All sets of $11$ quaternions satisfy some multilinear $11$th degree polynomial equation with fixed real coefficients

Question

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.

Answer 1

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.

Answer 2

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 $w$.