Just a simple question: is it correct to say that the Lie algebra of ${\rm U}(1)$ is formed by the generator $e^{i \pi}$?

Is it correct that ${\rm U}(1)=S^1 = T$ is the circle group?

  • 5
    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Jul 4, 2020 at 20:04

1 Answer 1


I will give a "physicist" pictorial explanation, very qualitative. For technical details, I agree that Math SE would be a better place.

Let's try to picture the group as a manifold (actually, a Lie group is a group that is also a smooth manifold, see the definition): each element of the group (in this case $U(1)$) is represented by a point of a manifold. The associated Lie algebra (in this case $u(1)$) is the tangent space to the point $1_{group}$ of the manifold that represents the identity element of the group.

Example: consider a group that is represented by a 3-sphere $S^3$, and choose to represent the identity element $1_{group}$ by the "North pole". Then, the associated Lie algebra is the tangent plane to the North pole. Both $S^3$ and the tangent space have the same dimension of 3 (but in the case of $U(1)$ and $u(1)$ the dimension is just 1). Note: $S^2$ can not be a Lie group, this is why I used $S^3$, see this.

This identification group-manifold is not such a "crazy thing": you already did it, since you wrote that the group $U(1)$ is "equivalent" to the manifold $S^1$, i.e., the group $U(1)$ is represented by a circle and every point of the circle is a group element. However, (remember the $S^2$ case above) not every manifold is a Lie Group.

The "exponential" is the operation that takes you from the tangent space (the algebra) to the manifold (the group): by taking the exponential of an element of the algebra you obtain an element of the group.

A tangent to the manifold $S^1$ is just a line, namely $\mathbb{R}$. So, the Lie algebra $u(1)$ is the real line (the 1-dimensional vector space $\mathbb{R}$). Note: the Lie algebra is a vector space (OK, it also has some algebraic structure, but from the "space point of view" it is a vector space). In fact, the Lie algebra is the tangent space to the group, and the tangent space in this context is a vector space.

Now, the generators are just the basis vectors of the tangent space (and you can make different choices!). Of course, they are elements of the algebra. If you know them, you can write down every element of the Lie algebra: every element (vector) of the algebra can be expressed as a linear combination of the generators.

In your case there is only one generator (the 1-dimensional vector $1$, or any other real number, because it is the only "base vector" of $\mathbb{R} \sim u(1)$). On the other hand, I would say that $e^{i \pi}$ is just an element of the group, not a generator of the group (generators are elements of the algebra, not elements of the group!).

A final note: very loosely speaking, you "multiply" group elements and "sum" elements of the algebra. The map that converts a sum into a product is exactly the exponential map. Obviously, the identity of the group is related to the "zero" of the algebra, $1_{group} = e^{0_{algebra}}$: this is the reason why the algebra is the tangent space at the point that represents the identity element $1_{group}$.


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