Symmetric Powers

Introduction

This article introduces the symmetric powers of a module. The treatment is introductory and purely mathematical. The aim is to construct the symmetric power, to exhibit its universal property, and to compute its basis and rank.

The base structure is a commutative ring $R$ with identity $1 \neq 0$, the default of the series. We say at each point whether a statement needs a field, characteristic zero, or the invertibility of a factorial. Tensor products and tensor powers of modules are taken from Modules (§13); the tensor algebra and its general quotients belong to category 05, and the anti-symmetric counterpart — exterior powers, the exterior algebra and the determinant — belongs to category 07 and is not treated here. The quotient studied in this article is specifically the one by the two-sided ideal generated by $x \otimes y - y \otimes x$.

The symmetric power is the commutative counterpart of the tensor power. The tensor power $M^{\otimes n}$ records an ordered $n$-tuple of vectors; the symmetric power $\operatorname{Sym}^n M$ records an unordered $n$-tuple, so that $x \otimes y$ and $y \otimes x$ become the same element. It is the module that receives the symmetric multilinear maps from $M^n$, just as the tensor power receives the multilinear ones.

An essential point, and one that is easy to miss over a general ring, is that there are two natural candidates for "the unordered $n$-tuple": the module of vectors fixed by the permutation action, and the module of coinvariants, in which the action is divided out. Over a field of characteristic zero, or more generally whenever $n!$ is invertible, the two coincide. Over a general commutative ring they do not, and the symmetric power is by definition the coinvariant one. The refined object that repairs the difference is the divided power, treated in Divided Powers.

Multilinear and Symmetric Maps

Multilinear Maps

Let $M$ and $P$ be $R$-modules. A map

$$ f : M^n = \underbrace{M \times \cdots \times M}_{n} \longrightarrow P $$

is $n$-multilinear if it is $R$-linear in each argument separately: for every index $i$ and every fixed choice of the remaining arguments, the map $x \mapsto f(x_1, \ldots, x_{i-1}, x, x_{i+1}, \ldots, x_n)$ is $R$-linear. For $n = 1$ this is ordinary $R$-linearity, and for $n = 0$ a $0$-multilinear map is just an element of $P$, that is, a constant.

Multilinearity is equivalent to the four families of identities of Modules §13 written out in each variable: additivity in each variable and compatibility with scalars in each variable.

Symmetric Multilinear Maps

The symmetric group $S_n$ acts on $M^n$ by permuting the arguments. A multilinear map $f$ is symmetric if

$$ f(x_{\sigma(1)}, \ldots, x_{\sigma(n)}) = f(x_1, \ldots, x_n) $$

for every $\sigma \in S_n$ and all $x_1, \ldots, x_n \in M$. Since the transpositions generate $S_n$, it suffices to require invariance under the interchange of any two arguments.

Example. The product of a polynomial algebra is symmetric multilinear in a suitable sense: the monomial $x_1 \cdots x_n$ does not depend on the order of its factors. The dot product $\langle \cdot, \cdot \rangle$ on a free module is a symmetric bilinear map $M \times M \to R$.

Example. Let $\mathbb{K}$ be $\mathbb{R}$ or $\mathbb{C}$ and let $M = \mathbb{K}^d$ with basis $e_1, \ldots, e_d$. The monomial map

$$ (x_1, \ldots, x_n) \longmapsto \prod_{j=1}^{n} \Bigl(\sum_{i=1}^{d} x_{j,i} \, e_i\Bigr) $$

is the product of the $n$ linear forms, and is visibly symmetric multilinear; read through the defining map $\pi_n$ constructed below, it produces the monomial basis of $\operatorname{Sym}^n M$.

