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Let $R$ be an arbitrary $n$-dimensional associative algebra, over a field $\mathbb F$. Denote by $P_k(\mathbb F)$ the space of formal $k$'th degree associative non-commutative polynomials, with coefficients in $\mathbb F$, that are linear in each of $k$ variables. For example,

$$P_3(\mathbb F)=\{c_1XYZ+c_2XZY+c_3YXZ+c_4YZX+c_5ZXY+c_6ZYX\mid\vec c\in\mathbb F^6\}.$$

The dimension of $P_k(\mathbb F)$ is $k!$.

Denote by $P_k(\mathbb F,R)$ the corresponding space of functions on $R^k$.

$$P_3(\mathbb F,R)=\{f:R^3\to R\;;\;(A,B,C)\mapsto c_1ABC+c_2ACB+\cdots+c_6CBA\mid\vec c\in\mathbb F^6\}.$$

This is a subspace of the space of all multilinear functions from $R^k$ to $R$. Such a function is uniquely determined by its action on basis vectors; there are $n$ of these to choose from, for each of $k$ input places, and thus a total of $n^k$ possible input combinations. And the output has $n$ components. So the space of all multilinear functions has dimension $n^k\cdot n=n^{k+1}$, and the subspace $P_k(\mathbb F,R)$ cannot exceed this.

For large enough $k$, $k!$ will be larger than $n^{k+1}$, so the natural evaluation map from $P_k(\mathbb F)$ to $P_k(\mathbb F,R)$ will have a non-trivial nullspace.

In other words, there must be some multilinear polynomial equation, $f(A,B,C,\cdots)=0$ for all $(A,B,C,\cdots)\in R^k$.


If the algebra is commutative, there is obviously $AB-BA=0$.

Now, let's take the first non-commutative algebra that comes to mind: the quaternions (others would say $m\times m$ matrices, etc.).

$$R=\mathbb H,\quad n=4,\quad\mathbb F=\mathbb R.$$

The sequence $k!$ first exceeds $4^{k+1}$ at $k=11$:

$$11!=39,916,800;\qquad4^{12}=16,777,216.$$

So there is some polynomial, in a 39-million-dimensional space, that vanishes identically on all sets of 11 quaternions. What is it?

Of course, it's not unique. And 11 is just an upper bound; for all I know, there could be a 3rd degree polynomial that works.

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  • $\begingroup$ The "evaluation map" actually goes from $P_k(\mathbb F)$ directly to $R$, not to $P_k(\mathbb F,R)$; but I don't know what else to call it. $\endgroup$
    – mr_e_man
    Commented Jun 15, 2020 at 19:30

2 Answers 2

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Take the $11$ variables to be $X_1,\ldots,X_{11}$. Consider $$F(X_1,\ldots,X_{11})=(XY-YX)^2Z-Z(XY-YX)^2$$ where $X=X_1$, $Y=X_2$ and $Z=X_3$ (I lost the eight other variables!). This is identically zero on the quaternions.

For quaternions $u$ and $v$, $uv-vu$ is a "pure" quaternion, with zero real part, and so $(uv-vu)^2$ is real, and then $(uv-vu)^2w=w(uv-vu)^2$ for any other quaternion $w$.

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  • $\begingroup$ This isn't multilinear, though you could polarize it: $$G(X,Y,U,Z)=F(X+U,Y,Z)-F(X,Y,Z)-F(U,Y,Z)$$ $$H(X,Y,U,V,Z)=G(X,Y+V,U,Z)-G(X,Y,U,Z)-G(X,V,U,Z)$$ Good answer, anyway. $\endgroup$
    – mr_e_man
    Commented Jun 15, 2020 at 19:02
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Using the usual embedding of $\mathbb H$ into $M_2(\mathbb C)$, by the Amitsur-Levitzki theorem, the elements satisfy the standard polynomial of degree $4$. This polynomial is linear in four noncommuting indeterminates.

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    $\begingroup$ After yoou brought this result to my attention it is actually easy to see that the quaternions satisfy the standard polynomial in four variables. The standard polynomial being multilinear, it suffices to show that it vanishes when the four variables come from the set $\{1,i,j,k\}$. If any value is repeated, the vanishing is automatic, because an odd permutation keeps the input fixed, but changes the signs of all the terms. If they are distinct, then the terms cancel pairwise because $1$ commutes with everything. $\endgroup$ Commented Jun 15, 2020 at 17:00
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    $\begingroup$ (cont'd) For example we can apply either $(12)$ or $(34)$ to the ordering according to which half $1$ is in. So $i1jk-1ijk=0$, $ji1k-jik1=0$ etc. OTOH, the standard polynomial in three variables does not vanish at permutations of $(i,j,k)$. $\endgroup$ Commented Jun 15, 2020 at 17:01

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