Talk:K-group, zeroth (Ex)

1 Solution of 1

Let us denote the monoid of isomorphism classes of finitely generated projective $R$$\newcommand{\Z}{\mathbb{Z}} \newcommand{\R}{\mathbb{R}} \newcommand{\C}{\mathbb{C}} \newcommand{\N}{\mathbb{N}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\F}{\mathbb{F}} \newcommand{\bZ}{\mathbb{Z}} \newcommand{\bR}{\mathbb{R}} \newcommand{\bC}{\mathbb{C}} \newcommand{\bH}{\mathbb{H}} \newcommand{\bQ}{\mathbb{Q}} \newcommand{\bF}{\mathbb{F}} \newcommand{\bN}{\mathbb{N}} \DeclareMathOperator\id{id} % identity map \DeclareMathOperator\Sq{Sq} % Steenrod squares \DeclareMathOperator\Homeo{Homeo} % group of homeomorphisms of a topoloical space \DeclareMathOperator\Diff{Diff} % group of diffeomorphisms of a smooth manifold \DeclareMathOperator\SDiff{SDiff} % diffeomorphism under some constraint \DeclareMathOperator\Hom{Hom} % homomrphism group \DeclareMathOperator\End{End} % endomorphism group \DeclareMathOperator\Aut{Aut} % automorphism group \DeclareMathOperator\Inn{Inn} % inner automorphisms \DeclareMathOperator\Out{Out} % outer automorphism group \DeclareMathOperator\vol{vol} % volume \newcommand{\GL}{\text{GL}} % general linear group \newcommand{\SL}{\text{SL}} % special linear group \newcommand{\SO}{\text{SO}} % special orthogonal group \newcommand{\O}{\text{O}} % orthogonal group \newcommand{\SU}{\text{SU}} % special unitary group \newcommand{\Spin}{\text{Spin}} % Spin group \newcommand{\RP}{\Rr\mathrm P} % real projective space \newcommand{\CP}{\Cc\mathrm P} % complex projective space \newcommand{\HP}{\Hh\mathrm P} % quaternionic projective space \newcommand{\Top}{\mathrm{Top}} % topological category \newcommand{\PL}{\mathrm{PL}} % piecewise linear category \newcommand{\Cat}{\mathrm{Cat}} % any category \newcommand{\KS}{\text{KS}} % Kirby-Siebenmann class \newcommand{\Hud}{\text{Hud}} % Hudson torus \newcommand{\Ker}{\text{Ker}} % Kernel \newcommand{\underbar}{\underline} %Classifying Spaces for Families of Subgroups \newcommand{\textup}{\text} \newcommand{\sp}{^}R$-modules by $P( R)$$P( R)$. We will show that the pair $\left(K_0( R), \iota: P( R) \to K_0( R)\right)$$\left(K_0( R), \iota: P( R) \to K_0( R)\right)$ has the appropriate universal property. So let $\varphi: P( R) \to A$$\varphi: P( R) \to A$ be a morphism of abelian monoids and $A$$A$ some abelian group. We need to show that there is a unique map $\psi$$\psi$ making the diagram commutative:

$\displaystyle \xymatrix{P( R) \ar[r]^-{\varphi} \ar[d]_\iota & A \\ K_0( R) \ar@/_1.2pc/[ur]_-{\psi} }$

Since $K_0( R)$$K_0( R)$ is a group defined by generators and relations and $\iota$$\iota$ is surjective on generators, we can only define the map $\psi$$\psi$ by the property

$\displaystyle \psi( P) = \varphi(\lbrack P\rbrack)$ $$\displaystyle \psi( P) = \varphi(\lbrack P\rbrack)$$

and need to show that it is well-defined. But this follows easily since short exact sequences of projective modules always split. i.e., if we have a short exact sequence

$\displaystyle \xymatrix{0 \ar[r] & Q \ar[r] & P \ar[r] & Q^\prime \ar[r] & 0}$

then it follows that $P \cong Q\oplus Q^\prime$$P \cong Q\oplus Q^\prime$.

2 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$$S_3 \to \C$, i.e. the functions that only depend on the conjugacy classes. Notice that there are three conjugacy classes in $S_3$$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

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

3 Solution of 3

That this module is finitely generated and projective follows from the fact that any bundle $E$$E$ over a compact space $X$$X$ (our case is $S^2$$S^2$) may be embedded into a trivial bundle. This tells us that there is a bundle $F$$F$ (the orthogonal complement for example) over the same space and some natural number $N$$N$ such that

$\displaystyle E \oplus F \cong \underline{\R}^N$ $$\displaystyle E \oplus F \cong \underline{\R}^N$$

where $\underline{\R}^N$$\underline{\R}^N$ denotes the trivial bundle of rank $N$$N$. Applying the functor of taking sections of these bundles tells us that as modules over $C(X)$$C(X)$ we have an isomorphism

$\displaystyle \Gamma(E) \oplus \Gamma(F) \cong C(X)^N$ $$\displaystyle \Gamma(E) \oplus \Gamma(F) \cong C(X)^N$$

hence the claim follows.

In order to show that the module is not finitely generated free we note that if it were a free module this would imply the bundle itself to be trivial. This cannot be the case, because a trivial bundle has vanishing euler class, but the euler class of $TS^2$$TS^2$ cannot be zero because

$\displaystyle 2 = \chi(S^2) = \langle e(TS^2),\lbrack S^2 \rbrack \rangle.$ $$\displaystyle 2 = \chi(S^2) = \langle e(TS^2),\lbrack S^2 \rbrack \rangle.$$