2 Notation and conventions

All vector spaces considered here are finite-dimensional over $\mathbb R$. The abstract index conventions in the sense of Penrose [Reference Penrose and Rindler30, chapter $2$] are used when convenient. Given a vector space $\mathbb {V}$, $\alpha _{i_{1}\dots i_{l}}^{j_{1}\dots j_{k}}$ indicates an element of $\otimes ^{k}\mathbb {V} \otimes \otimes ^{l}\mathbb {V}^{\ast }$. The indices are labels indicating tensor valencies and symmetries and do not refer to any choice of reference frame. Enclosure of indices in square brackets or parentheses indicates complete antisymmetrization or complete symmetrization over the enclosed indices; indices delimited by vertical bars are omitted from such (anti)symmetrizations. For example, $2a_{ijk} = a_{[i|j|k} + a_{(i|j|k)}$ is the decomposition of $a_{ijk}$ into its parts antisymmetric and symmetric in the first and last indices. The symmetric product $\alpha \odot \beta \in S^{k+l}\mathbb {V}^{\ast }$ of symmetric tensors $\alpha \in S^{k}\mathbb {V}^{\ast }$ and $\beta \in S^{l}\mathbb {V}^{\ast }$ is defined by complete symmetrization, $(\alpha \odot \beta )_{i_{1}\dots i_{k+l}} = \alpha _{(i_{1}\dots i_{k}}\beta _{i_{k+1}\dots i_{k+l})}$, whereas the wedge product $\alpha \wedge \beta $ of antisymmetric tensors $\alpha \in {\bigwedge }^{k}\mathbb {V}^{\ast }$ and $\beta \in {\bigwedge }^{l}\mathbb {V}^{\ast }$ is defined as a multiple of the complete antisymmetrization of their tensor product, by $(\alpha \wedge \beta )_{i_{1}\dots i_{k+l}} = \binom {k+l}{k}\alpha _{[i_{1}\dots i_{k}}\beta _{i_{k+1}\dots i_{k+l}]}$.

Indices are raised and lowered, respecting horizontal position, using a nondegenerate symmetric bilinear form $h_{ij}$ (called a metric) and the inverse symmetric bivector $h^{ij}$ satisfying $h^{ip}h_{pj} = \delta _{j}{\kern2pt}^{i}$. The pair $(\mathbb {V}, h)$ is called a metric vector space. The metric h is Euclidean if it is positive definite and in this case $(\mathbb {V}, h)$ is called a Euclidean vector space. Throughout the article the norms used on tensor modules are those given by complete contraction with the metric (and not those induced from the standard $O(h)$-representation). A subspace $\mathbb {M} \subset \otimes ^{k}\mathbb {V}^{\ast } \otimes \otimes ^{l}\mathbb {V}$ is a metric vector space with the metric $\langle \,\cdot \,\,, \,\cdot \,\, \rangle $ defined via complete contraction with $h_{ij}$ and $h^{ij}$ by $\langle \alpha , \beta \rangle = \alpha _{i_{1}\dots i_{k}}^{j_{1}\dots j_{l}}\beta _{a_{1}\dots a_{k}}^{b_{1}\dots b_{l}}h^{i_{1}a_{1}}\dots h^{i_{k}a_{k}}h_{j_{1}b_{1}}\dots h_{j_{l}b_{l}}$.

Let $(\mathbb {V}, h)$ be an n-dimensional metric vector space. When Euclidean h is fixed, the abstract orthogonal group $O(n)$ is identified with the orthogonal group $O(h)$ of linear automorphisms of $\mathbb {V}$ preserving h. The action of $GL(\mathbb {V})$ on $\mathbb {V}$ given by $(g\cdot x)^{i} = x^{p}g_{p}{}^{i}$ induces the cogredient action on $\mathbb {V}^{\ast }$ given by $(g\cdot \mu )_{i} = (g^{-1})_{i}{}^{p}\mu _{p}$ and these actions extend in the usual way to $\otimes ^{k}\mathbb {V} \otimes \otimes ^{l}\mathbb {V}^{\ast }$. By definition, $g_{i}{}^{j} \in GL(\mathbb {V})$ is in $O(h)$ if and only if $g_{i}{}^{p}g_{jp} = h_{ij}$ or, similarly, $(g^{-1})_{i}{}^{j} = g^{j}{}_{i}$. This implies the action of $O(h)$ commutes with taking traces. For example, for $\mathscr {X} \in \mathscr {MC}(\mathbb {V}^{\ast })$ and $g \in O(h)$, $\operatorname {\rho }(g\cdot \mathscr {X}) = g \cdot \operatorname {\rho }(\mathscr {X})$, so $\mathscr {MC}_{\mathscr {W}}(\mathbb {V}^{\ast })$ is an $O(h)$-submodule of $\mathscr {MC}(\mathbb {V}^{\ast })$.

The space $\operatorname {End}(\mathbb {V})$ of linear endomorphisms of $\mathbb {V}$ is regarded as an algebra with multiplication $\circ $ given by composition. The adjoint involution $\sigma _{h}:(\operatorname {End}(\mathbb {V}), \circ )\to (\operatorname {End}(\mathbb {V}), \circ )$ of the metric h on $\mathbb {V}$ is the real linear antiautomorphism defined by $h(\sigma _{h}(\phi )x, y) = h(x, \phi (y))$ for all $x, y \in \mathbb {V}$ and $\phi \in \operatorname {End}(\mathbb {V})$. For a metrized vector space $(\mathbb {V}, h)$, the subspace $\operatorname {Sym}(\mathbb {V}, h) = \operatorname {Sym}(\operatorname {End}(\mathbb {V}), \sigma _{h}) = \{\phi \in \operatorname {End}(\mathbb {V}): \sigma _{h}(\phi ) = \phi \}$ of h-self-adjoint endormorphisms is a Jordan algebra with the product $\phi \circledcirc \psi = \tfrac {1}{2}(\phi \circ \psi + \psi \circ \phi )$ for $\phi , \psi \in \operatorname {Sym}(\mathbb {V}, h)$. (When h is clear from context there is written simply $\operatorname {Sym}(\mathbb {V})$ for brevity.) There holds $\alpha \circ \beta = \tfrac {1}{2}[\alpha , \beta ] + \alpha \circledcirc \beta $, where $\alpha \circledcirc \beta = \tfrac {1}{2}(\alpha \circ \beta + \beta \circ \alpha )$.

Via metric duality, $\alpha _{ij} \in \otimes ^{2}(\mathbb {V}^{\ast })$ is identified with the endomorphism $x^{j} \to x^{i}\alpha _{i}{}^{j}$ of $\mathbb {V}$, and the composition of the endomorphisms of $\mathbb {V}$ determined by raising the second indices of $\alpha _{ij} , \beta _{ij} \in \otimes ^{2}\mathbb {V}^{\ast }$ is given by $(\alpha \circ \beta )_{i}{}^{j} = \alpha _{p}{}^{j}\beta _{i}{}^{p}$. These conventions are such that, for $x, y, z, w \in \mathbb {V}^{\ast }$,

(2.1)$$ \begin{align} &(x \otimes y)\circ (z \otimes w) = \langle w, x \rangle z \otimes y. \end{align} $$

The pullback to $\otimes ^{2}\mathbb {V}^{\ast }$ of the Lie bracket of endomorphisms yields the Lie bracket $[\,\cdot \,\,, \,\cdot \,\,]:\otimes ^{2}(\mathbb {V}^{\ast }) \times \otimes ^{2}(\mathbb {V}^{\ast }) \to \otimes ^{2}(\mathbb {V}^{\ast })$ given by $[\alpha , \beta ] = \alpha \circ \beta - \beta \circ \alpha $. Similarly, the multiplication induced via metric duality on $\otimes ^{2}\mathbb {V}^{\ast }$ by the usual Jordan product of endomorphisms is denoted by $\circledcirc $. The subspace ${\bigwedge }^{2}\mathbb {V}^{\ast }$ is a subalgebra of $(\otimes ^{2}\mathbb {V}^{\ast }, [\,\cdot \,\,, \,\cdot \,\,])$ and $({\bigwedge }^{2}\mathbb {V}^{\ast }, [\,\cdot \,\,, \,\cdot \,\,])$ is isomorphic to the Lie algebra $\mathfrak {so}(\mathbb {V}, h)$.

3 General results about commutative algebras

As explained in the introduction, the argument proving [Reference Popov31, Theorem $5$] adapts almost without change to the present setting to prove Theorem 3.1. For the reader’s convenience, it is reproduced here with modifications appropriate to the current setting.

4 Action of metric curvature tensors on tensors

Let $(\mathbb {V}, h)$ be a metric vector space. Note that $\omega _{ijkl} \in \overline {\mathscr {MC}}(\mathbb {V}^{\ast }) = \{\omega _{ijkl} \in \otimes ^{4}\mathbb {V}^{\ast }: \omega _{(ij)kl} = \omega _{ijkl} = \omega _{(kl)ij}, \omega _{(ijk)l} = 0\}$ satisfies $\omega _{klij} = \omega _{ijkl}$ and $\omega _{k[ij]l}$ is antisymmetric in k and l. A straightforward calculation proves that the linear map $\mathcal {T}:\overline {\mathscr {MC}}(\mathbb {V}^{\ast }) \to \mathscr {MC}(\mathbb {V}^{\ast })$ defined by $\mathcal {T}(\mu )_{ijkl} = \tfrac {2}{\sqrt {3}}\mu _{k[ij]l}$ is an isometric isomorphism with inverse given by $\mathcal {T}^{-1}(\nu )_{ijkl} = -\tfrac {2}{\sqrt {3}}\nu _{k(ij)l}$.

It is convenient to abuse notation by identifying $S^{2}({\bigwedge }^{2}\mathbb {V}^{\ast })$ and $S^{2}(S^{2}\mathbb {V}^{\ast })$ with the subspaces $\{\Psi _{ijkl} \in \otimes ^{4}\mathbb {V}^{\ast }: \Psi _{klij} = \Psi _{ijkl} = \Psi _{[ij]kl} = \Psi _{ij[kl]}\}$ and $\{\Phi _{ijkl} \in \otimes ^{4}\mathbb {V}^{\ast }: \Phi _{klij} = \Phi _{ijkl} = \Phi _{(ij)kl} = \Phi _{ij(kl)}\}$. As $\Psi _{ijkl} \in {\bigwedge }^{4}\mathbb {V}^{\ast }$ satisfies $\Psi _{klij} = \Psi _{ijkl}$, ${\bigwedge }^{4}\mathbb {V}^{\ast }$ can be viewed as a subspace of $S^{2}({\bigwedge }^{2}\mathbb {V}^{\ast })$ and, likewise, $S^{4}\mathbb {V}^{\ast }$ can be regarded as a subspace of $S^{2}(S^{2}\mathbb {V}^{\ast })$. The Bianchi identity implies ${\bigwedge }^{4}\mathbb {V}^{\ast }$ and $\mathscr {MC}(\mathbb {V}^{\ast })$ are orthogonal subspaces of $S^{2}({\bigwedge }^{2}\mathbb {V}^{\ast })$ with trivial intersection and, similarly, $S^{4}\mathbb {V}^{\ast }$ and $\overline {\mathscr {MC}}(\mathbb {V}^{\ast })$ are orthogonal subspaces of $S^{2}(S^{2}\mathbb {V}^{\ast })$ with trivial intersection. Comparing dimensions shows $S^{2}({\bigwedge }^{2}\mathbb {V}^{\ast }) \simeq \mathscr {MC}(\mathbb {V}^{\ast })\oplus {\bigwedge }^{4}\mathbb {V}^{\ast }$ and $S^{2}(S^{2}\mathbb {V}^{\ast }) \simeq \overline {\mathscr {MC}}(\mathbb {V}^{\ast })\oplus S^{4}\mathbb {V}^{\ast }$.

