## A Bijection Between A Subset of the Complex Reps of a Finite Group and the Real Reps (Pt. I)

**Point of post: **In this post we describe a certain subset of the set of all -representations of a finite group and show that this subset is in bijective correspondence with the set of all -representations of . Moreover, we shall show that this bijection naturally restricts to a subset of the set of all -reps of to the the -irreps of .

*Motivation*

In our last post we discussed the notion of representations for a finite group . Naturally our first desire would be to see if we could, in some way, connect representations of to the -representations which has held the center of our attention for so long. We begin this process in this post by showing that there is a natural place for which these -representations occur. Namely, we shall see that every -representation for which there is a complex conjugate for which for every naturally admits a -representation . We shall show then that in fact the reverse is true–namely that for every -representation there is a natural way to produce a -representation such that . Moreover, we’ll show that if is a -representation satisfying the aforementioned conditions and the added condition that there does not exist such that and is -invariant then we shall see that is an irreducible -representation.

**A Characterization of – Representations**

Our first goal is to show that a certain class of -representations naturally admits a representation. Indeed, say that a -representation satisfies the *real condition *if there exists a complex conjugate such that for every . We call such a the ‘realizer’ of . With this in mind:

**Theorem: ***Let be a finite group and be a -representation satisfying the real condition with realizer . Then, the space is a real vector space with and the same inner product as . Moreover, the map given by is a -representation of . Further more, is irreducible if and only if there does not exist a -invariant for which .*

** Proof: **We first show that is a real inner product space of real dimension . The fact that is a real vector space with the same addition as and scalar multiplication restricted to is clear, thus it remains to prove then that with this definition of operations . To see this recall from our past characterizations of complex conjugates that admits a basis such that for . We claim then that is similarly a basis for . Indeed, the fact that is linearly independent is evident and thus it suffices to prove that . To see this let be arbitrary. Then, since is a basis for there exists such that . Note though that by assumption

and thus and thus it clearly follows then that . Since was arbitrary it follows that is a real basis for and thus as desired. Thus, it remains to show that the restriction of the inner product to is really a real inner product. But, all the axioms are clear except for possibly that maps into (there’s not ostensible reason this must be true). To see this merely note that for any one has that

from where the result follows.

Now, we prove that as defined in the theorem statement is really a -representation. The first thing to show is that for every one has that really is an element of . The first, but crucial step in this procedure is to show that, while silly, since this isn’t absolutely clear. But, this is where the real condition comes into play. Namely, since we may reinterpret the real condition as saying for each . Thus, in particular for any and any one has that so that from where the fact that follows. Now, the fact that is a -homomorphism and unitary clearly follow from definition. Thus, to prove that is a -representation it merely suffices to prove that it is a homomorphism, but this is clear since is a representation. Thus, it follows that is a homomorphism from where the fact that it is a -representation of clearly follows.

Finally we prove that is irreducible if and only if doesn’t admit any non-trivial proper subspaces for which . Indeed, suppose that is reducible so that there exists such that is invariant. Define then . Clearly by definition one has that . Moreover, we claim that . Indeed, let then where and so

Conversely, if where then and evidently

and so from where the fact that follows. We now claim that is -invariant. Indeed, let and be arbitrary (where . Then, one has that

but by assumption one has that so that we may conclude that . Since and was arbitrary we may conclude that is -invariant as claimed. Lastly, noting that since we clearly have that which (recalling that contains a complex basis for ) implies that from where it follows that is a -invariant, non-trivial proper subspace of for which . It follows then that if does not admit such a subspace then must be irreducible.

Conversely, suppose that is -invariant and . Define then . Evidently, it’s true then that . We claim that is invariant. Indeed, let and and then and moreover so that in fact . Since and were arbitrary we have that is -invariant and thus is reducible. Thus, if is irreducible then cannot admit a non-trivial proper subspace which is invariant under both and .

Putting this all together the conclusion follows.

**References:**

1. Simon, Barry. *Representations of Finite and Compact Groups*. Providence, RI: American Mathematical Society, 1996. Print.

[…] Point of post: This post is a continuation of this one. […]

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