Simplifying CMB correlation function with spherical harmonics I originally asked this on the physics Stack Exchange site, but perhaps it could be more easily answered here.
Given the definition of the correlation function for CMB temperature fluctuations as
$$ C\left(\theta\right) = \left\langle \frac{\delta T}{T}\left(\hat{n}_1\right) \frac{\delta T}{T}\left(\hat{n}_2\right) \right\rangle_{\hat{n}_1\cdot \hat{n}_2 = \cos\theta} ,$$
I should be able to simplify it to
$$ C\left(\theta\right) = \frac{1}{4\pi} \sum_{l=0}^\infty (2l + 1) \, C_l \, P_l\left(\cos\theta\right) $$
(where $P_l \left(x\right)$ are the Legendre polynomials) by decomposing the temperature fluctuations into spherical harmonics like this
$$ \frac{\delta T}{T} = \sum_{l=0}^\infty \sum_{m=-l}^l a_{lm} Y_{lm}. $$
I think the first step of this procedure should look like this
$$ C\left(\theta\right) = \left\langle \sum_{l_1=0}^\infty \sum_{m_1=-l_1}^{l_1} a_{l_1 m_1} Y_{l_1 m_1}\left(\hat{n}_1\right) \sum_{l_2=0}^\infty \sum_{m_2=-l_2}^{l_2} a_{l_2 m_2} Y_{l_2 m_2}\left(\hat{n}_2\right) \right\rangle_{\hat{n}_1\cdot \hat{n}_2 = \cos\theta} .$$
I understand that the spherical harmonics can be written in the form
$$ Y_{lm}(\theta,\phi) \propto P_{lm} \left(\cos\theta\right) e^{i m \phi} $$
(where $P_{lm}(x)$ are the associated Legendre polynomials) and that $C_l$ should come out as
$$ C_l = \frac{1}{2l + 1} \sum_{m=-l}^l a_{lm} a_{l-m} $$
(though I could be off on this last piece). However, I am unsure of the mathematical steps involved in simplifying the four sums down to one. What identities, properties, or other insights will allow me to make this simplification?
Thanks!
 A: In order to arrive at the correlation function in terms of the Legendre polynomials you begin by noting that
\begin{align}
\left\langle a_{l_1 m_1}a^*_{l_2 m_2}\right\rangle = C_{l_1}\,\delta_{l_1l_2}\,\delta_{m_1m_2}
\end{align}
where $\delta_{xy}$ are Kronecker deltas.  (Note that, this only holds if the random fields described by the $a_{l m}$, i.e. in this case the CMB, is statistically homogeneous and isotropic!)
After using expansion into spherical harmonics (your third equation), you'll find the correlation function to read
\begin{align}
C(\theta) = \left\langle\frac{\delta T}{T}(\hat n_1)\frac{\delta T}{T}(\hat n_2)\right\rangle = \left\langle\sum_{l_1}\sum_{l_2}\sum_{m_1}\sum_{m_2}a_{l_1 m_1}Y_{l_1m_1}(\hat n_1)\,a_{l_2m_2}Y_{l_2m_2}(\hat n_2)\right\rangle
\end{align}
For real fields (and the CMB temperature is a real field) one can write
\begin{align}
a^*_{lm} = (-1)^ma_{l-m}
\quad\text{and}\quad
Y^*_{lm}(\hat n) = (-1)^mY_{l-m}(\hat n).
\end{align}
This allows you to use $\sum_{m_2} a_{l_2m_2}Y_{l_2m_2}(\hat n_2)=\sum_{m_2} a^*_{l_2m_2}Y^*_{l_2m_2}(\hat n_2)$ such that the correlation function reads
\begin{align}
C(\theta) &= \left\langle\sum_{l_1}\sum_{l_2}\sum_{m_1}\sum_{m_2}a_{l_1 m_1}a^*_{l_2m_2}\,Y_{l_1m_1}(\hat n_1)Y^*_{l_2m_2}(\hat n_2)\right\rangle =\\
&= \sum_{l_1}\sum_{l_2}\sum_{m_1}\sum_{m_2}\left\langle a_{l_1 m_1}a^*_{l_2m_2}\right\rangle\,Y_{l_1m_1}(\hat n_1)Y^*_{l_2m_2}(\hat n_2) =\\
&= \sum_{l_1}\sum_{l_2}\sum_{m_1}\sum_{m_2} C_{l_1}\,\delta_{l_1l_2}\,\delta_{m_1m_2}  \,Y_{l_1m_1}(\hat n_1)Y^*_{l_2m_2}(\hat n_2) =\\
&= \sum_{l_1}C_{l_1}\,\sum_{m_1} Y_{l_1m_1}(\hat n_1)Y^*_{l_2m_2}(\hat n_2)
\end{align}
Finally, use the relation
\begin{align}
P_l(\cos\theta) = \frac{4\pi}{2l+1}\sum_m Y_{lm}(\hat n_1)Y_{lm}(\hat n_2)
\end{align}
to find your quoted result (your 2nd equation).