The Universal Property

Definition of the Symmetric Power

The $n$-th symmetric power of $M$ is an $R$-module $\operatorname{Sym}^n M$ together with a symmetric multilinear map

$$ \pi_n : M^n \longrightarrow \operatorname{Sym}^n M $$

such that for every $R$-module $P$ and every symmetric multilinear map $f : M^n \to P$ there is a unique $R$-linear map $\tilde{f} : \operatorname{Sym}^n M \to P$ with

$$ \tilde{f} \circ \pi_n = f . $$

The map $\pi_n$ is the defining map, and the element $\pi_n(x_1, \ldots, x_n)$ is written $x_1 \cdots x_n$; it is a class modulo the submodule $J_n$ of relations constructed in the next section, not a tensor, and the notation with the product sign keeps the two apart. The universal property says that the symmetric powers corepresent symmetric multilinear maps:

$$ \operatorname{Hom}_R(\operatorname{Sym}^n M, P) \cong \operatorname{SymMult}_R(M^n; P), $$

naturally in $P$.

Uniqueness

The universal property determines the symmetric power up to a unique isomorphism. If $\pi_n : M^n \to S$ and $\pi_n' : M^n \to S'$ are two objects with the universal property, applying it to each with the other as target produces unique maps $u : S \to S'$ and $v : S' \to S$ with $u \pi_n = \pi_n'$ and $v \pi_n' = \pi_n$. Then $vu \pi_n = \pi_n = \operatorname{id}_S \pi_n$, and uniqueness applied to $\operatorname{id}_S$ gives $vu = \operatorname{id}_S$; symmetrically $uv = \operatorname{id}_{S'}$. So $u$ is an isomorphism.

Convention. The $0$-th symmetric power is $\operatorname{Sym}^0 M = R$ with defining map the constant $1 \in R$, and the first is $\operatorname{Sym}^1 M = M$ with $\pi_1 = \operatorname{id}_M$. These follow from the universal property: a $0$-multilinear map is a constant and an $R$-linear recipient of a constant is determined by the constant; a $1$-multilinear map is linear.

Construction as a Quotient of the Tensor Power

The Tensor Power

The $n$-th tensor power is the iterated tensor product

$$ M^{\otimes n} = M \otimes_R M \otimes_R \cdots \otimes_R M \qquad (n \text{ factors}), $$

with $M^{\otimes 0} = R$ and $M^{\otimes 1} = M$. Its elements are finite sums of elementary tensors $x_1 \otimes \cdots \otimes x_n$; by the associativity of the tensor product (Modules, §13), the parenthesisation is immaterial. The tensor power is characterised by the universal property that every $n$-multilinear map $M^n \to P$ factors uniquely through an $R$-linear map $M^{\otimes n} \to P$.

The Permutation Action

For each $\sigma \in S_n$ there is a unique $R$-linear automorphism of $M^{\otimes n}$ with

$$ \sigma \cdot (x_1 \otimes \cdots \otimes x_n) = x_{\sigma^{-1}(1)} \otimes \cdots \otimes x_{\sigma^{-1}(n)}. $$

Existence follows from the universal property applied to the multilinear map $(x_1, \ldots, x_n) \mapsto x_{\sigma^{-1}(1)} \otimes \cdots \otimes x_{\sigma^{-1}(n)}$; invertibility follows because the construction is reversed by $\sigma^{-1}$. This makes $M^{\otimes n}$ a module over the group ring $R[S_n]$, and the correspondence $\sigma \mapsto (\text{that automorphism})$ a representation of $S_n$ by $R$-linear maps.

The Quotient

Let $J_n \subseteq M^{\otimes n}$ be the submodule generated by all the elements

$$ x_1 \otimes \cdots \otimes x_n - x_{\sigma(1)} \otimes \cdots \otimes x_{\sigma(n)}, \qquad x_i \in M,\ \sigma \in S_n. $$

Equivalently, $J_n$ is spanned by the images of the operators $\sigma - 1$, $\sigma \in S_n$; since a permutation is a product of transpositions and $\sigma - 1$ is an $R[S_n]$-combination of the differences $\tau - 1$ for transpositions $\tau$, it suffices to take the elements $x \otimes y - y \otimes x$ in the second tensor power together with their products with the remaining factors. For $M = R^d$ free, the two-sided ideal of the tensor algebra generated by these elements is the one whose quotient is the polynomial algebra in $d$ variables; that graded statement is developed.

Theorem. $\operatorname{Sym}^n M \cong M^{\otimes n} / J_n$, with defining map the composite $M^n \to M^{\otimes n} \to M^{\otimes n}/J_n$.