For $\Psi \in S^{2}({\bigwedge }^{2}\mathbb {V}^{\ast })$, $\Psi _{[ijkl]} = \Psi _{[ijk]l}$ and the orthogonal projection $\mathscr {M}:S^{2}({\bigwedge }^{2}\mathbb {V}^{\ast }) \to \mathscr {MC}(\mathbb {V}^{\ast })$ is given by $\mathscr {M}(\Psi )_{ijkl} = \Psi _{ijkl} - \Psi _{[ijk]l}$, so $\mathscr {M} \oplus (\operatorname {Id}_{S^{2}({\bigwedge }^{2}\mathbb {V}^{\ast })} - \mathscr {M}): S^{2}({\bigwedge }^{2}\mathbb {V}^{\ast }) \to \mathscr {MC}(\mathbb {V}^{\ast })\oplus {\bigwedge }^{4}\mathbb {V}^{\ast }$ is an isometric isomorphism. Similarly, the orthogonal projection $\mathscr {S}:S^{2}(S^{2}\mathbb {V}^{\ast }) \to \overline {\mathscr {MC}}(\mathbb {V}^{\ast })$ is given by $\mathscr {S}(\Psi )_{ijkl} = \Psi _{ijkl} - \Psi _{(ijk)l}$, so $(\mathcal {T}\circ \mathscr {S}) \oplus (\operatorname {Id}_{S^{2}(S^{2}\mathbb {V}^{\ast })} - \mathscr {S}): S^{2}(S^{2}\mathbb {V}^{\ast }) \to \mathscr {MC}(\mathbb {V}^{\ast })\oplus S^{4}\mathbb {V}^{\ast }$ is an isometric isomorphism, where, for $\omega _{ijkl} \in S^{2}(S^{2}\mathbb {V}^{\ast })$,

(4.1)$$ \begin{align} (\mathcal{T} \circ \mathscr{S})(\omega)_{ijkl} = \tfrac{2}{\sqrt{3}}\omega_{k[ij]l}. \end{align} $$

Lemma 4.1. Let $(\mathbb {V}, h)$ be a finite-dimensional metric vector space and let $k \geq 2$. For $\alpha , \beta \in \otimes ^{k}\mathbb {V}^{\ast }$, setting $\varrho (\alpha , \beta )$ equal to the h-orthogonal projection $\mathscr {M}(\pi (\alpha , \beta )) \in \mathscr {MC}(\mathbb {V}^{\ast })$ of $\pi (\alpha , \beta )_{abcd} = \Psi (\alpha , \beta )_{[ab][cd]} \in S^{2}({\bigwedge }^{2}\mathbb {V}^{\ast })$ where

(4.2)$$ \begin{align} \Psi(\alpha, \beta)_{abcd} = \tfrac{1}{k}\sum_{r \neq s}\alpha_{i_{1}\dots i_{r-1}b i_{r+1}\dots i_{s-1} c i_{s+1}\dots i_{k}}\beta^{i_{1}\dots i_{r-1}}{}_{a}{}^{i_{r+1}\dots i_{s-1}}{}_{d}{}^{i_{s+1}\dots i_{k}} \end{align} $$

defines a symmetric $O(h)$-equivariant bilinear map $\varrho :\otimes ^{k}\mathbb {V}^{\ast } \times \otimes ^{k}\mathbb {V}^{\ast } \to \mathscr {MC}(\mathbb {V}^{\ast })$.

Proof. It needs to be checked only that $\varrho (\beta , \alpha ) = \varrho (\alpha , \beta )$. The definition (4.2) implies $\Psi (\beta , \alpha )_{abcd} = \Psi (\alpha , \beta )_{badc}$ and there follows

(4.3)$$ \begin{align} \begin{split} \pi(\beta, \alpha)_{abcd} &=\Psi(\beta, \alpha)_{[ab][cd]} \\ &= \tfrac{1}{4}\left(\Psi(\beta, \alpha)_{abcd} -\Psi(\beta, \alpha)_{bacd} -\Psi(\beta, \alpha)_{abdc} + \Psi(\beta, \alpha)_{badc}\right) \\ &= \tfrac{1}{4}\left(\Psi(\alpha, \beta)_{badc} -\Psi(\alpha, \beta)_{abdc} -\Psi(\alpha, \beta)_{bacd} + \Psi(\alpha, \beta)_{abcd}\right) \\ &= \Psi(\alpha, \beta)_{[ab][cd]} = \pi(\alpha, \beta)_{abcd}, \end{split} \end{align} $$

so that $\varrho (\beta , \alpha ) = \mathscr {M}(\pi (\beta , \alpha )) = \mathscr {M}(\pi (\alpha , \beta )) = \varrho (\alpha , \beta )$. That $\varrho $ is $O(h)$-equivariant means that $\varrho (g\cdot \alpha , g \cdot \beta ) = g \cdot \varrho (\alpha , \beta )$ for all $\alpha , \beta \in \otimes ^{k}\mathbb {V}^{\ast }$ and $g \in O(h)$, and this is apparent from the manifest $O(h)$-equivariance of the construction of $\varrho $.

Because $\varrho $ is $O(h)$-equivariant, its restriction to $\mathbb {W} \times \mathbb {W}$ where $\mathbb {W}$ is any $O(h)$-submodule $\mathbb {W} \subset \otimes ^{k}\mathbb {V}^{\ast }$ takes values in some $O(h)$-submodule of $\mathscr {MC}(\mathbb {V}^{\ast })$.

The symmetric group on k elements, $S_{k}$, acts on the left on $\otimes ^{k}\mathbb {V}^{\ast }$ by $(\sigma \cdot \alpha )_{i_{1}\dots i_{k}} = \alpha _{i_{\sigma ^{-1}(1)}\dots i_{\sigma ^{-1}(k)}}$ for $\sigma \in S_{k}$. A filling of a k box Young diagram by distinct indices $i_{1},\dots , i_{k}$ determines the submodule of $\otimes ^{k}\mathbb {V}^{\ast }$ comprising tensors antisymmetric in the indices in any column of the filled Young diagram and such that there vanishes the antisymmetrization over a subset of indices comprising the indices in any given column plus any index from any column to the right of the given column. The tensors in such a submodule are said to have the type given by the filled Young diagram. By [Reference Weyl40, Theorems $5.7.$A and $5.7.$C], an $O(n)$-module of covariant trace-free tensors on an n-dimensional vector space having symmetries corresponding to a Young diagram is nontrivial if and only if the sum of the lengths of the first two columns is no greater than n, and by [Reference Weyl40, Theorem $5.7$G] all irreducible finite-dimensional $O(n)$-modules correspond to some such Young diagram. For example, $\mathscr {MC}_{\mathscr {W}}(\mathbb {V}^{\ast }) = \ker \operatorname {\rho } \subset \mathscr {MC}(\mathbb {V}^{\ast })$ corresponds with a filling of a $2\times 2$ square array of boxes, so $\mathscr {MC}_{\mathscr {W}}(\mathbb {V}^{\ast }) = \{0\}$ if $\dim \mathbb {V} < 4$. If $n = \dim \mathbb {V} \geq 4$, $\mathscr {MC}_{\mathscr {W}}(\mathbb {V}^{\ast })$ is a nontrivial irreducible $O(n)$-module. When $\dim \mathbb {V} = 4$, it decomposes as an $SO(n)$-module into two 5-dimensional submodules (see Section 9). A linear endomorphism is said to preserve the type of tensors if it maps tensors with the symmetries determined by a given filled Young diagram into tensors with the same symmetries.

Lemma 4.2. Let $(\mathbb {V}, h)$ be a finite-dimensional metric vector space and let $k \geq 2$. The linear map $\mathscr {MC}(\mathbb {V}^{\ast }) \to \operatorname {End}(\otimes ^{k}\mathbb {V}^{\ast })$ associating with $\mathscr {X} \in \mathscr {MC}(\mathbb {V}^{\ast })$ the operator $\widehat {\mathscr {X}} \in \operatorname {End}(\otimes ^{k}\mathbb {V}^{\ast })$ defined by

(4.4)$$ \begin{align} \widehat{\mathscr{X}}(\alpha)_{i_{1}\dots i_{k}} = \tfrac{1}{k}\sum_{r \neq s}\mathscr{X}^{p}{}_{i_{r}i_{s}}{}^{q}\alpha_{i_{1}\dots i_{r-1}pi_{r+1}\dots i_{s-1} q i_{s+1}\dots i_{k}} \end{align} $$

commutes with the action of $S_{k}$ on $\otimes ^{k}\mathbb {V}^{\ast }$ and so preserves the type of tensors; is $O(h)$-equivariant, meaning $\widehat {g \cdot \mathscr {X}}(\alpha ) = g\cdot \widehat {\mathscr {X}}(g^{-1}\cdot \alpha )$ for all $\alpha \in \otimes ^{k}\mathbb {V}^{\ast }$ and $g \in O(h)$; and has image in the self-adjoint endomorphisms, meaning $\widehat {\mathscr {X}} \in \operatorname {Sym}(\otimes ^{k}\mathbb {V}^{\ast }, \langle \,\cdot \,\,, \,\cdot \,\, \rangle )$ for all $\mathscr {X} \in \mathscr {MC}(\mathbb {V}^{\ast })$.

Proof. That $\widehat {\mathscr {X}}$ preserves the type of tensors follows from $\widehat {\mathscr {X}}(\sigma \cdot \alpha ) = \sigma \cdot \widehat {\mathscr {X}}(\alpha )$ for $\sigma \in S_{k}$ and $\alpha \in \otimes ^{k}\mathbb {V}^{\ast }$. This is immediate from the definition (4.4) and the symmetries of $\mathscr {X}_{ijkl}$. The relation $\widehat {g \cdot \mathscr {X}}(\alpha ) = g\cdot \widehat {\mathscr {X}}(g^{-1}\cdot \alpha )$ likewise follows from a straightforward computation (for this relation to hold it is necessary that g be orthogonal because the metric is used in (4.4) where indices are contracted). It follows from (4.4) that

(4.5)$$ \begin{align} \begin{split} \langle \widehat{\mathscr{X}}(\alpha), \beta \rangle = \langle \mathscr{X}, \varrho(\alpha, \beta)\rangle \end{split} \end{align} $$

where $\varrho (\alpha , \beta )$ is as in Lemma 4.1. By Lemma 4.1, $\varrho (\alpha , \beta )$ is symmetric in $\alpha $ and $\beta $ and by (4.5) this implies that $\langle \widehat {\mathscr {X}}(\alpha ), \beta \rangle = \langle \alpha , \widehat {\mathscr {X}}(\beta )\rangle $ for all $\alpha , \beta \in \otimes ^{k}\mathbb {V}^{\ast }$, so $\widehat {\mathscr {X}}$ is self-adjoint.

When it is known that $\widehat {\mathscr {X}}$ preserves a subspace $\mathbb {W} \subset \otimes ^{k}\mathbb {V}^{\ast }$, $\widehat {\mathscr {X}}_{\mathbb {W}}$ is written instead of $\widehat {\mathscr {X}}$ when helpful for clarity. By Lemma 4.2, for $\mathscr {X} \in \mathscr {MC}(\mathbb {V}^{\ast })$ and $k \geq 2$, $\widehat {\mathscr {X}}$ preserves ${\bigwedge }^{k}\mathbb {V}^{\ast }$ and $S^{k}\mathbb {V}^{\ast }$, so $\widehat {\mathscr {X}}_{{\textstyle \wedge }^{k}\mathbb {V}^{\ast }} \in \operatorname {Sym}({\bigwedge }^{k}\mathbb {V}^{\ast })$ and $\widehat {\mathscr {X}}_{S^{k}\mathbb {V}^{\ast }} \in \operatorname {Sym}(S^{k}\mathbb {V}^{\ast })$ are written for the restrictions of $\widehat {\mathscr {X}}$ to ${\bigwedge }^{k}\mathbb {V}^{\ast }, S^{k}\mathbb {V}^{\ast } \subset \otimes ^{k}\mathbb {V}^{\ast }$. By Lemma 4.2, $\widehat {\mathscr {X}}_{S^{k}\mathbb {V}^{\ast }}(\alpha )_{i_{1}\dots i_{k}} = (k-1)\mathscr {X}^{p}{}_{(i_{1}i_{2}}{}^{q}\alpha _{i_{3}\dots i_{k)}pq}$ for $\alpha \in S^{k}\mathbb {V}^{\ast }$ and $\widehat {\mathscr {X}}_{{\textstyle \wedge }^{k}\mathbb {V}^{\ast }}(\alpha )_{i_{1}\dots i_{k}} = (k-1)\mathscr {X}^{p}{}_{[i_{1}i_{2}}{}^{q}\alpha _{i_{3}\dots i_{k]}pq} = -\tfrac {k-1}{2}\mathscr {X}^{pq}{}_{[i_{1}i_{2}}\alpha _{i_{3}\dots i_{k]}pq}$ for $\alpha \in {\bigwedge }^{k}\mathbb {V}^{\ast }$. In particular, when $k = 2$, $\widehat {\mathscr {X}}_{S^{2}\mathbb {V}^{\ast }}(\alpha )_{ij} = \alpha ^{pq}\mathscr {X}_{ipqj}$ and $\widehat {\mathscr {X}}_{{\textstyle \wedge }^{2}\mathbb {V}^{\ast }}(\alpha )_{ij} = \alpha ^{pq}\mathscr {X}_{ipqj} = -\tfrac {1}{2}\alpha ^{pq}\mathscr {X}_{ijpq}$. For example, $\widehat {\mathscr {X}}_{S^{2}\mathbb {V}^{\ast }}(h) = \operatorname {\rho }(\mathscr {X})$. For $\sigma _{ij} \in \otimes ^{2}\mathbb {V}^{\ast }$, write $(\mathsf {Sym}\, \sigma )_{ij} = \sigma _{(ij)}$ and $(\mathsf {Skew}\,\sigma )_{ij} = \sigma _{[ij]}$. For $\mathscr {X} \in \mathscr {MC}(\mathbb {V}^{\ast })$, by Lemma 4.2, $\widehat {\mathscr {X}}_{\otimes ^{2}\mathbb {V}^{\ast }}(\sigma )= \widehat {\mathscr {X}}_{S^{2}\mathbb {V}^{\ast }}(\mathsf {Sym}\, \sigma ) + \widehat {\mathscr {X}}_{{\textstyle \wedge }^{2}\mathbb {V}^{\ast }}(\mathsf {Skew}\, \sigma )$. This means that if $\otimes ^{2}\mathbb {V}^{\ast }$ is regarded as the orthogonal direct sum $S^{2}\mathbb {V}^{\ast } \oplus {\bigwedge }^{2}\mathbb {V}^{\ast }$ via $\sigma _{ij} = \sigma _{(ij)} + \sigma _{[ij]}$, then $\widehat {\mathscr {X}}_{\otimes ^{2}\mathbb {V}^{\ast }} = \widehat {\mathscr {X}}_{S^{2}\mathbb {V}^{\ast }}\oplus \widehat {\mathscr {X}}_{{\textstyle \wedge }^{2}\mathbb {V}^{\ast }}$ is an orthogonal direct sum too.

