# Abstract Nonsense

## Class Functions

Point of post: In this post we derive results about the set of class functions on a finite group $G$, in particular finding its dimension as a subspace of the group algebra and characterizing it as the center of the group algebra.

Motivation

In our last series of posts we saw an interesting technique. We saw the interesting idea that if we want to prove the cardinality of a set $X$ is equal to $\kappa$ it suffices to construct a vector space $\mathscr{V}$ of dimension $\kappa$ such that $X$ is a basis for $\mathscr{V}$. In particular, we saw that $\displaystyle \sum_{\alpha\in\widehat{G}}d_\alpha^2=|G|$ by considering the group algebra $\mathcal{A}(G)$ of dimension $|G|$ and then showing that $\left\{D^{(\alpha)}_{i,j}:\alpha\in\widehat{G}\text{ and }i,j\in[d_\alpha]\right\}$ is a basis for $\mathcal{A}(G)$ (where, as in the last post, the $D^{(\alpha)}_{i,j}$ are the matrix entry functions). We now wish to get our milage out of this technique by applying it again in a different context (different set and different cardinality). We don’t want to ruin the surprise of what precisely this will be, but we shall construct the vector space with the ‘proper dimension’ in this post. In particular, we will consider and study the set of class functions on a finite group $G$. Intuitively, these are functions which satisfy (of course we mean this loosely since we haven’t define the domain, range, etc.) the common ‘trace identity’ $\text{tr}\left(ABA^{-1}\right)=\text{tr}(B)$.

Class Functions

Let $G$ be a finite group. Then, if $f$ is a member of the group algebra $\mathcal{A}(G)$ which satisfies $f\left(xyx^{-1}\right)=f(x)$ for every $x,y\in G$ then we call $f$ a class function. We denote the set of all class functions by $\text{Cl}(G)$. We now derive some fairly elementary characterizations of the class functions.

Theorem: Let $G$ be a finite group. Then for some $f\in\mathcal{A}(G)$ the following are equivalent:

$\text{ }$

\begin{aligned}&(1)\quad f\textit{ is a class function}\\ &(2)\quad f(xy)=f(yx)\textit{ for every }x,y\in G\\ &(3)\quad f\textit{ is constant on the conjugacy classes of}\;G\\ &(4)\quad f\in\mathcal{Z}\left(\mathcal{A}(G)\right)\textit{ where }\mathcal{Z}\textit{ stands for the center of the algebra}\end{aligned}

$\text{ }$

(for those who need a reminder conjugacy classes and centers of algebras are defined here and here respectively)

Proof: Although inefficient we prove this by $(i)\Leftrightarrow (j)\;\; j=2,3$ and $(2)\Leftrightarrow (4)$ for maximum clarity.

$(1)\Leftrightarrow (2)$: Suppose first that $f$ is a class function. Then, one has that for any $f(xy)=f(xyxx^{-1})=f(yx)$. Conversely, if $f(yx)=f(xy)$ for every $x,y\in G$ then for any $x,y\in G$ one has that $f\left(xyx^{-1}\right)=f\left(yx^{-1}x\right)=f(y)$ and so $f\in\text{Cl}(G)$.

$\text{ }$

$\text{ }$

$(1)\Leftrightarrow (3)$: Suppose first $f$ is constant on conjugacy classes then clearly $f\in\text{Cl}(G)$ since for any $x,y\in G$ one has that $xyx^{-1}$ is, by definition, conjugate to $y$ and thus in the same conjugacy classes. Thus, by assumption $f\left(xyx^{-1}\right)=f(y)$ since $x,y\in G$ were arbitrary the conclusion follows. Conversely, suppose that $f\in\text{Cl}(G)$ and $x,y$ be in the same conjugacy class. Then, by definition there exists some $z\in G$ such that $zxz^{-1}=y$ and so $f(x)=f\left(zxz^{-1}\right)=f(y)$. Since $x,y$ in the conjugacy class of $x$ were arbitrary it follows that $f$ is constant on the conjugacy class of $x$. Since $x\in G$ was arbitrary it follows that $f$ is constant on all conjugacy classes of $G$.

$\text{ }$

$\text{ }$

$(2)\Leftrightarrow(4)$ This is by far the trickiest one. Suppose first that $f\in\mathcal{Z}\left(\mathcal{A}(G)\right)$ then we note that for any $x,y\in G$ one has that $\left(\delta_{y^{-1}}\ast f\right)(x)=\left(f\ast\delta_{y^{-1}}\right)(x)$. That said, $\left(\delta_{y^{-1}}\ast f\right)(x)=f\left(yx\right)$ and $\left(f\ast \delta_{y^{-1}}\right)=f(xy)$ from where the conclusion follows by the arbitrariness of $x,y$. Conversely, suppose that $f(xy)=f(yx)$ for every $x,y\in G$ then for any $a\in\mathcal{A}(G)$ and any $x\in G$ one has that

$\text{ }$

\displaystyle \begin{aligned}\left(a\ast f\right)(x) &= \sum_{g\in G}a\left(x g^{-1}\right)f\left(g\right)\\ &= \sum_{g\in G}a\left(xg^{-1}\right)f\left(gx^{-1}x\right)\\ &= \sum_{g\in G}a\left(xg^{-1}\right) f\left(x \left(x g^{-1}\right)^{-1}x\right)\\ &= \sum_{g\in G}a(g) f\left(g^{-1}x\right)\\ &= \sum_{g\in G}a(g) f\left(xg^{-1}\right)\\ &= \left(f\ast a\right)(x)\end{aligned}

$\text{ }$

and since $x\in G$ was arbitrary it follows that $a\ast f=f\ast a$ and since $a\in \mathcal{A}(G)$ was arbitrary it follows that $f\in\mathcal{Z}\left(\mathcal{A}\left(G\right)\right)$. $\blacksquare$

$\text{ }$

Corollary: $\text{Cl}(G)$ is a subalgebra of $\mathcal{A}(G)$

Proof: This follows from a previous theorem regarding the center of an algebra. $\blacksquare$

$\text{ }$

Remark: Note that this gives an alternate proof to our previous one that the group algebra is a commutative algebra if and only if $G$ is abelian. Indeed, $\mathcal{A}(G)$ is a commutative algebra if and only if $\mathcal{A}(G)=\mathcal{Z}\left(\mathcal{A}(G)\right)$ which (with a little justification is true) if and only if $\delta_g$ is constant on each conjugacy class for each $g\in G$ which is true  if and only if eachc conjugacy class has one element which is true if and only if $G$ is abelian.

$\text{ }$

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

February 24, 2011 -

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6. […] though from basic properties of -representations we have that for any (i.e. a class function) one has that for any .  Thus, by Schur’s lemma we may conclude that for some . Since this […]

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8. […] . We define the character table of to be the matrix with where is the single value , being a class function, takes on . So, in general the character table of a group should look […]

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9. […] Theorem: Let be defined as above. Then, is a class function. […]

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10. […] of containing from where the claim follows. Clearly then from this and the fact that each   is a class function we see […]

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11. […] we recall that since a character is a class function by and we mean the unique value each of those characters takes on the conjugacy […]

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12. […] one noticed that by extending every (irreducible) character on to a character (thus a class function) we’ve automatically defined a map by extending the map by linearity (recalling that the […]

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13. […] map for every where is the induced character. We saw dually there was a map which just took a class function on and restricted it to . In this post we prove the amazing fact that the maps and are adjoint, […]

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14. […] thus by a previous theorem. With this in mind we can state the following […]

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