# What is the dimension of this algebraic variety?

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 ?

-
That variety has in general several irreducible components of different dimensions. I'm not sure the maximal dimension of a component is known. – Mariano Suárez-Alvarez Nov 11 '11 at 12:51

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.
The Higman-Sims formula says that the number of groups of order $p^n$, $\#G(p^n)$, satisfies $\log_p(\#G(p^n)) = \left(\frac{2}{27}n^3+O(n^{8/3})\right)$; I presume these two results are closely coupled? – Steven Stadnicki Oct 28 '14 at 15:41