Proof. Write $Q = M^{\otimes n}/J_n$ and let $q : M^{\otimes n} \to Q$ be the quotient map. The composite $\pi = q \circ \otimes^n$ is symmetric multilinear: it is multilinear because $\otimes^n$ is and $q$ is linear, and it is symmetric because $q$ kills every $J_n$, so $q(x_1 \otimes \cdots \otimes x_n) = q(x_{\sigma(1)} \otimes \cdots \otimes x_{\sigma(n)})$.

Now let $f : M^n \to P$ be symmetric multilinear. By the universal property of the tensor power there is a unique $R$-linear $\bar f : M^{\otimes n} \to P$ with $\bar f(x_1 \otimes \cdots \otimes x_n) = f(x_1, \ldots, x_n)$. Symmetry of $f$ gives $\bar f(x_1 \otimes \cdots \otimes x_n) = \bar f(x_{\sigma(1)} \otimes \cdots \otimes x_{\sigma(n)})$, so $\bar f$ vanishes on the generators of $J_n$; hence $\bar f$ factors uniquely through a map $\tilde f : Q \to P$. This is the required factorisation, and uniqueness holds because the elementary tensors generate $M^{\otimes n}$, hence their images generate $Q$.

Thus the symmetric power is exactly the module of coinvariants of the permutation action,

$$ \operatorname{Sym}^n M \cong (M^{\otimes n})_{S_n} = M^{\otimes n} \big/ \langle \sigma v - v : v \in M^{\otimes n},\ \sigma \in S_n \rangle, $$

where $\langle - \rangle$ denotes the submodule generated.

Coinvariants and Invariants

The submodule of invariants is

$$ (M^{\otimes n})^{S_n} = \{v \in M^{\otimes n} : \sigma v = v \text{ for all } \sigma \in S_n\}, $$

the symmetric tensors in the strict sense. There is always a natural map

$$ (M^{\otimes n})^{S_n} \longrightarrow (M^{\otimes n})_{S_n} = \operatorname{Sym}^n M, $$

restricting the quotient map to the fixed submodule. Its kernel is the intersection $(M^{\otimes n})^{S_n} \cap J_n$.

When $n!$ is invertible in $R$, the symmetrisation operator

$$ s_n = \frac{1}{n!} \sum_{\sigma \in S_n} \sigma \ \in R[S_n] $$

is an idempotent projector whose image is $(M^{\otimes n})^{S_n}$ and whose kernel is $J_n$. Consequently

$$ M^{\otimes n} = (M^{\otimes n})^{S_n} \oplus J_n $$

and the natural map $(M^{\otimes n})^{S_n} \to \operatorname{Sym}^n M$ is an isomorphism. This is the precise sense in which the two candidates agree in characteristic zero. Over a field of characteristic dividing $n!$ they genuinely differ.

Example (the characteristic-2 failure). Take $R = \mathbb{F}_2$, $n = 2$, $M = R^2$ with basis $e_1, e_2$. The swap fixes $e_1 \otimes e_1$, $e_2 \otimes e_2$ and $e_1 \otimes e_2 + e_2 \otimes e_1$, so the invariants have dimension $3$. Since $-1 = +1$, the element $e_1 \otimes e_2 - e_2 \otimes e_1 = e_1 \otimes e_2 + e_2 \otimes e_1$ is fixed and also lies in $J_2$; the natural map therefore has a one-dimensional kernel and is neither injective nor surjective onto $\operatorname{Sym}^2 M$. The symmetric tensors and the symmetric power are distinct.

Remark. Over an integral domain in which $n! \neq 0$, the symmetrisation operator exists over the fraction field, and the failure above is invisible there. It reappears exactly when factorials cease to be units, which is the phenomenon the divided powers are designed to handle.

Basis, Rank and Dimension

The Monomial Basis

Assume now that $M$ is free of finite rank $d$ with basis $e_1, \ldots, e_d$. For a multi-index $a = (a_1, \ldots, a_d)$ with $a_i \geq 0$ and $|a| = a_1 + \cdots + a_d = n$, write

$$ e^a = e_1^{a_1} e_2^{a_2} \cdots e_d^{a_d} = \pi_n(\underbrace{e_1, \ldots, e_1}_{a_1}, \ldots, \underbrace{e_d, \ldots, e_d}_{a_d}) \in \operatorname{Sym}^n M. $$

Theorem. If $M$ is free with basis $e_1, \ldots, e_d$, then $\operatorname{Sym}^n M$ is free with basis $\{e^a : |a| = n\}$.

Proof. The elements $e^a$ span. Indeed every elementary tensor $x_1 \otimes \cdots \otimes x_n$ is a sum of elementary tensors in the basis elements, and $\pi_n$ is multilinear, so its image is a sum of terms $e^a$; the images of the elementary tensors span the quotient, hence the $e^a$ span.

For independence, let $P_n = R[x_1, \ldots, x_d]_n$ be the degree-$n$ part of the polynomial algebra, a free module with basis the monomials $x^a$, $|a| = n$. The map

$$ M^n \longrightarrow P_n, \qquad (u_1, \ldots, u_n) \longmapsto \prod_{j=1}^{n} \Bigl(\sum_{i=1}^{d} u_{j,i} x_i\Bigr), $$

where $u_j = \sum_i u_{j,i} e_i$, is symmetric multilinear, so by the universal property it induces an $R$-linear map $\Phi : \operatorname{Sym}^n M \to P_n$ with $\Phi(e^a) = x^a$. In the other direction let $\psi : P_n \to \operatorname{Sym}^n M$ be the unique $R$-linear map on the basis with $\psi(x^a) = e^a$. Then $\Phi\psi = \operatorname{id}_{P_n}$ because it fixes the basis, and $\psi\Phi(e^a) = \psi(x^a) = e^a$ for every $a$, so $\psi\Phi = \operatorname{id}_{\operatorname{Sym}^n M}$ because the $e^a$ span. Hence $\Phi$ is an isomorphism and the $e^a$ are a basis.

Equivalently, the map $R[x_1, \ldots, x_d] \to \operatorname{Sym}(R^d)$ sending $x_i \mapsto e_i$ is an isomorphism of graded algebras; this identification is developed.

Dimension Formula

The number of multi-indices $a$ with $a_i \geq 0$ and $|a| = n$ is the number of ways of distributing $n$ indistinguishable balls into $d$ boxes, namely

$$ \operatorname{rank}_R \operatorname{Sym}^n M = \binom{n + d - 1}{d - 1} = \binom{n + d - 1}{n}. $$

For $d = 1$ this is $1$ for every $n$; for $n = 1$ it is $d$; for $n = 2$ it is $\frac{d(d+1)}{2}$, the dimension of the space of symmetric $d \times d$ matrices. The values for small $d$ and $n$ are tabulated below.

Worked values.

The entries are the binomial coefficients $\binom{n+d-1}{d-1}$, the multiset coefficients.

Symmetric Tensors in Coordinates

When $M$ is free, an element of $\operatorname{Sym}^n M$ has a coordinate description. Fix a basis and write $x_j = \sum_i x_{j,i} e_i$ for $j = 1, \ldots, n$. Expanding the product and grouping the terms by the multiplicities of the basis elements,

$$ x_1 \cdots x_n = \sum_{|a| = n} \Bigl(\sum_{\substack{(i_1, \ldots, i_n) \in \{1, \ldots, d\}^n \\ \#\{j : i_j = i\} = a_i}} \ \prod_{j=1}^{n} x_{j,i_j}\Bigr) e^a . $$

The inner sum runs over the ordered $n$-tuples of basis indices with multiplicity vector $a$; such a tuple is determined by choosing, for each $i$, the $a_i$ positions carrying the index $i$, so there are $\binom{n}{a} = n!/\prod_i a_i!$ of them. This is the coordinate form of the monomial basis, and it is the reason symmetric powers are often called polynomial modules: the element $x_1 \cdots x_n$ is the degree-$n$ polynomial form in the $n$ vectors, and $e^a$ is the monomial $\prod_i y_i^{a_i}$ in the dual coordinates.

