Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New features!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

123 submissions , 104 unreviewed
3,547 questions , 1,198 unanswered
4,552 answers , 19,366 comments
1,470 users with positive rep
411 active unimported users
More ...

Do generators belong to the Lie group or the Lie algebra?

+ 5 like - 0 dislike
178 views

In Physics papers, would it be correct to say that when there is mention of generators, they really mean the generators of the Lie algebra rather than generators of the Lie group? For example I've seen sources that say that the $SU(N)$ group has $N^2-1$ generators, but actually these are generators for the Lie algebra aren't they?

Is this also true for representations? When we say a field is in the adjoint rep, does this typically mean the adjoint rep of the algebra rather than of the gauge group?


This post imported from StackExchange Physics at 2014-03-22 17:13 (UCT), posted by SE-user Siraj R Khan

asked May 19, 2013 in Mathematics by Siraj R Khan (105 points) [ revision history ]
retagged Mar 25, 2014 by dimension10

3 Answers

+ 6 like - 0 dislike

User twistor59 has addressed the part regarding the "generator" terminology, but let me give a bit more detail on the second part of the question. I'm going to restrict the discussion to matrix Lie groups for simplicity.

Some background.

Given a Lie group $G$ with Lie algebra $\mathfrak g$, there exist two mappings $\mathrm{Ad}$ and $\mathrm{ad}$, both are called "adjoint." In particular for all $g\in G$ and for all $X,Y\in\mathfrak g$, we define $\mathrm {Ad}_g:\mathfrak g\to \mathfrak g$ and $\mathrm{ad}_X$ by $$ \mathrm{Ad}_g(X) = gX g^{-1}, \qquad \mathrm{ad}_X(Y) = [X,Y] $$ The mapping $\mathrm{Ad}$ which takes an element $g\in G$ and maps it to $\mathrm{Ad}_g$ is a representation of $G$ acting on $\mathfrak g$, while the mapping $\mathrm{ad}$ which takes an element $X\in \mathfrak g$ and maps it to $\mathrm{ad}_X$ is a representation of $\mathfrak g$ acting on itself.

In other words, $\mathrm{Ad}$ is a Lie group representation while $\mathrm{ad}$ is a Lie algebra representation, but they both act on the Lie algebra which is a vector space.

Aside.

In response to user Christoph's comment below. Note that if we define the conjugation operation $\mathrm{conj}$ by $$ \mathrm{conj}_g(h) = g h g^{-1} $$ Then for matrix Lie groups (which I initially stated I was restricting the discussion to for simplicity) we have $$ \frac{d}{dt}\Big|_{t=0}\mathrm{conj}_g(e^{tX}) =\mathrm{Ad}_g X $$

Addressing the question.

Having said all of this, in my experience (in high energy theory), physicists usually are referring to $\mathrm{ad}$, the Lie algebra representation. In fact, you'll often see it written in physics texts that

generators $T_a$ of the Lie algebra furnish the adjoint representation provided $(T_a)_b^{\phantom bc} = f_{ab}^{\phantom{ab}c}$.

where the $f$'s are the structure constants of the Lie algebra with respect to the basis $T_a$; $$ [T_a,T_b] = f_{ab}^{\phantom{ab}c} T_c $$ But notice that $$ \mathrm{ad}_{T_a}(T_b) = [T_a,T_b] = f_{ab}^{\phantom{ab}c} T_c $$ which shows that the matrix representations of the generators in the Lie algebra representation $\mathrm{ad}$ precisely have entries given by the structure constants.

Addendum (May 22, 2013).

Let a Lie-algebra valued field $\phi$ on a manifold $M$ be given. If the field transforms under the representation $\mathrm{Ad}$ (which is a representation of the group acting on the algebra) then we have $$ \phi(x)\to \mathrm{Ad}_g(\phi(x)) = g\phi(x) g^{-1} $$ But recall that (see here) $\mathrm{Ad}$ is related to $\mathrm{ad}$ (a representation on the algebra acting on itself) as follows: Write an element of the Lie group as $g=e^X$ for some $X$ in the algebra (here we assume that $G$ is connected) then $$ \mathrm{Ad}_g(\phi(x)) = e^{\mathrm{ad}_X}\phi(x) = \phi(x) + \mathrm{ad}_X(\phi(x)) +\mathcal O(X^2) $$ so that the corresponding "infinitesimal" transformation law is $$ \delta\phi(x) = \mathrm{ad}_X(\phi(x)) $$ So when talking about a field transforming under the adjoint representation, $\mathrm{Ad}$ and $\mathrm{ad}$ in some sense have the same content; $\mathrm{ad}$ is the "infinitesimal" version of $\mathrm {Ad}$

This post imported from StackExchange Physics at 2014-03-22 17:13 (UCT), posted by SE-user joshphysics
answered May 19, 2013 by joshphysics (810 points) [ no revision ]
Most voted comments show all comments
shouldn't $\mathrm{Ad}$ be the differential of conjugation instead of conjugation itself, ie $\mathrm{Ad}_g=\mathrm{T}_e(\mathrm{conj}_g):\mathrm{T}_eG\to \mathrm{T}_eG$, whereas $\mathrm{conj}_g:G\to G$?

