Talk:Lie algebra

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Borderline Start/B-class. Could do with some work on the structure of the article and some illustrations. Several sections could do with a bit more explanation. --Salix alba (talk) 07:33, 15 May 2007 (UTC)

Our definition of Lie algebra currently appears to be slightly wrong. If the base field has characteristic 2, then xy+yx=0 and x²=0 are not equivalent identities, and it's surely x²=0 that is required. I haven't attempted to fix the article, since there appears to be a new Wikipedia bug that corrupts HTML character entity references. --Zundark, 2002 Feb 7

It seems this bug only affects the preview, so I've modified the article now. --Zundark, 2002 Feb 7

1 Lie algebra whose bracket is identically zero(Example 1)
2 Clarification requested
3 Vector fields as operators
4 Category-theoretic definition
5 Killing vector field?
6 Killing vectors of Lie algebras
- 6.1 Lie algebras from Lie groups
7 dual algebra
8 connection forms in coordinates
9 Check my work?
10 New To Advanced Math
11 The link to Weyl's theorem
12 Structure and classification
13 Do the Lie algebras form an abelian category?
14 Infinite dimensions and correspondence with "Lie" groups

[edit] Lie algebra whose bracket is identically zero(Example 1)

i think it is completely fair to call a Lie algebra whose bracket is identically zero "uninteresting". it is a trivial Lie algebra.

It was right to change "uninteresting" to nothing or to "trivial", because "uninteresting" on its own, thrown into an article on algebra, sounds like a schoolboyish comment on how boring it is.

Now who is going to change the german version of this page?

Just to be clear, i think you are right, the phrase "very uninteresting" doesn't sound very encyclopedic. "trivial" is better. on the other hand, it doesn't sound schoolboyish to me, i have heard methematicians use this very language to describe things like 0-dimensional vector spaces, various trivial things. so I am pretty sure it was not vandalism, which is why i initially reverted your change. whether or not it is the most appropriate choice of wording, well, that is a different matter. Lethe

I understand. I didn't mean to say it was necessarily vandalism, just that it looked like it.

Is the Lie group for this trivial algebra the nonzero elments of the base field?Thanks,Rich 20:24, 31 October 2006 (UTC)
Or maybe it would be just the one element group {1} of the field? (Since there are probably lots of less 'brutal' Lie brackets than the triv one here that still make for commutative Lie algebras and so give an abelian lie groups-I bet it's one of those less brutal brackets that gives rize to the nonzero scalars as the Lie group.)Rich 21:02, 31 October 2006 (UTC)

[edit] Clarification requested

Could someone clarify the following definition:

[X, Y] f = (XY − YX) f for every function f on the manifold ?

I don't see what XY is. PJ.de.Bruin 18:26, 12 Jul 2004 (UTC)

A vector in differential geometry is a directional derivative evaluated at a point, which therefore acts on a smooth function on the manifold and returns a number X_pf. A vector field is a directional derivative of a function X(f) which is also a smooth function (just like the derivative of sin x is also a smooth function).
Therefore you can take a second directional derivative, with respect to a different direction Y, and get Y(X(f)). The second order differential operator Y X is not a vector field (vector fields must be first order differential operators by definition), but the combination XY−YX is, because the second order terms cancel out, due to equality of mixed partial derivatives.

In local components, if X=X^a∂_a and Y=Y^b∂_b (using the Einstein summation notation), then

$[X,Y]f=(XY-YX)f=X(Y(f))-Y(X(f))=X^a(\partial_aY^b)(\partial_bf)+X^aY^b\partial_a\partial_bf-Y^bX^a\partial_b\partial_af-Y^b(\partial_bX^a)(\partial_af)$ .

The middle two terms there will cancel, leaving only first order derivative terms of f.

So in short, the notation XYf means Y acts on the function, taking the directional derivative, and returning a smooth function, and then we let X act on the resulting smooth function. This is more clearly brought out when you use the notation X(Y(f)), however people often leave off the parentheses in this context, because they become cumbersome.

One could also take the f out of the equation, switch the dummy index in one of the terms, and get

$[X,Y]^b=X^a\partial_aY^b-Y^a\partial_aX^b$ ;

this notation would be common in, for example, a GR textbook.

