Proof. Consider the free non-associative algebra on the set $\{b,\, b',\, x\}$. We quotient it by the ideal $I$ generated by $xb$, $bx$, $b'x$, $xb'$, $xx$, $bb$, $b'b'$ and all monomials of degree $3$ or higher in which one of the variables $b$, $b'$, $x$ is repeated. We reflect the result into the variety ${\mathcal {V}}$ by dividing out the ideal generated by the identities of ${\mathcal {V}}$ and obtain an algebra $Q$. Since ${\mathbb {K}}$ is an infinite field, by Theorem 1.4, the sub-vector space generated by the classes of the elements of the form $(bb')x$ (all possible permutations and bracketings) consists entirely of classes of linear combinations of such elements of degree $3$. We call this vector space $A$, and view it as an abelian ${\mathbb {K}}$-algebra.

We write $B$ for the abelian ${\mathcal {V}}$-algebra generated by $b$, $B'$ for the abelian ${\mathcal {V}}$-algebra generated by $b'$ and $C$ for the free vector space generated by $c$, viewed as an abelian ${\mathbb {K}}$-algebra. In ${\mathcal {V}}$, we then define a representation of $B + B'$ on the abelian algebra $V=A\times C$ by ${}^{b}a = 0 = a^{b}$, ${}^{b'}a = 0 = a^{b'}$, ${}^{bb'}a = {}^{b'b}a = 0 = a^{bb'} = a^{b'b}$ where $a$ is any element of $A$, ${}^{b}c = 0 = c^{b}$, ${}^{b'}c = 0 = c^{b'}$ and ${}^{bb'}c = (bb')x$, ${}^{b'b}c = (b'b)x$, $c^{bb'} = x(bb')$, $c^{b'b} = x(b'b)$. We need to verify that this does indeed determine an action. To do so, let us give an explicit description of the corresponding ${\mathcal {V}}$-algebra morphism $\xi \colon (B+B')\flat V\to V$. We first consider four ${\mathcal {V}}$-algebra morphisms $B+B'+A+C\to Q\times C$, one which sends $\theta (b,\,b',\,a,\,c)$ – where $\theta$ is a polynomial in $n+3$ variables and $a\in A^{n}$ for some $n\in {\mathbb {N}}$ – to $(\theta (b,\,b',\,0,\,x),\,0)$, and three others, which send it to $(\theta (0,\,0,\,a,\,0),\,0)$, $(\theta (0,\,0,\,0,\,x),\,0)$ and $(0,\,\theta (0,\,0,\,0,\,c))$, respectively. We combine them into the ${\mathbb {K}}$-linear map

\begin{align*} B+B'+A+C& \to Q\times C\colon\\ \theta(b,b',a,c)& \mapsto (\theta(b,b',0,x)-\theta(0,0,0,x)+\theta(0,0,a,0),\theta(0,0,0,c)). \end{align*}

Since this map vanishes on monomials in $a$ and $c$ that contain at least one element of $A$ and one $c$, it factors through the quotient

\[ {(B+B')+(A+C)\to (B+B')+(A\times C)=(B+B')+V} \]

to a ${\mathbb {K}}$-linear map $(B+B')+V\to Q\times C$, which in turn restricts to a ${\mathbb {K}}$-linear map $(B+B')\flat V\to Q\times C$. It is easy to check that this map factors over the inclusion of $A\times C=V$ into $Q\times C$. The resulting factorization is a ${\mathbb {K}}$-linear map $\xi \colon (B+B')\flat V\to V$, which is actually a morphism of ${\mathbb {K}}$-algebras, because it sends all binary products in $(B+B')\flat V$ to zero, as it should. We may now check that $\xi$ satisfies the axioms of a $B\flat (-)$-algebra, which follows immediately from the definitions – see 1.2.

Since ${\mathcal {V}}$ has representable representations in ${\mathcal {V}}$, the action $\xi$ is necessarily zero, because that is where it is sent by the canonical isomorphism

\[{\mathit{Rep}}(B+B',V)\to {\mathit{Rep}}(B,V)\times {\mathit{Rep}}(B',V) \]

defined by restricting an action of $B+B'$ on $V$ to an action of $B$ on $V$ together with an action of $B'$ on $V$. It follows that $(bb')x$, $(b'b)x$, $x(bb')$ and $x(b'b)$ vanish in $A$. This means that the identities of ${\mathcal {V}}$ allow us to write each of them as a linear combination of some degree $3$ elements of the ideal $I$, which can only mean that the $\lambda /\mu$-rules hold in ${\mathcal {V}}$. In other words, ${\mathcal {V}}$ is algebraically coherent.

