4
$\begingroup$

If Brauer characters are $\bar{\mathbb{Q}}$-linearly independent, why are they $\mathbb{C}$-linearly independent?

I think this is a linear algebra fact showing up when proving the irreducible Brauer characters on a finite group are linearly independent over $\mathbb{C}$. The proof I've seen observes that the characters take values in the ring of algebraic integers, and then proves linear independence over $\bar{\mathbb{Q}}$.

Why is it sufficient to only check linear independence over $\bar{\mathbb{Q}}$? It seems like something could go wrong when extending the field all the way up to $\mathbb{C}$.

The proof I'm reading is Theorem 15.5 in Isaacs' Character Theory of Finite Groups.

enter image description here

$\endgroup$
  • $\begingroup$ This does not hold in the generality you presume (e. g. the constant functions $1,\sqrt 2$ are linearly independent over $\mathbb Q$ but not over $\mathbb Q(\sqrt 2)$). One needs that $\mathbb C$ is transcendental over $\overline {\mathbb Q}$ (and that the values of your functions lie in $\overline {\mathbb Q}$). $\endgroup$ – Claudius Jul 29 '16 at 11:26
  • $\begingroup$ I might be easier if everyone had the same information as you: Where (give a source) did you stumble over this problem? $\endgroup$ – Claudius Jul 29 '16 at 11:34
  • $\begingroup$ @user218931 It is Theorem 15.5 of Isaacs' Character Theory. $\endgroup$ – Kally Jul 29 '16 at 17:07
4
$\begingroup$

If $E/F$ is a field extension, we have $F^n\subset E^n$, and if a subset of $F^n$ is $F$-linearly independent, then it is also $E$-linearly independent. A nice, super easy way to see it: extend the subset to a basis for $F^n$. Form the matrix whose columns are elements of this basis. Its determinant is nonzero. But this shows that the columns form a basis for $E^n$ since the determinant has the same formula regardless of the field you work over.

The space of class functions of a finite group can be identified with $F^n$ in an obvious way ($n$= number of conjugacy classes).

$\endgroup$
2
$\begingroup$

Fact. Extension of scalars preserves linear independence. That is, if $X$ is a $k$-linearly independent subset of a $k$-vector space $V$, and $K/k$ is any field extension, then it will also be a $K$-linearly independent subset of the $K$-vector space $K\otimes_k V$ (where we identify $x$ with $1\otimes x$).

Proof. Write $K=\bigoplus_{t\in T}kt$ as vector spaces for some $k$-basis $T$ of $K$. Then if $\sum c_i x_i=0$ for some $x_1,x_2,\cdots,x_n\in X$ and scalars $c_1,c_2,\cdots,c_n\in K$, we may write $c_i=\sum_{t\in T} c_{i,t}t$ for some restricted scalars $c_{i,t}\in k$, in which case the relation becomes $\sum_i c_{i,t}x_i=0$ for all $t$, but this implies each $c_{i,t}=0$ and hence every $c_i=0$.

It applies here with $V$ the vector space of functions spanned by characters and $\mathbb{C}/\overline{\mathbb{Q}}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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