Transferring the canonical isomorphisms $\operatorname {End}({\bigwedge }^{2}\mathbb {V}^{\ast }) \simeq {\bigwedge }^{2}\mathbb {V}^{\ast } \otimes ({\bigwedge }^{2}\mathbb {V}^{\ast })^{\ast }$ and $\operatorname {End}(S^{2}\mathbb {V}^{\ast }) \simeq S^{2}\mathbb {V}^{\ast } \otimes (S^{2}\mathbb {V}^{\ast })^{\ast }$ via metric duality yields linear isomorphisms

(4.6)$$ \begin{align} \begin{split} &\sharp:\otimes^{2}({\textstyle\bigwedge}^{2}\mathbb{V}^{\ast}) = \{\Psi_{ijkl} \in \otimes^{4}\mathbb{V}^{\ast}: \Psi_{ijkl} = \Psi_{[ij]kl} = \Psi_{ij[kl]}\} \to \operatorname{End}({\textstyle\bigwedge}^{2}\mathbb{V}^{\ast}),\\ &\sharp:\otimes^{2}(S^{2}\mathbb{V}^{\ast}) = \{\Phi_{ijkl} \in \otimes^{4}\mathbb{V}^{\ast}: \Phi_{ijkl} = \Phi_{(ij)kl} = \Phi_{ij(kl)}\} \to {\operatorname{End}}(S^{2}\mathbb{V}^{\ast}), \end{split} \end{align} $$

defined by sending $\Psi \in \otimes ^{2}({\bigwedge }^{2}\mathbb {V}^{\ast })$ to $\Psi ^{\sharp }(\alpha )_{ij} = \Psi _{ij}{}^{pq}\alpha _{pq}$ and $\Phi \in \otimes ^{2}(S^{2}\mathbb {V}^{\ast }) $ to $\Phi ^{\sharp }(\sigma )_{ij} = \Phi _{ij}{}^{pq}\sigma _{pq}$. Note that, for $\mathscr {X} \in \mathscr {MC}(\mathbb {V}^{\ast })$, the tensor identified in this way with $\widehat {\mathscr {X}}_{{\textstyle \wedge }^{2}\mathbb {V}^{\ast }}$ is $-\tfrac {1}{2}\mathscr {X}_{ijkl}$. Said otherwise, for $\mathscr {X}$ viewed as an element of $S^{2}({\bigwedge }^{2}\mathbb {V}^{\ast })$, $\mathscr {X}^{\sharp } = -2\widehat {\mathscr {X}}_{{\textstyle \wedge }^{2}\mathbb {V}^{\ast }}$.

Proof. For an orthonormal basis $\{E^{(1)}_{i}, \dots , E^{(n)}_{i}\}$ of $\mathbb {V}^{\ast }$, $\{\tfrac {1}{\sqrt {2}} E^{(a)}\wedge E^{(b)}: 1 \leq a < b \leq n\}$ is an orthonormal basis of ${\bigwedge }^{2}\mathbb {V}^{\ast }$ and $\{\sqrt {2} E^{(a)}\odot E^{(b)} : 1 \leq a < b \leq n\}\cup \{E^{(a)}\odot E^{(a)}:1 \leq a \leq n\}$ is an orthonormal basis of $S^{2}\mathbb {V}^{\ast }$, so

(4.7)$$ \begin{align} \begin{aligned} \operatorname{\mathsf{tr}} \Psi^{\sharp} & = \tfrac{1}{2}\sum_{1 \leq a < b \leq n}\langle\Psi^{\sharp}(E^{(a)}\wedge E^{(b)}), E^{(a)}\wedge E^{(b)}\rangle \\ &= \sum_{a = 1}^{n}\sum_{b = 1}^{n}\Psi^{ijkl}E^{(a)}_{i}E^{(b)}_{j}E^{(a)}_{k}E^{(b)}_{l} = \Psi_{pq}{}^{pq},\\ \operatorname{\mathsf{tr}} \Phi^{\sharp} & = 2\sum_{1 \leq a < b \leq n}\langle\Phi^{\sharp}(E^{(a)}\odot E^{(b)}), E^{(a)}\odot E^{(b)}\rangle + \sum_{1 \leq a \leq n}\langle\Phi^{\sharp}(E^{(a)}\odot E^{(a)}), E^{(a)}\odot E^{(a)}\rangle \\ &= \sum_{a = 1}^{n}\sum_{b = 1}^{n}\Phi^{ijkl}E^{(a)}_{i}E^{(b)}_{j}E^{(a)}_{k}E^{(b)}_{l} = \Phi_{pq}{}^{pq}. \end{aligned} \\[-2.2pc] \nonumber \end{align} $$

For example, Lemma 4.4 implies that $\operatorname {\mathsf {tr}}\widehat {\mathscr {X}}_{{\textstyle \wedge }^{2}\mathbb {V}^{\ast }} = \tfrac {1}{2}\operatorname {s}(\mathscr {X})$ for $\mathscr {X} \in \mathscr {MC}(\mathbb {V}^{\ast })$.

When $\widehat {\mathscr {MC}(\mathbb {V}^{\ast })}$ preserves $\mathbb {W} \subset \otimes ^{k}\mathbb {V}^{\ast }$, Lemma 4.2 does not affirm that the linear map $\mathscr {X} \to \widehat {\mathscr {X}}_{\mathbb {W}}$ is injective. From (4.5) it is apparent that this is true if and only if $\{\varrho (\alpha , \beta ): \alpha , \beta \in \mathbb {W}\}$ spans $\mathscr {MC}(\mathbb {V}^{\ast })$. Corollary 4.5 shows that $\mathscr {X} \to \widehat {\mathscr {X}}_{{\textstyle \wedge }^{2}\mathbb {V}^{\ast }}$ and $\mathscr {X} \to \widehat {\mathscr {X}}_{S^{2}\mathbb {V}^{\ast }}$ are injective. It is used in the proof of Theorem 1.5. Note that viewing $\mathscr {X} \in \mathscr {MC}(\mathbb {V}^{\ast })$ as an element of $\mathcal {T}(S^{2}(S^{2}\mathbb {V}^{\ast }))$ yields $\mathcal {T}^{-1}(\mathscr {X})^{\sharp } = -\tfrac {2}{\sqrt {3}}\widehat {\mathscr {X}}_{S^{2}\mathbb {V}^{\ast }}$.

Corollary 4.5. Let $(\mathbb {V},h)$ be an n-dimensional Euclidean vector space. For $\mathscr {X}, \mathscr {Y} \in \mathscr {MC}(\mathbb {V}^{\ast })$,

(4.8)$$ \begin{align} &\operatorname{\mathsf{tr}}(\widehat{\mathscr{X}}_{{\textstyle\wedge}^{2}\mathbb{V}^{\ast}}\circ \widehat{\mathscr{Y}}_{{\textstyle\wedge}^{2}\mathbb{V}^{\ast}}) = \tfrac{1}{4}\langle \mathscr{X}, \mathscr{Y}\rangle, && \operatorname{\mathsf{tr}}(\widehat{\mathscr{X}}_{S^{2}\mathbb{V}^{\ast}}\circ \widehat{\mathscr{Y}}_{S^{2}\mathbb{V}^{\ast}}) = \tfrac{3}{4}\langle \mathscr{X}, \mathscr{Y}\rangle. \end{align} $$

The $O(n)$-equivariant linear maps $\mathscr {MC}(\mathbb {V}^{\ast }) \to \operatorname {Sym}({\bigwedge }^{2}\mathbb {V}^{\ast })$, $\mathscr {MC}(\mathbb {V}^{\ast }) \to \operatorname {Sym}(S^{2}\mathbb {V}^{\ast })$, and $\mathscr {MC}(\mathbb {V}^{\ast }) \to \operatorname {Sym}(\otimes ^{2}\mathbb {V}^{\ast })$ given by $\mathscr {X} \to 2\widehat {\mathscr {X}}_{{\textstyle \wedge }^{2}\mathbb {V}^{\ast }} = -\mathscr {X}^{\sharp }$, $\mathscr {X} \to \tfrac {2}{\sqrt {3}}\widehat {\mathscr {X}}_{S^{2}\mathbb {V}^{\ast }} = - \mathcal {T}^{-1}(\mathscr {X})^{\sharp }$, and $\mathscr {X} \to \widehat {\mathscr {X}}_{\otimes ^{2}\mathbb {V}^{\ast }}$ are isometric with respect to the trace-norms on $\operatorname {Sym}({\bigwedge }^{2}\mathbb {V}^{\ast })$, $\operatorname {Sym}(S^{2}\mathbb {V}^{\ast })$, and $\operatorname {Sym}(\otimes ^{2}\mathbb {V}^{\ast })$ and injective.

5 Algebra structures on the space of metric curvature tensors

This section defines the multiplication $\ast $ and derives its basic properties.