Proof. Let us assume that ${\mathcal {V}}$ is a variety with representable representations, all of whose non-trivial identities are of degree three or higher. From Theorem 2.2, we already know that ${\mathcal {V}}$ satisfies the $\lambda / \mu$-rules. The strategy of this proof is to show that representability of representations forces the coefficients $\lambda _i$ and $\mu _i$ to satisfy a specific system of polynomial equations whose set of solutions is empty (unless some degree two identities are satisfied, but we assumed that it is not the case). This then means that algebraic coherence, and therefore representability of representations, cannot hold in ${\mathcal {V}}$ – which is a contradiction.

In order to do so, we let $X$ be the $79$-dimensional abelian algebra generated by the elements $\{x,\, y^{i}_\alpha, z^{ij}_{\alpha \beta } ,\,t^{ijk}_{\alpha \beta \gamma }\mid \alpha ,\beta ,\gamma \in \{ r,l\} \text{ and } (i\; j\; k) \in S_3\}$, where $r$ stands for right and $l$ for left, while $i$, $j$, $k\in \{1,\,2,\,3\}$ are pairwise non-equal. Furthermore, $B^{i}$ is the abelian algebra generated by $b^{i}$, for $i \in \{ 1,\,2,\,3\}$. Those algebras are in ${\mathcal {V}}$ because they are abelian. We now define (abelian) actions of the algebras $B^{i}$ on $X$:

\begin{align*} b^{i}x & =y^{i}_l & xb^{i} & =y^{i}_r \\ b^{i} y^{j}_\alpha & = \begin{cases} z^{ji}_{\alpha l} & \text{if }i \neq j\\ 0 & \text{otherwise} \end{cases} & y^{j}_\alpha b^{i} & = \begin{cases} z^{ji}_{\alpha r} & \text{if }i \neq j\\ 0 & \text{otherwise} \end{cases} \\ b^{i} z^{jk}_{\alpha\beta}& = \begin{cases} t^{jki}_{\alpha \beta l} & \text{if }i \neq j \text{ and }i\neq k\\ 0 & \text{otherwise} \end{cases} & z^{jk}_{\alpha\beta}b^{i}& = \begin{cases} t^{jki}_{\alpha \beta r} & \text{if } i \neq j \text{ and } i\neq k\\ 0 & \text{otherwise} \end{cases} \end{align*}

while $b^{i} t^{jkm}_{\alpha \beta \gamma }=t^{jkm}_{\alpha \beta \gamma }b^{i} =0$. Those actions are well defined. Indeed, the $\lambda / \mu$-rules are satisfied since any product of three elements in the semi-direct product $B^{i} \ltimes X$ vanishes. In order to familiarize ourselves with those actions which will appear again later, we first give some examples of their behaviour in $(B^{1}+B^{2}+B^{3})\ltimes X$:

\[ (b^{1}(b^{2}x))b^{3} = t^{213}_{ l l r} , \quad ((b^{2}x)b^{3})b^{2} = 0, \quad b^{1}(b^{2}y^{3}_r)=t^{321}_{rll}. \]

Next, as explained in § 1, by representability of representations the semi-direct product $(B^{1}+B^{2}+B^{3})\ltimes X$ is in ${\mathcal {V}}$ and thus all elements of this semi-direct product have to satisfy the $\lambda / \mu$-rules. We construct an equational system on the $\lambda _i$ and $\mu _i$ whose set of solutions is empty, by checking the $\lambda / \mu$-rules for some specific elements. For example, since $(xb^{1})b^{2}$ is in $(B^{1}+B^{2}+B^{3})\ltimes X$, decomposing it a first time gives:

\begin{align*} (xb^{1})b^{2} & = \mu_{1}(b^{2}x)b^{1}+\mu_{2}(xb^{2})b^{1}+ \mu_{3}b^{1}(b^{2}x)+\mu_{4}b^{1}(xb^{2}) \\& \quad +\mu_{5}(b^{2}b^{1})x+\mu_{6}(b^{1}b^{2})x+ \mu_{7}x(b^{2}b^{1})+\mu_{8}x(b^{1}b^{2}). \end{align*}

