Fast and stable computations with spherical harmonics Richard Mika¨elSlevinsky Joint work with Hadrien Montanelli and Qiang Du Department of Mathematics, University of Manitoba [email protected] PDEs on the Sphere 2019 April 30, 2019 Spherical Harmonics Let µ be a positive Borel measure on D ⊂ Rn. The inner product: Z hf ; gi = f (x)g(x) dµ(x); D p 2 induces the norm kf k2 = hf ; f i and the Hilbert space L (D; dµ(x)). Let S2 ⊂ R3 denote the unit 2-sphere and let dΩ = sin θ dθ d'. Then any function f 2 L2(S2; dΩ) may be expanded in spherical harmonics: +1 +` +1 +1 X X m m X X m m f (θ; ') = f` Y` (θ; ') = f` Y` (θ; '); `=0 m=−` m=−∞ `=jmj where the expansion coefficients are: m m hY` ; f i f` = m m hY` ; Y` i 2 of 24 Spherical Harmonics For ` 2 N0 and jmj ≤ `, orthonormal spherical harmonics are defined by: s eim' (` − jmj)! Y m(x) = Y m(θ; ') = p (−1)jmj (` + 1 ) Pjmj(cos θ) : ` ` 2π 2 (` + jmj)! ` | {z } ~jmj P` (cos θ) Associated Legendre functions are defined by ultraspherical polynomials: 1 m m 1 m (m+ 2 ) P` (cos θ) = (−2) ( 2 )m sin θC`−m (cos θ): ~m The notation P` is used to denote orthonormality, and: Γ(x + n) (x) = n Γ(x) is the Pochhammer symbol for the rising factorial. 2 of 24 Synthesis and Analysis A band-limited function of degree-n on the sphere has no nonzero spherical harmonic expansion coefficient of degree-n or greater: n−1 +` X X m m fn−1(θ; ') = f` Y` (θ; '): `=0 m=−` m For a spherical harmonic Y` , ` is the degree and m is the order. The transforms of synthesis and analysis convert between representations of a band-limited function in momentum and physical spaces: Synthesis Sample a band-limited function at a set of points on S2. Analysis Convert samples on S2 to spherical harmonic expansion coefficients. Normally, an appropriate set of points is chosen to be able to perfectly reconstruct a band-limited function of degree-n. 3 of 24 Synthesis and Analysis Point sets include: Equiangular θk and 'j are equispaced and their Cartesian product is taken. These include the celebrated Driscoll{Healy (2n)(2n − 1) and McEwen{Wiaux (n − 1)(2n − 1) + 1 sampling theorems. Gaussian cos θk are Gauss{Legendre points and 'j are equispaced and their Cartesian product is taken. HEALPix procedure to generate points that correspond to a hierarchical equal area isolatitude pixelization of S2. Random with common distributions (e.g. uniformly distributed on S2). Data-driven may be treated similar to random. The na¨ıvecost of synthesis and analysis is O(n4) but it can be trivially reorganized to O(n3) for isolatitudinal point sets. 3 of 24 The Connection Problem Complementary to synthesis and analysis is the connection problem: the exact expansion of spherical harmonics in another basis. The sphere is doubly periodic and supports a bivariate Fourier series. The (N)FFT solves synthesis and analysis with bivariate