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Associated Legendre Polynomials and Legendre Polynomials are strongly connected. From what I read, Legendre polynomials is a special case of Associated Legendre polynomials where $m=0$. Thus the Orthogonality and parity condition followed from Associated Legendre Polynomial.

From alexjo's answer in this post Associated Legendre Polynomials!? , Associated Legendre polynomial was the solution with "a separation ansatz", and Legendre polynomial was the solution with azimuthal symmetry. Spherical Harmonic was then introduced with Associated Legendre polynomial multiply by a complex phase.

  1. Is the above summary correct? or have I missed something?

  2. From this post Proof that Legendre Polynomials are Complete, Legendre Polynomial itself had already formed a complete basis, so why do we need Associated Legendre Polynomial? Isn't it too excessive?

  3. Further, I still have a hard time in dealing with the the geometrical picture of those functional basis. I think I saw somewhere Spherical Harmonic was represented with a vector on a sphere aiming at $m,(m-1),...,-m$. However, how could we visualize associated Legendre polynomials? Does those $m$ represented the similar thing?

Could you in help me to distinguish the usage of those function basis, please?

ShoutOutAndCalculate
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    You should start from the $x=\sin\theta,y+iz=e^{i\phi}\cos\theta$ parametrization of the sphere $x^2+y^2+z^2=1$, find the natural inner product (invariant under rotations), the Fourier series and Legendre polynomials inner products appear, thus obtaining the $L^2$ orthogonal basis $e^{im\phi}P_l(\cos(\theta))$, multiply with $\sin(\theta)^m$ it becomes an homogeneous polynomial $Y_l^m$ of $x,y,z$ and (from Legendre polynomials differential equation) it satisfies $(\partial_x^2+\partial_y^2+\partial_z^2)Y_l^m=0$, from there it suffices to show it is $L^2$-complete in the harmonic functions. – reuns Nov 20 '19 at 05:53

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