Talk:Lie algebra

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

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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.

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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

<math>[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)<math> .

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

<math>[X,Y]^b=X^a\partial_aY^b-Y^a\partial_aX^b<math> ;

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)

Contents

1 Vector fields as operators 2 Category-theoretic definition 3 Killing vector field? 4 Killing vectors of Lie algebras 4.1 Lie algebras from Lie groups

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

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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 <math>[\cdot,\cdot]:A\otimes A\rightarrow A<math> such that

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

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

(It should probably also be added that the morphism <math>\tau_{A,A}:A\otimes A\rightarrow A\otimes A<math> 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 <math>f\otimes g:A\otimes B\rightarrow A\otimes B<math>, 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.

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)

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)

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.