Angular momentum and Noether's theorem Studying the lagrangian formulation of Noether's theorem and came upon how the invariance under rotations gives conservation of angular momentum.
Whilst setting up the problem the notes state that if a potential only depends on the distance between 2 points, namely $V(|r_i-r_j|)$, then you can apply the transformation:
$$\textbf{r}\rightarrow \textbf{r}+\epsilon T\textbf{r}$$
where $\epsilon$ is a small variation, $\textbf{r}$ is just a vector and $T$ is a rotation matrix. I'm confused about the fact that the notes state that $T$ is an anti-symmetric matrix, I thought rotation matrices where orthogonal.
 A: You are absolutely right, a rotation matrix should be orthogonal. However, when we investigate symmetry transformations, we often like to consider infinitesimal variations, as they are often much easier to investigate. 
Consider then a orthogonal matrix $R$, the rotation matrix. The rotation transformation is written as follows :
$$\vec{r}\rightarrow R\vec{r}$$
Now, let us consider an infinitesimal rotation. Since it is infinitesimal, it must be very close to the "$0$" rotation, or in other words, the identity $I$. Thus, we must write it in some way as :
$$R = I+\epsilon T$$
Where $\epsilon$ is a small parameter. Now, we want to find which $T$ we should use. Obviously, a random matrix $T$ wouldn't work in general. We must enforce the fact that $R$ is a rotation, i.e. that $R^T R = I$ :
$$I = R^T R = (I+\epsilon T^T)(I+\epsilon T) = I+\epsilon(T+T^T)+O(\epsilon^2)$$
Here, we don't care about terms in $\epsilon^2$, as they are much smaller than terms in $\epsilon$. So, to enforce $R^T R = I$, we must have simply $T+T^T = 0 \Leftrightarrow T = -T^T$. This is exactly the condition of anti-symmetricity you were looking for !
Thus, for an infinitesimal rotation, we can conveniently write $R = I +\epsilon T$ with T antisymmetric. Note that we often use an even more compact notation, $R = e^{\epsilon T}$, which you can check is equivalent to the one you gave for small $\epsilon$.
P.S. : Physicist use yet another notation, where they write $R = I -i\epsilon (iT) = I -i\epsilon T'$, so we would have purely imaginary antisymmetric matrices. This is mainly because in this way, when you consider unitary matrices $U^\dagger U =I$, then the associated $T'$ is hermitian $T'^\dagger = T'$
