Mode weights

On the previous page, we introduced the eigenfunctions of the differential operators defined on $\mathrm{Spin}(3)$. These eigenfunctions are the spin-weighted spherical harmonics (SWSHs) ${}_{s}Y_{ℓ,m}(R)$, or equivalently Wigner's 𝔇 matrices.

Now that we have introduced the spin-weighted spherical harmonics (SWSHs) as eigenfunctions of the relevant differential operators, we can define mode weights of a general spin-weighted function in terms of these harmonics.

These eigenfunctions form a complete orthogonal basis for the space of square-integrable functions defined on $\mathrm{Spin}(3)$. They are orthogonal rather than orthonormal, because they are normalized (but not orthogonal) on $𝕊²$ instead, which leaves $\left\| {}_{s}Y_{ℓ,m} \right\|²_{\mathrm{Spin}(3)} = π/2$ (see the previous page). Thus, any square-integrable function $f(R)$ can be expressed as a linear combination of these harmonics:

\[f(R) = \sum_{ℓ=0}^{∞} \sum_{s=-ℓ}^{ℓ} \sum_{m=-ℓ}^{ℓ} {}_{s}f_{ℓ,m}\, {}_{s}Y_{ℓ,m}(R),\]

where the coefficients ${}_{s}f_{ℓ,m}$ are called the mode weights of the function $f$. These mode weights can be computed from the function using the orthogonality of the SWSHs:

\[{}_{s}f_{ℓ,m} = \frac{2}{π} \int_{\mathrm{Spin}(3)} f(R)\, {}_{s}\bar{Y}_{ℓ,m}(R)\, dR,\]

where the factor in front of the integral is the reciprocal of $\left\| {}_{s}Y_{ℓ,m} \right\|²_{\mathrm{Spin}(3)} = π/2$, which is explained here. Note that we have not restricted the spin weight $s$ of the function $f$; a general function on $\mathrm{Spin}(3)$ can have contributions from SWSHs of any spin weight, so $s$ was included in the sum above. In fact, those spin weights may have half-integral values as well, in which case the sum over $ℓ$ must include all positive half-integral values as well as the integral values.

However, if we restrict to functions with a specific spin weight $s$, only the SWSHs with that spin weight will contribute to the expansion, and we can simplify the expressions above to

\[f(R) = \sum_{ℓ=|s|}^{∞} \sum_{m=-ℓ}^{ℓ} {}_{s}f_{ℓ,m}\, {}_{s}Y_{ℓ,m}(R)\]

and

\[{}_{s}f_{ℓ,m} = \frac{2}{π} \int_{\mathrm{Spin}(3)} f(R)\, {}_{s}\bar{Y}_{ℓ,m}(R)\, dR.\]

In practice the integral is almost always written over the sphere $𝕊²$ instead, but the relationship is simple. The integrand $f\, {}_{s}\bar{Y}_{ℓ,m}$ is the product of a function of spin weight $s$ with the conjugate of another, so that product has spin weight $0.$ A function of spin weight 0 does not depend on the third Euler angle at all, so it pushes forward to an ordinary function on $𝕊²$, and — by exactly the argument that relates the two norms above — its integral over $\mathrm{Spin}(3)$ is $π/2$ times its integral over $𝕊²$. The factor $2/π$ therefore cancels, leaving the familiar expression

\[{}_{s}f_{ℓ,m} = \int_{𝕊²} f(θ, ϕ)\, {}_{s}\bar{Y}_{ℓ,m}(θ, ϕ)\, dΩ, \qquad dΩ = \sin θ\, dθ\, dϕ,\]

in which $f(θ, ϕ)$ means $f$ evaluated at the rotor that spherical coordinates assign to the point $(θ, ϕ)$. Note that this step relies on $f$ having a definite spin weight; for a function with contributions from several spin weights — where the third Euler angle really matters — the integral over $\mathrm{Spin}(3)$ in the previous expression is the one to use. Functions whose domain is really $I × 𝕊¹$ rather than $𝕊²$ are discussed here.

The same result holds for half-integer spins: the cancellation above only needs $f$ and ${}_{s}Y_{ℓ,m}$ to have the same spin weight, so the product is $γ$-independent — and in particular unchanged under $𝐑 → -𝐑$ — whatever the parity of $s$ may be. What is peculiar to half-integer $s$ is that each factor separately is double-valued on $𝕊²$, changing sign under $ϕ → ϕ + 2π$; the integral must therefore be taken with one fixed rotor assignment $(θ, ϕ) ↦ 𝐐$, under which the integrand is single-valued and the identity holds unchanged.