Then we apply the $\lambda / \mu$-rules again on the monomials where $x$ is isolated and write everything on the same side. This gives us a linear combination $\sum \nolimits f_k(\lambda _i,\,\mu _j) z^{ij}_{\alpha \beta }$ which must be equal to zero. Since the $z^{ij}_{\alpha \beta }$ are linearly independent, all the polynomials $f_k\in {\mathbb {Q}}[\lambda _1,\, \ldots ,\, \lambda _8,\, \mu _1,\, \ldots ,\, \mu _8]$ vanish. Following this procedure for $(xb^{1})b^{2}$ gives us the eight first polynomials of Figure 1 (lines 2–9, written in Singular [Reference Decker, Greuel, Pfister and Schönemann12] code). To compute the $f_k$ for $k=9$, …, $32$, we repeat this procedure on the elements $(b^{1}x)b^{2}$, $b^{2}(b^{1}x)$ and $b^{2}(xb^{1})$. The details of the computations are done and explained in [Reference García-Martínez, Tsishyn, Van der Linden and Vienne14].

Next, we use the fact that $(b^{1}b^{2})b^{3}$ and $b^{1}(b^{2}b^{3})$ should also satisfy the $\lambda /\mu$-rules since they are elements of $(B^{1}+B^{2}+B^{3})\ltimes X$. Making the decomposition work on $(b^{1}(b^{2}b^{3}))x$, $x(b^{1}(b^{2}b^{3}))$, $((b^{1}b^{2})b^{3})x$ and $x((b^{1}b^{2})b^{3})$ – again, a procedure explained in [Reference García-Martínez, Tsishyn, Van der Linden and Vienne14] – we find the last 192 polynomials in the system of equations.

Finally, we employ the computer algebra system Singular [Reference Decker, Greuel, Pfister and Schönemann12], which uses Gröbner bases, to look for a common root of the polynomials. Based on the code in Figure 1, the computer system tells us that $1$ is a linear combination in ${\mathbb {Q}}[\lambda _1,\, \ldots ,\, \lambda _8,\, \mu _1,\, \ldots ,\, \mu _8]$ of the polynomials $f_{1}$, …, $f_{224}$. The arXiv version of this paper includes an ancillary file containing an explicit way to write $1$ as a linear combination of the $(f_i)_i$ in ${\mathbb {Q}}[\lambda _1,\, \ldots ,\, \lambda _8,\, \mu _1,\, \ldots ,\, \mu _8]$. Hence the set of solutions of the polynomial system is empty, which completes the proof.

Proof. In this proof, we mimic the trick used in Theorem 4.2 of [Reference García-Martínez and Van der Linden16]. Indeed, in the proof of Proposition 3.1, we wrote $1$ as a linear combination of the $(f_i)_{1\leq i\leq 224}$ in ${\mathbb {Q}}[\lambda _1,\, \ldots ,\, \lambda _8,\, \mu _1,\, \ldots ,\, \mu _8]$. Instead of doing this, we will write some $m\in {\mathbb {N}}$ as a linear combination in ${\mathbb {Z}}[\lambda _1,\, \ldots ,\, \lambda _8,\, \mu _1,\, \ldots ,\, \mu _8]$ of these 224 polynomials because, then, we will just have to check that the system has an empty set of solutions for all infinite fields whose characteristic divides $m$. The problem is that the natural number $m$, equal to ${145679959084559802430969530553780449546315662406534685329857227}$ ${0548680472045472116250386006867468944668940571897397172623649920632890267296075654343504}$ ${20878447841877721000590295768558884307124148778177377876830064916666592523291}$ ${59304174905496937087738581344349487066880186556945585175775695576209957462932}$ ${78480812431412260574404024455598320441859722042018260900473969846828608645627}$ ${87183059873566162910333424228212906065894343670540539725147842861513488173278}$ ${22021774577694196501188227813156363270426620366150688257342489840683094034209}$ ${50318}$, which we find with a first computation in Singular is 541 digits long. This makes finding its prime divisors very difficult. The trick is to compute the Gröbner basis with a different monomial order. This will give us another natural number $m'$ and we will only need to check the inconsistency of the system of equations for the common prime divisors of $m$ and $m'$. A different monomial order may be chosen in line 1 of the code in Figure 1: for instance, swapping $\mu _7$ and $\mu _8$ is expressed as

