Exercise about symmetry in the Lagrange equations This question was asked during a classical mechanics exam (no solutions were given afterwards). 

Suppose a free particle in $\mathbb{R}^n$ with the following Lagrangian:
  $$L = \frac{m}{2}\sum_{i=1}^n \dot{x}_i^2$$
  1) What are the symmetries of this system? Hint: there are at least $n +\frac{n(n-1)}{2} $
2) Use Noether's theorem to write down all the conserved quantities

Now for the first question, I don't understand how to write down these symmetries mathematically. Also, how does one know you didn't miss one ? For the case $n = 3$ I suspect there are 6 symmetries ( 3 translational and 3 rotational) but for the general case when $n > 3$ , I don't reach the minimum amount of symmetries they gave me as a hint. Concerning the second question, I understand Noether's theorem but I don't know how to use it. How does one find all the possible transformations that leave the Lagrangian invariant + getting all the conserved quantities?
 A: (I assume that the sum starts from $1$.) That  lagrangian is evidently invariant  under the Lie group of isometries of $\mathbb R^n$. 
This group is the semidirect product of $\mathbb R^n$ viewed as group of translations and $O(n)$.
$$\vec{x} \to \vec{v}+ R\vec{x}\:, \quad \dot{\vec{x}} \to  R\dot{\vec{x}}$$
The Lie algebra of this group has $n$ dimensions due to the translations plus $n(n-1)/2$ due to rotations (that is the dimension of the antisymmetric real matrices of dimension $n$). Each generator induces a corresponding one-parameter subgroup and an associted conserved quantity as a consequence of Noether's theorem. Actually,  there is a further independent conserved quantity arising from the invariance under time translations of the Lagrangian. 
Applying Noether theorem the said conserved quantities turn out to be 
$$p_k := m \dot{x}_k\quad k=1,2, \cdots, n$$
due to transaltions, in addition to
$$J_k := m\sum_{i,j=1}^n \dot{x_i} A^{(k)}_{ij}x_j\quad k= 1,2,\ldots, n(n-1)/2$$
due to rotations, 
where the matrices $A^{(k)}$ form a basis of the real vector space of antisymmetric $n\times n$ matrices. The final conserved quantity due to temporal translational invariance  is $L$ itself.
A: The $n(n-1)/2$ symmetries are the rotations in $n$ dimensions. Assuming you haven't studied $O(n)$ in much detail here is a way to do it using elementary methods (in the following $\vec{x}$ is treated as a column vector so:
Consider the coordinate change $\vec{x} \mapsto (\hat{1}+\epsilon \hat{M})\vec{x}$ where $M$ is a matrix. Under this change the velocities become:
$$\dot{\vec{x}} \mapsto \dot{\vec{x}}+\epsilon \hat{M}\dot{\vec{x}}$$
We can then plug this into the Lagrangian to find the condition that this is a symmetry, we find:
$$L' = L + \epsilon (\hat{M}^T+\hat{M})\dot{\vec{x}} + O(\epsilon^2) $$
So we need our infinitesimal transformation to correspond to anti-symmetric $M$ for this method to give us a symmetry (and its easy to count the number of anti-symmetric matrices).
The conserved charge is:
$$Q = \frac{\partial L}{\partial \dot{\vec{x}}}\delta \vec{x} $$
The first factor on the right is the canonical momentum, $\vec{p}^T$ whilst the second is $\hat{M}\vec{x}$. (The reason for the transpose is because canonical momentum is a covector so is represented by a row vector not a column vector.)
This gives us our conserved charges (really the angular momentum) as the quadratic forms $Q = \vec{p}^T\hat{M}\vec{x}$. There's one of these for each anti-symmetric matrix you can write down.
