is a finite group, and is a field. Assume all modules are finite-dimensional over (for simplicity). We showed that
If then .
Suppose now that is invertible in , i.e., . is also a -vector space. Therefore we have , i.e., . So under this assumption,
By the dimension shifting argument, for all .
Also last time
where is the trivial -dimensional module. By definition of ,
for all under the same assumption that .
can still be non-zero.
This proves:
Theorem (Semisimplicity of -representations)
#TODO2
Let be a finite group and be a field with invertible in . Then for any (finite-dimensional, although not needed) -modules and ,
In other words, (finite-dimensional) is semisimple. This means that any short exact sequence splits:
Example: from representation theory
If in , then is not semisimple. To show this, we just map injectively into via
If were semisimple, then we must have
for some complementary submodule . In particular, there exists a projection such that . Let’s see what it actually does: let . Then for all , we have
since is acted on by trivially. Therefore
This is a contradiction.
Twists and descent
Setup: Let be a field extension.
Suppose is some object “over ” (e.g., -vector space, -algebra, variety/scheme over ). In all of these cases, we can extend scalars to , and therefore get an object over .
Q: If is some other object over with , is ?
Example: vector spaces
If and are vector spaces over , then
is a vector space over . Now if , then . But now,
Therefore (by writing down a basis).
Example: quaternions
Consider the setting of -algebras. Let and . Let , the Hamiltonian quaternions, and . Here, , because for example every non-zero has an inverse, but does not. Now,
To see the isomorphism , consider the two actions . Since is a -dimensional -vector space (by the Cayley-Dickson construction), out of the action by we can get a -linear map . Specifically,
Example: varieties
Let and . Consider the varieties
However,
under the invertible substitution , since
Galois things
Assume is is (finite) Galois and take
So . Let and suppose that we have an isomorphism
What gives us is a map .
For example, for varieties, we’re acting on the set of solutions in to equations with coefficient in , while for vector spaces, we’re acting on the coefficients by : .
Q: What if is non-commutative? When we discussed cocycles, the cocycle had to take values in a module of the group, and the module is an abelian group itself. However, is not necessarily abelian. A: Later/soon.
More Galois things
Suppose we have , , and isomorphisms
i.e.,
We want to know how many different s there can be which become the same over . Out of this data, we can build two -cocycles: one from and one from . We want to compare them. We have
Conjugating by ,
since is defined over . This is a $1$-coboundary condition-coboundary condition: if we pretend we are in an abelian context, the above equation is saying that the difference of the two cocycles is a coboundary:
The two -cocycles are cohomologous.
Non-abelian group cohomology
Suppose is a group acting on a group by automorphisms. Define (c.f. $1$-cocycles-cocycles)