In my opinion, explanations of these terms should not necessarily be included in an article on Lie algebra. Lie algebra is an application of linear algebra, and discussions of differential topology are somewhat out of context. It is interesting that the space of vector fields from differential topology is also a Lie algebra, so it is good that that is mentioned as an important example of a Lie algebra in this article, but further explanations of these terms should probably be in differential topology or vector field (although I don't see this explanation in either of those two articles. It should be added somewhere).

-Lethe 21:21, Jul 12, 2004 (UTC)

[edit] Vector fields as operators

[X, Y] f = (XY − YX) f for every function f on the manifold

(here we view vector fields as operators that turn functions on a manifold into other functions).

Could someone remind the naive reader (me) how this is done? I have a vague recollection: X f = the directional derivative of f in the direction of X, so that X is in effect a partial differential operator; then XY means the composition of partial differential operators. Is that right? Michael Hardy 23:24, 24 Sep 2004 (UTC)

yes, XY is the composition of the two derivative operators, making it a second order derivative, and therefore itself not a vector. only the difference XY-YX is a first order derivative operator -Lethe | Talk

also, i noticed you added the word "smooth" there. I added another instance; we usually require the vector fields to be smooth as well. however, strictly speaking, we only need f to be C² and the vector fields to be C¹. I'm not sure it matters enough to change it though, and most textbooks usually assume smoothness for simplicity. -Lethe | Talk

[edit] Category-theoretic definition

The category theoretic definition provided doesn't work over a field of characteristic 2. The correct definition is as follows:

A Lie algebra is an object A in the category of vector spaces together with a morphism $[\cdot,\cdot]:A\otimes A\rightarrow A$ such that

$[\cdot,\cdot]\circ \Delta=0$ where $\Delta:A\rightarrow A\otimes A$ is the diagonal morphism, and

$[\cdot,\cdot]\circ ([\cdot,\cdot]\otimes id)\circ(id+\sigma+\sigma^2)=0$ where σ is the cyclic permutation braiding $(id\otimes \tau_{A,A})\circ(\tau_{A,A}\otimes id)$ .

(It should probably also be added that the morphism $\tau_{A,A}:A\otimes A\rightarrow A\otimes A$ is the interchange morphism rather than leave that as assumed knowledge.)

I didn't want to edit the given definition because I don't want to get rid of the diagram, which is correct apart from the case of characteristic 2.

Could we put this category theoretic definition towards the end? To me this notation seems almost as bad as Bourbaki's notation for quantifiers. (Yeah I know it has a glorious tradition — Frege and all that) CSTAR 00:08, 30 Dec 2004 (UTC)

I think it also needs some indication of which id is which here. Kuratowski's Ghost 15:09, 7 Apr 2005 (UTC)

My impression is that whenever you write something like $f\otimes g:A\otimes B\rightarrow A\otimes B$ , the implicit meaning is essentially that f acts on things in A and g acts on things in B. So I don't share your confusion with this aspect of the notation, although I do agree with the great-grandparent in this thread that the notation is rediculously unclear.

[edit] Killing vector field?

Good god, this section is incomprehensible! You are calling a left invariant vector field a Killing vector field? This is going to generate an enormous amount of confusion with Killing forms. Also the explanation is ridiculusly complicated. True the left action of G on itself allows to transport the tangent space at to any tanget space on the group. (and we don't have to say without holonomy). Somebody with time please rewrite this!

And what's this conversational style? Let's say ...blah.CSTAR 15:33, 18 Jan 2005 (UTC)

Is this a mixup between LIVFs and Killing vectors? LIVF are already mentioned early in the article, so...? -Lethe | Talk 15:42, Jan 18, 2005 (UTC)

[edit] Killing vectors of Lie algebras

User:Phys, can you help me understand this recent addition of yours?