The injectivity of $\widehat {\,\cdot \,\,}_{{\textstyle \wedge }^{2}\mathbb {V}^{\ast }}$ and $\widehat {\,\cdot \,\,}_{S^{2}\mathbb {V}^{\ast }}$ shown in Corollary 4.5 means that commutative algebra structures $\ast _{A}$ and $\ast _{S}$ on $\mathscr {MC}(\mathbb {V}^{\ast })$ can be constructed as follows. By the injectivity there are unique $\mathscr {X} \ast _{A} \mathscr {Y} \in \mathscr {MC}(\mathbb {V}^{\ast })$ and $\mathscr {X} \ast _{S} \mathscr {Y} \in \mathscr {MC}(\mathbb {V}^{\ast })$ such that $(\mathscr {X}\ast _{A} \mathscr {Y})^{\sharp } = -2\widehat {\mathscr {X} \ast _{A} \mathscr {Y} }_{{\textstyle \wedge }^{2}\mathbb {V}^{\ast }}$ equals the orthogonal projection on $\widehat {\mathscr {MC}(\mathbb {V}^{\ast })}_{{\textstyle \wedge }^{2}\mathbb {V}^{\ast }}$ of the Jordan product $(-2\widehat {\mathscr {X}}_{{\textstyle \wedge }^{2}\mathbb {V}^{\ast }})\circledcirc (-2\widehat {\mathscr {Y}}_{{\textstyle \wedge }^{2}\mathbb {V}^{\ast }}) = \mathscr {X}^{\sharp }\circledcirc \mathscr {Y}^{\sharp }\in \operatorname {Sym}({\bigwedge }^{2}\mathbb {V}^{\ast }) $ and $\mathcal {T}^{-1}(\mathscr {X}\ast _{S} \mathscr {Y})^{\sharp } = -\tfrac {2}{\sqrt {3}}\widehat {\mathscr {X} \ast _{S} \mathscr {Y} }_{S^{2}\mathbb {V}^{\ast }}$ equals the orthogonal projection on $\widehat {\mathscr {MC}(\mathbb {V}^{\ast })}_{S^{2}\mathbb {V}^{\ast }}$ of the Jordan product $(-\tfrac {2}{\sqrt {3}}\widehat {\mathscr {X}}_{S^{2}\mathbb {V}^{\ast }})\circledcirc (-\tfrac {2}{\sqrt {3}}\widehat {\mathscr {Y}}_{S^{2}\mathbb {V}^{\ast }}) = (\mathcal {T}^{-1}(\mathscr {X})^{\sharp }\circledcirc \mathcal {T}^{-1}(\mathscr {Y})^{\sharp } \in \operatorname {Sym}(S^{2}\mathbb {V}^{\ast }))$. More precisely:

(5.1)$$ \begin{align} &\mathscr{X}\ast_{A} \mathscr{Y} = -2\mathscr{M}\left(\left(\widehat{\mathscr{X}}_{{\textstyle\wedge}^{2}\mathbb{V}^{\ast}}\circledcirc \widehat{\mathscr{Y}}_{{\textstyle\wedge}^{2}\mathbb{V}^{\ast}}\right)^{\flat}\right),&& &\mathscr{X}\ast_{S} \mathscr{Y} = -\tfrac{2}{\sqrt{3}}\mathcal{T}\circ \mathscr{S}\left(\left(\widehat{\mathscr{X}}_{S^{2}\mathbb{V}^{\ast}}\circledcirc \widehat{\mathscr{Y}}_{S^{2}\mathbb{V}^{\ast}}\right)^{\flat}\right), \end{align} $$

where $\flat $ is the inverse of $\sharp $.

Lemma 5.1. Let $(\mathbb {V}, h)$ be a metric vector space. For $\mathscr {X}, \mathscr {Y} \in \mathscr {MC}(\mathbb {V}^{\ast })$, the commutative multiplications $\ast _{A}$ and $\ast _{S}$ on $\mathscr {MC}(\mathbb {V}^{\ast })$ are given by

(5.2)$$ \begin{align} (\mathscr{X} \ast_{A} \mathscr{Y})_{ijkl} & =- \tfrac{1}{4}\left(\mathscr{X}_{ij}{}^{pq}\mathscr{Y}_{klpq} + \mathscr{Y}_{ij}{}^{pq}\mathscr{X}_{klpq} - 2\mathscr{X}_{[ij}{}^{pq}\mathscr{Y}_{kl]pq}\right), \end{align} $$

(5.3)$$ \begin{align} (\mathscr{X} \ast_{S} \mathscr{Y})_{ijkl} & =-\tfrac{1}{3}\left(\mathscr{X}_{k(pq)i}\mathscr{Y}_{j}{}^{(pq)}{}_{l}-\mathscr{X}_{k(pq)j}\mathscr{Y}_{i}{}^{(pq)}{}_{l} + \mathscr{X}_{j(pq)l}\mathscr{Y}_{k}{}^{(pq)}{}_{i}- \mathscr{X}_{i(pq)l}\mathscr{Y}_{k}{}^{(pq)}{}_{j}\right). \end{align} $$

The Ricci and scalar traces are

(5.4)$$ \begin{align} \begin{aligned} {} &\operatorname{\rho}(\mathscr{X} \ast_{A} \mathscr{Y})_{ij} = \tfrac{1}{2}\mathscr{X}_{(i}{}^{abc}\mathscr{Y}_{j)abc}, \qquad \operatorname{s}(\mathscr{X} \ast_{A} \mathscr{Y}) = \tfrac{1}{2}\langle \mathscr{X}, \mathscr{Y}\rangle, &\\ &\operatorname{\rho}(\mathscr{X} \ast_{S} \mathscr{Y})_{ij} = -\tfrac{1}{2}\mathscr{X}_{(i}{}^{abc}\mathscr{Y}_{j)abc} + \tfrac{1}{3}\left(\widehat{\mathscr{X}}_{S^{2}\mathbb{V}^{\ast}}(\operatorname{\rho}(\mathscr{Y}))_{ij} + \widehat{\mathscr{Y}}_{S^{2}\mathbb{V}^{\ast}}(\operatorname{\rho}(\mathscr{X}))_{ij} \right), \\ &\operatorname{s}(\mathscr{X} \ast_{S} \mathscr{Y}) = -\tfrac{1}{2}\langle \mathscr{X}, \mathscr{Y}\rangle + \tfrac{2}{3}\langle \operatorname{\rho}(\mathscr{X}), \operatorname{\rho}(\mathscr{Y})\rangle. \end{aligned} \end{align} $$

Proof. By (5.1), Lemma 4.5, and using $\mathscr {Z}^{\sharp } = -2\widehat {\mathscr {Z}}_{{\textstyle \wedge }^{2}\mathbb {V}^{\ast }}$ and $\mathcal {T}^{-1}(\mathscr {Z})^{\sharp } = -\tfrac {2}{\sqrt {3}}\widehat {\mathscr {Z}}_{S^{2}\mathbb {V}^{\ast }}$,

(5.5)$$ \begin{align} \begin{aligned} \langle \mathscr{X} \ast_{A} \mathscr{Y}, \mathscr{Z}\rangle &= -2\left\langle\mathscr{M}\left(\left(\widehat{\mathscr{X}}_{{\textstyle\wedge}^{2}\mathbb{V}^{\ast}}\circledcirc \widehat{\mathscr{Y}}_{{\textstyle\wedge}^{2}\mathbb{V}^{\ast}}\right)^{\flat}\right), \mathscr{Z} \right\rangle = 4\left\langle \left(\widehat{\mathscr{X}}_{{\textstyle\wedge}^{2}\mathbb{V}^{\ast}}\circledcirc \widehat{\mathscr{Y}}_{{\textstyle\wedge}^{2}\mathbb{V}^{\ast}}\right)^{\flat}, \widehat{\mathscr{Z}}_{{\textstyle\wedge}^{2}\mathbb{V}^{\ast}}^{\flat} \right\rangle\\ &= 2\operatorname{\mathsf{tr}}\left(\widehat{\mathscr{X}}_{{\textstyle\wedge}^{2}\mathbb{V}^{\ast}}\circ \widehat{\mathscr{Y}}_{{\textstyle\wedge}^{2}\mathbb{V}^{\ast}}\circ \widehat{\mathscr{Z}}_{{\textstyle\wedge}^{2}\mathbb{V}^{\ast}} + \widehat{\mathscr{Y}}_{{\textstyle\wedge}^{2}\mathbb{V}^{\ast}}\circ \widehat{\mathscr{X}}_{{\textstyle\wedge}^{2}\mathbb{V}^{\ast}}\circ \widehat{\mathscr{Z}}_{{\textstyle\wedge}^{2}\mathbb{V}^{\ast}} \right),\\ \langle \mathscr{X} \ast_{S} \mathscr{Y}, \mathscr{Z}\rangle &=-\tfrac{2}{\sqrt{3}}\left\langle\mathcal{T}\circ \mathscr{S}\left(\left(\widehat{\mathscr{X}}_{S^{2}\mathbb{V}^{\ast}}\circledcirc \widehat{\mathscr{Y}}_{S^{2}\mathbb{V}^{\ast}}\right)^{\flat}\right), \mathscr{Z} \right\rangle = \tfrac{4}{3}\left\langle \left(\widehat{\mathscr{X}}_{S^{2}\mathbb{V}^{\ast}}\circledcirc \widehat{\mathscr{Y}}_{S^{2}\mathbb{V}^{\ast}}\right)^{\flat}, \widehat{\mathscr{Z}}_{S^{2}\mathbb{V}^{\ast}}^{\flat} \right\rangle \\ &= \tfrac{2}{3}\operatorname{\mathsf{tr}}\left(\widehat{\mathscr{X}}_{S^{2}\mathbb{V}^{\ast}}\circ \widehat{\mathscr{Y}}_{S^{2}\mathbb{V}^{\ast}}\circ \widehat{\mathscr{Z}}_{S^{2}\mathbb{V}^{\ast}} + \widehat{\mathscr{Y}}_{S^{2}\mathbb{V}^{\ast}}\circ \widehat{\mathscr{X}}_{S^{2}\mathbb{V}^{\ast}}\circ \widehat{\mathscr{Z}}_{S^{2}\mathbb{V}^{\ast}} \right). \end{aligned} \end{align} $$

The right-hand sides of (5.5) are completely symmetric in $\mathscr {X}$, $\mathscr {Y}$, and $\mathscr {Z}$, and this shows the invariance with respect to $\langle \,\cdot \,\,, \,\cdot \,\,\rangle $ of $\ast _{A}$ and $\ast _{S}$ and so also of any linear combination $s\ast _{A} + t \ast _{S}$. That $O(h)$ acts on $(\mathscr {MC}(\mathbb {V}^{\ast }), \ast _{s, t})$ by isometric algebra automorphisms follows from the manifest $O(h)$ invariance of the construction of $\ast _{A}$ and $\ast _{S}$.

Because rescaling the multiplication of a commutative algebra yields an isomorphic algebra, the products $\ast _{s, t}$ are best viewed as parametrized by $[s:t] \in \mathbb {P}(\mathbb R)$.

For $\mathscr {X}, \mathscr {Y} \in \mathscr {MC}(\mathbb {V}^{\ast })$ define $B_{ijkl} = B(\mathscr {X}, \mathscr {Y})_{ijkl} \in S^{2}(\otimes ^{2}\mathbb {V}^{\ast })$ by

(5.6)$$ \begin{align} B(\mathscr{X}, \mathscr{Y})^{\sharp} = \widehat{\mathscr{X}}_{\otimes^{2}\mathbb{V}^{\ast}}\circledcirc \widehat{\mathscr{Y}}_{\otimes^{2}\mathbb{V}^{\ast}} = \widehat{\mathscr{X}}_{{\textstyle\wedge}^{2}\mathbb{V}^{\ast}}\circledcirc \widehat{\mathscr{Y}}_{{\textstyle\wedge}^{2}\mathbb{V}^{\ast}} + \widehat{\mathscr{X}}_{S^{2}\mathbb{V}^{\ast}}\circledcirc \widehat{\mathscr{Y}}_{S^{2}\mathbb{V}^{\ast}}, \end{align} $$

the second equality by Lemma 4.2, so that

(5.7)$$ \begin{align} B_{ijkl} = B(\mathscr{X}, \mathscr{Y})_{ijkl} &= \tfrac{1}{2}\left(\mathscr{X}_{ipjq}\mathscr{Y}_{k}{}^{p}{}_{l}{}^{q} + \mathscr{Y}_{ipjq}\mathscr{X}_{k}{}^{p}{}_{l}{}^{q} \right) = \left(\widehat{\mathscr{X}}_{\otimes^{2}\mathbb{V}^{\ast}}\circledcirc \widehat{\mathscr{Y}}_{\otimes^{2}\mathbb{V}^{\ast}}\right)^{\flat}_{ijkl}. \end{align} $$

Lemma 5.3. Let $(\mathbb {V}, h)$ be a metric vector space. For $\mathscr {X}, \mathscr {Y} \in \mathscr {MC}(\mathbb {V}^{\ast })$, the commutative multiplications $\ast _{A}$ and $\ast _{S}$ on $\mathscr {MC}(\mathbb {V}^{\ast })$ are given by

(5.8)$$ \begin{align} (\mathscr{X} \ast_{A} \mathscr{Y})_{ijkl} & = -2(B_{[ij]kl} - B_{[ijkl]}) = -\tfrac{2}{3}\left(2B_{[ij]kl} - B_{k[ijl]} + B_{[i|k|j]l}\right), \end{align} $$

(5.9)$$ \begin{align} (\mathscr{X} \ast_{S} \mathscr{Y})_{ijkl} & = -\tfrac{2}{3}\left(B_{k[ij]l} + B_{[i|k|j]l} \right) = -\tfrac{2}{3}\left(B_{[ij]kl} + 2B_{k[ij]l} - 3B_{[ijkl]} \right). \end{align} $$

