$3$-linear functions on $\mathbb{R}^3$

Question

I know that any bilinear function $\Phi$ can be presented in a unique way as a sum $$\Phi = S + A,$$ where $S$ is a symmetric and $A$ is skew-symmetric bilinear functions. Does a similar statement hold for $3$-linear functions on $\mathbb{R}^3$?

Answer 1

For readability denote $\Bbb V := \Bbb R^3$.

The space $\bigotimes^3 \Bbb V^*$ of trilinear forms on $\Bbb V$ has dimension $3^3 = 27$, but the subspace $\bigwedge^3 \Bbb V^*$ of skew-symmetric symmetric forms has dimension ${3\choose 3} = 1$ (spanned by the map $(a, b, c) \mapsto \det[\,a\,\,\,b\,\,\,c\,]$) and the subspace $\bigodot^3 \Bbb V^*$ of symmetric trilinear forms has dimension ${{3 + 3 - 1}\choose 3} = 10$, so the space of trilinear forms that can be decomposed into the sum of a symmetric form and a skew form has dimension $10 + 1 = 11$. Thus, not all trilinear forms can be so decomposed.

There is, however, a direct analogue of the decomposition you mention for trilinear forms, and the key is to add two additional subspaces whose elements have particular combinations of symmetry and skew-symmetry. Consider the following maps $$\bigotimes{}^3 \Bbb V^* \to \bigotimes{}^3 \Bbb V^*:$$ \begin{align*} \operatorname{Sym} &: T_{abc} \mapsto T_{(abc)} \\ \operatorname{Alt} &: T_{abc} \mapsto T_{[abc]} \\ A &: T_{abc} \mapsto \frac{1}{3}(2 T_{(ab)c} - 2 T_{(bc)a}) \\ B &: T_{abc} \mapsto \frac{1}{3}(2 T_{[ab]c} - 2 T_{c[ab]}) . \end{align*} By convention, indices in parentheses are symmetrized over, and those in brackets are skewed over. One can check that these maps are projections and together define a direct sum decomposition of $\bigotimes{}^3 \Bbb V^*$: $$\underbrace{\bigotimes{}^3 \Bbb V^*}_{27} = \underbrace{\bigodot{}^3 \Bbb V^*}_{10} \oplus \underbrace{\bigwedge\!{}^3\Bbb V^*}_1 \oplus \underbrace{\operatorname{im} A}_8 \oplus \underbrace{\operatorname{im} B}_8 .$$ (The numbers underneath each vector space is its dimension.) Put another way, any trilinear form on $\Bbb V$ uniquely decomposes as a sum of four trilinear forms, one in each summand. Explicitly, $$\Phi = \operatorname{Sym}(\Phi) + \operatorname{Alt}(\Phi) + A(\Phi) + B(\Phi) .$$ This is, by the way, exactly the decomposition of $\bigotimes^3 \Bbb V^*$ into irreducible representations of the symmetric group $S_3$.

Example For distinct $\alpha, \beta \in (\Bbb R^3)^* - \{{\bf 0}\}$ the trilinear form $$3 A(\alpha \otimes \alpha \otimes \beta) = 2 \alpha \otimes \alpha \otimes \beta - \alpha \otimes \beta \otimes \alpha - \beta \otimes \alpha \otimes \alpha$$ is not the sum of a symmetric form and a skew form.

In general, for an $n$-dimensional vector space (at least over fields of characteristic $0$), one gets one irreducible summand for each standard Young tableau with $n$ cells; these counts are recorded in OEIS A000085.

Answer 2

Let $$T = x_1 \otimes x_1 \otimes x_2, \text{ }u = (1, 0, 0), \text{ }v = (0, 1, 0).$$ Then $T(u, u, v) = 1$ and $T(u, u, u) = 0$. Say that $T = A + B$ where $A$ is symmetric and $B$ is anti-symmetric. Then $A(u, u, v) = A(u, v, u)$, whereas $B(u, u, v) = 0 = B(u, v, u)$, a contradiction.

This shows that not all $3$-linear functions on $\mathbb{R}^3$ can be expressed as a sum of a symmetric function and a skew-symmetric function.

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We give a better response in the interests of completeness. Write $V = \mathbb{R}^3$. Then $\left(V^*\right)^{\otimes3}$ is a real vector space of dimension $3^3 = 27$. The subspaces $S^3V^*$ and $\wedge^3 V^*$ have dimensions $10$ and $1$ respectively. So$$S^3V^* + \wedge^3V^*$$is a subspace in $\left(V^*\right)^{\otimes3}$ of dimension at most $11$, and therefore cannot be all of $\left(V^*\right)^{\otimes3}$. In fact the sum is orthogonal direct, since if $S \in S^3V^*$ and $A \in \wedge^3 V^*$, then$$\langle S, A\rangle = \sum_{i_1, i_2, i_3} S_{i_1i_2i_3}A_{i_1i_2i_3} = S_{123}A_{123}\sum_{\sigma \in S_3} (-1)^{\text{sign}(\sigma)} = 0.$$If $\phi = S + A$ for $S$ symmetric and $A$ antisymmetric, then$$\phi^\text{sym} = 3!S,\text{ }\phi^\text{asym} = 3!A.$$In fact ${1\over{3!}}(\cdot)^\text{sym}$, ${1\over{3!}}(\cdot)^\text{asym}$ are orthogonal projections onto the mutually orthogonal spaces of symmetric, antisymmetric $3$-tensors in $\mathbb{R}^3$.

Let $\phi = x_3 \otimes x_1 \otimes x_2$. Then$$\phi^\text{sym}(e_1, e_2, e_3) = 1, \text{ }\phi^\text{asym}(e_1, e_2, e_3) = 1$$but $\phi(e_1, e_2, e_3) = 0 \neq 1/3$.