"for each element of the tangent space of G at the identity e, there naturally corresponds a Killing vector field over G generated by the regular representation of G upon itself (Take a differentiable parametrized path passing through the identity and take the derivative at the identity)."

firstly, as far as I am aware, Killing vectors depend on the existence of a metric. I know that the Killing form of a semisimple Lie group gives rise to a Riemann metric, but the non-semi-simple case? I didn't follow how the regular representation gives you a metric...? -Lethe | Talk 15:37, Jan 18, 2005 (UTC)

The new text follows here for reference. I have removed it from the article, until its validity can be established. Lethe | Talk 03:24, Jan 21, 2005 (UTC)

[edit] Lie algebras from Lie groups

Let's say we have a Lie group G. for each element of the tangent space of G at the identity e, there naturally corresponds a Killing vector field over G generated by the regular representation of G upon itself (Take a differentiable parametrized path passing through the identity and take the derivative at the identity). From differential geometry, we have the Lie bracket (see Lie derivative) between any two vector fields. It turns out the Lie bracket of the two Killing vector fields generated by any two elements of the tangent space at the identity is another Killing vector field generated by another element of the tangent space at the identity. It turns out this has the structure of a Lie algebra.

[edit] dual algebra

Is the vector space dual of a Lie algebra naturally a Lie algebra? And the tensor product of two Lie algebras? If so then Hom(g, g) would also be a Lie algebra if g is and the connection form could take values therein. --MarSch 11:48, 19 October 2005 (UTC)

No, the dual and tensor of Lie algebras are not natually Lie algebras. -Lethe | Talk 13:29, 19 October 2005 (UTC)

So what about the automorphisms of a Lie algebra? --MarSch 14:22, 19 October 2005 (UTC)

That's a group. If the Lie algebra is a finite-dimensional Lie algebra over R, then it's a closed subgroup G of GL and hence a Lie group. The Lie algebra of the group G is the Lie algebra of derivations of the original Lie algebra (Umm I'm giving a quick answer so the last statment might require some modification).--CSTAR 14:27, 19 October 2005 (UTC)

Okay. What I want to know is what Lie algebra a connection form for a G-bundle is valued in, since that article says it is Hom(F, F) for fiber F, which makes sense but perhaps the algebra product generates Aut(F)? Or perhaps a connection form is not a Lie algebra form?--MarSch 15:30, 19 October 2005 (UTC)

The connection form of a G-bundle is g-valued, where g is the Lie algebra of G. The part you mention is not for G-bundles though, it's for the associated vector bundle. A vector bundle can be associated with a GL(n)-bundle. The Lie algebra of GL(n) is gl(n). gl(n) is another way of writing Hom(F,F), where the fibre F is an n-dimensional vector space, not a Lie group. Incidently, I guess Hom(g,g) is a Lie algebra. Simply define [S,T](X) to be ST(X)-TS(X). I don't think that construction is too useful though , and it's still true that the dual and tensor are not Lie algebras, and it's not what's meant in the connection form article, where F is a vector space, not a Lie algebra. -Lethe | Talk 19:58, 19 October 2005 (UTC)

I suspect the question is based on a misunderstanding, and so has a flawed answer. Lets say we have a connnection on a fiber bundle P whose fiber is a Lie group G, over a base manifold M. Now pick a vector field X on M, and pick a point p in P. Then the connection A for vector field X at point p will just be an element of the lie algebra g of G. There is no dual. The "form" part of the "connection form" refers only to the base manifold, and not to the lie group. Does that answer the qeustion? linas 23:34, 19 October 2005 (UTC)

To answer his question, you should probably address how Hom(F,F) is supposed to be a Lie algebra. I think that's why he's bringing up duals and tensors. He's thinking of the isomorphism Hom(F,F) = F*xF. His real problem is that he is confusing the vector space F on which a rep of g acts with the Lie algebra g itself. It becomes clear once you distinguish between a principle G-bundle and an associated vector bundle. -Lethe | Talk 00:23, 20 October 2005 (UTC)

Here's a way to think about it: suppose M is two-dimensional, and suppose G is 9-dimensional. Then the connection form has one index that runs over 1,2 and two more indecies that run over 1..9. The two indecies running over 1..9 are a matrix, and that matrix is an element of the Lie algebra. The index that runs over 1,2 is the "form" part of the connection form, its the part that is to be contracted with a vector field X on M. (the tangent space to G is g, and one does not need to use forms to move around on G, it is enough to use the group action). linas 23:47, 19 October 2005 (UTC)

Of course, you only get a matrix once you choose a representation (and thereby move to the associated vector bundle). And then it need not share the dimensionality of the Lie group. For example, an associated SU(2) vector bundle using the fundamental rep will be a 2x2 matrix valued form, even though SU(2) is 3-dimensional. -Lethe | Talk 00:23, 20 October 2005 (UTC)

