Symmetric Tensors and Spherical Harmonics

Introduction

This article is the second application of the category, and it identifies the symmetric powers of Symmetric Powers with the harmonic layers of the polynomial algebra. The symmetric-tensor statements hold over a field $F$ of characteristic $0$, so that all factorials are invertible and the symmetric power is the module of invariants; the polynomial algebra is that of The Symmetric Algebra, and the symmetric powers are its graded pieces $\operatorname{Sym}^k$.

Let $V = \mathbb{R}^n$ with its standard inner product and let $P_k$ be the space of homogeneous polynomials of degree $k$ in $x_1, \ldots, x_n$. The Laplacian $\Delta = \sum_i \partial_i^2$ maps $P_k$ to $P_{k-2}$, and the harmonic polynomials of degree $k$ are

$$ \mathcal{H}_k = \{f \in P_k : \Delta f = 0\}. $$

The two structures are the same up to a trace: symmetric powers of the dual are polynomials, and the harmonic polynomials are the symmetric tensors that are trace-free. The restriction of $\mathcal{H}_k$ to the unit sphere $S^{n-1}$ consists of eigenfunctions of the spherical Laplacian, and the resulting spherical harmonics decompose the functions on the sphere into orthogonal layers.

The article defines symmetric tensors and their contraction with the metric, develops the decomposition $P_k = \mathcal{H}_k \oplus r^2 P_{k-2}$, proves the orthogonality and eigenvalue properties of the spherical harmonics, and computes the dimension of each layer together with the trace-free interpretation. All mathematics is real; no physical interpretation of the harmonics is used.

Symmetric Tensors

Symmetric Powers and Symmetric Tensors

Let $V$ be a finite-dimensional vector space over a field $F$ of characteristic $0$ and let $V^*$ be its dual. By Symmetric Powers, the $k$-fold symmetric power $\operatorname{Sym}^k(V^*)$ is the quotient of the $k$-fold tensor power $(V^*)^{\otimes k}$ by the subspace spanned by the tensors $\varphi_1 \otimes \cdots \otimes \varphi_k - \varphi_{\sigma(1)}\otimes\cdots\otimes\varphi_{\sigma(k)}$ for $\sigma$ in the symmetric group $S_k$. An element of $\operatorname{Sym}^k(V^*)$ is a symmetric tensor of rank $k$, that is, a $k$-linear form $\varphi : V^k \to F$ invariant under permutation of its arguments. In characteristic $0$ the symmetric power is canonically the module of $S_k$-invariant elements of $(V^*)^{\otimes k}$, so this use of the term agrees with the invariants that Symmetric Powers calls symmetric tensors in the strict sense.

Proposition. If $V$ has basis $e_1, \ldots, e_n$ with dual basis $e^1, \ldots, e^n$ defined by $e^i(e_j) = \delta^i_j$, then the products $e^{i_1}\cdots e^{i_k}$ with $i_1 \leq \cdots \leq i_k$ are a basis of $\operatorname{Sym}^k(V^*)$. Hence

$$ \dim \operatorname{Sym}^k(V^*) = \binom{n+k-1}{k}. $$

Proof. The monomials in the $e^i$ of degree $k$ span the symmetric power under the multiplication $\operatorname{Sym}^i\times\operatorname{Sym}^j\to\operatorname{Sym}^{i+j}$, and the commutative law identifies all orderings of the factors; the nondecreasing multi-indices parametrise the distinct monomials. Independence follows by evaluating on the symmetric tensors $e_{i_1}\cdots e_{i_k}$, whose coordinates are $\delta$-symbols.

Example. $\operatorname{Sym}^2(V^*)$ has dimension $\binom{n+1}{2}$ and is the space of quadratic forms; its basis is $e^ie^j$ for $i\leq j$. The space $\operatorname{Sym}^k(V^*)$ is canonically the space $P_k$ of homogeneous polynomials of degree $k$ in the coordinate functions, and this identification is used throughout.

Symmetric Tensors under the Orthogonal Group

Assume $F = \mathbb{R}$ and let $V = \mathbb{R}^n$ carry the standard positive definite inner product $g$, with $g_{ij} = \delta_{ij}$. The orthogonal group $O(n)$ acts on $\operatorname{Sym}^k(V^*)$ by $(g\varphi)(v_1,\ldots,v_k) = \varphi(g^{-1}v_1,\ldots,g^{-1}v_k)$, and this action determines the structure of the layer. Inside $\operatorname{Sym}^k(V^*)$ sits the image of the contraction with the metric,

$$ \operatorname{Sym}^{k-2}(V^*) \longrightarrow \operatorname{Sym}^k(V^*), \qquad \varphi \longmapsto g\,\varphi, $$

which lowers the rank by two by summing over a pair of indices with the metric and its inverse. The trace-free or harmonic symmetric tensors are the kernel of a dual contraction; when the metric is the identity the contraction is the Laplacian and the trace-free tensors are exactly the harmonic polynomials defined below.