Differential operators

Mode weights and functions transform in related but different ways under the action of the differential operators.

One important point to note is that mode weights transform "oppositely" (loosely speaking) relative to the spin-weighted spherical functions under some operators. For example, take the action of the $L_+$ operator, which acts on a SWSH as

\[L_+ \left\{{}_{s}Y_{ℓ,m}\right\} (R) = \sqrt{(ℓ-m)(ℓ+m+1)}\ {}_{s}Y_{ℓ,m+1}(R).\]

We can use this to derive mode weights of a general spin-weighted function $f$ under the action of this operator:

\[\begin{aligned} \left\{L_+ f\right\}_{ℓ,m} &= \frac{2}{\pi} \int \left\{L_+ f(R)\right\}\, {}_{s}\bar{Y}_{ℓ,m}(R)\, dR \\ &= \frac{2}{\pi} \int \left\{L_+ \sum_{ℓ',m'}f_{ℓ',m'}\, {}_{s}Y_{ℓ',m'}(R)\right\}\, {}_{s}\bar{Y}_{ℓ,m}(R)\, dR \\ &= \frac{2}{\pi} \int \sum_{ℓ',m'} f_{ℓ',m'}\, \left\{L_+ {}_{s}Y_{ℓ',m'}(R)\right\}\, {}_{s}\bar{Y}_{ℓ,m}(R)\, dR \\ &= \sum_{ℓ',m'} f_{ℓ',m'}\, \frac{2}{\pi} \int \left\{L_+ {}_{s}Y_{ℓ',m'}(R)\right\}\, {}_{s}\bar{Y}_{ℓ,m}(R)\, dR \\ &= \sum_{ℓ',m'} f_{ℓ',m'}\, \frac{2}{\pi} \int \left\{\sqrt{(ℓ'-m')(ℓ'+m'+1)} {}_{s}Y_{ℓ',m'+1}(R)\right\}\, {}_{s}\bar{Y}_{ℓ,m}(R)\, dR \\ &= \sum_{ℓ',m'} f_{ℓ',m'}\, \sqrt{(ℓ'-m')(ℓ'+m'+1)} \frac{2}{\pi} \int {}_{s}Y_{ℓ',m'+1}(R)\, {}_{s}\bar{Y}_{ℓ,m}(R)\, dR \\ &= \sum_{ℓ',m'} f_{ℓ',m'}\, \sqrt{(ℓ'-m')(ℓ'+m'+1)} δ_{ℓ,ℓ'} δ_{m,m'+1} \\ &= f_{ℓ,m-1}\, \sqrt{(ℓ-m+1)(ℓ+m)} \end{aligned}\]

Note that this expression (and in particular its signs) more resembles the expression for $L_- \left\{{}_{s}Y_{ℓ,m}\right\}$ than for $L_+ \left\{{}_{s}Y_{ℓ,m}\right\}$. Similar relations hold for the action of $L_-$:

\[\begin{gathered} L_- \left\{{}_{s}Y_{ℓ,m}\right\} (R) = \sqrt{(ℓ+m)(ℓ-m+1)}\ {}_{s}Y_{ℓ,m-1}(R), \\ \left\{L_- f\right\}_{ℓ,m} = f_{ℓ,m+1}\, \sqrt{(ℓ+m+1)(ℓ-m)}. \end{gathered}\]

Seeing this, it is obvious that this "opposite" transformation is just down to the fact that $L_+$ and $L_-$ are dual operators, and we have

\[\left\langle {}_{s}Y_{ℓ,m} \middle| L_+ f \right\rangle = \left\langle L_- {}_{s}Y_{ℓ,m} \middle| f \right\rangle.\]

However, it is important to note that the same duality is not apparent for the spin-raising and -lowering operator laws:

\[\begin{aligned} \left\{\eth f\right\}_{s+1,ℓ,m} &= \frac{2}{\pi} \int \left\{\eth f(R)\right\}\, {}_{s+1}\bar{Y}_{ℓ,m}(R)\, dR \\ &= \frac{2}{\pi} \int \left\{\eth \sum_{ℓ',m'}f_{ℓ',m'}\, {}_{s}Y_{ℓ',m'}(R)\right\}\, {}_{s+1}\bar{Y}_{ℓ,m}(R)\, dR \\ &= \sum_{ℓ',m'} f_{ℓ',m'}\, \frac{2}{\pi} \int \left\{\eth {}_{s}Y_{ℓ',m'}(R)\right\}\, {}_{s+1}\bar{Y}_{ℓ,m}(R)\, dR \\ &= \sum_{ℓ',m'} f_{ℓ',m'}\, \sqrt{(ℓ'-s)(ℓ'+s+1)} \frac{2}{\pi} \int {}_{s+1}Y_{ℓ',m'}(R)\, {}_{s+1}\bar{Y}_{ℓ,m}(R)\, dR \\ &= \sum_{ℓ',m'} f_{ℓ',m'}\, \sqrt{(ℓ'-s)(ℓ'+s+1)} δ_{ℓ,ℓ'} δ_{m,m'} \\ &= \left\{f\right\}_{s,ℓ,m}\, \sqrt{(ℓ-s)(ℓ+s+1)}. \end{aligned}\]

The reason for this apparent asymmetry between $L_\pm$ and $\eth$ is actually just an asymmetry in the way we treat modes as they vary over $m$ and over $s$, belied by a sleight of hand we played in the notation above. We generally assemble mode weights for a single spin weight $s$, while varying the index $m$ over its full range. So the expression for $\left\{L_+ f\right\}_{ℓ,m}$ has $m$ on the left-hand side and $m-1$ on the right-hand side, whereas the expression for $\left\{\eth f\right\}_{s+1,ℓ,m}$ has $s+1$ on the left-hand side and $s$ on the right-hand side. That is, we assume that $f$ has a fixed spin weight $s$, so we don't care about the $s$ component of $\eth f$ — it is automatically zero. On the other hand, we assume that $f$ has a range of $m$ components, so we do care about the $m$ component of $L_+ f$. Perhaps a simpler way to see this is to write the modes as vectors, and look at the operators in matrix form, as we do in the next section.

Vector/matrix forms

We conventionally assemble the mode weights of a function $f$ with a fixed spin weight $s$ into a single vector of data, with $ℓ$ increasing, but $m$ running from $-ℓ$ to $ℓ$ between each increment of $ℓ$. Schematically, this looks like

\[f_{ℓ,m} \leftrightarrow [f_{0,0}, f_{1,-1}, f_{1,0}, f_{1,1}, f_{2,-2}, f_{2,-1}, f_{2,0}, f_{2,1}, f_{2,2}, \ldots]^T.\]

This is the form used throughout this package — and in particular the ModeWeights type. (Optionally, the $ℓ < |s|$ modes can be omitted since they are automatically zero, but on this page we will assume they are present, for simplicity.)

The differential operators can then be represented as matrices acting on these vectors. For example, $L_z$ is diagonal in this basis, with the $m$ values along the diagonal:

\[L_z = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 0 & -1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & -2 & 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & 0 & -1 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & \cdots \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}.\]

The $L_+$ operator is represented by a matrix that is not diagonal, but rather has nonzero entries just below the diagonal, with the square-root factors from the previous section:

\[L_+ = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 0 & \sqrt{2} & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & \sqrt{2} & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & 0 & \sqrt{6} & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & 0 & 0 & \sqrt{6} & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}.\]

Similarly, $L_-$ has nonzero entries just above the diagonal. Note that these matrices are independent of $s$ (though entries with $ℓ < |s|$ will never be used).

The $R$ matrices are different: they are all diagonal, and they all depend on $s$. $R_z$ is simply $s$ times the identity (up to irrelevant factors in the $ℓ < |s|$ entries):

\[R_z = \begin{pmatrix} s & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 0 & s & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & s & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & s & 0 & 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & s & 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & 0 & s & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & 0 & 0 & s & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & s & 0 & \cdots \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & s & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}.\]

But now, $R_+$ is also diagonal. For $s=1$, for example:

\[R_+ = \begin{pmatrix} 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 & 0 & \cdots \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & 0 & \cdots \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 2 & \cdots \\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{pmatrix}.\]

The contrast with the sub-diagonal $L_+$ is striking. The difference is that the matrix-vector product $L_+ f$ represents modes of a function with the same spin weight as $f$, while $R_+ f$ implicitly represents modes of a function with different spin weight — a fact that is not readily apparent in matrix-vector notation.

Because of this subtlety, we cannot represent $R_x = (R_+ + R_-)/2$ and $R_y = (R_+ - R_-)/(2i)$ as matrices in the same way. These operators result in functions with indefinite spin weight, so they cannot be represented as a single vector of mode weights. This package defines functions for L₊, L₋, R₊, and R₋, but only Lx and Ly — not Rx or Ry.