Respons to the 2-dim M, 9-dim G example. This means that g is also 9-dim, so why would the g-valued connection form have two indices running from 1 to 9? Perhaps we can choose an example to work out and add to connection form which does a very bad job of making things clear which is why I added the expert tag. --MarSch 10:21, 20 October 2005 (UTC)

Actually remembering that A is really an array of LAv-forms I can understand this, but the example would still be appreciated. --MarSch 10:32, 20 October 2005 (UTC)

And the representation of the Lie algebra is the isomorphism g ~=~ Aut(V) = Hom(V, V), which explains another way of turning one index into two. So why do christoffel symbols only have 3 indices. One form-index and 2 from Aut(TM) I guess, since they all run from 1 to 4 (0 to 3) or is the form-index suppressed? What happened to the array index? --MarSch 10:50, 20 October 2005 (UTC)

Aut(V) is the group of isomorphisms, Hom(V,V) is the algebra of homomorphisms. These things are never isomorphic. (For example, End(V) always contains 0, Aut(V) never does). Elements of Aut(V) and End(V) will both have 2 indices when written in coordinates, so changing from one to the other won't get rid of indices. The Christoffel symbols have 3 indices like all connection forms. 2 indices indices for g (so for Christoffel), and one for the one-form. -Lethe | Talk 20:19, 20 October 2005 (UTC)

As I found Lethe's remarks below confusing, my best recommendation is to cite a book that is quite readable: Jurgen Jost "Riemannian Geoemtry and Applicatons" (do not confuse with his similarly named book "Compact Riemann Surfaces"). I've been reading this recently, and (to me at least) its the clearest, best-constructed and most approachable book on the topic that I've seen. (And yes, it seems that gauge theory is now a part of reimannian geometry these days, so he explains Yang-Mills at the same time, without any big to-do.). linas 23:37, 20 October 2005 (UTC)

Are you talking about ISBN 3-540-42627-2, Riemannian Geometry and Geometric Analysis? I couldn't find another Riemannian geometry book by Jost on Amazon, so I'll assume that's what you're talking about. Anyway, looking through the table of contents of the book online, I see that he only treats vector bundles, makes no mention of principle bundles or more general bundles. I also don't see Yang-Mills in the table of contents, so maybe I've got the wrong book. Anyway, I'm disappointed that you didn't find my list elucidating. I thought by expressing everything in local coordinates, I was being as down-to-earth and concrete as one can get. -Lethe | Talk 00:15, 21 October 2005 (UTC)

Sorry, I didn't mean to dis your work; it looks good but my eyes glazed over. Yes, that is the book. The chapter titles are misleading, the book is a bit deeper than the titles would indicate; but its not a book about fiber bundles per-se. Mostly I just thought the treatment was refreshing in that it used the general language of bundles, introducing F and A first (even on vector bundles) and then showing how the riemann curvature R is related to F when the fiber is a vector bundle. When I first learned this stuff, the textbooks didn't do this, so it was very hard to figure out how these concepts were similar/the same, as superficially, they seemed to be so different (i.e. using F for lie groups, using R for geometry). linas 10:34, 21 October 2005 (UTC)

I know what you mean. I just figured out that they are related too. Unfortunately GR has a different Lagrangian than Yang-Mills. Do you happen to know what happens when you choose the same Yang-Mills Lagrangian for GR? Is it in accord with experiment? Is there an article yet? --MarSch 11:35, 21 October 2005 (UTC)

I think the YM langrangian and the EH lagrangian are as close to being the same as they can be, or at least as they should be. YM is tr(F ^ *F) , while EH is tr(R ^ *(θ^θ). The difference is in that second term, θ^θ (here θ is a bundle morphism from a Lorentz bundle to our Riemannian spacetime aka vielbein). We can include it here because it makes the action into a guage invariant scalar. We do not have that corresponding structure available in YM theory (which is one way that makes YM theory different from GR in a basic way). Including that term has an important effect: it makes the connection not an independent dynamical variable. If you used R^*R as your action, you'd have YM theory on your connection, which would mean that your connection is no longer the Levi-Civita connection, and your vector bosons would be spin-3. I believe its known that there cannot be interacting theories of spin 3 massless particles. The EH action is known as the only action in the curvature which satisfies some physical requirements like the equivalence principle and vanishing of torsion in the vacuum. But actions higher order in the curvature (which still do not have the connection as independent) have been considered in the literature. String theory, for example, predicts higher order terms in the curvature, so models based on string theory generally have more general actions. -Lethe | Talk 17:29, 21 October 2005 (UTC)

