Which are the right configurations in the Markov chain of a Hamiltonian Monte Carlo algorithm?

The algorithm for perfoming a single HMC step is as follows:

Input: Some initial configuration $vec{y}_i$ and momentum $vec{p}_i$.
Output: Next configuration $vec{y}_{i+1}$ and momentum $vec{p}_{i+1}$

1. Draw a random momentum $vec{p}_*$ from a Gaussian distribution.
2. Numerically solve Hamilton's equations of motion for some time (i.e., perform some Leapfrog steps) with initial conditions $(vec{y}_i, vec{p}_*)$. The end point of the trajectory is denoted by $(vec{y}_c, vec{p}_c)$.
3. Compute the acceptance probability $rho_i in [0,1]$ of the proposal / candidate $(vec{y}_c, vec{p}_c)$: $$rho_i = min{expleft( - beta left( mathcal{H}(vec{y}_c, vec{p}_c) - mathcal{H}(vec{y}_i, vec{p}_i) right) right), 1}.$$ Here, $mathcal{H}$ denotes the Hamiltonian and $beta > 0$ is the inverse temperature.
4. Draw a random number $alpha_i sim U([0,1])$ from the uniform distribution on $[0,1]$ and set $$(vec{y}_{i+1}, vec{p}_{i+1}) = begin{cases} (vec{y}_c, vec{p}_c) & text{if } alpha_i leq rho_i, \ (vec{y}_i, vec{p}_i) & text{otherwise}. end{cases}$$

Now, to answer your questions (and to give some additional remarks):

• Usually, the degrees of freedom are not treated separately. This is especially problematic if the Hamiltonian $mathcal{H}$ does not decompose into multiplicative factors for each DOF. In the algorithm above, the full state $vec{y}$ is drawn (i.e., all DOFs are drawn simultaneously).
• If a proposal is rejected, the initial state is taken ($vec{y}_{i+1} = vec{y}_i$).
• If two proposals are rejected in a row, the chain is still in the original state ($vec{y}_{i+1} = vec{y}_i$ and $vec{y}_{i+2} = vec{y}_{i+1}$, which implies $vec{y}_{i+2} = vec{y}_i$).