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Representation theory (at least the origin of this terminology) aims to exhibit a model (a represetative) in the group of matrices for an abstract group which is known by only its group law. So complex representations is a satisfying beautiful theory with Schur's theorem, Frobenius reciprocity, Artin, Brauer theorems and various results about irreducible degrees for finite groups.

But my concern is that the field of complex numbers is extraneous to groups. Do we have theorems like $\sum_j d_j^2 = |G|$ when $d_j$ runs though all irreducible degrees of $G$ over an unspecified algebraically closed field. Because the complex case uses character theory with Hermitian structure on the vector space of complex-valued class functions I am not sure if this result is available for a general field.

I have not studied what is known as modular representation theory. Are all non-modular cases no different from the theory over complex numbers? (always working over algebraically closed field). Do the degrees of irreducible representations divide the order of the group whatever the base field is?

Instead of asking for questions individually of this nature let me ask:

Is there an analogue of Lefschetz principle of Algebraic Geometry that is valid for finite group representations? so that to study non-modular representations it suffices to study complex representations.

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    $\begingroup$ Certainly any algebraically closed field of characteristic 0 will reproduce the classical results (in fact, a much smaller extension of $\mathbb{Q}$ suffices). But in prime characteristic $p$ dividing $|G|$ everything tends to change. It's true however that when $p$ doesn't divide $|G|$, you get back the same picture as in characteristic 0. All of this can be found in standard sources. $\endgroup$
    – Jim Humphreys
    Commented Feb 2, 2015 at 15:59
  • $\begingroup$ I am uncomfortable about using the property of complex conjugation. This seems to play a crucial role in the proof of Schur orthogonality relations. $\endgroup$
    – P Vanchinathan
    Commented Feb 2, 2015 at 16:12
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    $\begingroup$ Character theory provides a very elegant tool when studying representations of finite groups but many statements can be proved without using characters. For instance, the identity you mention follows from the fact that the group algebra $\mathbb{K}G$ is semisimple when $\mathbb{K}$ is an algebraically closed field whose characteristic doesn't divide $|G|$. In particular $\mathbb{K}G$ is isomorphic to a direct sum of matrix algebras $M_{d_i}(\mathbb{K})$. Counting dimensions as $\mathbb{K}$-vector spaces gives you the numerical identity. This can be proved without character theory. $\endgroup$
    – Jay Taylor
    Commented Feb 2, 2015 at 16:40
  • $\begingroup$ See for instance (27.21) on pg. 186 of Curtis and Reiner's fantastic book "Representation theory of finite groups and associative algebras". $\endgroup$
    – Jay Taylor
    Commented Feb 2, 2015 at 16:41
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    $\begingroup$ There's no need to mention complex conjugation in the theory. The conjugate of a character $\chi(g)$, over any field, is $\chi(g^{-1})$. $\endgroup$
    – Qiaochu Yuan
    Commented Feb 3, 2015 at 9:40

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I was encouraged to make my comment an answer, so will do so.

If $G$ is a finite group and $\mathbb{K}$ is a field then many interesting results that can be proved using character theory can also be proved by analysing the structure of the group algebra $\mathbb{K}G$. For instance by (27.20) of Curtis and Reiner's "Representation theory of finite groups and associative algebras" we have

$$\mathbb{K}G \cong \bigoplus_i M_{d_i}(\mathbb{K})$$

is a direct sum of matrix algebras whenever $\mathbb{K}$ is an algebraically closed field whose characteristic does not divide the order of the group. Counting dimensions this gives the desired numerical identity

$$|G| = \sum_i d_i^2.$$

Note however that dropping either of these conditions can be problematic. It is well known that if the characteristic of $\mathbb{K}$ divides the order of the group then $\mathbb{K}G$ is no longer semisimple. However, if the field is too small then the irreducible representations of $G$ may not be realisable over that field. For instance, take $\mathbb{K} = \mathbb{Q}$ and $G$ to be a cyclic group.

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    $\begingroup$ To supplement this: note by the way that when $p$ is odd, Eisenstein's criterion tells us that the cyclic group of order $p$ has an irreducible representation of degree $p-1$ over $\mathbb{Q}$, so the degree of an irreducible characteristic $0$ representation need not divide the group order in general. Also, the degrees of irreducible characteristic $p$ representations (even over algebraically closed fields) need not divide the group order. $\endgroup$
    – Geoff Robinson
    Commented Feb 2, 2015 at 17:11
  • $\begingroup$ That's a nice example for $C_p$, I'd never thought about that before. $\endgroup$
    – Jay Taylor
    Commented Feb 2, 2015 at 17:16
  • $\begingroup$ @Geoff Robinson: Thanks for your valuable comment connecting Eisenstein criterion. Learnt something new. I am curious about irreducible degrees not dividing the order of the group. Surprised that such a fact is not mentioned in textbooks. Can you direct me to a reference for an example of that? $\endgroup$
    – P Vanchinathan
    Commented Feb 3, 2015 at 2:38
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    $\begingroup$ I have given a characteristic 0 example. In characteristic 3, the simple group PSU(3,3) has an absolutely irreducible representation of degree 15, but has order 6048. There are many more examples in the "Atlas of Brauer Characters". $\endgroup$
    – Geoff Robinson
    Commented Feb 3, 2015 at 4:25

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