Gaack. For better or worse, there is now a brand-new article called Gauge covariant derivative that attempts some sort of low-brow review of this topic. Its maybe an over-simplified presnetation but thought I'd mention it. linas 23:50, 20 October 2005 (UTC)

I think reading this just caused me to lose my mind. —Preceding unsigned comment added by 68.144.103.61 (talk) 03:56, 15 April 2008 (UTC)

[edit] connection forms in coordinates

I'll use A for the connection form components of a principle bundle in local coordinates, and Γ for the covariant derivative components of a vector bundle.

[edit] principle bundle

on a principle G-bundle, given a local trivialisation, a connection form in local coordinates can be written as a g-valued one form on the base space. Therefore, it will look like

$A=T^r\otimes A_{r\mu}dx^\mu$

where {T^r} is a basis for g, and {x^μ} are local coordinates for the base space. So if the base space is m-dimensional and the bundle group is n-dimensional, then μ ranges to m and r ranges to n. Here A can be any functions (locally, at least).

[edit] vector bundle

on a vector bundle, given a local trivialisation, a connection is described by a exterior covariant derivative, which is an element of End(V)⊗Ω(M) which satisfies the Leibniz law, so in local coordinates, you have

$Dv=(\partial_\mu v^b+v^a\Gamma_{\mu a}{}^b)e_b\otimes dx^\mu$

where v is a section of the vector bundle. So if the vector space is k-dimensional, then a and b range up to k (these are End(V) indices). Here Γ can be any functions (at least locally).

[edit] associated bundle

if the vector bundle is the ρ-associated bundle to the principle bundle, then ρ takes elements of g to End(V) (End(V) is the same thing as Hom(V,V), but distinct from Aut(V). End(V) is the Lie algebra of Aut(V). if ρ is a faithful rep, it will actually map G to a subgroup of Aut(V) and g to a Lie subalgebra of End(V). So ρ(T) is actually an element of the image of the Lie algebra of g in End(V)). Thus we may write

$\Gamma_{\mu a}{}^b=\rho(T^r)^b{}_aA_{r\mu}\;$

and so you can see how the covariant derivative on a vector bundle associated to a principle bundle explicitly depends on the connection form of the principle bundle. Here, a and b are k-dimensional ρ rep of g indices.

Lethe | Talk 22:17, 20 October 2005 (UTC)

Thanks Lethe, that is pretty clear. Could you also explain why A can represent the connection? --MarSch 10:45, 21 October 2005 (UTC)

Well, the connection form on the total manifold determines a connection by decreeing that the kernel of the form are "horizontal" spaces. A path is parallel transported if its tangent vector remains horizontal. The local connection form is just the pullback of the connection form by a local trivialisation. Most of my textbooks define a connection to be a splitting of the tangent space of the total space into horizontal and vertical components, so once you see how you can use a one form to split the tangent space into parts, you see why it can represent a connection. Some textbooks simply define the connection one form to be the connection, in which case it is called the Ehresmann connection. -Lethe | Talk 17:02, 21 October 2005 (UTC)

It does not seem like the connection form A depends on the local trivialization. This probably has to do with the isomorphism between the tangent bundle of a Lie group and the product of that Lie group with its Lie algebra. See also the discussion on this at http://en.wikipedia.org/wiki/Talk:Gauge_theory#Frame-dependent_claims --MarSch 14:41, 24 May 2006 (UTC)

[edit] Check my work?