Proposition. The trace-free symmetric tensors of rank $k$ on $\mathbb{R}^n$ form a subspace of dimension $\binom{n+k-1}{k} - \binom{n+k-3}{k-2}$ for $k \geq 2$, and this is the dimension of $\mathcal{H}_k$.

The equality of dimensions is established in the last section together with the decomposition; it is the tensor form of the isomorphism between harmonic polynomials and trace-free symmetric tensors.

Harmonic Polynomials

The Laplacian

Let $P_k$ be the space of homogeneous polynomials of degree $k$ in $x_1, \ldots, x_n$ over $\mathbb{R}$, and let

$$ \Delta = \sum_{i=1}^{n}\frac{\partial^2}{\partial x_i^2} : P_k \longrightarrow P_{k-2} $$

be the Laplacian; it is a linear map decreasing the degree by two. A polynomial is harmonic if $\Delta f = 0$, and

$$ \mathcal{H}_k = \ker\bigl(\Delta : P_k \to P_{k-2}\bigr) . $$

Example. Every element of $P_0$ and of $P_1$ is harmonic, so $\mathcal{H}_0 = P_0$ and $\mathcal{H}_1 = P_1$. In $P_2$ the harmonic polynomials are the traceless quadratic forms; for $n = 3$ a basis is $xy$, $yz$, $zx$, $x^2 - y^2$, $x^2 - z^2$, so $\dim\mathcal{H}_2 = 5$.

The Fundamental Decomposition

Write $r^2 = x_1^2 + \cdots + x_n^2$ for the squared distance from the origin.

Lemma. For any polynomial $h$, the Laplacian of $r^2 h$ is

$$ \Delta(r^2 h) = 4\,x\cdot\nabla h + 2n\,h + r^2\,\Delta h . $$

Proof. The product rule gives $\Delta(r^2 h) = (\Delta r^2)h + 2\nabla r^2 \cdot \nabla h + r^2 \Delta h$, and $\nabla r^2 = 2x$, $\Delta r^2 = 2n$, so $2\nabla r^2\cdot\nabla h = 4\,x\cdot\nabla h$.

If $h$ is homogeneous of degree $k - 2$, then $x \cdot \nabla h = (k-2)h$ by Euler's relation, and the lemma reads

$$ \Delta(r^2 h) = \bigl(2n + 4(k-2)\bigr) h + r^2 \Delta h , \qquad h \in P_{k-2} . $$

Theorem (harmonic decomposition). For each $k \geq 0$,

$$ P_k = \mathcal{H}_k \oplus r^2 P_{k-2}, $$

and iterating,

$$ P_k = \bigoplus_{j=0}^{\lfloor k/2\rfloor} r^{2j}\,\mathcal{H}_{k-2j} . $$

Proof. Put $m = k - 2$ and filter $P_m$ by the submodules $F_j = r^{2j}P_{m-2j}$, $j \geq 0$, a descending chain with $F_0 = P_m$ and $F_j = 0$ for $2j > m$. The operator

$$ T = \Delta \circ (r^2\,\cdot) : P_{m} \longrightarrow P_{m}, \qquad T(h) = c_k h + r^2\,\Delta h, \qquad c_k = 2n + 4(k-2), $$

preserves the filtration. Indeed, for $g \in P_{m-2j}$ and $u = r^{2j}g$, the product rule applied to $r^{2(j+1)}g$ gives

$$ \Delta\bigl(r^{2j+2}g\bigr) = \beta_j\, r^{2j} g + r^{2j+2}\Delta g, \qquad \beta_j = 2(j+1)\bigl(n + 2k - 4 - 2j\bigr), $$

using $x\cdot\nabla g = (m-2j)g$; the second summand lies in $F_{j+1}$, so $T$ induces the scalar $\beta_j$ on each graded piece $F_j/F_{j+1}$. For $0 \leq j \leq m/2$ one has $n + 2k - 4 - 2j \geq n + k - 2 \geq 1$, so every $\beta_j$ is nonzero over $\mathbb{R}$, and a triangular operator with nonzero diagonal is invertible. Hence $T$ is invertible, so

$$ \Delta\bigl(r^2 P_{k-2}\bigr) = P_{k-2} $$

and $\Delta : P_k \to P_{k-2}$ is surjective; therefore $\dim\mathcal{H}_k = \dim P_k - \dim P_{k-2}$. The intersection $r^2P_{k-2}\cap\mathcal{H}_k$ is zero, because $r^2h\in\mathcal{H}_k$ means $T(h) = \Delta(r^2h) = 0$, and $T$ is invertible. Hence the sum $\mathcal{H}_k + r^2P_{k-2}$ is direct of dimension $\dim P_k$, so it equals $P_k$. Iterating the identity gives the second formula.

Remark. The first graded diagonal is $\beta_0 = c_k$, recovering the lemma; the later ones differ from it, since $\Delta$ does not respect the decomposition by harmonic degree but only the filtration by powers of $r^2$.

Corollary. $\dim\mathcal{H}_k = \dim P_k - \dim P_{k-2}$ for $k \geq 2$, and $\mathcal{H}_0 = P_0$, $\mathcal{H}_1 = P_1$.

Spherical Harmonics

Restriction to the Sphere

Let $S^{n-1} = \{x \in \mathbb{R}^n : r^2 = 1\}$ be the unit sphere with surface measure $d\sigma$. The restriction of a polynomial to the sphere is not injective on $P_k$ if $k \geq 2$, because $r^2 - 1$ vanishes there; the following theorem shows that it is injective on each harmonic layer and identifies the images.

Theorem. The restriction map $\mathcal{H}_k \to C^{\infty}(S^{n-1})$, $f \mapsto f|_{S^{n-1}}$, is injective.

Proof. Suppose $f \in \mathcal{H}_k$ vanishes on the sphere. Since $f$ is homogeneous of degree $k$, the identity $f(x) = r^k f(x/r)$ holds for $x \neq 0$, and the right hand side vanishes whenever $r = 1$; by homogeneity $f$ vanishes on every sphere $r = \rho$, hence on all of $\mathbb{R}^n$ except possibly the origin, and by continuity everywhere.

The images of the layers are the spherical harmonics. The elements of $\mathcal{H}_k|_{S^{n-1}}$ are eigenfunctions of the spherical Laplacian.

Theorem. Let $\Delta_{S}$ be the Laplace–Beltrami operator of $S^{n-1}$. For $f \in \mathcal{H}_k$,

$$ \Delta_S f = -k(k + n - 2)\, f . $$

Proof. In polar coordinates the Euclidean Laplacian is $\Delta = \partial_r^2 + \frac{n-1}{r}\partial_r + \frac{1}{r^2}\Delta_S$. For a homogeneous harmonic $f$ of degree $k$ one has $\partial_r^2 f = k(k-1)r^{k-2}\tilde f$ and $\frac{n-1}{r}\partial_r f = (n-1)k r^{k-2}\tilde f$, where $\tilde f = f|_{S^{n-1}}$. Hence

$$ 0 = \Delta f = r^{k-2}\Bigl(k(k-1) + (n-1)k\Bigr)\tilde f + r^{k-2}\Delta_S \tilde f , $$

and $\Delta_S\tilde f = -k(k-1) - (n-1)k\,\tilde f = -k(k+n-2)\tilde f$.

Example. For $n = 3$ the eigenvalue is $-k(k+1)$, so the spherical harmonics of degrees $0, 1, 2$ have eigenvalues $0, -2, -6$. The polynomial $f = x^2 - y^2$ restricts to the sphere as $\sin^2\theta\cos2\varphi$, and a direct computation of $\Delta_{S^2}$ gives $\Delta_{S^2} f = -6f$, in agreement with the theorem.

Orthogonality

Theorem. For $j \neq k$, the layers $\mathcal{H}_j$ and $\mathcal{H}_k$ are orthogonal in $L^2(S^{n-1}, d\sigma)$.

Proof. Let $f \in \mathcal{H}_j$, $h \in \mathcal{H}_k$ and consider the vector field $F = h\,\nabla f - f\,\nabla h$. Its divergence is

$$ \nabla\cdot F = h\,\Delta f - f\,\Delta h = 0 . $$

Integrating over the unit ball $B$ and applying the divergence theorem,

$$ 0 = \int_B \nabla\cdot F\,dV = \int_{S^{n-1}} \langle F, x\rangle\,d\sigma = \int_{S^{n-1}}\bigl(h\,\partial_r f - f\,\partial_r h\bigr)d\sigma , $$

because the outward normal to the ball on the sphere is $x$. Since $f$ and $h$ are homogeneous, Euler's relation gives $\partial_r f = j f$ and $\partial_r h = k h$, so

$$ 0 = (j - k)\int_{S^{n-1}} f h\,d\sigma , $$

and $j \neq k$ gives the orthogonality.

