Carrier Frequency Offset, Rotating Constellation and Correlator I have been further reading up on carrier recovery methods and something is not clear to me:

It's written that a carrier frequency offset results in a rotating constellation diagram but if you are sampling at the output of a correlator such as RRC it's not clear to my why this is the case.

For an all digital receiver, the ADC samples will be rotated prior to going into the correlator but it appears the correlation process in conjunction with the frequency offset will do more than rotate the constellation diagram. 

Am I wrong?
The correlator is just a filtering operation. So in the case of passband pulse-amplitude modulation (PAM), such as QPSK or QAM, the (noiseless) received analytic signal before demodulation and sampling can be written as
$$r(t)=e^{j((\omega_c+\Delta\omega) t+\phi))}\sum_{k}A_kp(t-kT)\tag{1}$$
where $\Delta\omega$ is a frequency offset, $\phi$ is a phase offset, $\omega_c$ is the carrier frequency as estimated by the receiver, $A_k$ are the complex data symbols, and $T$ is the symbol period. The pulse $p(t)$ represents the combination of the transmit filter, linear distortions on the channel, and the receive filter (correlator).
Demodulation with $e^{-j\omega_ct}$ and sampling at the symbol rate $1/T$ gives
$$\tilde{r}[n]=e^{j(\Delta\omega nT +\phi)}\sum_kA_kp((n-k)T)\tag{2}$$
If for simplicity we assume that the combined pulse $p(t)$ satisfies the Nyquist criterion $p(kT)=\delta[k]$ (i.e., there is no inter-symbol interference), then $(2)$ simplifies to
$$\tilde{r}[n]=e^{j(\Delta\omega nT +\phi)}A_n\tag{3}$$
which should make it clear that at time $t=nT$ the received symbol is a rotated version of the original symbol, where the rotation has a constant component $\phi$ (the phase offset), and a time-dependent component $\Delta\omega nT$ caused by the frequency offset. Consequently, if $\Delta\omega\neq 0$, the constellation rotates.