Is this algebra simple?

Question

I have this algebra $A$, defined on the elements of the basis $ \left\{ L_{k}\right\} _{k\in\mathbb{Z}} $ with a bilinear product such that $$L_{a}\circ L_{b}=\frac{1}{2}\left(L_{\phi(a,b)+2a-b}+L_{-\phi(a,b)+2b-a}\right),$$

where $\phi(a,b)=-\phi(b,a)$ is an integer number such that $\phi(a,b)\sim -2(a-b)/3$. Unfortunately $\phi(a,b)$ does not have a definite law, the only thing that is known is that the integer is bounded by $$−2(a-b)/3−3≤\phi(a,b)≤−2(a-b)/3+3 $$ and that letting $a,b$ grow to infinity we have $\phi(a,b)\sim -2(a-b)/3$. Essentially you can think of $\phi(a,b)$ as $\phi(a,b)=2(b−a)/3$ plus or minus something that makes it an integer

Now, I have reasons to say that this algebra is simple since suppose you have an ideal $I$ then $L_{a}\circ I$ has to be contained in $I$ for all $L_{a}$ and it seems to me that this will always end up implying $I=A$. Nevertheless I cannot make this formulation precise. Does anybody have an idea on how to proof that?

Remark: note that if $$L_{a}\circ L_{b}=\frac{1}{2}\left(L_{n}+L_{m}\right).$$ Then $a+b=n+m$. This identity was useful for other things, so even is almost trivial I wanted to pointed out.

Answer 1

The conditions on $\phi$ stated are NOT enough to show that the algebra is simple. We can pick, see below, our own $\phi$ (within the constraints) for which $A$ has an ideal. Now I get the impression that you are dealing with a concrete (though hard to compute) $\phi$ coming from an application, so below results do not rule out that $A$ is simple in your case. However it means that you would need to extract more of its secrets before you can start hoping for a simple $A$ again.

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Following the notation in your useful formula we define

$$n(a, b) = \phi(a, b) + 2a - b; \quad m(a, b) = -\phi(a, b) + 2b - a$$

so that

$$L_a \circ L_b = 1/2(L_{n(a, b)} + L_{m(a,b)}).$$

As remarked in the comments we have that $m(a, b) = n(b, a)$ so that

$$L_a \circ L_b = L_b \circ L_a$$

and $A$ is commutative.

This in turn means that the right ideal $L_0A$ is in fact a two sided ideal.

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If $A$ is simple the two-sided ideal $L_0A$ must be all of $A$, hence contain the element $L_{17}$. In other words, there must be a solution $X$ of the equation

$$L_0 \circ X = L_{17} \qquad \qquad {(1)}$$

Any $X$ is of the form $\sum_{i = 1}^k \lambda_i L_{b_i}$ for a finite set of $b_i \in \mathbb{Z}$ and $\lambda_i \in \mathbb{C}$. The left hand side of $(1)$ is then a finite linear combination of the basis elements $L_{m(0, b_i)}$ and $L_{n(0, b_i)}$ and so in order to have a solution of (*) we at the very least must have that one $m(0, b_1), \ldots, m(0, b_k)$, $n(0, b_1), \ldots, n(0, b_k)$ equals 17.

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Hence, in order to create a counter-example, all we need to do is device a $\phi$ so that no $m(0, b)$ and no $n(0, b)$ equals 17. Since there are at most 7 (positive) values of $b$ for which there is any hope anyway that $m(0, b) = 17$ and at most 7 (negative) values of $b$ for which we might dream that $n(0, b)$ is 17, this is rather easily achieved.

We choose the values of these 14 $\phi(0, b)$ carefully so that no $n(0, b)$ and no $m(0, b)$ equals 17 and fill in the other values of $\phi(a, b)$ as we please. Then the $\phi$ we constructed yields an algebra $A$ in which the ideal $L_0A$ is proper.

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The above suggests an easy necessary condition for $A$ to be simple: for each $z$ (playing the role of 17 in the above) there must be a $b \in \mathbb{Z}$ so that $m(0, b) = z$ or $n(0, b) = z$. However, this condition only guarantees that the ideal $L_0A$ contains an element of the form

$$L_{17} + \textrm{ something.}$$

If we want it to contain $L_{17}$ itself we get a more conditions. Staring a bit at the situation we conclude that in order for $X \mapsto L_0 \circ X$ to be surjective, we need that each $z \in \mathbb{Z}$ appears as $m(0, b)$ or $n(0, b)$ in at least two ways.

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Moreover, the above only talks about the ideal $L_0A$. But we also need $L_1A, L_2A$ etc to be all of $A$ in order for $A$ to be simple. Packing everything together we conclude that in order for $A$ to be simple we must at least satisfy the following

NECESSARY CONDITION:

For each $a, z \in \mathbb{Z}$ there are $b, c \in \mathbb{Z}$ (possibly equal) such that at least one of the following three scenarios is fulfilled:

• $m(a, b) = m(a, c) = z$
• $m(a, b) = n(a, c) = z$
• $n(a, b) = n(a, c) = z$

I do not claim this necessary condition is also sufficient.