Corollary. The restrictions of distinct harmonic layers are orthogonal on the sphere, and within each layer the $O(n)$-action is irreducible; the spaces $\mathcal{H}_k$ are the irreducible components of the natural action of $O(n)$ on the polynomial algebra.

The completeness of the spherical harmonics is the statement of Peter–Weyl for the compact group $O(n)$, or equivalently of the density of the polynomial algebra in $C(S^{n-1})$: every continuous function on the sphere can be uniformly approximated by polynomials, and the harmonic decomposition of the polynomials gives the orthogonal expansion. This is standard and is quoted.

Zonal Harmonics and the Addition Theorem

Fix a unit vector $y \in S^{n-1}$. By the irreducibility of the $O(n)$-action on $\mathcal{H}_k$, the subspace fixed by the stabiliser $\mathrm{Stab}(y) \cong O(n-1)$ is one-dimensional; its generator is the zonal harmonic $Z_k(\cdot, y)$, characterised by the reproducing property

$$ f(y) = \int_{S^{n-1}} f(x)\, Z_k(x, y)\, d\sigma(x), \qquad f \in \mathcal{H}_k . $$

The zonal harmonic depends only on the inner product $\langle x, y\rangle$, so it is a function $Z_k(t)$ of one variable $t = \cos\gamma$, where $\gamma$ is the geodesic distance between $x$ and $y$; up to normalisation it is the Gegenbauer or ultraspherical polynomial $C_k^{(n/2 - 1)}(t)$. For $n = 3$ this is the Legendre polynomial, and the reproducing kernel is $(2k+1)P_k(t)/(4\pi)$.

The reproducing property for an orthonormal basis $Y_{k1}, \ldots, Y_{kd}$ of $\mathcal{H}_k$ reads $Z_k(x,y) = \sum_{j} Y_{kj}(x)\overline{Y_{kj}(y)}$, and taking $n = 3$, where $d = 2k+1$, gives the classical addition theorem

$$ P_k(\cos\gamma) = \frac{4\pi}{2k+1}\sum_{m=-k}^{k} Y_{km}(\theta, \phi)\,\overline{Y_{km}(\theta', \phi')} , $$

with $\cos\gamma = \cos\theta\cos\theta' + \sin\theta\sin\theta'\cos(\phi - \phi')$. The rewriting of the kernel in the single variable $t$ is the general addition theorem for the harmonic layers; it expresses the reproducing kernel of each irreducible summand of the polynomial algebra in terms of a single classical special function.

Dimensions of the Layers

The Dimension Formula

Theorem. For $k \geq 2$,

$$ \dim\mathcal{H}_k = \binom{n+k-1}{k} - \binom{n+k-3}{k-2} = \frac{(2k+n-2)\,(k+n-3)!}{k!\,(n-2)!} , $$

while $\dim\mathcal{H}_0 = 1$ and $\dim\mathcal{H}_1 = n$.

Proof. The first equality is the corollary of the decomposition theorem, and the second is the algebraic simplification of the difference of the two binomial coefficients, using $\binom{n+k-3}{k-2} = \frac{(n+k-3)!}{(k-2)!\,(n-1)!}$.

Example. For $n = 3$, $\dim\mathcal{H}_k = 2k+1$, the classical dimension of the degree-$k$ spherical harmonics on the two-sphere. For $n = 4$, $\dim\mathcal{H}_k = (k+1)^2$. For $n = 2$, $\dim\mathcal{H}_k = 2$ for every $k \geq 1$: the harmonic homogeneous polynomials of degree $k$ in two variables are spanned by the real and imaginary parts of $z^k$. The dimensions for small $n$ and $k$ are

$k$	$n = 2$	$n = 3$	$n = 4$	$n = 5$
0	1	1	1	1
1	2	3	4	5
2	2	5	9	14
3	2	7	16	30
4	2	9	25	55

Corollary. The restrictions to the sphere of the polynomials of degree at most $K$ span $\bigoplus_{k \leq K}\mathcal{H}_k|_{S^{n-1}}$, of dimension

$$ \sum_{k=0}^{K}\dim\mathcal{H}_k = \dim P_K + \dim P_{K-1} = \binom{n+K-1}{K} + \binom{n+K-2}{K-1}, $$

by telescoping, with the second term omitted for $K = 0$. For $n = 3$ this dimension is $(K+1)^2$, the classical count of the spherical harmonics of degree at most $K$ on the two-sphere.

Trace-Free Tensors

The harmonic layers and the trace-free symmetric tensors are the same object, read through the metric.

