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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.

Turion
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  • $\Bbb R[\sqrt[4]{2}]=\Bbb R$ because $\sqrt[4]{2}\in\Bbb R$. – anon Feb 11 '14 at 19:33
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    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. – anon Feb 11 '14 at 19:40
  • Sorry, my notation was flawed. See the updated example. – Turion Feb 11 '14 at 20:26
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    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. – anon Feb 11 '14 at 21:03
  • 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. – Turion Feb 11 '14 at 23:35
  • Why do you think your example is 2-dimensional? I did not miss the relation $ab-b$. The first isomorphism I wrote down is from deleting the last relation in the denominator while simultaneously replacing $a$ with $b^2/\sqrt{2}$ (in place of the last relation). – anon Feb 11 '14 at 23:47
  • @anon, sorry, it's a polynomial ring of course. I guess I was actually thinking about a noncommutative algebra, but I just failed several times at writing it down, but I'll not try again since the algebra I wanted to think of was 2d anyways ;) – Turion Feb 11 '14 at 23:59
  • You should be able to extract a $7$-dimensional counterexample from section 2 of this paper: http://math.mit.edu/~poonen/papers/dimension6.pdf – Qiaochu Yuan Feb 12 '14 at 04:28
  • @Qiaochu, this extends your argument, but how does it give a counterexample? I see that I can choose two of these matrices to get an algebra and there are infinitely many different choices, but I don't see why such an example doesn't ever have all structure constants rational. – Turion Feb 12 '14 at 10:51
  • @Turion: if an algebra of the form described in Proposition 2.1 has rational structure constants, the cross ratio you get at the end of the proof of Proposition 2.1 must be an algebraic number. So it suffices to find an example where this is not the case. – Qiaochu Yuan Feb 12 '14 at 18:15

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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.

Qiaochu Yuan
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  • 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. – Turion Feb 11 '14 at 23:39
  • @Turion: I totally agree that this is not a complete answer! – Qiaochu Yuan Feb 11 '14 at 23:41
  • 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 $n=2$. – anon Feb 11 '14 at 23:41
  • @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 $1$). – Qiaochu Yuan Feb 11 '14 at 23:43
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    @anon You missed $\mathbb{R} [x] / (x^2)$; $\mathbb{R} \times \mathbb{R}$ is reduced. – Zhen Lin Feb 11 '14 at 23:48
  • @Zhen Oops, yes. Forgot zeros could have multiplicity. – anon Feb 11 '14 at 23:50
  • Associativity is $n^3$ constraints on the structure constants: $$(x_ix_j)x_k = x_i(x_jx_k) \implies \sum_l c^{ij}_l c^{lk}_m = \sum_l c^{il}_m c^{jk}_l \forall i,j,k$$. The constraints are not independent, though, for example interchanging $i$ and $k$ gives the same constraint. So probably the dimension is still bigger 0 for sufficiently large $n$. – Turion Feb 12 '14 at 00:08
  • @Turion: right, that's the problem with the naive counting argument with associativity, which is why I ignored it. – Qiaochu Yuan Feb 12 '14 at 00:43