# Can anyone help me derive and/or define the infinitesimal generators of a Lie group?

I'm doing a project involving Lie groups in Physics, and a part of the project involves generators. I initially used Robert Gilmore's "Lie Groups, Lie Algebras, and some of their Applications" which gives a very detailed account ending with the definition

$$X_\mu(x')=-\frac{\partial}{\partial q^\mu}(f^j(q^\mu,x'(p))) \Big|_{q=0}\frac{\partial}{\partial x'^j}$$

where $$X$$ is the infinitesimal generator of a Lie group, with $$q$$ being some dummy variable, with $$X$$ going in

$$\delta F=\delta^\mu X_\mu(x')F^S(x'(p))$$

where $$F$$ is a transformation in the geometric space. However, according to my thesis supervisor, my explanation inspired by this is not completely correct (for example, it treats a Lie group as its own geometric space, even though the geometric space is a vector space, whereas Lie groups cannot be assumed to be vector spaces).

I also tried to instead explain things in terms of Killing vectors, but he noted that Killing vectors are representations of generators, rather than generators themselves, and therefore that this is not entirely correct.

Can anyone help me give a better account of what exactly the infinitesimal generators of Lie groups are, and perhaps also direct me to a text that explains this more clearly than Robert Gilmore's textbook?

• Hello! I have edited your question using MathJax (LaTeX) math typesetting. For future questions, you can refer to MathJax basic tutorial and quick reference. Thanks! Jul 1, 2021 at 16:42
• Would Mathematics be a better home for this question? Jul 1, 2021 at 17:13
• Qmechanic you might be right. It's a Mathematical Physics project so I wasn't sure whether to put the question here or there, so in the end I put it in both. Jul 2, 2021 at 12:47

Infinitesimal generators are associated with an action of a Lie group $$G$$ on a manifold $$M$$, and not to the Lie group itself (so it makes more sense to speak of infinitesimal generators of an action). They are a map $$a : \mathfrak{g} \to \mathfrak{X}(M)$$ (where $$\mathfrak{g}$$ is the Lie algebra of $$G$$ and $$\mathfrak{X}(M)$$ are vector fields on $$M$$) usually defined by

$$a(\xi)_p = \left. \frac{d}{dt} \right|_{t=0} \exp(-t\xi)p$$

The minus sign ensures that $$a$$ is a morphism of Lie algebras, i.e. $$a([\xi, \eta]) = [a(\xi), a(\eta)]$$.

As for a reference, I recommend looking at chapter 2 in these notes, specifically pages 57-58.

• This might just be a difference in language, but I was under the impression that the "infinitesimal generators of a Lie group" were simply the elements of its Lie algebra (or a basis thereof), because with the exponential map the elements of the Lie algebra generate elements of the Lie group. This is a concept that makes perfect sense without requiring the Lie group to act on some other object.
– HTNW
Jul 1, 2021 at 17:53
• Indeed I wasn't familiar with that possibility of terminology (I believe it's not used by mathematicians). If in that case the question is how to define the Lie algebra of a Lie group, it's even simpler: it's the tangent space at the identity element (which can be identified with left (or right) invariant vector fields on $G$). The notes I linked also explain this. Jul 1, 2021 at 19:52
• Though it should be noted that, especially in quantum mechanical contexts, physicists generally want the complexified Lie algebra; so take what I said above and tensor it with $\mathbb{C}$. Jul 1, 2021 at 19:55
• @Pedro Not necessarily. There are nice and multiple applications of real forms such as so(3,1), e(n), and sp(2n,R) where unitarity is essential Jul 1, 2021 at 21:52
• Thanks Pedro and HTNW both, for your help. Gimore's book discusses generators by talking about the action of a Lie group on a geometric space, almost exactly what you write here, Pedro (although what you've written makes it a lot clearer than Gilmore's book), but HTNW, I think you're also right that you can talk about generators without needing to talk about the action of the Lie group on a geometric space. Am I right in thinking that X(M) here refers to killing vectors in the tangent space of M, or literally to vectors in M? And could you clarify what you mean by the greek letter xi, Pedro? Jul 2, 2021 at 12:54

Lie groups are differentiable manifolds, and thus have a notion of tangent (vector) spaces. As a group, there is also a special point on the manifold that is singled out as the identity. The Lie algebra of a Lie group is the tangent space to the identity. While the group is not necessarily a vector space, the algebra is. The infinitesimal generators of the Lie group are simply the elements of this vector space. In particular, we're usually interested in finding a convenient, finite basis of generators. The exponential map then extends any vector in the Lie algebra along the manifold to produce an element of the group itself.

