Finite Averaging and Character-Theory Prerequisites: Examples

For $v=(x,y,z)$, $v^{\mathrm{\prime }}=(x^{\mathrm{\prime }},y^{\mathrm{\prime }},z^{\mathrm{\prime }})$ and $a,b\in \mathbb{C}$, the formula gives $P(av+b{v}^{\mathrm{\prime }})=(ax+b{x}^{\mathrm{\prime }},a(x+z)+b(x^{\mathrm{\prime }}+z^{\mathrm{\prime }}),az+b{z}^{\mathrm{\prime }})=aP(v)+bP(v^{\mathrm{\prime }})$. Moreover $P^2(x,y,z)=P(x,x+z,z)=(x,x+z,z)=P(x,y,z)$.

For every $(x,y,z)$, $P(x,y,z)=x(1,1,0)+z(0,1,1)\in U$. Conversely any $u=s(1,1,0)+t(0,1,1)=(s,s+t,t)$ satisfies $P(u)=(s,s+t,t)=u$, so $\mathrm{im}P=U$ and $P{\mathrm{\mid }}_U={\mathrm{id}}_U$. This explicitly realizes the projection supplied by F1.

The equation $P(x,y,z)=0$ is equivalent to $x=0$ and $z=0$, with $y$ arbitrary. Hence $\ker P=\{(0,y,0):y\in \mathbb{C}\}=\mathrm{span}\{(0,1,0)\}$. Every vector decomposes as $(x,y,z)=(x,x+z,z)+(0,y-x-z,0)$ with the first summand in $U$ and the second in $\ker P$. If $(s,s+t,t)=(0,y,0)$, then $s=t=y=0$, so the intersection is zero and the decomposition is unique. Thus the stated kernel is exactly the direction along which $P$ projects.

Conjugation sends a transposition $(ab)$ to $(\tau (a)\tau (b))$: at $\tau (a)$ the composite $\tau (ab){\tau }^{-1}$ sends $\tau (a)$ to $\tau (b)$, at $\tau (b)$ it sends it to $\tau (a)$, and it fixes every other point. In particular $(23)(12)(23)=(13)$ and $(13)(12)(13)=(23)$. Thus all three transpositions, and no other permutations, constitute one conjugacy class.

Similarly, $\tau (123){\tau }^{-1}$ sends $\tau (1)$ to $\tau (2)$, then to $\tau (3)$, then to $\tau (1)$, so it is a three-cycle. Directly $(12)(123)(12)=(132)$, so the two three-cycles form one class. Every conjugate of $e$ is $e$. A permutation of three points is either the identity, a transposition, or a three-cycle: fixing two points forces the third to be fixed; fixing exactly one exchanges the other two; fixing none forces a three-cycle by following any point’s successive images. Hence the three listed classes exhaust $S_3$.

The group is finite and these are exactly its three classes by steps 1.1–1.2, so F1 gives the asserted indicator basis over $k$. More explicitly, for every class function $f$ one has $f=f(e){\mathbf{1}}_{C_1}+f((12)){\mathbf{1}}_{C_2}+f((123)){\mathbf{1}}_{C_3}$, because evaluation on each of the three classes picks out its constant value. For example the coefficients $(0,1,-1)$ give the function with values $0$ at $e$, $1_k$ on the transpositions, and $-1_k$ on the three-cycles. This calculation remains valid in characteristic two, where the last two values coincide, while the indicator basis itself stays independent by evaluation on the three disjoint nonempty classes.

The set $X$ is nonempty of cardinality three, so F1 applies with normalization $1\mathrm{/}3$. The point indicators are a basis: $v=v(1)e_1+v(2)e_2+v(3)e_3$ for every function $v$, and evaluating a zero combination at each point makes all coefficients zero.

For $j=k$, only the value at point $j$ contributes to $⟨e_j,e_k⟩$, giving $1\mathrm{/}3$. For $j\mathrm{\ne }k$, the two indicators never both have nonzero value at the same point, so the pairing is zero. Thus $⟨e_j,e_k⟩={\delta }_{jk}\mathrm{/}3$, which is the matrix $I_3\mathrm{/}3$.

Since $\bar{i}=-i$, $⟨f,h⟩=(1(-i)+i(1)+0(1))\mathrm{/}3=0$. Also $⟨f,f⟩=(1\cdot 1+i(-i)+0)\mathrm{/}3=(1+1)\mathrm{/}3=2\mathrm{/}3$. In particular this nonzero vector has strictly positive diagonal value.

Put $u_j=\sqrt{3}{e}_j$, where $\sqrt{3}$ is the positive real square root. Direct substitution gives $⟨u_j,u_k⟩=3⟨e_j,e_k⟩={\delta }_{jk}$. Scaling each basis vector by the nonzero scalar $\sqrt{3}$ preserves spanning and independence (divide the coefficients by $\sqrt{3}$), so $(u_1,u_2,u_3)$ is an orthonormal basis.

Take $n=2$, $z_1=1$ and $z_2=-1$ in $\mathbb{C}$. Their moduli are $\sqrt{1^2+0^2}=1$ and $\sqrt{(-1{)}^2+0^2}=1$, so both satisfy the unit-modulus hypothesis and $n>0$.

Their sum is $1+(-1)=0$, whose modulus is zero. Consequently $\mathrm{\mid }{z}_1+z_2\mathrm{\mid }=0<2=n$, so this instance fails the asserted equality and refutes its universal quantifier. The summands are distinct, in agreement with F1.