## Complex Conjugate Representation (Characterization of Complex Conjugates)

**Point of post: **This post is a continuation of this one.

We’d now like to find necessary and sufficient conditions for an irrep to be a self-conjugate and when this is true whether it is real or quaternionic. But, first we need a theorem which can be used to show that all complex conjugates, given the right basis, have a particularly nice form and which will be integral in this aforementioned characterization of quaternionic and real self-conjugate irreps.

**Theorem: ***Let be a complex conjugate on a pre-Hilbert space of dimension . Then, admits an orthonormal basis such that .*

**Proof: **Consider the vector space which is but considered as a vector space over instead of a vector space over . It’s an elementary fact that . We can define an inner product on , denoted by

where is the given inner product on . This is indeed a symmetric inner product on . To see that it’s linear in the first column note that (recall that this is a -space) since

It’s symmetric since

Lastly, to see that it’s positive definite we merely note that since for any inner product and the conclusion follows since is positive definite.

Now, with respect to this new pre-Hilbert space we claim that the map is linear, involutory, and self-adjoint. This first part of this statement is clear since

(once again recalling that we are now dealing with an -space). The second is also clear by our initial assumptions. The last statement follows since for any we have that

With this, define and . We claim that . Indeed, to see that we note that for any one has that

Indeed, expansion merely gives

from where the rest is trivial. To see that is trivial, assume that then for some . But,

from where it follows that (note we made use of the fact that ).

Next note that if then since for some and thus

similarly one shows that if then . With this we claim that the map by is an isomorphism. Indeed, is clearly linear. To see that is bijective we note that and thus is invertible. It follows then that is an isomorphism and thus by a counting argument . Let be an orthonormal basis for . But, recalling the simple identity one sees that

So, note then that if then by assumption of the way the basis was chosen . That said, this implies that so that for some real number . It follows then that and thus from where it follows that and thus . That said, note that since we have that and thus

by assumption. It follows that is orthonormal in and since it follows that is an orthonormal basis for .

That said, since by prior discussion we know that from where the theorem follows.

From this we get the following interesting corollary:

**Corollary: ***Let be a pre-Hilbert space and a complex conjugate on . Then, there exists some orthonormal basis such that*

**References:**

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

2. Serre, Jean Pierre. *Linear Representations of Finite Groups*. New York: Springer-Verlag, 1977. Print

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

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