Fourier series. eim' Y m(θ; ') = p P~m(cos θ) ) in longitude, we are done. ` 2π ` ~m The problem is to convert P` (cos θ) to Fourier series. (m+ 1 ) ~m m 2 Since P` (cos θ) / sin θC`−m (cos θ), the intuition is that ~m even-ordered P` are trigonometric polynomials in cos θ; and, ~m odd-ordered P` are trigonometric polynomials in sin θ. Conversions are not one-to-one. The connection problem is an analogue of the McEwen{Wiaux sampling theorem. 4 of 24 Connection vs. Synth. & Anal. 10 10 2 SHTns SHTns 10 11 2 FT FT 10 12 10 13 Relative Error 10 14 10 15 102 103 104 n + 1 5 of 24 Fast Transforms Fast transforms should: have an O(n2 logO(1) n) run-time with a similar pre-computation for synthesis and analysis or the connection problem; be (backward) stable; and, be fast in practice. Algorithms may be classified: as either exact in exact arithmetic and unstable in floating-point arithmetic, and numerically stable algorithms in fixed precision that are approximate to some arbitrarily small tolerance ; as either acceleration of synthesis and analysis (>10), or acceleration of the connection problem (3); or, by analytical apparatus: split-Legendre functions, WKB approximation, the fast multipole method (FMM), and the butterfly algorithm. 6 of 24 The SH Connection Problem Definition Let fφn(x)gn≥0 be a family of orthogonal functions with respect to L2(D^; dµ^(x)); and, let f n(x)gn≥0 be another family of orthogonal functions with respect to L2(D; dµ(x)). The connection coefficients: h `; φnidµ c`;n = ; h `; `idµ allow for the expansion: 1 X φn(x) = c`;n `(x): `=0 7 of 24 The SH Connection Problem Theorem Let fφn(x)gn≥0 and f n(x)gn≥0 be two families of orthonormal functions with respect to L2(D; dµ(x)). Then the connection coefficients satisfy: 1 X c`;mc`;n = δm;n: `=0 Any matrix A 2 Rm×n, m ≥ n, with orthonormal columns is well-conditioned and Moore{Penrose pseudo-invertible A+ = A>. ~m For every m, the P` (x) are a family of orthonormal functions for the same Hilbert space L2([−1; 1]; dx). 7 of 24 The SH Connection Problem Definition Let Gn denote the real Givens rotation: 01 ··· 0 0 0 ··· 01 B: :: : : : :C B: : : : : :C B C B0 ··· cn 0 sn ··· 0C B C Gn = B0 ··· 0 1 0 ··· 0C ; B C B0 · · · −sn 0 cn ··· 0C B C B: : : : :: :C @: : : : : :A 0 ··· 0 0 0 ··· 1 where the sines and the cosines are in the intersections of the nth and n + 2nd rows and columns, embedded in the identity of a conformable size. 7 of 24 The SH Connection Problem Theorem (Slevinsky 2017a) ~m+2 ~m The connection coefficients between Pn+m+2(cos θ) and P`+m(cos θ) are: s 8 (` + 2m)! (n + m + 5 )n! > (2` + 2m + 1)(2m + 2) 2 ; for ` ≤ n; ` + n even; > 1 > (` + m + 2 )`! (n + 2m + 4)! m < s c`;n = (n + 1)(n + 2) > − ; for ` = n + 2; > > (n + 2m + 3)(n + 2m + 4) : 0; otherwise: Furthermore, the matrix of connection coefficients (m) (m) (m) (m) (m) C = G0 G1 ··· Gn−1Gn I(n+3)×(n+1), where the sines and cosines are: s s (n + 1)(n + 2) (2m + 2)(2n + 2m + 5) sm = ; and cm = : n (n + 2m + 3)(n + 2m + 4) n (n + 2m + 3)(n + 2m + 4) 7 of 24 The SH Connection Problem \Proof." W.l.o.g., consider m = 0 and n = 5. 