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.)

$\endgroup$

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.