TITLE: What is the dimension of this algebraic variety?
QUESTION [8 upvotes]: Let $\mathbb K$ be a number field of degree $n$ over $\mathbb Q$, and let
$\alpha_1,\alpha_2, \ldots ,\alpha_n$ be a $\mathbb Q$-basis of $\mathbb K$. Then there
are coefficients $(c^{ij}_k)$ (where $i,j,k$ are independent indices between $1$ and $n$)
such that
$$
\alpha_i \times \alpha_j = \sum_{k=1}^{n} c^{ij}_k \alpha_k
$$ 
and we have for any indices $i,j,k,l$,
$$
(1) c^{ij}_k=c^{ji}_k \ (\ {\rm commutativity})
$$
$$
(2) \sum_{y=1}^{n}c^{iy}_lc^{jk}_y=\sum_{y=1}^{n}c^{ij}_yc^{yk}_l \ (\ {\rm associativity})
$$
If we take $\alpha_n$ to be $1-\sum_{y=1}^{n-1} \alpha_y$, we also have
$$
(3) \sum_{y=1}^{n}c^{yi}_j=\delta_{ij} 
$$
where $\delta_{ij}$ is the Kronecker symbol.
Now, let $V$ be the algebraic variety in the variables $( c^{ij}_k)$ 
defined as the subset of ${\mathbb C}^{n^3}$ satisfying equations (1) to (3). Is the dimension of $V$ (in the sense of algebraic geometry) known ?

REPLY [2 votes]: As Mariano says, this variety has multiple components for $n \geq 8$. There is no known formula for the dimension of the largest component, and none is likely to be known, but it is known to be $\frac{2}{27} n^3 + O(n^{8/3})$. For proofs of these facts and much more, see Poonen, The moduli space of commutative algebras of finite rank.