Example. For $d = 2$, $n = 2$, and $M = R^2$ with basis $e_1, e_2$, a basis of $\operatorname{Sym}^2 M$ is $e_1^2, e_1 e_2, e_2^2$, and $(c e_1 + d e_2)^2 = c^2 e_1^2 + 2cd\, e_1 e_2 + d^2 e_2^2$, matching the binomial expansion.

Functoriality

The Functor $\operatorname{Sym}^n$

Let $u : M \to N$ be $R$-linear. The map

$$ M^n \to \operatorname{Sym}^n N, \qquad (x_1, \ldots, x_n) \mapsto u(x_1) \cdots u(x_n), $$

is symmetric multilinear, so by the universal property it factors uniquely through an $R$-linear map

$$ \operatorname{Sym}^n u : \operatorname{Sym}^n M \to \operatorname{Sym}^n N, \qquad x_1 \cdots x_n \mapsto u(x_1) \cdots u(x_n). $$

This assignment is functorial: $\operatorname{Sym}^n(\operatorname{id}_M) = \operatorname{id}_{\operatorname{Sym}^n M}$ and $\operatorname{Sym}^n(v \circ u) = \operatorname{Sym}^n v \circ \operatorname{Sym}^n u$, both by uniqueness. So $\operatorname{Sym}^n$ is a functor from $R$-modules to $R$-modules. For $n \geq 2$ it is not additive: it does not send $M \oplus N$ to $\operatorname{Sym}^n M \oplus \operatorname{Sym}^n N$ but to the tensor-product decomposition of the next theorem, and it does not preserve sums of maps either, since for $M = N = R$ with $u = v = \operatorname{id}_R$ the map $\operatorname{Sym}^2(u+v)$ is multiplication by $4$ on $\operatorname{Sym}^2 R \cong R$ while $\operatorname{Sym}^2 u + \operatorname{Sym}^2 v$ is multiplication by $2$. Nor is it left or right exact in general. On free modules of finite rank it is given on bases by the rule $e^a \mapsto u(e_1)^{a_1} \cdots u(e_d)^{a_d}$.

Direct Sums and the Symmetric Binomial Theorem

Theorem. For all $R$-modules $M, N$ there is a natural isomorphism

$$ \operatorname{Sym}^n (M \oplus N) \cong \bigoplus_{k=0}^{n} \operatorname{Sym}^k M \otimes_R \operatorname{Sym}^{n-k} N. $$

Proof. Both sides represent the same functor of $P$. Giving a symmetric multilinear map $f : (M \oplus N)^n \to P$ is the same as giving, for each $k$, a map on $M^k \times N^{n-k}$ that is symmetric multilinear in the $M$-variables and in the $N$-variables separately, the blocks being independent; a symmetric multilinear $f$ is determined by these restrictions, and any such family combines uniquely to a symmetric multilinear $f$ because the symmetry only permutes variables within each block. Hence

$$ \operatorname{SymMult}_R\bigl((M \oplus N)^n; P\bigr) \cong \prod_{k=0}^{n} \operatorname{Hom}_R\bigl(\operatorname{Sym}^k M \otimes_R \operatorname{Sym}^{n-k} N,\, P\bigr), $$

where the right-hand side uses the tensor–hom adjunction of Modules §13. Both sides are natural in $P$, so by the Yoneda lemma the representing objects are isomorphic, which is the displayed isomorphism.

Consistency check. Over a field, taking $M = K^d$, $N = K^e$ and $P = K$ reduces the isomorphism to the dimension identity

$$ \binom{n + d + e - 1}{d + e - 1} = \sum_{k=0}^{n} \binom{k + d - 1}{d - 1} \binom{n - k + e - 1}{e - 1}, $$

which is the Vandermonde convolution, recorded in the table above for small values.

Corollary (the symmetric binomial theorem). For a free module of rank $d$ there is a natural isomorphism

$$ \operatorname{Sym}^n(R^d) \cong \bigoplus_{a_1 + \cdots + a_d = n} R \cdot x_1^{a_1} \cdots x_d^{a_d}, $$

the degree-$n$ part of the polynomial algebra in $d$ variables, and the product of the algebra is the symmetric product on the direct sum over $n$.