Note that this definition makes no reference whatsoever to the action of the Lie group on any objects other than itself. This was your supervisor's objection to the explanation of the generators in terms of Killing vector fields: those arise by extending the action of the Lie group on a manifold to the generators, but they aren't the generators themselves.

Let's try to use our definition on the Lie group $$SO(3),$$ following this Math.SE answer. We choose to present the elements of this group as 3-by-3 rotation matrices $$A\in SO(3)\subset\mathbb{R}^{3\times3}.$$ These matrices are defined by the condition $$A^T\!A=I$$ (and $$\det A=1$$ for the "S"). Now consider a (smooth) curve $$A(t):\mathbb{R}\to SO(3)$$ that passes through the identity as $$A(0)=I.$$ Then for $$t$$ near 0, we have the approximation $$A(t)\approx I+t\frac{dA(0)}{dt},$$ where the derivative $$G=\frac{dA(0)}{dt}$$ belongs to the Lie algebra $$\mathfrak{so}(3).$$

Plug $$A=I+tG$$ into the condition $$A^T\!A=I$$ and you find $$tG+tG^T+t^2G^T\!G=0.$$ Cut off the term with the square of the infinitesimal $$t$$ and you're left with an explicit presentation of $$\mathfrak{so}(3)$$: $$G=-G^T.$$ That is, the infinitesimal generators of $$SO(3)$$ are given by the 3-dimensional space of antisymmetric 3x3 matrices, isomorphic to just $$\mathbb{R}^3.$$ As a language note, "generators" can either refer to all of the elements of the Lie algebra, or just to those of some preferred basis, but I don't think anyone really cares too much about it.

The way that the generators $$G\in\mathfrak{so}(3)$$ actually do the "generating" is through the map $$\exp:\mathfrak{so}(3)\to SO(3).$$ I think here is a good point to note that "the rotation group" is not fundamentally related to rotation matrices; there are other presentations of the elements of $$SO(3)$$ and they lead to other presentations of $$\mathfrak{so}(3)$$ and the map $$\exp$$ that relates them. In this case we get lucky and $$\exp$$ is literally the matrix exponential, but e.g. in the axis-angle presentation, $$\exp$$ is basically the identity.

Without context, I'm not sure what exactly Gilmore is trying to do, but I can guess that it's computing the action of some generator on some other object. In a simpler setting, consider that (with the rotation matrix presentation) $$SO(3)$$ acts on vectors $$\mathbb{R}^3$$ with the map $$\cdot:SO(3)\times\mathbb{R}^3\to\mathbb{R}^3.$$ Take the derivative with respect to the first argument (and evaluated at the identity) and you get that $$\cdot:\mathfrak{so}(3)\times\mathbb{R}^3\to\mathbb{R}^3$$ tells you the direction a vector will be (infinitesimally) moved under a (infinitesimal) rotation. If you consider that (as a differentiable manifold) a Lie group is endowed with coordinates $$q^\mu$$, then you can describe the action of the group on some object as a function $$f(q^\mu,v)$$ thereof. Then the derivative $$X_\mu=\left.\frac{d}{dq^\mu}f(q^\mu,v)\right|_{q=0}$$ extends that action to the infinitesimal generators of the group, which live in the Lie algebra and themselves have coordinates, so that $$X_\mu\delta^\mu$$ gives the infinitesimal transformation of the object $$v$$ (whatever it is) by whatever element of the Lie algebra has coordinates $$\delta^\mu.$$ Note that this has strayed far from actually defining the generators. We are now in the weeds of figuring out how to apply them to things, which is a very different thing.

• HTNW thanks a lot, this really helps to make things a lot clearer. I think the Gilmore book tried to talk in terms of Lie group action on a manifold, but got a bit confusing in terms of distinguishing generators in the tangent space of the Lie group from Killing vectors in the tangent space of the other manifold Jul 2, 2021 at 17:13