This leads us to the number $m'$ which is equal to ${525717631958791658273542822932875536277944482117059047298355615522}$ ${4447303866073034964879226451745451233297544838}$ ${18964761098115453534593150454043600868151586630528071290312031929148817936277}$ ${58784552281330948466297460892213124611259557307705714671224295558125970110748}$ ${22233317864875426854341643972018163355893297334149632686212253120055566461120}$ ${4869818358}$ and only 353 digits long. Obviously, $2$ is a common prime divisor. Using the Euclidean algorithm, it is rapidly seen that it is the only one. In order to conclude, we prove as before, using Singular, that $I = {\mathbb {K}}[\lambda _1,\, \ldots ,\, \lambda _8,\, \mu _1,\, \ldots ,\, \mu _8]$ for any field ${\mathbb {K}}$ of characteristic $2$, where $I$ is the ideal of ${\mathbb {K}} [\lambda _1 ,\, \ldots ,\, \lambda _8,\, \mu _1,\, \ldots ,\, \mu _8]$ generated by the polynomials $f_k$ for $k=1$, …, $224$. This is done in Singular by using

in line 1 of the code in Figure 1. Again, explicit linear combinations are given in the arXiv version of this paper as an ancillary file.

Proof. Combine the above with Proposition 1.3.

Proof. First, since ${\mathcal {V}}$ has representable representations, it is algebraically coherent by Theorem 2.2. Hence the $\lambda / \mu$-rules hold. By commutativity, these can be rewritten as

(†)\begin{equation} x(yz) = \lambda (xy)z + \mu y (xz) \end{equation}

for some $\lambda$, $\mu \in {\mathbb {K}}$.

Now we need to use essentially the same algebras and actions as in the proof of Proposition 3.1, but without considerations for left and right, so that the actions satisfy the commutativity rule. In other words, for all $b^{i}$ in $B^{i}$ and $w$ in $X$ we define the actions such that $bw=wb$. Such actions are well defined, and thus the semi-direct product $(B^{1}+B^{2}+B^{3})\ltimes X$ is an object of ${\mathcal {V}}$ by representability of representations. Therefore, its elements should satisfy the identity (†). Let us check this on $b^{1}(b^{2}x)$: indeed, $b^{1}(b^{2}x) = \lambda (b^{1}b^{2})x + \mu b^{2}(b^{1}x) = \lambda ^{2} (xb^{1})b^{2} +\lambda \mu b^{1}(xb^{2})$, so that $0=(\lambda ^{2} + \mu ) z^{12} + (\lambda \mu -1) z^{21}$. By linear independence of $\{z^{12},\,z^{21}\}$, either ${\mathcal {V}}$ is abelian or $\lambda = \mu = -1$.

The second case would mean that the identity (†) is now

(‡)\begin{equation} x(yz) ={-} (xy)z - y (xz). \end{equation}

Therefore, since $b^{1}(b^{2}b^{3})$ is an element of the semi-direct product we have that $b^{1}(b^{2}b^{3})= -(b^{1}b^{2})b^{3} - b^{2}(b^{1}b^{3})$ as an action, and in particular on $x$ this implies

\[ (b^{1}(b^{2}b^{3}))x={-}((b^{1}b^{2})b^{3})x - (b^{2}(b^{1}b^{3}))x. \]

Decomposing each element on the left side and on the right side twice by applying (‡) gives us $t^{123} + t^{132} + t^{231} + t^{321} = - t^{312} - t^{321} - t^{123} - t^{213} - t^{213} - t^{231} - t^{132} - t^{312}$, which is equivalent to $2t^{312} + 2t^{321} + 2t^{123} + 2t^{213} + 2t^{132} + 2t^{231}= 0$. Since ${\mathit {char}}({\mathbb {K}}) \neq 2$ and all these elements are linearly independent, we encounter a contradiction. Hence ${\mathcal {V}}$ is abelian.

