can be considered to span a "root space" \(\displaystyle \Phi\).

Here's the question. The text says that if we look at the \(\displaystyle \mathfrak{g}_0\) portion of the killing form we have the equation: \(\displaystyle \kappa ^{ij} = \frac{1}{I_{ad}} tr (ad_{H^i} \circ ad_{H^j} ) = \sum_{\gamma \in \Phi}\gamma (H^i) \gamma (H^j)\), where i, j = 1, 2. I can verify that the trace part of the formula, but I don't know what they mean by the RHS. How do we do the sum?

-Dan

 

[topsquark]

Ha! I figured out two things.

First, the text's notation is "bad" but not as bad as I had thought. It uses \(\displaystyle \alpha\) as both a vector and an index. But it is nothing worse than the \(\displaystyle i \epsilon_{ijk}\) notation where we have to untuit that the the symbol "i" is used both for the complex number and an index.

So \(\displaystyle \mathfrak{sl}(3, \mathbb{C})\) has 6 1 x 2 roots each corresponding to a specific generator. Specifically a root \(\displaystyle \alpha\) associated with each \(\displaystyle E^{\alpha}\). That pretty much clears up the roadblock.

The second thing is that the generators the text is using to define \(\displaystyle \mathfrak{sl}(3, \mathbb{C})\) is not, in fact, a Cartan-Weyl basis even though the notation looks almost exactly the same. Two of the generators get switched going from one basis to another.

I think I pretty much have this. I'm going to keep working on \(\displaystyle \mathfrak{sl}(3, \mathbb{C})\) and then see if I can't find another example to check things out. Possibly \(\displaystyle \mathfrak{so}(5, \mathbb{C})\) since it isn't all that large.

Thanks for all the help, Euge. I owe you big for this. (Bow)

-Dan

 

[Euge]

Here's the question. The text says that if we look at the \(\displaystyle \mathfrak{g}_0\) portion of the killing form we have the equation: \(\displaystyle \kappa ^{ij} = \frac{1}{I_{ad}} tr (ad_{H^i} \circ ad_{H^j} ) = \sum_{\gamma \in \Phi}\gamma (H^i) \gamma (H^j)\), where i, j = 1, 2. I can verify that the trace part of the formula, but I don't know what they mean by the RHS. How do we do the sum?

-Dan

Hi Dan,

Sorry for the late response. To evaluate $\sum\limits_{\gamma \in \Phi} \gamma(H^i)\gamma(H^j)$, suppose $\Phi = \{\gamma_1,\gamma_2,\ldots, \gamma_r\}$. Then

$$\sum_{\gamma\in \Phi} \gamma(H^i)\gamma(H^j) = \gamma_1(H^i)\gamma_1(H^j) + \gamma_2(H^j)\gamma_2(H^j) + \cdots + \gamma_r(H^i)\gamma_r(H^j).$$

As you can see, you compute each summand $\gamma_\ell(H^i)\gamma_\ell(H^j)$, $1 \le \ell \le r$, then add them all up to get the desired sum.