Getting new irreducible representations from old ones

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SUMMARY

This discussion focuses on the classification of irreducible representations of a group G across different fields K, specifically real (R), complex (C), and quaternionic (H) representations. It highlights that while one can construct complex representations from real representations using the tensor product, the resulting representations are not guaranteed to be irreducible. The example of the cyclic group of order 3 illustrates that the only irreducible real representation is trivial, whereas it possesses three irreducible complex representations, demonstrating the complexity of representation theory.

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hideelo
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Suppose I had some group G, and I could classify all of its irreducible K-representations for some K = R,C, or H. Given that information (how) can I classify its irreducible K-representations for all K.

i.e. suppose I knew all the irreducible real representations of G, (how) could I then get all the irreducible complex and quaternionic representations?
 
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A standard method in case of linear representations on a real vectorspace ##V_\mathbb{R}## is to construct ##V_\mathbb{C} = V_\mathbb{R} \otimes_\mathbb{R} \mathbb{C}## and define for a given ##\varphi_\mathbb{R} \, : \,G \longrightarrow GL(V_\mathbb{R})##
$$
\varphi_\mathbb{C} \, : \,G \longrightarrow GL(V_\mathbb{C}) \quad \text{ by } \quad \varphi_\mathbb{C}(g)(\lambda\cdot v) = \varphi_\mathbb{R}(g)(v) \otimes \lambda
$$
I'm not sure, however, whether they automatically will be irreducible again, as there are simply more eigenvalues available, so I doubt it.
 
hideelo said:
suppose I knew all the irreducible real representations of G, (how) could I then get all the irreducible complex and quaternionic representations?

Unfortunately, you can't. Consider the cyclic group of order 3. Its only irreducible real representation is the trivial one, but it has three irreducible complex representations.
fresh_42 said:
I'm not sure, however, whether they automatically will be irreducible again, as there are simply more eigenvalues available, so I doubt it.
Indeed, let the cyclic group of order 3 act on \mathbb{R}^2 with a generator corresponding to rotation by 2\pi/3. This representation is irreducible but its complexification isn't.
 

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