What are the elements of Lie algebra in the case of a single generator?

[LagrangeEuler]

Elements of Lie algebra are generators. So for example Pauli matrices are generators of rotation and the elements of Lie algebra. And multiplication in Lie algebra is commutator. Right?

What about if there is only one generator. As in case in rotation in plane. What is Lie algebra product in that case?

 

[WWGD]

I don't understand, elements of Lie Algebra are generators of what? And the " multiplication" , meaning the binary operation associated is the bracket. In some cases, if the Lie group is a matrix lie group, then the bracket is given by XY-YX : https://en.wikipedia.org/wiki/Lie_algebra#Generators_and_dimension . I am not an expert in the area, but I don't think the bracket is always of the form [X,Y]=XY-YX .

 

[fresh_42]

Elements of Lie algebra are generators. So for example Pauli matrices are generators of rotation and the elements of Lie algebra. And multiplication in Lie algebra is commutator. Right?

What about if there is only one generator. As in case in rotation in plane. What is Lie algebra product in that case?

In that case you get a one-dimensional Lie Algebra L = F*X which is abelian: [cX,dX] = 0 (c,d constants of the underlying field F, X your generator). By the way: elements of a Lie Algebra are firstly vectors. Calling them generators implies a lot of theory in between.

 

[fresh_42]

I am not an expert in the area, but I don't think the bracket is always of the form [X,Y]=XY-YX .

Correct. In physics however, Lie algebras usually arise from transformations / symmetries and you'll therefore have a natural ordinary multiplication.

EDIT: In case Lie Groups as analytic manifolds are the origin (left invariant vector fields), you get the above bracket multiplication by considering the group multiplication.

 

[LagrangeEuler]

I don't understand, elements of Lie Algebra are generators of what? And the " multiplication" , meaning the binary operation associated is the bracket. In some cases, if the Lie group is a matrix lie group, then the bracket is given by XY-YX : https://en.wikipedia.org/wiki/Lie_algebra#Generators_and_dimension . I am not an expert in the area, but I don't think the bracket is always of the form [X,Y]=XY-YX .

It depends on the group. Lie algebra elements could be generators of translations or generator of rotation... It is very important in physics that for infinite number of elements one has finite number of generators.
$$exp(i\theta generator)=group \quad element$$
where $\theta$ is continuous parameter.

 

[LagrangeEuler]

In that case you get a one-dimensional Lie Algebra L = F*X which is abelian: [cX,dX] = 0 (c,d constants of the underlying field F, X your generator). By the way: elements of a Lie Algebra are firstly vectors. Calling them generators implies a lot of theory in between.

Thanks. And for example in case of translation in 3d generators are impulse operators $\hat{p}_x, \hat{p}_y,\hat{p}_z$ and in quantum mechanics those operators commute. So I also can tell that this is Abelian Lie algebra? Right?

 

[fresh_42]

Thanks. And for example in case of translation in 3d generators are impulse operators $\hat{p}_x, \hat{p}_y,\hat{p}_z$ and in quantum mechanics those operators commute. So I also can tell that this is Abelian Lie algebra? Right?

Commuting operators as elements of Lie algebras is equivalent to the Lie algebra multiplication being zero. The Lie algebra itself is only abelian if all elements commute, i.e. if you don't embed it elsewhere, e.g. considering extensions.

EDIT: For example: Heisenberg and Poincaré algebras are neither abelian nor semisimple.

EDIT2: The impulse operators commute with each other but not with the position operators!

 

[LagrangeEuler]

EDIT2: The impulse operators commute with each other but not with the position operators!

This is interesting. Element of group of translation in one dimension $T(x)=e^{ix\hat{p}}$. Generator of translation is $\hat{p}=-i\frac{d}{dx}$. Now I am confused. Is $x$ also part of Lie algebra. As far as I understand this is not case.

 

[fresh_42]

This is interesting. Element of group of translation in one dimension $T(x)=e^{ix\hat{p}}$. Generator of translation is $\hat{p}=-i\frac{d}{dx}$. Now I am confused. Is $x$ also part of Lie algebra. As far as I understand this is not case.