I just added a section called "relation to Lie groups". I was unsure about some of the points, I only thought they were true. Irresponsibly, I added them anyway. I welcome comments and corrections.

the functor is full, faithful, and exact. Does the fact that the canonical projection SU(2) --> SO(3) goes to the identity prevent the functor from being faithful? I'm not sure. I'm positive it's full, but as for exactness, I feel like it should hold, but can't convince myself why (or why not).
The connected simply connected Lie group associated to a Lie algbra is isomorphic to the group of units of the completion of the universal enveloping algebra. I know that it can be embedded in this completion, but I don't recall hearing that it is the group of units. I just thought it ought to be.
Umm... I guess that's it. The rest is OK (I hope).

thanks -13:29, 30 October 2005 (UTC)

Re: the completion of the universal enveloping algebra. I'm at the limits of what I know here (I'm travelling so I'm without my favorite reference which is Hochschild's book) but you need to specify a ~~topology~~(correction: uniform structure --CSTAR) for this to make sense. The enveloping algebra is a graded algebra so has an ultrametric, but this obviously won't do since ultrametric topologies are totally disconnected. The Enveloping algebra is an algebra of differential operators so maybe you can assign a topology in that way. But in any case, somebody more knowledgeable better comment on your completion remarks. Maybe Serre's ancient notes deal with this. --CSTAR 16:19, 30 October 2005 (UTC)

I'm not familiar with Hochschild, but I think I read about this in Barut. Which I don't have access to presently. Anyway, now I'm thinking this can't be right: if x is a unit, so is 2x. In general, they won't both be in the Lie group. I think there ought to be a contruction where you arrive at the Lie group from the enveloping algebra, but until I can find something in the literature, I should probably remove the remark. Anyway, regarding a topology for the enveloping algebra, can't we somehow get one from the Lie algebra? Like uh... well I don't know how. Nevermind. -Lethe | Talk 16:41, 30 October 2005 (UTC)

I'm interested in learning more about complete reducibility.. Can anyone help me find a completely reducible gl_n module? This question is possibly lame..

[edit] New To Advanced Math

Hi; I'm trying desperately to understand many of these advanced principals of mathematics, such as Lie algebras, but no matter how many times I review the material, it doesn't sink in. Could someone please provide examples, problems to solve (with their solutions) and/or ways to visualize this? beno 26 Jan 2006

You posted the same comments at Talk:Lattice (group), and received a reply there. I can only suggest that whenever you find a term that you don't know, click on the link until you reach an article that you do understand. Then work forwards again, and backwards, as necessary. Google can help as well. linas 04:18, 1 February 2006 (UTC)

This guy posted the same exact message (with the corresponding article title swapped in) at about 20 math pages. I've replied to 2 or 3 of them, as have a few others. What's the deal with this guy? -lethe ^{talk +} 05:25, 1 February 2006 (UTC)

[edit] The link to Weyl's theorem

The link to Weyl's theorem currently leads to a disambiguation page. Someone with more knowledge of the subject should correct the link to lead to a specific article.--Bill 19:56, 3 February 2006 (UTC)

I've always known this as the unitary trick, so I've changed the link accordingly. -lethe ^{talk +} 20:52, 3 February 2006 (UTC)

[edit] Structure and classification

I've corrected multiple inaccuracies in that section (and renamed it). Since Lie theory part of Wikipedia has developed quite a bit in recent months, fairly large clean-up of this article may be in order (there are also internal redundancies). In particular, I can envisage moving out the structure theory to a separate article, with only a summary on this page. Arcfrk 05:58, 15 May 2007 (UTC)

[edit] Do the Lie algebras form an abelian category?

No.

In fact, the morphisms between two Lie algebras don't form a abelian group, so the category of Lie algebras is not even an additive category. —Preceding unsigned comment added by 59.66.121.105 (talk) 15:03, July 9, 2007

But modules over a fixed Lie algebra do form an abelian category, in analogy with the associative case (Rings is non-abelian; R-mod for a fixed R is abelian). Arcfrk 16:30, 9 July 2007 (UTC)

It is stated every Lie Alegra has a representation via matrices which I guess correspond to multiplications of the elements of the matices but the lie group corresponding to the algebra is not unique as previously stated so how is this possible.If this can be simply explained . Is it a gobal versus a local thing and is a lie group only a local correspondence

[edit] Infinite dimensions and correspondence with "Lie" groups

I'd like a reference to the statement "some infinite-dimensional Lie algebras are not the Lie algebra of any group". I guess this holds for a specific definition of an infinite-dimensional Lie group, otherwise it would be quite remarkable to me. 147.122.53.27 (talk) 18:32, 12 May 2008 (UTC)