Symmetric Powers over a General Ring

Non-Free Modules

Over a general commutative ring a module need not be free, and then $\operatorname{Sym}^n M$ need not be free; it need not have a basis and need not have a well-defined rank. The quotient construction of the theorem above nevertheless applies verbatim, and all the structural statements — the universal property, functoriality, the direct-sum decomposition — remain valid. What is lost is the dimension formula, which is a statement about free modules.

Example. Let $R = \mathbb{Z}$ and $M = \mathbb{Z}/m\mathbb{Z}$, with $e$ the image of $1$. Tensoring the presentation $\mathbb{Z} \xrightarrow{m} \mathbb{Z} \to \mathbb{Z}/m\mathbb{Z} \to 0$ with itself gives $M^{\otimes 2} \cong \mathbb{Z}/m\mathbb{Z}$, generated by $e \otimes e$; the swap fixes $e\otimes e$ and therefore acts as the identity on this cyclic module, so $J_2 = 0$ and $\operatorname{Sym}^2(\mathbb{Z}/m\mathbb{Z}) \cong \mathbb{Z}/m\mathbb{Z}$. In general $\operatorname{Sym}^n(\mathbb{Z}/m\mathbb{Z}) \cong \mathbb{Z}/m\mathbb{Z}$ for $n \geq 1$, since every permutation fixes $e^{\otimes n}$.

The Canonical Maps

The universal property gives two canonical maps that are worth keeping apart.

• The quotient map $\otimes^n : M^{\otimes n} \to \operatorname{Sym}^n M$, always defined and surjective.
• The symmetrisation map $s_n : M^{\otimes n} \to M^{\otimes n}$, defined when $n!$ is invertible, idempotent, with image the symmetric tensors and kernel $J_n$.

When $n!$ is invertible the second splits the first: the composite $M^{\otimes n} \xrightarrow{s_n} (M^{\otimes n})^{S_n} \xrightarrow{\ \cong\ } \operatorname{Sym}^n M$ is the quotient map, and the symmetric power is a direct summand of the tensor power. When $n!$ is not invertible no such splitting exists in general, and the symmetric power is genuinely a quotient, not a submodule.

Summary

The $n$-th symmetric power $\operatorname{Sym}^n M$ is the $R$-module corepresenting symmetric $n$-multilinear maps $M^n \to P$:

$$ \operatorname{Hom}_R(\operatorname{Sym}^n M, P) \cong \operatorname{SymMult}_R(M^n; P). $$

It is unique up to a unique isomorphism, with $\operatorname{Sym}^0 M = R$ and $\operatorname{Sym}^1 M = M$. It is constructed as the quotient of the tensor power by the submodule generated by the differences $x_1 \otimes \cdots \otimes x_n - x_{\sigma(1)} \otimes \cdots \otimes x_{\sigma(n)}$, that is, as the module of coinvariants of the permutation action of $S_n$ on $M^{\otimes n}$. It agrees with the module of invariants (the symmetric tensors) precisely when $n!$ is invertible in $R$, via the symmetrisation operator $\frac{1}{n!}\sum_\sigma \sigma$; when $n!$ fails to be invertible the two differ, and the characteristic-2 example $M = \mathbb{F}_2^2$, $n = 2$ exhibits the failure. The symmetric powers form a functor, and they convert direct sums into tensor products by the symmetric binomial theorem. If $M$ is free of rank $d$ then $\operatorname{Sym}^n M$ is free of rank $\binom{n+d-1}{d-1}$ with monomial basis $e^a$, $|a| = n$; over a general ring this basis may not exist.

Summary of Notation

Further Reading

• Nicolas Bourbaki, Algebra I: Chapters 1–3 (Springer, 1989), for symmetric and alternating powers of modules.
• Serge Lang, Algebra (Springer, 3rd ed. 2002), for the multilinear algebra and the symmetric algebra.
• Charles A. Weibel, An Introduction to Homological Algebra (Cambridge University Press, 1994), for tensor and symmetric powers as functors.
• Nicolas Bourbaki, Éléments de mathématique: Algèbre commutative (Hermann, 1961), for the symmetric algebra over a commutative ring.