Theorem. Under the identification of $P_k$ with $\operatorname{Sym}^k(V^*)$, the harmonic polynomials of degree $k$ correspond exactly to the symmetric tensors of rank $k$ that are trace-free with respect to $g$, and the trace-free symmetric tensors of rank $k$ form a subspace of dimension $\dim\mathcal{H}_k$.

Proof. The Laplacian on polynomials corresponds to the metric contraction on symmetric tensors: with the standard form, $\Delta$ is the contraction of two indices by $\delta^{ij}$, so $\Delta f = 0$ is exactly the trace-free condition. The dimensions agree by the formula above.

Corollary. The dimension of the space of quadratic forms on $\mathbb{R}^n$ is $\binom{n+1}{2} = \tfrac{n(n+1)}{2}$, and the trace-free part has dimension $\binom{n+1}{2} - 1 = \tfrac{n(n+1)}{2} - 1$, which for $n = 3$ is $5$, the number of independent harmonic quadratics. The decomposition $P_2 = \mathcal{H}_2 \oplus r^2 P_0$ separates the quadratic form into its trace part $\tfrac{\operatorname{tr} Q}{n}r^2$ and its trace-free part, and this is the spectral decomposition of a quadratic form into its mean and its traceless component.

Summary

Let $V = \mathbb{R}^n$ with its standard inner product. The symmetric tensors of rank $k$ on $V^*$ are the symmetric powers $\operatorname{Sym}^k(V^*)$, of dimension $\binom{n+k-1}{k}$, and they are the homogeneous polynomials of degree $k$. The Laplacian $\Delta : P_k \to P_{k-2}$ defines the harmonic polynomials $\mathcal{H}_k = \ker\Delta$, and the fundamental decomposition is

$$ P_k = \mathcal{H}_k \oplus r^2 P_{k-2}, \qquad P_k = \bigoplus_{j=0}^{\lfloor k/2\rfloor} r^{2j}\mathcal{H}_{k-2j}, $$

proved from the identity $\Delta(r^2 h) = (2n+4(k-2))h + r^2\Delta h$ on $P_{k-2}$ and the downward induction on the degree. The restriction of $\mathcal{H}_k$ to the unit sphere is injective, its image consists of eigenfunctions of the spherical Laplacian with eigenvalue $-k(k+n-2)$, and distinct layers are orthogonal in $L^2(S^{n-1})$, proved by the divergence theorem applied to $h\nabla f - f\nabla h$ and Euler's relation. The dimension of the layer is

$$ \dim\mathcal{H}_k = \binom{n+k-1}{k} - \binom{n+k-3}{k-2} = \frac{(2k+n-2)(k+n-3)!}{k!(n-2)!}, $$

equal to $2k+1$ for $n=3$ and to $(k+1)^2$ for $n=4$, and the harmonic layers are exactly the trace-free symmetric tensors. The spherical harmonics are the images of the layers, and their orthogonality and completeness are the Fourier expansion on the sphere.

Summary of Notation

Symbol	Meaning
$F$	Base field of characteristic $0$
$V = \mathbb{R}^n$, $V^*$	Euclidean space and its dual
$e_1, \ldots, e_n$; $e^1, \ldots, e^n$	Basis of $V$ and its dual basis, $e^i(e_j) = \delta^i_j$
$g$, $g_{ij} = \delta_{ij}$	Standard positive definite inner product
$\operatorname{Sym}^k(V^*)$	Symmetric tensors of rank $k$, dimension $\binom{n+k-1}{k}$
$P_k$	Homogeneous polynomials of degree $k$
$\Delta = \sum_i\partial_i^2$	Laplacian, $P_k \to P_{k-2}$
$\mathcal{H}_k = \ker\Delta \cap P_k$	Harmonic polynomials of degree $k$
$r^2 = x_1^2 + \cdots + x_n^2$	Squared norm
$S^{n-1}$, $d\sigma$	Unit sphere and surface measure
$\Delta_S$	Laplace–Beltrami operator; eigenvalue $-k(k+n-2)$
$O(n)$	Orthogonal group acting on each layer
$\binom{n+k-1}{k} - \binom{n+k-3}{k-2}$	Dimension of $\mathcal{H}_k$
Trace-free	Symmetric tensors in the kernel of metric contraction
$Z_k(x,y)$	Zonal harmonic, reproducing kernel of $\mathcal{H}_k$
$Y_{km}$	Orthonormal basis of $\mathcal{H}_k$ on $S^2$
$C_k^{(n/2-1)}(t)$	Gegenbauer polynomial of the zonal harmonic