Talk:K-group, zeroth (Ex)

Solution of 2

We will use the facts that every representation is uniquely determined by its character and that every representation decomposes as the direct sum of irreducible representations (this holds at least for representations of finite groups over algebraically closed fields). We begin by characterizing the class functions $S_3 \to \C$$<wikitex>; == Solution of 2 == We will use the facts that every representation is uniquely determined by its character and that every representation decomposes as the direct sum of irreducible representations (this holds at least for representations of finite groups over algebraically closed fields). We begin by characterizing the class functions $S_3 \to \C$, i.e. the functions that only depend on the conjugacy classes. Notice that there are three conjugacy classes in $S_3$, one for each cycle type. Indeed, since each element of $S_3$ may be represented by a cycle of maximal length $ and the cycle length does not change under conjugation, we see that there are at least $ conjugacy classes. It is easy to check that all transpositions and the two $-cycles are conjugated to each other. Thus, a class function is uniquely determined by its values on, say, $\mathrm{id}$, $(1 2)$, $(1 2 3)$ and, conversely, an element of $\C^3$ uniquely determines a class function. Consequently, the space of class functions is $-dimensional, which is the same as saying that there are, up to isomorphism, three irreducible representations on $S_3$ since the irreducible characters span the space of class functions. The first two are found quickly: Of course, the trivial representation $S_3 \to \C^\times, \, \sigma \mapsto 1$ and the sign-representation $S_3 \to \C^\times, \, \sigma \mapsto \mathrm{sign}(\sigma)$ are irreducible (they are both one-dimensional!) and they cannot be isomorphic because their characters differ. The natural representation of $S_3$ as permutation matrices on $\C^3$ is not irreducible, as the hyperplane $E := \left\{ (z_1, z_2, z_3) \mid z_1 + z_2 + z_3 = 0 \right\}$ is invariant under permutation of coordinates. But the induced representation on $E$ is irreducible: A one-dimensional subspace $L \subset E$ cannot be invariant under permutation of coordinates because if it was, it would be spanned by, say, $(z_1, z_2, -z_1 - z_2)$ and then it would follow that $z_1 = z_2$ (because $L$ is invariant under $(1 2)$) which implies $z_1 = -2z_1$ (because $L$ is invariant under $(1 3)$) so that $L$ would be spanned by zero, a contradiction (I would like to see a quicker argument here but could not come up with one.) Now we represent the three characters by elements of $\C^3$ (i.e. their values on $\mathrm{id}$, $(1 2)$, $(1 2 3)$): The trivial representation corresponds to $\mathbf{1} = (1, 1, 1)$, the sign representation corresponds to $\mathbf{sign} = (1, -1, 1)$ and the permutation representation corresponds to $\mathbf{perm} = (2, 0, -1)$. Recalling that the multiplication in $R_\C(S_3)$ is given by the tensor product which corresponds to pointwise multiplication in the space of class funcions, i.e to componentwise multiplication on $\C^3$, we get the following multiplication table: {| class="wikitable" |- ! $\cdot$ !! $\mathbf{1}$ !! $\mathbf{sign}$ !! $\mathbf{perm}$ |- ! scope="row" | $\mathbf{1}$ | $\mathbf{1}$ | $\mathbf{sign}$ | $\mathbf{perm}$ |- ! scope="row" | $\mathbf{sign}$ |$\mathbf{sign}$ | $\mathbf{1}$ | $\mathbf{perm}$ |- ! scope="row" | $\mathbf{perm}$ | $\mathbf{perm}$ | $\mathbf{perm}$ | $\mathbf{1} +\mathbf{sign} + \mathbf{perm}$ |} This shows that sending $[\mathbf{sign}] \mapsto X$ and $[\mathbf{perm}] \mapsto Y$ induces a well-defined isomorphism $$ K_0(\C[S_3]) \simeq R_\C(S_3) \to \Z[X,Y] / (X^2 - 1, XY - Y, Y^2 - X - Y - 1). $$ </wikitex>S_3 \to \C$, i.e. the functions that only depend on the conjugacy classes. Notice that there are three conjugacy classes in $S_3$$S_3$, one for each cycle type. Indeed, since each element of $S_3$$S_3$ may be represented by a cycle of maximal length $3$$3$ and the cycle length does not change under conjugation, we see that there are at least $3$$3$ conjugacy classes. It is easy to check that all transpositions and the two $3$$3$-cycles are conjugated to each other. Thus, a class function is uniquely determined by its values on, say, $\mathrm{id}$$\mathrm{id}$, $(1 2)$$(1 2)$, $(1 2 3)$$(1 2 3)$ and, conversely, an element of $\C^3$$\C^3$ uniquely determines a class function. Consequently, the space of class functions is $3$$3$-dimensional, which is the same as saying that there are, up to isomorphism, three irreducible representations on $S_3$$S_3$ since the irreducible characters span the space of class functions.

The first two are found quickly: Of course, the trivial representation $S_3 \to \C^\times, \, \sigma \mapsto 1$$S_3 \to \C^\times, \, \sigma \mapsto 1$ and the sign-representation $S_3 \to \C^\times, \, \sigma \mapsto \mathrm{sign}(\sigma)$$S_3 \to \C^\times, \, \sigma \mapsto \mathrm{sign}(\sigma)$ are irreducible (they are both one-dimensional!) and they cannot be isomorphic because their characters differ.

The natural representation of $S_3$$S_3$ as permutation matrices on $\C^3$$\C^3$ is not irreducible, as the hyperplane $E := \left\{ (z_1, z_2, z_3) \mid z_1 + z_2 + z_3 = 0 \right\}$$E := \left\{ (z_1, z_2, z_3) \mid z_1 + z_2 + z_3 = 0 \right\}$ is invariant under permutation of coordinates. But the induced representation on $E$$E$ is irreducible: A one-dimensional subspace $L \subset E$$L \subset E$ cannot be invariant under permutation of coordinates because if it was, it would be spanned by, say, $(z_1, z_2, -z_1 - z_2)$$(z_1, z_2, -z_1 - z_2)$ and then it would follow that $z_1 = z_2$$z_1 = z_2$ (because $L$$L$ is invariant under $(1 2)$$(1 2)$) which implies $z_1 = -2z_1$$z_1 = -2z_1$ (because $L$$L$ is invariant under $(1 3)$$(1 3)$) so that $L$$L$ would be spanned by zero, a contradiction (I would like to see a quicker argument here but could not come up with one.)

Now we represent the three characters by elements of $\C^3$$\C^3$ (i.e. their values on $\mathrm{id}$$\mathrm{id}$, $(1 2)$$(1 2)$, $(1 2 3)$$(1 2 3)$): The trivial representation corresponds to $\mathbf{1} = (1, 1, 1)$$\mathbf{1} = (1, 1, 1)$, the sign representation corresponds to $\mathbf{sign} = (1, -1, 1)$$\mathbf{sign} = (1, -1, 1)$ and the permutation representation corresponds to $\mathbf{perm} = (2, 0, -1)$$\mathbf{perm} = (2, 0, -1)$. Recalling that the multiplication in $R_\C(S_3)$$R_\C(S_3)$ is given by the tensor product which corresponds to pointwise multiplication in the space of class funcions, i.e to componentwise multiplication on $\C^3$$\C^3$, we get the following multiplication table:

$\cdot$$\cdot$	$\mathbf{1}$$\mathbf{1}$	$\mathbf{sign}$$\mathbf{sign}$	$\mathbf{perm}$$\mathbf{perm}$
$\mathbf{1}$$\mathbf{1}$	$\mathbf{1}$$\mathbf{1}$	$\mathbf{sign}$$\mathbf{sign}$	$\mathbf{perm}$$\mathbf{perm}$
$\mathbf{sign}$$\mathbf{sign}$	$\mathbf{sign}$$\mathbf{sign}$	$\mathbf{1}$$\mathbf{1}$	$\mathbf{perm}$$\mathbf{perm}$
$\mathbf{perm}$$\mathbf{perm}$	$\mathbf{perm}$$\mathbf{perm}$	$\mathbf{perm}$$\mathbf{perm}$	$\mathbf{1} +\mathbf{sign} + \mathbf{perm}$$\mathbf{1} +\mathbf{sign} + \mathbf{perm}$

This shows that sending $[\mathbf{sign}] \mapsto X$$[\mathbf{sign}] \mapsto X$ and $[\mathbf{perm}] \mapsto Y$$[\mathbf{perm}] \mapsto Y$ induces a well-defined isomorphism

$\displaystyle K_0(\C[S_3]) \simeq R_\C(S_3) \to \Z[X,Y] / (X^2 - 1, XY - Y, Y^2 - X - Y - 1).$ $$\displaystyle K_0(\C[S_3]) \simeq R_\C(S_3) \to \Z[X,Y] / (X^2 - 1, XY - Y, Y^2 - X - Y - 1).$$