# Do “$K/k$ twisted” representations exist?

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.

• 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). – Watson Dec 24 '18 at 12:56
• You may want to look at Galois cohomology, namely 1.3.6 here : the twists of $(\rho, V)$ should be classified by $$H^1( \mathrm{Gal}(K/k) ; \mathrm{Aut}_K(\rho \otimes_k K) ).$$ – Watson Dec 24 '18 at 12:56

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$$M\cong N$$.
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.)