= Solution
Primordial <density contrasts> perturb the intrinsic photon <temperature>, the <gravitational potential>, and the <velocity> of the last-scattering material. Intrinsic photon-density fluctuations obey $\delta T/T=\delta_\gamma/4$ because $\rho_\gamma\propto T^4$; adiabatic initial conditions relate these to matter <density contrasts>. Subsequent acoustic evolution, <Doppler effect> and gravitational frequency shifts determine the observed <Cosmic microwave background anisotropy>. There is no scale-independent identification of the final sky map with matter <density contrast> alone.
The <spherical harmonic> coefficients are random variables inherited from these perturbations. After subtracting the mean sky <temperature>, $\langle a_{lm}\rangle=0$ for the anisotropy modes. Gaussian primordial perturbations give approximately <Gaussian random variables>; this extra Gaussian assumption is not needed for diagonal covariance. Reality of the sky map implies $a_{l,-m}=(-1)^ma_{lm}^*$.
For a statistically isotropic sky, its two-point function depends only on $\hat n\cdot\hat m$. Rotations mix the $m$ components at fixed $l$ and do not mix different irreducible $l$ spaces. A rotation-invariant <covariance matrix> is therefore a multiple of the identity in each $l$ block:
$$
\boxed{\langle a_{lm}a_{l'm'}^*\rangle=C_l\delta_{ll'}\delta_{mm'}.}
$$
On our one sky, average the $2l+1$ angular components at fixed $l$, or equivalently suitable rotated sky pairs, rather than treating different universes as observed samples. The <CMB angular power spectrum> is estimated by
$$
\boxed{\widehat C_l=\frac{1}{2l+1}\sum_{m=-l}^l|a_{lm}|^2.}
$$
Under the Gaussian, full-sky, noiseless assumptions, there are $2l+1$ independent real degrees of freedom and $\operatorname{Var}(\widehat C_l)=2C_l^2/(2l+1)$. This <cosmic variance> remains even with perfect detectors; a finite sky mask and detector noise add further uncertainty. The estimator is unbiased, $\langle\widehat C_l\rangle=C_l$, but does not turn one realization into the exact ensemble mean.
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