Suppose $G$ is a finite group, and $l$ is a prime, with $l$ coprime to the order $|G|$. (Thus we have complete reducibility for $G$ representations.) Is there a straightforward condition on $l$ which ensures that every irreducible representation of $G$ is liftable to a characteristic zero representation? (For instance, does the fact that we assume $l$ coprime to $|G|$ suffice?)

  • $\begingroup$ Do we need a new tag, characteristic-l? :) $\endgroup$ – Pete L. Clark Jan 24 '10 at 22:16

Yes, $\operatorname{gcd}(l,|G|) = 1$ is sufficient. This is an easy consequence of Brauer's modular representation theory. See Serre's Linear Representations of Finite Groups, especially Chapter 18.

  • $\begingroup$ By the way, this reference is somehow overkill. It would be nice to see a more direct argument in this much easier special case, e.g. starting with a version of Maschke's Theorem over $\mathbb{Z}_{\ell}[G]$. My instruments predict a response coming in from a northwesternly direction... $\endgroup$ – Pete L. Clark Jan 24 '10 at 22:41
  • $\begingroup$ Use the splitting of $A \rightarrow G \rightarrow B$ whenever the orders of $A$ and $B$ are relatively prime. $\endgroup$ – moonface Jan 25 '10 at 1:19
  • 3
    $\begingroup$ One possible argument is via deformation theory, which will be very natural for some (but maybe not all). Namely, given the irrep. $\rho$ in char. $\ell$, the obstructions to deformation are computed by ${\mathrm Ext}^2_{{\mathbb F}\_{\ell}}(\rho,\rho),$ and the mod $\ell$ tangent space to the deformation space is computed by ${\mathrm Ext}^1_{{\mathbb F}\_{\ell}}(\rho,\rho).$ Since both vanish when $\ell$ is prime to $|G|$, we see that each irrep. $\rho$ mod $\ell$ lifts uniquely to an irrep. in char. 0. $\endgroup$ – Emerton Jan 25 '10 at 2:29
  • $\begingroup$ Typo: in the above, the subscript ${\mathbb F}_{\ell}$ on the Ext spaces should be ${\mathbb F}_{\ell}[G]$. $\endgroup$ – Emerton Jan 25 '10 at 2:30

Another sufficient condition is that if $G$ is solvable, then for every prime $l$, every absolutely irreducible characteristic $l$ representation can be lifted to the complex numbers. In fact, solvability is not really necessary; $l$-solvability suffices. This is the Fong-Swan theorem.

Added later: Since groups with order not divisible by $l$ are trivially $l$-solvable, this sufficient condition includes, and is more general than the the condition stated by Pete L. Clark.


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.