Givens rotation with Eigendecomposition Setup
Let $\Sigma_t$ be the time $t$ positive-definite covariance matrix of some $N$-dimensional random vector $X_t$. Denote $\Sigma_t = P_t \Lambda_t P_t$ be the spectral/eigen-decomposition of this covariance matrix. There are $N + {N \choose 2} = N(N+1)/2$ free elements in this matrix. There are $N$ positive eigenvalues, and the rest are in the eigenvectors.
Questions
I am wondering how a Givens rotation matrices will "isolate out" the ${N \choose2}$ "free" elements of the eigenvectors. A paper I am reading reads:

We further write each $P_t$ as a product of $N(N-1)/2$ Givens rotation
  matrices $P_t = \Pi_{i<j}G_{ij}(\omega_{i,j,t})$...



*

*How should I visualize or understand the $N(N-1)/2$ free elements of $P_t$ before the Givens transformation. Probably something to do with orthogonality restrictions, right?

*How do I visualize or understand what the Givens rotations are doing? I imagine that plotting each $w_{i,j,t}$ versus time $t$ leads to something nice looking.

 A: @amoeba's comment answers the first thing question. This is all for the second. Let's suppress all the mentioning of $t$. Element $(k,l)$ of a Givens matrix $G_{i,j} \in \mathbb{R}^{N \times N}$ can be written as
$$
G_{i,j}[k,l] = 
\begin{cases} 
      \cos(\omega_{i,j}) & i=j \text{ and } (i=k \text{ or } i = l) \\
      \sin(\omega_{i,j}) &  k=i \text{ and } l = j\\
      -\sin(\omega_{i,j}) & k=j \text{ and } l = i \\
      1 & k = l \text{ and } k \neq i \text { and } k \neq j \\
      0 & \text{ otherwise } .
\end{cases}
$$
Rotating a Vector
Wikipedia has a pretty good description about how the transpose of this matrix takes vectors and rotates them in the $i-j$ plane in a counterclockwise fashion. 
Let $v \in \mathbb{R}^N$ be some vector. Picking the suitable angle $-2\pi \le \omega_{i,j} \le 2 \pi$ will zero out the $j$th element of this vector. It will also map the $i$th element of this vector to $\sqrt{v_i^2 + v_j^2}$. In other words:
$$
G_{i,j}^Tv = (v_1,\ldots,v_{i-1},\sqrt{v_i^2 + v_j^2},v_{i+1},\ldots,v_{j-1},0,v_{j+1},\ldots,v_n)^T.
$$
Rotating a Matrix
Think about rotating our orthogonal matrix $P = [P_1,\ldots,P_N]$ (remember no more $t$). The index $i$ will correspond with the column of $P$ we're talking about. Set $i=1$ for a second. Letting $j$ go from $2,\ldots,N$ we have
$$
  G_{1,2}^T \cdots G_{1,N}^T P_1 = (1,0,\ldots,0)^T = e_1
$$
where $e_n \in \mathbb{R}^N$ corresponds to a matrix of all zeros, except a $1$ in the $n$th spot. This is because this column is of unit length.
$G_{1,2}^T \ldots G_{1,N}^T$ affected $P_2$, but none of the other columns. By the same reasoning as above
$$
G_{2,3}^T\cdots G_{2,N}^T \cdot G_{1,2}^T \cdots G_{1,N}^T P_2 = e_2.
$$
Also, $G_{2,3}^T\cdots G_{2,N}^T \cdot G_{1,2}^T \cdots G_{1,N}^T P_1 = e_1$ still, because rotating $(0,0)^T$ yields the same thing.
So you can proceed inductively and get 
$$
\prod_{i<j}G_{i,j}^T P = [e_1,\ldots,e_N]= I,
$$
and we can solve this and get
$$
P = [\prod_{i<j}G_{i,j}^T]^{-1} = \prod_{i<j}G_{i,j},
$$
which is true as long as we're careful about ordering the $G_{i,j}$s correctly (recall that $G_{i,j}^T = G_{i,j}^{-1}$).
Remaining Questions
I'm not still confident in interpreting individual $\omega_{i,j}$s. Also, these matrices don't commute with one another, so how do we choose the "right" order. There's no point incentive for accepting my own answer, so anyone, feel free to chip in here.
