3

The algebra has this multiplication table:

$$\begin{array}{c|ccc}\odot&E_1&E_2&E_3\\\hline E_1&0&E_3&-E_2\\E_2&E_3&0&E_1\\E_3&-E_2&E_1&0\end{array}$$

Equivalently, any vector $aE_1+bE_2+cE_3=\begin{bmatrix}a\\b\\c\end{bmatrix}$ acts on other vectors by the matrix

$$\begin{bmatrix}0&c&b\\-c&0&-a\\b&a&0\end{bmatrix}.$$

Are there any (commutative, non-associative) polynomial equations that are satisfied by all elements of this algebra?

The base field is $\mathbb Q$ or $\mathbb R$ or $\mathbb C$, or maybe anything of characteristic $0$. We want formal polynomials to be equivalent to polynomial functions. (So, no finite fields.)

We only need to consider polynomials that are homogeneous in each variable. If $p(A,B,\cdots)=0$ is an equation with different degrees of $A$, then replacing $A$ with $tA$ gives a polynomial $\mathbb R\to\mathbb R^3:\;q(t)=p(tA,B,\cdots)=0$, all of whose coefficients must vanish; and each coefficient is homogeneous in $A$.

In fact, we only need to consider polynomials that are linear in each variable. For example, if $q(A,C,\cdots)$ is quadratic in $A$, then the polarization

$$p(A,B,C,\cdots)=\frac{q(A+B,C,\cdots)-q(A,C,\cdots)-q(B,C,\cdots)}{2}$$

is linear in $A$ and $B$, and vanishes identically if and only if $q$ does.

This section is wrong, but I'm still trying to figure out where I made my mistake. The first identity polarizes to something involving $(AB)(CD)$, but I didn't find such an identity before considering polarization.

This algebra contains two 2D subalgebras, spanned by $E_2$ and $E_1\pm E_3$. I was quite surprised to find (after a few pages of work) that they satisfy several quartic equations, which are not satisfied in the larger, 3D algebra. The first:

$$(AA)(BB)-(AB)(AB)=((AA)B)B-((AB)A)B-((AB)B)A+((BB)A)A.$$

The second:

$$((AA)B)C-((AB)A)C+((AB)C)A=((AA)C)B-((AC)A)B+((AC)B)A,$$

which is a special case of the third:

$$((AB)C)D+((AC)D)B+((AD)B)C-((AB)D)C-((AD)C)B-((AC)B)D=0.$$

Furthermore, these are the only identities of degree 4 or less in these subalgebras. So any identity in the 3D algebra must be quintic or higher.

(By the way, has anyone seen this algebra before?)

Do you know of any general theorems which would help here? For example, do we only need to consider polynomials of degree bounded by some function of the dimension?

mr_e_man
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  • If the field has characteristic $2$, then the equation $A^2=0$ works. – Anonymous Jun 09 '20 at 03:40
  • It is commutative then it must be Flexible. – AHandsomeAlien Jun 09 '20 at 12:42
  • It's commutative, and satisfies $XY=YX$. Does that count? – rschwieb Jun 09 '20 at 13:59
  • @AHandsomeAlien - As a commutative non-associative polynomial, $(AB)A-A(BA)$ is the trivial polynomial $0$. – mr_e_man Jun 10 '20 at 03:08
  • @rschwieb - As a commutative non-associative polynomial, $AB-BA$ is the trivial polynomial $0$. So it doesn't count. The polynomial must be identically $0$ as a function on this algebra, but not identically $0$ in the space of formal commutative non-associative polynomials. – mr_e_man Jun 10 '20 at 03:12
  • In fact, that gives us another way to think about this problem: Is the natural mapping, from formal commutative non-associative polynomials to polynomial functions on this algebra, an isomorphism? – mr_e_man Jun 10 '20 at 03:55

1 Answers1

1

I found my mistake: while considering polynomials of the form $A^2B^1C^1$, I missed 1 monomial out of 9, $((BC)A)A$.

All quartic identities on the 2D subalgebras can be derived from

$$((AB)C)D-((AD)C)B-((CB)A)D+((CD)A)B-(AB)(CD)+(AD)(CB)=0$$

and

$$((AB)C)D+((AC)D)B+((AD)B)C-((AB)D)C-((AD)C)B-((AC)B)D=0$$

by permutations (like $(A,B,C,D)\mapsto(B,A,C,D)$), specializations (like $(A,B,C,D)\mapsto(A,A,C,D)$), and linear combinations.

And that last polynomial (which is the standard polynomial in the adjoints $[B]=(X\mapsto BX)$ of $B,C,D$, applied to $A$) actually is satisfied in the 3D algebra.

To prove this, write $A=a_1E_1+a_2E_2+a_3E_3$ etc., expand $((AB)C)D$ in terms of these scalar components, factor out the $a$ terms, and note that the remaining polynomials in $b,c,d$ are symmetric with respect to $b\leftrightarrow c$ or $b\leftrightarrow d$ or $c\leftrightarrow d$.

(To simplify calculations, you can get rid of some minus signs by complexifying: $E_1'=iE_1,\,E_2'=E_2,\,E_3'=iE_3$.)

In general, any $n$-dimensional algebra must satisfy some non-trivial identities (not following from commutativity or associativity), assuming it's not both commutative and associative. This is because the dimension of the space of formal $k$-linear polynomials grows factorially, while the space of $k$-linear functions on the algebra (which contains the space of $k$-linear polynomial functions) only grows exponentially.

$$\begin{align}\text{commutative, associative:}\qquad\qquad1&\\ \text{non-commutative, associative:}\qquad\qquad k!&=1\cdot2\cdot3\cdots k\\ \text{commutative, non-associative:}\qquad(2k-3)!!&=1\cdot3\cdot5\cdots(2k-3)\\ \text{non-commutative, non-associative:}\qquad k!\,C_{k-1}&=k\cdot(k+1)\cdot(k+2)\cdots(2k-2)\\ \text{multilinear functions:}\qquad\qquad n^{k+1}&=n\cdot n\cdot n\cdots n\end{align}$$

So, for a large enough number of variables $k$, the natural mapping from formal polynomials to polynomial functions has a non-trivial nullspace.

mr_e_man
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