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I have been looking at the Laplace equation $\nabla^2 f = 0$ in various dimensions. In 3 dimensions, the angular equation leads to the well-known spherical harmonics, defined up to normalisation as \begin{align} Y_{lm}(\theta, \phi) = P^{|m|}_l(\cos\theta)e^{\imath n\phi}. \end{align} Here, the $P^{|m|}_l$ are the associated Legendre polynomials, defined by \begin{align} P^{|m|}_l(t) = (1-t^2)^{\frac{|m|}{2}}\frac{d^{|m|}}{dt^{|m|}}P_l(t), \end{align} where the $P_l$ are the Legendre polynomials.

This wikipedia article gives the Legendre polynomials by the Rodrigues formula as \begin{align} P_l(t) = \frac{1}{2^ll!}\frac{d^l}{dt^l}(t^2-1)^l. \end{align}

This paper I found, which discusses the Laplace equation in $p$ dimensions, gives the Legendre polynomials in $p$ dimensions by the Rodrigues formula as (equation 4.31 on page 69) \begin{align} P_l(t) = \frac{(-1)^l}{2^l(l+\frac{p-3}{2})_l}(1-t^2)^{\frac{3-p}{2}}\frac{d^l}{dt^l}(1-t^2)^{l-\frac{p-3}{2}}. \end{align}

$(l+\frac{p-3}{2})_l$ denotes the truncated factorial, i.e. $(n)_m = n(n-1)\cdot\cdot\cdot(n-m+1)$. For the case of $p = 3$, this is identical to the other formula, so I then assume that the wikipedia formula is implicitly for the 3-dimensional case (is this correct?).


I should also mention that I am a physicist, so I am familiar with a lot of the geometric reasoning in the case of 3 dimensions particularly, and the way I ask my questions may reflect this. Now this is what I am unsure about:

I have been through the paper I linked, and the $p$-dimensional case derivation makes sense to me. In this case, the argument $t$ of the polynomial is then interpreted as the cosine of the angle between points on the unit sphere, and we typically choose one of those points to be fixed as the 'North Pole'. So it makes sense to me how the spherical harmonics are associated with the sphere of a certain dimensionality.

However, the Legendre polynomials themselves seem '1-dimensional' to me, as in they take a single argument $t\in[-1, 1]$. Furthermore, there are alternative derivations of the Legendre polynomials that have nothing to do with Laplace's equation, and its associated dimensionality. For example, at the top of the wikipedia article linked above, there are 3 different definitions which I cannot see how to relate to some 'dimension'.

So my question is really what is the dimension of a Legendre polynomial?

If my question is unclear, or perhaps ill-defined, I am trying to understand why wikipedia says one thing, which when I compare to the $p$-dimensional case, is the $p=3$ case, yet there is no discussion of dimension on the wikipedia page - I cannot see where this additional information comes from.

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  • $\begingroup$ The "Legendre polynomials" from the arxiv paper you linked are not Legendre polynomials as conventionally understood, they are a generalization. This is why Wikipedia does not discuss $p$. Perhaps the authors should call them "$p$-Legendre polynomials" but since they keep $p$ fixed they opted for a shorthand. Of course, $p$-Legendre polynomials are 1D just like the ordinary Legendre polynomials for $p=3$, but they are different families of 1D polynomials for different $p$. $\endgroup$
    – Conifold
    Commented Aug 16, 2021 at 13:15
  • $\begingroup$ Thank you. So is my association of the Legendre polynomials to the Laplacian more of a 'coincidence', by which I mean the associated Legendre polynomials simply solve the differential equation for the $\theta$ part of Laplace's equation, but that does not mean the Legendre polynomials themselves have anything to do with the space on which the Laplacian is acting? $\endgroup$
    – Bedge
    Commented Aug 16, 2021 at 13:33
  • $\begingroup$ I wouldn't call it a coincidence, more of a natural generalization. Classical Legendre polynomials are associated to the spherical harmonics in 3D in a way analogous to $p$-Legendre polynomials are in $p$D. It is true that they are always one variable polynomials, and in this sense are not reflecting the Laplacian dimension explicitly, but implicitly it is still wrapped into the differential equation they satisfy. In the same way solutions to the radial part of the Laplacian are always 1D functions of $r$ but the dimension manifests in the power of $r$ in that function. $\endgroup$
    – Conifold
    Commented Aug 16, 2021 at 19:37

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