Proof. That $B_{ijkl}$ is contained in $S^{2}(\otimes ^{2}\mathbb {V}^{\ast })$ is equivalent to the symmetry $B_{klij} = B_{ijkl}$, which is apparent from (5.7). There holds

(5.10)$$ \begin{align} 2B_{jilk} = \mathscr{X}_{jpiq}\mathscr{Y}_{l}{}^{p}{}_{k}{}^{q} + \mathscr{Y}_{jpiq}\mathscr{X}_{l}{}^{p}{}_{k}{}^{q} = \mathscr{X}_{iqjp}\mathscr{Y}_{k}{}^{q}{}_{l}{}^{p} + \mathscr{Y}_{iqjp}\mathscr{X}_{k}{}^{q}{}_{l}{}^{p} = 2B_{ijkl}. \end{align} $$

Hence, $2B_{[ij]kl} = B_{ijkl} - B_{jikl} = B_{ijkl} - B_{ijlk} = 2B_{ij[kl]}$. It follows that $2B_{[ij][kl]} = B_{ij[kl]} - B_{ji[kl]} = B_{[ij]kl} - B_{[ji]kl} = 2B_{[ij]kl}$. This means that $B_{[ij]kl} = B_{[ij][kl]} = B_{ij[kl]}$ is the orthogonal projection of $B_{ijkl}$ onto $S^{2}({\bigwedge }^{2}\mathbb {V}^{\ast })$. It follows that $B_{ij(kl)} = B_{(ij)kl} = B_{(ij)(kl)} = B_{(kl)(ij)}$ is the orthogonal projection of $B_{ijkl}$ onto $S^{2}(S^{2}\mathbb {V}^{\ast })$. From the second equality of (5.7) it follows that $(\mathscr {X} \ast _{A} \mathscr {Y})_{ijkl}$ and $(\mathscr {X} \ast _{S} \mathscr {Y})_{ijkl}$ are given by applying $-2\mathscr {M}$ to $B_{[ij]kl}$ and $-\tfrac {2}{\sqrt {3}} \mathcal {T}\circ \mathscr {S}$ to $B_{(ij)kl}$, respectively, and doing so yields the first equalities of (5.8) and (5.9). The second equalities of (5.2) and (5.9) follow from $3B_{[ijkl]} = B_{[ij]kl} + B_{k[ij]l} - B_{[i|k|j]l}$.

Since, by Lemma 4.2, the operator $\widehat {\mathscr {X}}$ preserves type, if $\mathscr {X}, \mathscr {Y} \in \mathscr {MC}(\mathbb {V}^{\ast })$, then $\widehat {\mathscr {X}}(\mathscr {Y}), \widehat {\mathscr {Y}}(\mathscr {X}) \in \mathscr {MC}(\mathbb {V}^{\ast })$. Theorem 5.4 shows that, remarkably, $\widehat {\mathscr {X}}(\mathscr {Y}) = \widehat {\mathscr {Y}}(\mathscr {X})$, so that a commutative multiplication of curvature tensors can be defined by $\mathscr {X} \ast \mathscr {Y} = \widehat {\mathscr {X}}(\mathscr {Y})$, and, moreover, $\ast = \tfrac {3}{2}(\ast _{A} + \ast _{S})$.

Proof. Evaluation of $\widehat {\mathscr {X}}(\mathscr {Y})_{ijkl}$ using (4.4) and $\varrho (\mathscr {X}, \mathscr {Y})_{ijkl}$ using (4.2) and (4.3) yields

(5.11)$$ \begin{align} 4\widehat{\mathscr{X}}(\mathscr{Y})_{ijkl} &= - 2\mathscr{X}_{i}{}^{p}{}_{k}{}^{q}\mathscr{Y}_{jplq}- 2\mathscr{Y}_{i}{}^{p}{}_{k}{}^{q}\mathscr{X}_{jplq} + 2\mathscr{X}_{i}{}^{p}{}_{l}{}^{q}\mathscr{Y}_{jpkq} + 2\mathscr{Y}_{i}{}^{p}{}_{k}{}^{q}\mathscr{X}_{jpkq} \nonumber\\ &\qquad - \mathscr{X}_{ij}{}^{pq}\mathscr{Y}_{pqkl} - \mathscr{X}_{kl}{}^{pq}\mathscr{Y}_{pqij} = 4\varrho(\mathscr{X}, \mathscr{Y})_{ijkl}, \end{align} $$

which is evidently symmetric in $\mathscr {X}$ and $\mathscr {Y}$, showing that $\mathscr {X}\ast \mathscr {Y} = \widehat {\mathscr {X}}(\mathscr {Y}) = \widehat {\mathscr {Y}}(\mathscr {X})= \mathscr {Y} \ast \mathscr {X}$. By  Lemma 4.2, $\widehat {\mathscr {Y}}$ is self-adjoint, so, for $\mathscr {X}, \mathscr {Y}, \mathscr {Z} \in \mathscr {MC}(\mathbb {V}^{\ast })$,

(5.12)$$ \begin{align} \langle \mathscr{X} &\ast \mathscr{Y}, \mathscr{Z}\rangle =\langle \widehat{\mathscr{Y}}(\mathscr{X}), \mathscr{Z}\rangle = \langle \mathscr{X}, \widehat{\mathscr{Y}}(\mathscr{Z})\rangle = \langle \mathscr{X}, \mathscr{Y} \ast \mathscr{Z}\rangle. \end{align} $$

This shows the complete symmetry of $\langle \mathscr {X} \ast \mathscr {Y}, \mathscr {Z}\rangle $. That $O(h)$ acts by algebra automorphisms follows from (5.11) and the $O(h)$-equivariance of $\varrho $ established in Lemma 4.1.

From the symmetries of $B_{ijkl} = B(\mathscr {X}, \mathscr {Y})_{ijkl}$ it follows that

(5.13)$$ \begin{align} \tfrac{1}{2}\left(\mathscr{X}_{ij}{}^{pq}\mathscr{Y}_{pqkl} \right.&\left. + \mathscr{Y}_{ij}{}^{pq}\mathscr{X}_{pqkl}\right) \nonumber\\ &= \left(\mathscr{X}_{jpiq} + \mathscr{X}_{pijq} \right)\left(\mathscr{Y}^{q}{}_{k}{}^{p}{}_{l} + \mathscr{Y}_{k}{}^{pq}{}_{l} \right) + \left(\mathscr{Y}_{jpiq} + \mathscr{Y}_{pijq} \right)\left(\mathscr{X}^{q}{}_{k}{}^{p}{}_{l} + \mathscr{X}_{k}{}^{pq}{}_{l} \right) \nonumber\\ & = B_{ijkl} + B_{jilk} - B_{ijlk} - B_{jikl} = 4B_{[ij][kl]} = 4B_{ij[kl]} = 4B_{[ij]kl}. \end{align} $$

Combining (5.11) and (5.13) shows

(5.14)$$ \begin{align} -(\mathscr{X} \ast \mathscr{Y})_{ijkl} &= B_{ijkl} - B_{ijlk} + B_{ikjl} - B_{iljk}= B_{ijkl} - B_{jikl} + B_{ikjl} - B_{iljk}. \end{align} $$

That $\ast = \tfrac {3}{2}(\ast _{A} + \ast _{S})$ follows from (5.8), (5.9), and (5.14). By [Reference Hamilton18, section $7$] the polarization of the quadratic term of the expression (1.2) for the evolution of the curvature tensor under the Ricci flow equals the right-hand side of (5.14).

A metrized commutative $\mathbb R$-algebra $(\mathbb {A}, \circ , h)$ is determined up to isometric isomorphism by the $O(h)$-orbit of the associated homogeneous cubic polynomial, $P_{(\mathbb {A}, \circ )}(x)= (1/6)h(x\circ x, x)$, because P and the metric h determine the multiplication $\circ $ via polarization.

Corollary 5.6. The cubic polynomial of $(\mathscr {MC}(\mathbb {V}^{\ast }), \ast )$ has the form

(5.15)$$ \begin{align} P_{(\mathscr{MC}(\mathbb{V}^{\ast}), \ast)}(\mathscr{X}) = \operatorname{\mathsf{tr}} \widehat{\mathscr{X}}_{{\textstyle\wedge}^{2}\mathbb{V}^{\ast}}^{3} + \tfrac{1}{3}\operatorname{\mathsf{tr}} \widehat{\mathscr{X}}_{S^{2}\mathbb{V}^{\ast}}^{3}. \end{align} $$

Lemma 5.7. Let $(\mathbb {V}, h)$ be a metric vector space. For $\mathscr {X}, \mathscr {Y} \in \mathscr {MC}(\mathbb {V}^{\ast })$,

(5.16)$$ \begin{align} & \operatorname{\rho}(\mathscr{X} \ast \mathscr{Y}) = \tfrac{1}{2}\left( \widehat{\mathscr{X}}_{S^{2}\mathbb{V}^{\ast}}(\operatorname{\rho}(\mathscr{Y})) + \widehat{\mathscr{Y}}_{S^{2}\mathbb{V}^{\ast}}(\operatorname{\rho}(\mathscr{X}))\right),&& \operatorname{s}(\mathscr{X} \ast \mathscr{Y}) = \langle \operatorname{\rho}(\mathscr{X}), \operatorname{\rho}(\mathscr{Y})\rangle. \end{align} $$

In particular, the subspace $\mathscr {MC}_{\mathscr {W}}(\mathbb {V}^{\ast }) = \ker \operatorname {\rho } \subset \mathscr {MC}(\mathbb {V}^{\ast })$ is a subalgebra of $(\mathscr {MC}(\mathbb {V}^{\ast }),\ast )$.

In particular, Lemma 5.8 shows that, in Euclidean signature, if $\mathscr {X} \ast \mathscr {X} = 0$, then $\mathscr {X} \in \mathscr {MC}_{\mathscr {W}}(\mathbb {V}^{\ast })$. That is, a square-zero element of $(\mathscr {MC}(\mathbb {V}^{\ast }), \ast )$ must be contained in $\mathscr {MC}_{\mathscr {W}}(\mathbb {V}^{\ast })$. The first claim of Lemma 5.8 is an example of a claim that depends on the assumption that h have definite signature. In other signatures the same proof shows only that $\operatorname {\rho }(\mathscr {X})$ is null.

6 Calculation of products in $(\mathscr {MC}(\mathbb {V}^{\ast }), \ast )$

There are two ways to construct curvature tensors from simpler objects, a commutative product of symmetric two-tensors and a commutative product $\cdot $ of antisymmetric two-tensors. Lemma 6.1 records the $\ast $ products of curvature tensors obtained in these ways. They are used later in the construction of idempotents in $(\mathscr {MC}_{\mathscr {W}}(\mathbb {V}^{\ast }), \ast )$. Although the full strength of these computations is not needed in this article, their proofs illustrate the use of $\ast _{A}$ and $\ast _{S}$ in the computation of $\ast $.

By straightforward computations using (4.2), Lemma 4.1, and (4.1),

(6.1)

defines a symmetric bilinear map . When $k = 2$, is half what is usually called the Kulkarni–Nomizu product. Again by (4.2), for $\alpha , \beta \in {\bigwedge }^{2}\mathbb {V}^{\ast }$, $\pi (\alpha , \beta )_{ijkl} = \tfrac {1}{2}\left ( \alpha _{k[i}\beta _{j]l} - \alpha _{l[i}\beta _{j]k}\right )$, so that, by Lemma 4.1, an $O(h)$-equivariant symmetric bilinear map $\cdot :{\bigwedge }^{2}(\mathbb {V}^{\ast }) \times {\bigwedge }^{2}(\mathbb {V}^{\ast }) \to \mathscr {MC}(\mathbb {V}^{\ast })$ is defined by

(6.2)$$ \begin{align} \alpha \cdot \beta &= -6\varrho(\alpha, \beta)_{ijkl} = -6\mathscr{M}(\pi(\alpha, \beta))_{ijkl} = -3\left( \alpha_{k[i}\beta_{j]l} - \alpha_{l[i}\beta_{j]k} - 2\alpha_{[ij}\beta_{kl]}\right)\nonumber\\ &= \tfrac{3}{2}(\alpha_{ij}\beta_{kl} + \alpha_{kl}\beta_{ij}) - \tfrac{1}{2}(\alpha \wedge \beta)_{ijkl} = \tfrac{3}{2}\mathscr{M}(\alpha \otimes \beta + \beta \otimes \alpha)_{ijkl}. \end{align} $$

For a commutative algebra $(\mathbb {A}, \circ )$ define

(6.3)$$ \begin{align} &Q_{\circ}(x, y) = 2L_{\circ}(x)\circledcirc L_{\circ}(y) - L_{\circ}(x\circ y), & &Q_{\circ}(x) = Q_{\circ}(x, x) = 2L_{\circ}(x)^{2} - L_{\circ}(x\circ x). \end{align} $$

For a Jordan algebra, $Q_{\circ }(x)$ is what is usually called the quadratic representation [Reference Jacobson23, Reference Koecher24]. For the special case of $\alpha , \beta , \gamma \in \otimes ^{2}\mathbb {V}^{\ast }$,

(6.4)$$ \begin{align} &Q_{\circledcirc}(\alpha, \beta)(\gamma) = \tfrac{1}{2}(\alpha \circ \gamma \circ \beta + \beta \circ \gamma \circ \alpha), && Q_{\circledcirc}(\alpha)(\gamma) = \alpha \circ \gamma \circ \alpha. \end{align} $$

The following identities relate $Q_{\circledcirc }$ with $\circledcirc $ and $[\,\cdot \,\,, \,\cdot \,\,]$.

(6.5)$$ \begin{align} \begin{aligned} Q_{\circledcirc}(\alpha \circledcirc \beta) &+ \tfrac{1}{4}Q_{\circledcirc}([\alpha, \beta]) = \tfrac{1}{2}Q_{\circledcirc}(\alpha\circ \beta) + \tfrac{1}{2}Q_{\circledcirc}(\beta \circ\alpha),\\ Q_{\circledcirc}(\alpha \circledcirc \beta) &- \tfrac{1}{4}Q_{\circledcirc}([\alpha, \beta]) =Q_{\circledcirc}(\alpha) \circledcirc Q_{\circledcirc}(\beta). \end{aligned} \end{align} $$

Define $\operatorname {\mathsf {tr}}:S^{k}\mathbb {V}^{\ast } \to S^{k-2}\mathbb {V}^{\ast }$ by $\operatorname {\mathsf {tr}}(\omega )_{i_{1}\dots i_{k-2}} = \omega _{i_{1}\dots i_{k-2}p}{}^{p}$. For $\alpha \in \otimes ^{2}\mathbb {V}^{\ast }$, let $\alpha ^{\sharp } = \langle \alpha , \,\cdot \,\,\rangle \in \otimes ^{2}\mathbb {V}$, so that, in the notation of (4.6), $2(\alpha \odot \beta )^{\sharp } = \alpha \otimes \beta ^{\sharp } + \beta \otimes \alpha ^{\sharp }$, where $\odot $ is the symmetrized tensor product of elements of $\otimes ^{2}\mathbb {V}^{\ast }$. From the definitions there follow

(6.6)

(6.7)

(6.8)

By (4.5), (6.1), and (6.2), for $\mathscr {X} \in \mathscr {MC}(\mathbb {V}^{\ast })$, $\gamma , \sigma \in S^{2}\mathbb {V}^{\ast }$, and $\alpha , \beta \in {\bigwedge }^{2}\mathbb {V}^{\ast }$,

(6.9)

(6.10)$$ \begin{align} \langle \alpha \cdot \beta , \mathscr{X}\rangle & = -6 \langle \varrho(\alpha, \beta), \mathscr{X}\rangle = -6\langle \widehat{\mathscr{X}}(\alpha), \beta \rangle = -6\langle \alpha, \widehat{\mathscr{X}}(\beta)\rangle. \end{align} $$

For $\alpha , \beta , \gamma , \sigma \in {\bigwedge }^{2}(\mathbb {V}^{\ast })$, by (6.10) and (6.7),

(6.11)$$ \begin{align} \langle \alpha \cdot \beta, \gamma \cdot \sigma \rangle& = - 6\langle \widehat{\alpha\cdot \beta}_{{\textstyle\wedge}^{2}\mathbb{V}^{\ast}}(\gamma), \sigma\rangle \nonumber\\ & = 3\left(\langle \alpha, \gamma \rangle\langle \beta, \sigma \rangle + \langle \alpha, \sigma \rangle \langle \beta, \gamma \rangle + \operatorname{\mathsf{tr}}(\alpha\circ \gamma \circ \beta \circ \sigma+ \alpha\circ \sigma \circ \beta \circ \gamma)\right). \end{align} $$

For $\alpha , \beta , \gamma , \sigma \in S^{2}\mathbb {V}^{\ast }$, by (6.9) and (6.6),

(6.12)

For $\alpha , \beta \in {\bigwedge }^{2}\mathbb {V}^{\ast }$ and $\gamma , \sigma \in S^{2}\mathbb {V}^{\ast }$, by (6.8) together with (6.9) or (6.10),

(6.13)

Using (4.1) it is straightforward to check that, for $\sigma \in S^{2}\mathbb {V}^{\ast }$ and $\tau \in {\bigwedge }^{2}\mathbb {V}^{\ast }$, there hold

(6.14)

(6.15)$$ \begin{align} \hspace{-12pt}\tau_{i[k}\tau_{l]j} - \tau_{[ik}\tau_{lj]} = -\tfrac{1}{6}(\tau\cdot \tau)_{ijkl}, \phantom{sssss.ss}&\implies -2\mathscr{M}((Q_{\circledcirc}(\tau)_{{\textstyle\wedge}^{2}\mathbb{V}^{\ast}})^{\flat}) = \tfrac{1}{3}\tau \cdot \tau, \quad \ \, \end{align} $$

(6.16)

(6.17)$$ \begin{align} \hspace{-30pt}\tau_{i(k}\tau_{l)j} =-\tfrac{\sqrt{3}}{6}\mathcal{T}^{-1}(\tau \cdot \tau)_{ijkl},\phantom{ssssss.sssss}&\implies -\tfrac{2}{\sqrt{3}}\mathcal{T} \circ \mathscr{S}((Q_{\circledcirc}(\tau)_{S^{2}\mathbb{V}^{\ast}})^{\flat}) =\tfrac{1}{3}\tau \cdot \tau. \end{align} $$

Lemma 6.1. Let $(\mathbb {V}, h)$ be a metric vector space. For $\alpha , \beta , \gamma , \sigma \in S^{2}(\mathbb {V}^{\ast })$,

(6.18)

For $\alpha , \beta \in {\bigwedge }^{2}\mathbb {V}^{\ast }$ and $\gamma , \sigma \in S^{2}(\mathbb {V}^{\ast })$,

(6.19)

For $\alpha , \beta , \gamma , \sigma \in {\bigwedge }^{2}(\mathbb {V}^{\ast })$,

(6.20)

Proof. For $\alpha , \beta \in S^{2}\mathbb {V}^{\ast }$, by (6.8), (6.6), and (6.5),

(6.21)

(6.22)

By (5.1), applying $-2\mathscr {M}$ to (6.21) and $-\tfrac {2}{\sqrt {3}}\mathcal {T} \circ \mathscr {S}$ to (6.22), and using (6.2) and (6.14)–(6.17) with $\sigma = \alpha \circledcirc \beta $ and $\tau = [\alpha , \beta ]$ to simplify the results yields

(6.23)

Using $\ast = \tfrac {3}{2}(\ast _{A} + \ast _{S})$ and (6.23) to evaluate yields

(6.24)

Polarizing (6.24) first in $\alpha $ and then in $\beta $ yields (6.18).

For $\alpha \in {\bigwedge }^{2}\mathbb {V}^{\ast }$ and $\gamma \in S^{2}\mathbb {V}^{\ast }$, by (6.7), (6.8), (6.6), and (6.5),

(6.25)

(6.26)

By (5.1), applying $-2\mathscr {M}$ to (6.25) and $-\tfrac {2}{\sqrt {3}}\mathcal {T} \circ \mathscr {S}$ to (6.26), and using (6.2), (6.1), and (6.14)–(6.17) with $\sigma = [\alpha , \gamma ]$ and $\tau = \alpha \circledcirc \gamma $ to simplify the results yields

(6.27)

Using $\ast = \tfrac {3}{2}(\ast _{A} + \ast _{S})$ and (6.27) to evaluate yields

(6.28)

Polarizing (6.28) first in $\alpha $ and then in $\gamma $ yields (6.19).

For $\alpha , \beta \in {\bigwedge }^{2}\mathbb {V}^{\ast }$, by (6.7), (6.8), and (6.5),

(6.29)$$ \begin{align} \begin{split} &\widehat{\alpha \cdot \alpha}_{{\textstyle\wedge}^{2}\mathbb{V}^{\ast}}\circledcirc \widehat{\beta \cdot \beta}_{{\textstyle\wedge}^{2}\mathbb{V}^{\ast}} = (Q_{\circledcirc}(\alpha) - (\alpha \otimes \alpha)^{\sharp})\circledcirc(Q_{\circledcirc}(\beta) - (\beta \otimes \beta)^{\sharp})\\ & = Q_{\circledcirc}(\alpha \circledcirc \beta) - \tfrac{1}{4}Q_{\circledcirc}([\alpha, \beta]) + \langle \alpha, \beta \rangle \left(\alpha \odot \beta\right)^{\sharp} -\left( Q_{\circledcirc}(\beta)(\alpha) \odot \alpha\right)^{\sharp} -\left( Q_{\circledcirc}(\alpha)(\beta) \odot \beta\right)^{\sharp} , \end{split} \end{align} $$

(6.30)$$ \begin{align} \begin{split} &\widehat{\alpha \cdot \alpha}_{S^{2}\mathbb{V}^{\ast}}\circledcirc \widehat{\beta \cdot \beta}_{S^{2}\mathbb{V}^{\ast}} = 9Q_{\circledcirc}(\alpha)\circledcirc Q_{\circledcirc}(\beta) = 9Q_{\circledcirc}(\alpha \circledcirc \beta) - \tfrac{9}{4}Q_{\circledcirc}([\alpha, \beta]). \end{split} \end{align} $$

By (5.1), applying $-2\mathscr {M}$ to (6.29) and $-\tfrac {2}{\sqrt {3}}\mathcal {T} \circ \mathscr {S}$ to (6.30), and using (6.2) and (6.14)–(6.17) with $\tau = [\alpha , \beta ]$ and $\sigma = \alpha \circledcirc \beta $ to simplify the results yields

(6.31)

Using $\ast = \tfrac {3}{2}(\ast _{A} + \ast _{S})$ and (6.31) to evaluate $(\alpha \cdot \alpha ) \ast (\beta \cdot \beta )$ yields

(6.32)

Polarizing (6.32) first in $\alpha $ and then in $\beta $ yields (6.20).

Remark 6.2. Taking $\beta = \alpha \in S^{2}\mathbb {V}^{\ast }$ in (6.24) yields

(6.33)

Taking $\gamma = \alpha \in S^{2}\mathbb {V}^{\ast }$ and $\sigma = \beta \in S^{2}\mathbb {V}^{\ast }$ in (6.18) yields

(6.34)

The identities (6.33) and (6.34) suggest that, with further conditions on $\alpha $ and $\beta $, $\ast $-idempotents can be constructed from , , and . This is shown to be the case in Lemma 7.4. Similarly, taking $\alpha = \beta = \gamma = \sigma \in {\bigwedge }^{2}\mathbb {V}^{\ast }$ in (6.20) yields

(6.35)

The identity (6.35) suggests that with some further condition on $\alpha $, the element $\alpha \cdot \alpha $ or its trace-free part $\operatorname {\mathsf {tf}}(\alpha \cdot \alpha )$ might be a $\ast $-idempotent. Corollary 7.5 and Lemma 10.1 show that this works. In this regard, see also Lemma 7.3, which uses (6.32) and (6.35) to construct square-zero elements in $(\mathscr {MC}_{\mathscr {W}}(\mathbb {V}^{\ast }), \ast )$ when h has indefinite signature.

7 Simplicity of $(\mathscr {MC}_{\mathscr {W}}(\mathbb {V}^{\ast }), \ast )$ and fusion rules for $(\mathscr {MC}(\mathbb {V}^{\ast }), \ast )$

Lemma 5.7 shows that the Weyl curvature tensors constitute a subalgebra of $(\mathscr {MC}(\mathbb {V}^{\ast }), \ast )$. The first part of the section is dedicated to constructing idempotents in $(\mathscr {MC}_{\mathscr {W}}(\mathbb {V}^{\ast }), \ast )$. This is used in the proof of Theorem 1.3 and to show the nontriviality of the subalgebra $(\mathscr {MC}_{\mathscr {W}}(\mathbb {V}^{\ast }), \ast )$, which in turn yields Corollary 7.6, showing the simplicity of $(\mathscr {MC}_{\mathscr {W}}(\mathbb {V}^{\ast }), \ast )$ when $\dim \mathbb {V}> 4$, but is also interesting in its own right, as experience with Jordan and axial algebras suggests that detailed information about idempotents and the spectra of their multiplication endomorphisms is useful for understanding the internal structure of an algebra such as $(\mathscr {MC}_{\mathscr {W}}(\mathbb {V}^{\ast }), \ast )$.

Next, explicit formulas for products in $(\mathscr {MC}(\mathbb {V}^{\ast }), \ast )$ are deduced and these are used to deduce the main result of this section, Theorem 7.12, that describes the interaction of the subspaces $\mathscr {MC}_{\mathscr {S}}(\mathbb {V}^{\ast })$, $\mathscr {MC}_{\mathscr {R}}(\mathbb {V}^{\ast })$, and $\mathscr {MC}_{\mathscr {W}}(\mathbb {V}^{\ast })$ with respect to $\ast $.

Lemma 7.1. Let $(\mathbb {V}, h)$ be an n-dimensional Euclidean vector space. For an $O(n)$-module of tensors $\mathbb {W}$, let $\operatorname {\mathsf {tf}}\in \operatorname {End}(\mathbb {W})$ denote the orthogonal projection onto the $O(n)$-submodule $\mathbb {W}_{0} \subset \mathbb {W}$ comprising trace-free elements. Define $S^{2}_{0}(\mathbb {V}^{\ast }) = \{\alpha \in S^{2}(\mathbb {V}^{\ast }): \operatorname {\mathsf {tr}}\alpha = 0\}$, , and . The orthogonal projections $\mathscr {P}_{\mathscr {R}}, \mathscr {P}_{\mathscr {S}} \in \operatorname {End}(\mathscr {MC}(\mathbb {V}^{\ast }))$ on $\mathscr {MC}_{\mathscr {R}}(\mathbb {V}^{\ast })$ and $\mathscr {MC}_{\mathscr {S}}(\mathbb {V}^{\ast })$ are given by

(7.1)

where $\operatorname {\rho }_{\circ }(\mathscr {X}) = \operatorname {\rho }(\mathscr {X}) - \tfrac {1}{n}\operatorname {s}(\mathscr {X})h$, and the trace-free part $\operatorname {\mathsf {tf}}(\mathscr {X})$ of $\mathscr {X} \in \mathscr {MC}(\mathbb {V}^{\ast })$ is given by

(7.2)

Proof. For $\alpha , \beta \in S^{2}(\mathbb {V}^{\ast })$, computations using the definitions show

(7.3)

(7.4)

(7.5)

When $n>2$, $\mathscr {MC}(\mathbb {V}^{\ast }) = \mathscr {MC}_{\mathscr {W}}(\mathbb {V}^{\ast }) \oplus \mathscr {MC}_{\mathscr {R}}(\mathbb {V}^{\ast }) \oplus \mathscr {MC}_{\mathscr {S}}(\mathbb {V}^{\ast })$ is an orthogonal decomposition into irreducible $O(n)$-modules (although $\mathscr {MC}_{\mathscr {W}}(\mathbb {V}^{\ast })$ is trivial if $\dim \mathbb {V} = 3$). By (7.4), if $n> 2$, the map is a linear isomorphism from $S^{2}_{0}(\mathbb {V}^{\ast })$ onto its image in $\mathscr {MC}(\mathbb {V}^{\ast })$, which is $\mathscr {MC}_{\mathscr {R}}(\mathbb {V}^{\ast })$. The expressions (7.1) and (7.2) follow from (7.4) and (7.5).

Example 7.2. For $\alpha , \beta \in S^{2}\mathbb {V}^{\ast }$, taking in (7.2) and using (7.3) yields

(7.6)

Similarly, by (7.2) and (7.7), for $\alpha , \beta \in {\bigwedge }^{2}(\mathbb {V}^{\ast })$,

(7.7)$$ \begin{align} &\operatorname{\rho}(\alpha \cdot \beta)= 3 \alpha \circledcirc \beta,\qquad \operatorname{s}(\alpha \cdot \beta) = -3\langle \alpha, \beta \rangle, \end{align} $$

(7.8)

Since, by (6.9), for any $\mathscr {X} \in \mathscr {MC}(\mathbb {V}^{\ast })$, by (7.7) there holds . Alternatively, this is a special case of (6.10) or a special case of (6.13).

Let denote the set of idempotent elements in the algebra $(\mathbb {A}, \circ )$. Two idempotents are orthogonal if $e\circ f = 0 = f\circ e$.

If , then $\operatorname {\mathsf {tr}} \alpha = |\alpha |^{2}$ is the rank of the orthogonal proejction $\alpha _{i}{}^{j}$, so $\operatorname {\mathsf {tr}} \alpha $ is said to be the rank of $\alpha $. If are orthogonal idempotents, then $\alpha \circ \beta = 0 = \beta \circ \alpha $, for $\alpha \circ \beta = \alpha \circ \alpha \circ \beta \circ \beta = \beta \circ \beta \circ \alpha \circ \alpha = \beta \circ \alpha = -\alpha \circ \beta $.

Lemma 7.4. Let $(\mathbb {V}, h)$ be an n-dimensional Euclidean vector space. For orthogonal idempotents with $a = \operatorname {\mathsf {tr}} \alpha $ and $b =\operatorname {\mathsf {tr}} \beta$ there hold

(7.9)

Moreover, , so if and only if $b \neq 1$. In this case . If $a \neq 1$ and $b \neq 1$, then and are orthogonal idempotents in $(\mathscr {MC}(\mathbb {V}^{\ast }), \ast )$.

Corollary 7.5. Let $(\mathbb {V}, h)$ be a Euclidean vector space of dimension $n> 2$. If orthogonal idempotents satisfy $a = \operatorname {\mathsf {tr}} \alpha \neq 1$ and $b = \operatorname {\mathsf {tr}} \beta \neq 1$, then

(7.10)

is an idempotent in $(\mathscr {MC}_{\mathscr {W}}(\mathbb {V}^{\ast }), \ast )$ satisfying $|\mathscr {B}(\alpha , \beta )|^{2} = \tfrac {2(a + b -2)ab}{(a + b -1)(a - 1)(b - 1)}$. In particular, if $b \notin \{1, n-1\}$, $\hat {\beta } = h - \beta $, and $\hat {b} = \operatorname {\mathsf {tr}} \hat {\beta }$, then

(7.11)

is an idempotent in $(\mathscr {MC}_{\mathscr {W}}(\mathbb {V}^{\ast }), \ast )$ satisfying $|\mathscr {B}(\beta )|^{2} = \tfrac {2(n-2)(n-b)b}{(n-1)(n-1 - b)(b - 1)}$.

Lemma 7.7. Let $(\mathbb {V}, h)$ be a metric vector space. For $\mathscr {X} \in \mathscr {MC}(\mathbb {V}^{\ast })$ and $\alpha \in S^{2}(\mathbb {V}^{\ast })$,

(7.12)

Proof. For $\alpha \in S^{2}\mathbb {V}^{\ast }$ and $\mathscr {X}, \mathscr {Y} \in \mathscr {MC}(\mathbb {V}^{\ast })$,

(7.13)

the first equality by the invariance of $\langle \,\cdot \,\,, \,\cdot \,\,\rangle $, the second equality by (6.9), the third and fourth equalities because $\widehat {\mathscr {Z}}(h) = \operatorname {\rho }(\mathscr {Z})$ for any $\mathscr {Z} \in \mathscr {MC}(\mathbb {V}^{\ast })$, the fifth equality again by (6.9), and the last equality by (5.16). By the nondegeneracy of $\langle \,\cdot \,\,, \,\cdot \,\, \rangle $, (7.13) implies the first equality of (7.12). Because $\widehat {\mathscr {X}}(h) = \operatorname {\rho }(\mathscr {X})$, taking $\alpha = h$ in the first equality of (7.12) yields its second equality.

Lemma 7.8. Let $(\mathbb {V}, h)$ be an n-dimensional metric vector space. For $\alpha , \beta \in S^{2}(\mathbb {V}^{\ast })$,

(7.14)

(7.15)

(7.16)

Lemma 7.9. Let $(\mathbb {V}, h)$ be a Euclidean vector space of dimension $n \geq 4$. Let

(7.17)

Then $\mathscr {MC}_{\mathscr {W}}(\mathbb {V}^{\ast }) = \mathbb {W}_{1} = \mathbb {W}_{2} = \mathbb {W}_{3} = \mathbb {W}_{4}$.

Lemma 7.11. Let $(\mathbb {V}, h)$ be an n-dimensional Euclidean vector space. The projections onto the $O(h)$-irreducible summands of $\mathscr {MC}(\mathbb {V}^{\ast })$ of $\mathscr {X}, \mathscr {Y} \in \mathscr {MC}(\mathbb {V}^{\ast })$ satisfy

(7.18)

(7.19)

(7.20)$$ \begin{align} \operatorname{\mathsf{tf}}(\mathscr{X})\ast \mathscr{P}_{\mathscr{S}}(\mathscr{Y}) &= 0, \hspace{270pt} \end{align} $$

(7.21)$$ \begin{align} \operatorname{\mathsf{tf}}(\mathscr{X})\ast \mathscr{P}_{\mathscr{R}}(\mathscr{Y}) &= \mathscr{P}_{\mathscr{R}}(\operatorname{\mathsf{tf}}(\mathscr{X})\ast \mathscr{Y}), \hspace{220pt} \end{align} $$

(7.22)

An irreducible $O(n)$-submodule of $\otimes ^{k}\mathbb {V}^{\ast }$ comprises the completely trace-free tensors of a given type (this statement is false for even $k = n$ if $O(n)$ is replaced by $SO(n)$). By Lemma 4.2, if $G \subset O(n)$ is a Lie subgroup and $\mathbb {U} \subset \mathscr {MC}(\mathbb {V}^{\ast })$ and $\mathbb {W} \subset \otimes ^{k}\mathbb {V}^{\ast }$ are G-submodules, then $\widehat {\mathbb {U}}(\mathbb {W})$ is a G-submodule of $\otimes ^{k}\mathbb {V}^{\ast }$, for if $g \in O(n)$, $\mathscr {X} \in \mathbb {U}$ and $\mathscr {Y} \in \mathbb {W}$, $g\cdot \widehat {\mathscr {X}}(\mathscr {Y}) = \widehat {g\cdot \mathscr {X}}(g\cdot \mathscr {Y}) \in \widehat {\mathbb {U}}(\mathbb {W})$. In particular, because $\widehat {\mathscr {X}}$ preserves type, if $\mathbb {U}_{1}, \mathbb {U}_{2} \subset \mathscr {MC}(\mathbb {V}^{\ast })$ are G-submodules, then $\widehat {\mathbb {U}_{1}}(\mathbb {U}_{2})$ is a G-submodule of $\mathscr {MC}(\mathbb {V}^{\ast })$. By Theorem 5.4, $\widehat {\mathbb {U}_{2}}(\mathbb {U}_{1}) = \mathbb {U}_{2}\ast \mathbb {U}_{1} = \mathbb {U}_{1}\ast \mathbb {U}_{2}= \widehat {\mathbb {U}_{1}}(\mathbb {U}_{2})$. For example, Corollary 7.6 shows that $\widehat {\mathbb {U}}(\mathbb {U}) = \mathbb {U} \ast \mathbb {U} = \mathbb {U}$ for $\mathbb {U} = \mathscr {MC}_{\mathscr {W}}(\mathbb {V}^{\ast })$. Theorem 7.12 describes completely the submodules $\mathbb {U}_{1}\ast \mathbb {U}_{2}$ for $\mathbb {U}_{1}, \mathbb {U}_{2}$ among the $O(n)$-irreducible summands of $\mathscr {MC}(\mathbb {V}^{\ast })$.

