5
$\begingroup$

Given $k$-representations $V,W$ of a group $G$, where $k$ is a field, $K/k$ a field extension, if we have $V\otimes_k K\cong W\otimes_k K$ as $K$-representations, do we have that $V\cong W$?

Being more specific, what about in the case of $V,W$ irreps, $G$ a finite group, with $K/k$ finite and galois?

In characteristic $0$, with character theory, the question can be rephrased as: if the characters of $V$ and $W$ agree, then are $V$ and $W$ isomorphic over their field of definition?

Any reference for these questions would be much appreciated.

$\endgroup$
  • 1
    $\begingroup$ In general, the functor $$- \otimes_k K : k[G]\mathrm{-Mod} \to K[G]\mathrm{-Mod}$$ doesn't reflect isomorphisms (even if this is not explained by the existence of irreps that may not be absolutely irreducible). $$ $$ For algebras instead of modules, see math.stackexchange.com/questions/2578592. Notice that for commutative rings, $- \otimes_R S : R\text{-Mod} \to S\text{-Mod}$ reflects isomorphisms if $R \to S$ is faithfully flat (see here). $\endgroup$ – Watson Dec 24 '18 at 12:56
  • 2
    $\begingroup$ You may want to look at Galois cohomology, namely 1.3.6 here : the twists of $(\rho, V)$ should be classified by $$H^1( \mathrm{Gal}(K/k) ; \mathrm{Aut}_K(\rho \otimes_k K) ).$$ $\endgroup$ – Watson Dec 24 '18 at 12:56
3
$\begingroup$

In fact, for any field extension $K/k$ and any $k$-algebra $A$, if $M$ and $N$ are finite dimensional $A$-modules such that $M\otimes_kK\cong N\otimes_kK$ as $A\otimes_kK$-modules, then $M\cong N$ as $A$-modules. This is the Noether-Deuring Theorem (see, for example, (19.25) in Lam's "First Course in Noncommutative Rings"; he assumes $A$ is finite-dimensional, but that's not essential).

For finite field extensions the proof is very short. If $M\otimes_kK\cong N\otimes_kK$ as $A\otimes_kK$-modules, then, restricting to $A$, $M\otimes_kK\cong N\otimes_kK$ as $A$-modules. But if $[K:k]=n$, then as $A$-modules $M\otimes_kK\cong M^n$, the direct sum of $n$ copies of $M$, and $N\otimes_kK\cong N^n$. So $M^n\cong N^n$. Now just apply the Krull-Schmidt Theorem to deduce that$M\cong N$.

$\endgroup$
1
$\begingroup$

In the case of finite group with semisimple group algebra (i.e. $\mathrm{char}\, k = 0$ or $\mathrm{char}\, k$ is coprime to $|G|$), the claim should be true, i.e. $V\otimes_k K \simeq W\otimes_k K$ implies $V \simeq W$.

In the case $\mathrm{char}\, k = 0$, the reason is character theory: Namely, the orthogonality of characters still works (see e.g. here the version for non-alg. closed field). So after decomposing $V$ and $W$ into irreps, comparing the decomposition comes down to computing $\langle \chi_i, \chi \rangle$ where $\chi$ is the common character of $V, W$ (which is invariant under extension of scalars) and $\chi_i$ runs over irreducible chracters, as in the case $k=\mathbb{C}$.

Note that this argument does not suffice on its own in positive characteristic coprime to $|G|$, i.e. the other semisimple case, since the character orthogonality formula contains the number $d$ of absolutely irreducible components of an irrep, and it's not obvious that this is necessarily nonzero modulo $\mathrm{char}\, k$.

However, there is a Brauer-Nesbitt Theorem that can be used: It states that two semisimple representations of $G$ are isomorphic whenever their characteristic polynomials (meaning: functions taking $g$ to the char. polynomial of $\rho(g)$ when $\rho$ is a representation) agree. Since the char. polynomials are invariant under extension of scalars, the same conclusion as in the previous case follows. (See e.g. these notes of Gabor Wiese, Thm 2.4.6 for a more-general version of the Brauer-Nesbitt theorem.)

$\endgroup$

This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .