Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 ?

share|improve this question
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

1 Answer 1

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.

share|improve this answer
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 2 days ago
There is a really strong analogy but not, to my knowledge, a theorem to make it precise. –  David Speyer 2 days ago

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.