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.

Can anyone help me to understand the equation $$|G|=\sum^s_{i=1} d_i n_i^2$$ please?

The context is representation theory of finite groups over a field $\mathbb{K}$ of characteristic zero. I know that for complex representations the is a formula $|G|= \sum_in_i^2$ , where $n_i$'s are the dimensions of the representations. The $d_i$ here are defined as the index $d_i=[D_i : K]$, but I cannot find the book, I saw this, so I don't know what the $D_i$'s are.

Maybe someone knows this equation?

share|improve this question

1 Answer 1

up vote 7 down vote accepted

If $\mathbb{K}$ is algebraically closed (e.g. $\mathbb{K}=\mathbb{C}$), then the Artin-Wedderburn theorem says that the group algebra $\mathbb{K}G$ splits as a direct sum of matrix algebras over $\mathbb{K}$: $$ \mathbb{K}G \cong \bigoplus_i M_{n_i}(\mathbb{K}). $$ Each of the matrix algebras, the so-called blocks, consists of $n_i$ copies of isomorphic irreducible representations of dimension $n_i$, corresponding to sub-algebras consisting of matrices that are 0 outside a given column (my modules are left modules). For $i\neq j$, the associated irreducible representations are mutually non-isomorphic, since they are annihilated by different blocks. That is one way of deriving the familiar formula $|G|=\sum_i n_i^2$.

But when $\mathbb{K}$ is not algebraically closed, this statement needs to be modified. Now, Artin-Wedderburn says that the group algebra $\mathbb{K}G$ is isomorphic to a direct sum of matrix algebras over division algebras: $$ \mathbb{K}G \cong \bigoplus_i M_{n_i}(D_i), $$ where each $D_i$ is a division algebra over $\mathbb{K}$, and each block consists of $n_i$ isomorphic copies of irreducible representations, again corresponding to sub-algebras of matrices that are 0 outside a given column. But now, each of these is $n_i$-dimensional over $D_i$, which in turn has dimension $d_i$ over $\mathbb{K}$. Thus, each irreducible representation has dimension $d_in_i$ over $\mathbb{K}$. Hence the formula $$ \dim_{\mathbb{K}} \mathbb{K}G = \dim_{\mathbb{K}}\left(\bigoplus_i M_{n_i}(D_i)\right)= \sum_i d_in_i^2. $$

This will become clearer if you consider a specific example. Take $G=Q_8$, the quaternion group of order 8. Let's take a non-algebraically closed field of characteristic 0, e.g. $\mathbb{R}$. Recall that the only division algebras over $\mathbb{R}$ are $\mathbb{R}$, $\mathbb{C}$, and $\mathbb{H}$, the quaternion algebra. It turns out that $$ \mathbb{R}Q_8 \cong \mathbb{R}\oplus \mathbb{R}\oplus \mathbb{R}\oplus \mathbb{R}\oplus\mathbb{H}, $$ whence $8 = 1+1+1+1+4$. Note that the block corresponding to $\mathbb{H}$ becomes $M_2(\mathbb{C})$ over $\mathbb{C}$, meaning that the irreducible 4-dimensional representation over $\mathbb{R}$ splits as a direct sum of two 2-dimensional irreducible representations over $\mathbb{C}$.

share|improve this answer
    
Thanks a lot !Grace –  Grace Apr 21 '12 at 11:22

Your Answer

 
discard

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.