Theorem 7.12. Let $(\mathbb {V}, h)$ be an n-dimensional Euclidean vector space. The products of the $O(n)$-irreducible submodules of $\mathscr {MC}(\mathbb {V}^{\ast })$ satisfy:

(7.23)$$ \begin{align} \mathscr{MC}_{\mathscr{S}}(\mathbb{V}^{\ast}) &\ast \mathscr{MC}_{\mathscr{S}}(\mathbb{V}^{\ast}) = \mathscr{MC}_{\mathscr{S}}(\mathbb{V}^{\ast}), \quad \text{if} \,\,n> 1, \end{align} $$

(7.24)$$ \begin{align} \mathscr{MC}_{\mathscr{R}}(\mathbb{V}^{\ast}) &\ast \mathscr{MC}_{\mathscr{S}}(\mathbb{V}^{\ast}) = \mathscr{MC}_{\mathscr{R}}(\mathbb{V}^{\ast}), \quad \text{if} \,\,n> 2, \end{align} $$

(7.25)$$ \begin{align} \mathscr{MC}_{\mathscr{W}}(\mathbb{V}^{\ast}) &\ast \mathscr{MC}_{\mathscr{S}}(\mathbb{V}^{\ast}) = \{0\}, \ \, \quad\qquad\qquad\qquad \end{align} $$

(7.26)$$ \begin{align} \mathscr{MC}_{\mathscr{W}}(\mathbb{V}^{\ast}) &\ast \mathscr{MC}_{\mathscr{R}}(\mathbb{V}^{\ast}) = \mathscr{MC}_{\mathscr{R}}(\mathbb{V}^{\ast}), \quad \text{if} \,\,n> 3, \ \end{align} $$

(7.27)$$ \begin{align} \operatorname{\mathsf{tf}}(\mathscr{MC}_{\mathscr{R}}(\mathbb{V}^{\ast}) &\ast \mathscr{MC}_{\mathscr{R}}(\mathbb{V}^{\ast})) = \mathscr{MC}_{\mathscr{W}}(\mathbb{V}^{\ast}), \quad \text{if} \,\,n> 3, \ \ \ \, \end{align} $$

(7.28)$$ \begin{align} \mathscr{MC}_{\mathscr{W}}(\mathbb{V}^{\ast}) &\ast \mathscr{MC}_{\mathscr{W}}(\mathbb{V}^{\ast}) = \mathscr{MC}_{\mathscr{W}}(\mathbb{V}^{\ast}), \quad \text{if} \,\,n> 3. \ \end{align} $$

Proof. The containments in (7.23)–(7.27) of subspaces are consequences of polarizing (7.18)–(7.22). The equalities require further justification. The product (7.25) is immediate from (7.20). By (7.15), multiplication by , which spans $\mathscr {MC}_{\mathscr {S}}(\mathbb {V}^{\ast })$, is invertible on $\mathscr {MC}_{\mathscr {R}}(\mathbb {V}^{\ast })$ when $n> 2$ and on $\mathscr {MC}_{\mathscr {S}}(\mathbb {V}^{\ast })$ when $n> 1$, and this suffices to show the equalities (7.23) and (7.24).

Let $\alpha \in S^{2}(\mathbb {V}^{\ast })$. If $\mathscr {X} \in \mathscr {MC}_{\mathscr {W}}(\mathbb {V}^{\ast })$, then $\operatorname {\mathsf {tr}} \widehat {\mathscr {X}}(\alpha ) = \langle \operatorname {\rho }(\mathscr {X}), \alpha \rangle = 0$, so, by (7.12), . This shows the containment of $O(n)$-modules, $\mathscr {MC}_{\mathscr {W}}(\mathbb {V}^{\ast }) \ast \mathscr {MC}_{\mathscr {R}}(\mathbb {V}^{\ast }) \subset \mathscr {MC}_{\mathscr {R}}(\mathbb {V}^{\ast })$. By the irreducibility of $\mathscr {MC}_{\mathscr {R}}(\mathbb {V}^{\ast })$, to show equality it suffices to exhibit a nonzero element of $\mathscr {MC}_{\mathscr {W}}(\mathbb {V}^{\ast }) \ast \mathscr {MC}_{\mathscr {R}}(\mathbb {V}^{\ast })$. Let $u, v \in \mathbb {V}$ be such that $|u|^{2} = 2 = |v|^{2}$ and $\langle u, v \rangle = 0$. Then $\alpha = u\odot v \in S^{2}_{0}\mathbb {V}^{\ast }$ satisfies $|\alpha |^{2} = 2$, $2 \alpha \circ \alpha = u \otimes u + v \otimes v$, $\langle \alpha , \alpha \circ \alpha \rangle = 0$, and $\alpha \circ \alpha \circ \alpha = \alpha $. From these and there follow , , , and . By (7.6), , and by (6.6) and the preceding observations, , so, by (7.12),

(7.29)

which shows that $\mathscr {MC}_{\mathscr {W}}(\mathbb {V}^{\ast }) \ast \mathscr {MC}_{\mathscr {R}}(\mathbb {V}^{\ast })$ is nontrivial if $n> 3$ and so proves the equality in (7.26).

Suppose $\dim \mathbb {V}^{\ast }> 3$. Let $x, y, z, w \in \mathbb {V}^{\ast }$ be pairwise orthogonal unit norm vectors. Then $\alpha = x \odot y$ and $\beta = z \odot w$ are in $S^{2}_{0}\mathbb {V}^{\ast }$, so . Because $\alpha \circledcirc \beta = 0$ and $\langle \alpha , \beta \rangle = 0$, by (7.14), . By (7.3), , so . Since, by Lemma 7.9, $\mathscr {MC}_{\mathscr {W}}(\mathbb {V}^{\ast })$ is spanned by elements of the form , this shows the equality in (7.27). The equality (7.28) is Corollary 7.6.

As an application of Lemma 7.11 there is given a simple proof of [Reference Böhm and Wilking2, Theorem $2$]. For $\alpha , \beta \in \mathbb R$ define an $O(n)$-equivariant endomorphism $\Phi _{\alpha , \beta } \in \operatorname {End}(\mathscr {MC}(\mathbb {V}^{\ast }))$ by

(7.30)$$ \begin{align} \Phi_{\alpha, \beta}(\mathscr{X}) = \operatorname{\mathsf{tf}}(\mathscr{X}) + \beta \mathscr{P}_{\mathscr{R}}(\mathscr{X}) + \alpha \mathscr{P}_{\mathscr{S}}(\mathscr{X}) = \mathscr{X} + (\beta - 1)\mathscr{P}_{\mathscr{R}}(\mathscr{X}) + (\alpha - 1)\mathscr{P}_{\mathscr{S}}(\mathscr{X}). \end{align} $$

For example, (7.22) can be rewritten as .

Note that $\Phi _{1, 1} = \operatorname {Id}_{\mathscr {MC}(\mathbb {V}^{\ast })}$. Because $\Phi _{\alpha , \beta }\circ \Phi _{\bar {\alpha }, \bar {\beta }} = \Phi _{\alpha \bar {\alpha }, \beta \bar {\beta }}$, $\Phi _{\alpha , \beta }$ is invertible if and only if $\alpha \neq 0$ and $\beta \neq 0$, in which case $\Phi _{\alpha , \beta }^{-1} = \Phi _{\alpha ^{-1}, \beta ^{-1}}$. If $\alpha = 1 + 2(n-1)a$ and $\beta = 1 + (n-2)b$, then $\Phi _{\alpha , \beta }$ equals the endomorphism called $l_{a, b}$ introduced by C. Böhm and B. Wilking in [Reference Böhm and Wilking2]. The reason for working with the parameters $\alpha $ and $\beta $ is that the map $(\alpha , \beta ) \in \mathbb {R}^{\times } \times \mathbb {R}^{\times } \to \Phi _{\alpha , \beta } \in \operatorname {End}(\mathscr {MC}(\mathbb {V}^{\ast }))$ is an injective group homomorphism.

The key point of Theorem 7.14 for its applications is that (7.31) does not depend on $\operatorname {\mathsf {tf}}(\mathscr {X})$.

Theorem 7.14 C. Böhm and B. Wilking [Reference Böhm and Wilking2, Theorem $2$].

Let $(\mathbb {V}, h)$ be an n-dimensional Euclidean vector space. If $\alpha , \beta \in \mathbb R\setminus \{0\}$ and $D_{\alpha , \beta }(\mathscr {X}) = \Phi _{\alpha , \beta }^{-1}\left (\Phi _{\alpha , \beta }(\mathscr {X}) \ast \Phi _{\alpha , \beta }(\mathscr {X}) \right ) - \mathscr {X} \ast \mathscr {X}$, where $\Phi _{\alpha , \beta } \in \operatorname {End}(\mathscr {MC}(\mathbb {V}^{\ast }))$ is defined in (7.30), then

(7.31)

Proof. Straightforward calculations using (7.18)–(7.22) yield

(7.32)

The special case $\alpha = 1 = \beta $ yields

(7.33)

Combining (7.32) and (7.33) yields (7.31). After rewriting (7.31) in terms of the parameters a and b and a bit of computation it can be seen that (7.31) recovers the conclusion of [Reference Böhm and Wilking2, Theorem $2$].

8 Characterization of $(\mathscr {MC}(\mathbb {V}^{\ast }), \ast )$ when $\dim \mathbb {V} = 3$

If $\dim \mathbb {V} = 2$, the $1$-dimensional algebra $(\mathscr {MC}(\mathbb {V}^{\ast }), \ast )$ is generated by and, by (7.16), it is isomorphic to the field $\mathbb R$ of real numbers by the map sending to $1 \in \mathbb R$. This section identifies $(\mathscr {MC}(\mathbb {V}^{\ast }), \ast )$ in a similarly explicit manner when $(\mathbb {V}, h)$ is a $3$-dimensional Euclidean vector space. Pulling the multiplication $\ast $ back via an $O(3)$-equivariant linear isomorphisms $\Psi : S^{2}\mathbb {V}^{\ast } \to \mathscr {MC}(\mathbb {V}^{\ast })$ yields an $O(3)$-equivariant commutative multiplication $\diamond $ on $S^{2}\mathbb {V}^{\ast }$. By Lemma 8.1, requiring that the associated map $\alpha \in S^{2}\mathbb {V}^{\ast } \to 2\widehat {\Psi (\alpha )}_{{\textstyle \wedge }^{2}\mathbb {V}^{\ast }} \in \operatorname {Sym}({\bigwedge }^{2}\mathbb {V}^{\ast }, \langle \,\cdot \,\,, \,\cdot \,\, \rangle )$ be a Jordan algebra isomorphism determines $\Psi $ uniquely, and this determines a standard model $(S^{2}\mathbb {V}^{\ast }, \diamond )$ of $(\mathscr {MC}(\mathbb {V}^{\ast }), \ast )$ realizing it as a deformation of Jordan product $\circledcirc $ on $S^{2}\mathbb {V}^{\ast }$ by terms built from the metric and the trace. Most of the section is devoted to formulating and proving Theorem 8.5, which specifies algebraic conditions that characterize the resulting algebra $(S^{2}\mathbb {V}^{\ast }, \diamond )$ up to isomorphism.

Because $S^{2}\mathbb {V}^{\ast } = S^{2}_{0}\mathbb {V}^{\ast } \oplus \text {Span}\, \{h\}$ and $\mathscr {MC}(\mathbb {V}^{\ast }) = \mathscr {MC}_{\mathscr {R}}(\mathbb {V}^{\ast })\oplus \mathscr {MC}_{\mathscr {S}}(\mathbb {V}^{\ast })$ are decompositions into $O(3)$-irreducible submodules, by the Schur Lemma the most general $O(3)$-equivariant linear map $S^{2}\mathbb {V}^{\ast } \to \mathscr {MC}(\mathbb {V}^{\ast })$ has the form for some $p,\tau \in \mathbb R$ and $\psi _{p, \tau } \in \operatorname {End}(S^{2}\mathbb {V}^{\ast })$ defined by $\psi _{p, \tau }(\alpha )= p(\alpha +\tfrac {\tau - 1}{3} (\operatorname {\mathsf {tr}} \alpha )h)$. Because $\psi _{p, \tau }\circ \psi _{\bar {p}, \bar {\tau }} = \psi _{p\bar {p}, \tau \bar {\tau }}$, $\psi _{p, \tau }$ is invertible if and only if $p\tau \neq 0$, in which case $\psi _{p, \tau }^{-1} = \psi _{p^{-1}, \tau ^{-1}}$, and $\Psi _{p, \tau }\circ \psi _{\bar {p}, \bar {\tau }} = \Psi _{p\bar {p}, \tau \bar {\tau }}$. By (7.3),