The rules are $[\hat{x_i},\hat{x_j}] = 0 = [\hat{p_i},\hat{p_j}]$ for $i,j = 1,2,3$ but $[\hat{x_i},\hat{p_j}] = i ħ δ_{ij}$.
So the position operators by themselves as well as the impulse operators by themselves define each a three-dimensional abelian Lie algebra.
If you consider the linear span of all six you don't have a Lie algebra anymore, since $iħ \cdot I = [\hat{x_i},\hat{p_i}]$ is missing.
If you add $I$, e.g. to the position operators, and build the linear spans $P = span \{I,\hat{x_i}| i=1,2,3\}$ and $S = span\{\hat{p_i}| i =1,2,3\}$ then you get a 7-dimensional Lie algebra which is a semi-direct product of abelian subalgebras $P$ and $S$, $P$ being an ideal, and $span \{I\}$ its center.
Guess it's a Heisenberg algebra.

 

[LagrangeEuler]

The rules are $[\hat{x_i},\hat{x_j}] = 0 = [\hat{p_i},\hat{p_j}]$ for $i,j = 1,2,3$ but $[\hat{x_i},\hat{p_j}] = i ħ δ_{ij}$.
So the position operators by themselves as well as the impulse operators by themselves define each a three-dimensional abelian Lie algebra.
If you consider the linear span of all six you don't have a Lie algebra anymore, since $iħ \cdot I = [\hat{x_i},\hat{p_i}]$ is missing.
If you add $I$, e.g. to the position operators, and build the linear spans $P = span \{I,\hat{x_i}| i=1,2,3\}$ and $S = span\{\hat{p_i}| i =1,2,3\}$ then you get a 7-dimensional Lie algebra which is a semi-direct product of abelian subalgebras $P$ and $S$, $P$ being an ideal, and $span \{I\}$ its center.
Guess it's a Heisenberg algebra.

Ok. If I understand you well Lie algebra, as a Lie group needs to have closure property. My question is. I have group of translation in one dimension. I want to now what are Lie algebra elements in that case? $\hat{p}$ and $0$? Or $\hat{p}$, $\hat{x}$, $0$, $1$. For given Lie group, Lie algebra is unique?

 

[fresh_42]

Ok. If I understand you well Lie algebra, as a Lie group needs to have closure property. My question is. I have group of translation in one dimension. I want to now what are Lie algebra elements in that case? $\hat{p}$ and $0$? Or $\hat{p}$, $\hat{x}$, $0$, $1$. For given Lie group, Lie algebra is unique?

I'm not quite sure what you mean. For (strong) simplicity: given a Lie group (curved) you can consider it's Lie algebra (flat) as its tangent space at 1. The connection between both is how this varies when you consider other points of the group and its effect on the tangent vectors.
If you have only one generator, the translation, then things become easy. Let's say $G=(ℝ,+)$. Then the Lie algebra $g$ of $G$ is also one dimensional (and therefore abelian) and generated by the vector field $D = \frac{d}{dt}$, i.e. for $τ ∈ ℝ$ you have $D_τ = \frac{d}{dt}|_{t=τ}$ as tangent vector at the point $τ ∈ G$.
Yes, the Lie algebra of a Lie Group is unique. (The details however are a bit more complicated.)

 

[samalkhaiat]

I want to now what are Lie algebra elements in that case? $\hat{p}$ and $0$? Or $\hat{p}$, $\hat{x}$, $0$, $1$. For given Lie group, Lie algebra is unique?

$X = ix\hat{p}$ and $0$.
Regarded as real vector space, the element $A$ of a n-dimentional Lie algebra can be expanded in a basis (generators) $X_{i}$
$$A = \alpha^{i}X_{i}$$
where $\alpha^{i}$ are a set of n real parameters (local coordinates on the group manifold). So in the case of $e^{ix\hat{p}}$, $x$ is the real parameter and $i\hat{p}$ is the generator.