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

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?

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.

• 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? – Siraj R Khan May 19 '13 at 21:22
• 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. – twistor59 May 20 '13 at 6:24

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 $$\frac{d}{dt}\Big|_{t=0}\mathrm{conj}_g(e^{tX}) =\mathrm{Ad}_g X$$

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.

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 $\mathrm {Ad}$
• 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 $\mathrm{conj}_g:G\to G$? – Christoph May 19 '13 at 20:15
• ah... so I was missing something obvious, in particular that for matrix groups, $\mathrm{conj}_g$ is a linear map and thus $\frac{d}{dt}\big|_{t=0}\mathrm{conj}_g(\exp tX)=\mathrm{conj}_g(\frac{d}{dt}\big|_{t=0}\exp tX)=\mathrm{conj}_g(X)$; anyway, +1 – Christoph May 19 '13 at 21:23