This post imported from StackExchange Physics at 2014-03-22 17:13 (UCT), posted by SE-user Christoph
@Christoph Yeah actually I don't think it's obvious at all. When you asked that I got pretty confused for a moment; thanks for pointing that out.

This post imported from StackExchange Physics at 2014-03-22 17:13 (UCT), posted by SE-user joshphysics
@joshphysics Thank you josh, I really appreciate it. So when we say 'a field, $\phi$, is in the adjoint rep of SU(2)' (as an arbitrary example), does this mean that matrices belonging to the adjoint rep of the Lie algebra (ad) are the matrices that matrix multiply the field $\phi$?

This post imported from StackExchange Physics at 2014-03-22 17:13 (UCT), posted by SE-user Siraj R Khan
Incidentally, I was torn about who to click for the accepted answer. Both really helped me out but twistor59 got there first.

This post imported from StackExchange Physics at 2014-03-22 17:13 (UCT), posted by SE-user Siraj R Khan
@SirajRKhan No prob. Yeah that's right.

This post imported from StackExchange Physics at 2014-03-22 17:13 (UCT), posted by SE-user joshphysics
Most recent comments show all comments
@SirajRKhan I added an addendum that might help in this regard. And yeah, then physicists refer to generators in the context of Lie groups, they usually mean elements of a basis for the Lie algebra of the group.

This post imported from StackExchange Physics at 2014-03-22 17:13 (UCT), posted by SE-user joshphysics
Ah I see. That helps a lot, thanks!

This post imported from StackExchange Physics at 2014-03-22 17:13 (UCT), posted by SE-user Siraj R Khan
+ 4 like - 0 dislike

If you have a basis for the Lie algebra, you can talk of these basis vectors as being "generators for the Lie group". This is true in the sense that, by using the exponential map on linear combinations of them, you generate (at least locally) a copy of the Lie group. So they're sort of "primitive infinitesimal elements" that you can use to build the local structure of the Lie group from.

Re your second point, yes, fields in gauge theories are generally Lie algebra-valued entities.

This post imported from StackExchange Physics at 2014-03-22 17:13 (UCT), posted by SE-user twistor59
answered May 19, 2013 by twistor59 (2,490 points) [ no revision ]
Thank you very much for the speedy response to my question. So I guess that, in the strictest sense, the Pauli matrices aren't generators of the SU(2) group (they don't combine via the group action to generate the group). However, as you say, the SU(2) group can be obtained from them via the exponential of their linear combinations - so we call them generators. Technically, they are the generators of su(2) (the Lie algebra). Do you think this is a good way to view it?

This post imported from StackExchange Physics at 2014-03-22 17:13 (UCT), posted by SE-user Siraj R Khan
I guess they "generate" the Lie algebra in the sense that any basis "generates" the vector space it spans. You're right that they don't combine via the group action to generate the group, they combine via the exponential map to generate it.

This post imported from StackExchange Physics at 2014-03-22 17:13 (UCT), posted by SE-user twistor59
+ 1 like - 0 dislike

If $G$ is a simply connected Lie group with associated Lie algebra $g$, any basis of $g$ is referred to as a (minimal) set of generators of both $G$ and $g$. Indeed, the elements of $g$ are generated by taking linear combinations of generators, while the elements of $G$ are generated by taking products of exponentials $e^{\alpha A_i}$ with real or complex $\alpha$ and a generator $A_i$. In the compact case, the elements $G$ are also generated by taking all exponentials $e^{\sum_i \alpha_i A_i}$.

The number of generators of the group or Lie algebra is the dimension of the Lie algebra.

A field $\phi$ transforms under a symmetry group always in a particular group representation. This means in particular that at every point $x$, $\phi(x)$ belongs to the vector space on which the particular representation is represented. For a field in the adjoint representation, this vector space is the Lie algbra, but the adjoint action is still the group action.

answered Mar 31, 2014 by Arnold Neumaier (11,395 points) [ no revision ]

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOverfl$\varnothing$w
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...