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Take a finite dimensional associative algebra $A$ over the reals. Fix a basis $\{x_1, x_2, \ldots x_n\}$. The multiplication is completely specified by specifying structure constants $c^{ij}_k$ defined by the following equation: $$x_i \cdot x_j = \sum_k c^{ij}_k x_k \quad\forall i, j$$ Of course, the structure constants depend on the choice of basis.

My question is: Are there algebras such that no choice of basis leads to structure constants that are all rational?

My conjectured example would be the algebra spanned by two variables $a$ and $b$, and relations $a^2 = a, ab = ba = b, b^2 = \sqrt{2}a$, but I couldn't prove it yet.

Note: One could also ask "integer" instead of "rational", this is equivalent.

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  • $\begingroup$ $\Bbb R[\sqrt[4]{2}]=\Bbb R$ because $\sqrt[4]{2}\in\Bbb R$. $\endgroup$ – anon Feb 11 '14 at 19:33
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    $\begingroup$ For fields $L/K$, an $L$-algebra $A$ admits structure constants from $K$ if and only if there is an $L$-spanning $K$-subalgebra $B$ of $A$. Not sure if that's helpful or not, but it's a coordinate-free equivalent. $\endgroup$ – anon Feb 11 '14 at 19:40
  • $\begingroup$ Sorry, my notation was flawed. See the updated example. $\endgroup$ – Turion Feb 11 '14 at 20:26
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    $\begingroup$ Well $$\frac{\Bbb R[a,b]}{(a^2-a,ba-b,b^2-\sqrt{2}a)}\cong\frac{\Bbb R[b]}{((b^2/\sqrt{2})^2-b^2/\sqrt{2},b^3/\sqrt{2}-b)}\cong\frac{\Bbb R[b]}{(b^3-\sqrt{2}b)} $$ is isomorphic by CRT to $$\frac{\Bbb R[b]}{(b)}\times\frac{\Bbb R[b]}{(b-\sqrt[4]{2})}\times\frac{\Bbb R[b]}{(b+\sqrt[4]{2})}\cong \Bbb R\times\Bbb R\times\Bbb R$$ which certainly admits a basis with rational structure constants. Presumably, using finite-dimensionality one can proceed inductively on simple algebra extensions. $\endgroup$ – anon Feb 11 '14 at 21:03
  • $\begingroup$ How can you end up with a three-dimensional algebra at the end if my example is 2-dimensional (I think)? You're missing out on a relation, $ab-b$. Not sure if that changes anything. It probably leads to $$\frac{\mathbb{R}[b]}{(b^2-\sqrt{2})}$$, not much of a difference, then. $\endgroup$ – Turion Feb 11 '14 at 23:35
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Nice question! Here is a coarse heuristic argument suggesting that the answer should be yes: the structure constants of a multiplication on a finite-dimensional real vector space $V$ of dimension $n$ form an $n \times n \times n$ table of real numbers, so the space of these should have dimension $n^3$ (ignoring associativity, since it's hard to tell how many independent constraints this adds). The action of change of basis on these structure constants is the action of $\text{GL}(V)$, which has dimension $n^2$, hence the quotient of structure constants by change of basis should have dimension at most $n^3 - n^2$, which in particular is positive as soon as $n \ge 2$.

On the other hand, there are only countably many choices of rational structure constants even before quotienting by change of basis. So as soon as $n \ge 2$ the "generic" algebra should be a counterexample (although again it's hard to tell how strong a constraint associativity is; perhaps you need to go a few dimensions higher).

Edit: Including units, the dimension count is as follows. If we always include the unit as the first element of our basis, then a multiplication (still not assumed to be associative) on an $(n+1)$-dimensional real vector space $V$ is a table of $n \times n \times (n+1)$ numbers, and the action of change of basis is a group of dimension $n^2 + n$ (we also want to be able to add the unit to other elements of our basis). So now the dimension count is

$$n^2 (n + 1) - n^2 - n = n^3 - n$$

which is positive as soon as $n \ge 2$, hence as soon as $n+1 \ge 3$, so this is the dimension where we should start seeing generic counterexamples.

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  • $\begingroup$ I find your point very convincing, but I still fail to find an example where I can actually prove that its structure constants are never rational. That would be a requirement for an answer that I can accept, I'm afraid. $\endgroup$ – Turion Feb 11 '14 at 23:39
  • $\begingroup$ @Turion: I totally agree that this is not a complete answer! $\endgroup$ – Qiaochu Yuan Feb 11 '14 at 23:41
  • $\begingroup$ The only two-dimensional real algebras up to isomorphism are $\Bbb R\times\Bbb R$ and $\Bbb C$, so we at least have to skip $n=2$. $\endgroup$ – anon Feb 11 '14 at 23:41
  • $\begingroup$ @anon: oh, I see, I didn't properly take into account the influence of units. What I wrote above is for non-unital, non-associative multiplications (but we can add in units by adjoining them so this just involves shifting dimensions up by $1$). $\endgroup$ – Qiaochu Yuan Feb 11 '14 at 23:43
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    $\begingroup$ @anon You missed $\mathbb{R} [x] / (x^2)$; $\mathbb{R} \times \mathbb{R}$ is reduced. $\endgroup$ – Zhen Lin Feb 11 '14 at 23:48

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