0 0:91287 0:0 0:31623 0:0 0:17593 0:0 1 B 0:0 0:83666 0:0 0:39641 0:0 0:24398 C B C B−0:40825 0:0 0:70711 0:0 0:3934 0:0 C B C (0) B 0:0 −0:54772 0:0 0:60553 0:0 0:37268 C C = B C B 0:0 0:0 −0:63246 0:0 0:5278 0:0 C B C B 0:0 0:0 0:0 −0:69007 0:0 0:46718 C B C @ 0:0 0:0 0:0 0:0 −0:73193 0:0 A 0:0 0:0 0:0 0:0 0:0 −0:76376 7 of 24 The SH Connection Problem (0)> \Proof." Apply the Givens rotation G0 : 01:0 0:0 0:0 0:0 0:0 0:0 1 B0:0 0:83666 0:0 0:39641 0:0 0:24398 C B C B0:0 0:0 0:7746 0:0 0:43095 0:0 C B C (0)> (0) B0:0 −0:54772 0:0 0:60553 0:0 0:37268 C G C = B C 0 B0:0 0:0 −0:63246 0:0 0:5278 0:0 C B C B0:0 0:0 0:0 −0:69007 0:0 0:46718 C B C @0:0 0:0 0:0 0:0 −0:73193 0:0 A 0:0 0:0 0:0 0:0 0:0 −0:76376 7 of 24 The SH Connection Problem \Proof." And again: 01:0 0:0 0:0 0:0 0:0 0:0 1 B0:0 1:0 0:0 0:0 0:0 0:0 C B C B0:0 0:0 0:7746 0:0 0:43095 0:0 C B C (0)> (0)> (0) B0:0 0:0 0:0 0:72375 0:0 0:44544 C G G C = B C 1 0 B0:0 0:0 −0:63246 0:0 0:5278 0:0 C B C B0:0 0:0 0:0 −0:69007 0:0 0:46718 C B C @0:0 0:0 0:0 0:0 −0:73193 0:0 A 0:0 0:0 0:0 0:0 0:0 −0:76376 7 of 24 The SH Connection Problem \Proof." And again: 01:0 0:0 0:0 0:0 0:0 0:0 1 B0:0 1:0 0:0 0:0 0:0 0:0 C B C B0:0 0:0 1:0 0:0 0:0 0:0 C B C (0)> (0)> (0)> (0) B0:0 0:0 0:0 0:72375 0:0 0:44544 C G G G C = B C 2 1 0 B0:0 0:0 0:0 0:0 0:68139 0:0 C B C B0:0 0:0 0:0 −0:69007 0:0 0:46718 C B C @0:0 0:0 0:0 0:0 −0:73193 0:0 A 0:0 0:0 0:0 0:0 0:0 −0:76376 7 of 24 The SH Connection Problem \Proof." And again: 01:0 0:0 0:0 0:0 0:0 0:0 1 B0:0 1:0 0:0 0:0 0:0 0:0 C B C B0:0 0:0 1:0 0:0 0:0 0:0 C B C (0)> (0)> (0)> (0)> (0) B0:0 0:0 0:0 1:0 0:0 0:0 C G G G G C = B C 3 2 1 0 B0:0 0:0 0:0 0:0 0:68139 0:0 C B C B0:0 0:0 0:0 0:0 0:0 0:6455 C B C @0:0 0:0 0:0 0:0 −0:73193 0:0 A 0:0 0:0 0:0 0:0 0:0 −0:76376 7 of 24 The SH Connection Problem \Proof." And again: 01:0 0:0 0:0 0:0 0:0 0:0 1 B0:0 1:0 0:0 0:0 0:0 0:0 C B C B0:0 0:0 1:0 0:0 0:0 0:0 C B C (0)> (0)> (0)> (0)> (0)> (0) B0:0 0:0 0:0 1:0 0:0 0:0 C G G G G G C = B C 4 3 2 1 0 B0:0 0:0 0:0 0:0 1:0 0:0 C B C B0:0 0:0 0:0 0:0 0:0 0:6455 C B C @0:0 0:0 0:0 0:0 0:0 0:0 A 0:0 0:0 0:0 0:0 0:0 −0:76376 7 of 24 The SH Connection Problem \Proof." And finally: 01:0 0:0 0:0 0:0 0:0 0:01 B0:0 1:0 0:0 0:0 0:0 0:0C B C B0:0 0:0 1:0 0:0 0:0 0:0C B C (0)> (0)> (0)> (0)> (0)> (0)> (0) B0:0 0:0 0:0 1:0 0:0 0:0C G G G G G G C = B C 5 4 3 2 1 0 B0:0 0:0 0:0 0:0 1:0 0:0C B C B0:0 0:0 0:0 0:0 0:0 1:0C B C @0:0 0:0 0:0 0:0 0:0 0:0A 0:0 0:0 0:0 0:0 0:0 0:0 7 of 24 The SH Connection Problem \Proof." Schematically: (m) C : I = : : (m) (m) G0 ··· Gn Conversion between neighbouring layers is O(n) flops and storage.