Proof. Because of anticommutativity, the $\lambda / \mu$-rules can be rewritten as

(§)\begin{equation} x(yz)= \lambda (xy)z + \mu y(xz) \end{equation}

for some $\lambda$, $\mu \in {\mathbb {K}}$. We consider the actions we defined in the proof of Proposition 4.1, corrected for anticommutativity. So we decree that an element $b^{i}$ acting on the left is the same as $b^{i}$ acting on the right times $-1$. For instance, $b^{1}y^{2}=-y^{2} b^{1}$ and $b^{2}x = -xb^{2}$. In order to check the identity (§) on $b^{1}(b^{2}x)$, we compute:

\[ b^{1}(b^{2}x) =\lambda (b^{1}b^{2})x + \mu b^{2}(b^{1}x) ={-} \lambda^{2} (xb^{1})b^{2} -\lambda\mu b^{1}(xb^{2}) + \mu b^{2}(b^{1}x). \]

By linear independence, this yields the system $\mu - \lambda ^{2} =0$, $1-\lambda \mu =0$ whose solution is $\lambda =\mu =1$. Thus (§) becomes the Jacobi identity.

Proof. Let ${\mathcal {V}}$ be a non-abelian variety of ${\mathbb {K}}$-algebras with representable representations. In the previous sections, it was explained that there are no identities of degree two besides $xy=-yx$ or $xx=0$, and that the Jacobi identity holds. The idea here is to first prove that there is no identity of degree three besides Jacobi, and then we show that no other identities (of degree $n > 3$) can hold.

First, suppose that ${\mathcal {V}}$ satisfies some degree three identities. Because of anticommutativity and Theorem 1.5, we may assume this identity to be $x(yz)= \lambda (xy)z + \mu y(xz)$ for certain $\lambda$, $\mu \in {\mathbb {K}}$. Then we may recycle the ideas of the proof of Proposition 4.3 to reduce it to the Jacobi identity.

Now, the goal is to prove by induction that the existence of an identity of degree $n>3$ is in contradiction with Theorem 1.5. Let us assume that no other identities of degree lower than or equal to $n>3$ are satisfied. Let $\psi$ be a homogeneous identity of degree $n+1$ with $n>3$ which holds in ${\mathcal {V}}$. We consider a non-trivial multilinear consequence $\varphi (x_1 ,\,\ldots ,\, x_n,\,x_0)=0$ of $\psi$, whose existence is assured by Theorem 1.5. Using anticommutativity and the Jacobi identity, we can rewrite $\varphi$ in the shape $0= \sum \nolimits _{i=1} \lambda _i \varphi _i (x_1 ,\, \ldots ,\, x_i x_0,\, \ldots ,\, x_n)$. We observe that if $\lambda _i=0$ for all $i$, then $\varphi$ is just a consequence of $xy=-yx$ and $x(yz)+y(zx)+z(xy)=0$, or in other words a trivial identity. Note that it is impossible to be a consequence of other identities by the induction hypothesis. Proving that the $\lambda _i$'s are zero will then bring us to a contradiction.

We consider the same actions we used in Proposition 4.3 but extended to $n$ algebras $B^{i}$ acting on $X$ (which is also enlarged to more elements). Those are again well defined and thus we have that $(B^{1} + \ldots + B^{n}) \ltimes X$ lies in ${\mathcal {V}}$ as already explained. Therefore the element $\psi ((b^{1},\,0),\,(b^{2},\,0),\, \ldots ,\, (b^{n},\,0),\,(0,\, x))=\psi (b^{1},\, \ldots ,\, b^{n},\,x)$ of this semi-direct product has to vanish and the multilinear consequence $\varphi$ as well, which allows us to write $0= \sum \nolimits _{i=1} \lambda _i \varphi _i (b^{1} ,\, \ldots ,\, b^{i} x,\, \ldots ,\, b^{n})$. Again, by construction of the actions, the elements $\varphi _i (b^{1} ,\, \ldots ,\, b^{i} x,\, \ldots ,\, b^{n})$ of $X$ are linearly independent and thus $\lambda _i=0$ for all $i$ – which completes the proof.