My doubt is this:

I saw in a paper that the Lie Algebra is the relation between the commutator of the generators and the generators multiplied by structure constants.


Does these generators belong to the group that they "generate"

This is because $SU(2)$ for example is generated by Pauli's matrices but these have a determinant -1. I thought that $SU(2)$ is the group of unitary 2 by 2 matrices with determinant 1. So, my guess is that the ganerator not necessarily belongs to the group itself, am I wrong? Are these generators unique?


1 Answer 1


You're confusing Lie groups with Lie algebras; they're closely related but not the same thing. For example, the $S_i$ are generators of a Lie algebra, and therefore elements of a Lie algebra, but they aren't generators or elements of a Lie group. The relationship is that the algebra expresses the composition of group elements that are infinitesimally close to the identity. You get the group (or at least the part connected to the identity) back by exponentiating the algebra.

The first clue that the $S_i$ aren't group generators or group elements is that in the commutator they are being multiplied together, then one of the products is being multiplied by -1, and then the two products are being added. Groups have only one operation: composition of group elements (often called "multiplication", especially when the group can be represented as matrices). By contrast, algebras have three operations: vector addition (which produces a vector), multiplication of a vector by a scalar (which produces a vector), and vector multiplication (which produces a vector). They are basically vector spaces with a vector multiplication that obeys the distributive law.

In this case the $S_i$ are the basis vectors of the Lie algebra, even though they happen to be matrices. They belong to the Lie algebra su(2) that they generate. They can be exponentiated to form elements of the Lie group SU(2).

The generators are not unique. They are just an arbitrary basis for the vector space of the algebra. You can use other linear combinations of them as a basis as long as the combinations are all independent vectors.

  • $\begingroup$ Thanks a lot, you answer really clarify my ideas but I have a new question; what happen when a algebra generator is multiplied by another matrix? It creates another generator? A group member? And what are the constrictions for these matrices? Must they be a group element? Or is this product defined for any matrix?. $\endgroup$
    – Amadeus
    Commented Jul 14, 2018 at 14:44
  • $\begingroup$ An algebra generator multiplied by another algebra generator produces some element of the algebra, but not necessarily one of the generators. Every algebra element can be expressed as some linear combination of the algebra generators. $\endgroup$
    – G. Smith
    Commented Jul 14, 2018 at 16:49
  • $\begingroup$ The generators of the algebra are NOT group elements. Taking the exponential of an algebra element, such as $\exp(a_1 S_1 + b_1 S_2)$, gives you some group element. This constricts the generators. $\endgroup$
    – G. Smith
    Commented Jul 14, 2018 at 16:53
  • $\begingroup$ Consider the 3D rotation group SO(3). A rotation around the z-axis is $$R_z(\theta)=\begin{bmatrix}\cos\theta & \sin\theta & 0 \\ -\sin\theta & \cos\theta & 0 \\ 0 & 0 & 1\end{bmatrix}. $$ For infinitesimal $d\theta$ this is $$R_z(d\theta)=I + d\theta \begin{bmatrix}0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}=I+d\theta S_z.$$ You can get a finite rotation by doing an infinite number of infinitesimal rotations so $$R_z(\theta)=\lim_{n->\infty}R(\theta/n)^n=\lim_{n->\infty}(I+\frac{\theta}{n} S_z)^n=\exp(\theta S_z).$$ $\endgroup$
    – G. Smith
    Commented Jul 14, 2018 at 17:10
  • $\begingroup$ So you can see how to go from infinitesimal group elements to algebra generators and then back again. $\endgroup$
    – G. Smith
    Commented Jul 14, 2018 at 17:15

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