# Noether's theorem, dimension of the group of symmetry and dimension of conserved quantity

I previously asked on Mathematics Stack Exchange about the relation between $\bigwedge^2(\Bbb R^n)$ and $\text{SO}(n)$. See this link to the post. I noticed that they have equal dimensions, that I think they are related. It happens that the question I have in mind may actually be related to physics. So I post the question here.

First, I noticed the following things. Hopefully my understandings are correct:

Time translational symmetry is described by $\Bbb R$, which is a one-dimensional Lie group. The corresponding conserved quantity (energy) lies in a one-dimensional vector space.

Spatial translational symmetry is described by $\Bbb R^3$, which is a three-dimensional Lie group. The corresponding conserved quantity (momentum) lies in a three-dimensional vector space.

Rotational symmetry is described by $\text{SO}(3)$, which is a three-dimensional Lie group. The corresponding conserved quantity (angular momentum) lies in a three-dimensional vector space.

The reason I have thought that $\bigwedge^2(\Bbb R^n)$ and $\text{SO}(n)$ are related is actually about a hypothetical angular momentum in a hypothetical $n$-dimensional space. If there is an $n$-dimensional space, and there is a Lagrangian which is invariant under the group action of $\text{SO}(n)$, I guess that the quantity $r\wedge p$ (angular momentum) of a particle is conserved along the allowed paths, where $r$ is the position of the particle from the origin, and $p$ is the momentum. The quantity $r\wedge p$ lies in the vector space $\bigwedge^2(\Bbb R^n)$, which has the same dimension as $\text{SO}(n)$.

From the above observations, I have made the following conjecture:

If a continuous symmetry of the Lagrangian is given by a Lie group of dimension $n$, then the corresponding conserved quantity lies in an $n$-dimensional vector space.

Is it really the case?

• They are related simply because $\Lambda^2(\mathbb R^n)$, when seen as vector space of antisymmetric real $n\times n$ matrices and equipped with the standard commutator, is the Lie algebra of $SO(n)$. So if $R\in SO(n)$ since $SO(n)$ is compact, we have $R = e^{A}$ for some $A \in \Lambda^2(\mathbb R^n)$. – Valter Moretti Oct 15 '16 at 12:57
• The answer to your final question is YES, but it is independent from the case of $SO(n)$ and $\Lambda^2(\mathbb R^n)$. It is just the application of Noether theorem for every one-parameter subgroup of the Lie group (assuming that the representation of the group in terms of symmetries is faithful). – Valter Moretti Oct 15 '16 at 13:01