6
$\begingroup$

I am trying to do a problem from artin's algebra 2nd ed (Chapter 10, Exercise 2.3) but having trouble:

Let $(\rho , V)$ be a representation of the symmetric group $S_3$. Let $x=(123), y=(12)$ be the usual generators of $S_3$.

(a) Let $u\neq 0$ be a vector in $V$. Let $v=u+xu+x^2u$ and $w=u+yu$. By analyzing the $S_3$-orbits of $v,w$, show that $V$ contains a nonzero invariant subspace of dimension $\leq 2$.

(b) Prove that all 2-dimensional irreducible representations of $S_3$ are isomorphic, and determine them all.

After some calculations, i found that $xv=v$ and $yw=w$. But I dont know what to do next. Also, we are not allowed to use character theory, thanks in advance.

$\endgroup$
  • $\begingroup$ (a) If $v \neq 0$, then show that the span of $v$ and $yv$ is a nonzero invariant subspace. If $v = 0$, then show that the span of $w$, $xw$ and $x^2w$ is a nonzero invariant subspace and already spanned by $w$ and $xw$. $\endgroup$ – darij grinberg Nov 18 '14 at 5:03
4
$\begingroup$

We have $$x^3 = y^2 = (xy)^2 = 1,$$ so $$yx = (yx)^{-1} = x^2y,\text{ }yx^2 = x^2yx = x^4y = xy.$$

a.

Fix a nonzero $u\in V$, and let $$v = u+xu+x^2u = (1+x+x^2)u,\text{ }w=u+yu = (1+y)u.$$ Since $\{1,x,x^2\},\{1,y\}$ are subgroups of $G = S_3$, we have $xv = x^2v = v$ and $yw = w$.

We have the operator identities $$y(1+x+x^2) = y+yx+yx^2 = y+x^2y+xy = (1+x+x^2)y,$$ $$x(1+y) = x+xy = x+yx^2,$$$$x^2(1+y) = x^2+x^2y = x^2+yx.$$ It follows that $yv = (1+x+x^2)(yu)$, so $x,x^2$ fix $yv$ $($just as they fix $v$$)$. Of course, $y(yv) = y^2v = v$, so $\text{span}(\text{Orb}(v)) = \text{span}(v,yv)$ is a $G$-invariant subspace of dimension at most $2$. This gives us a nonzero invariant subspace of dimension at most $2$ if $v\ne0$.

Otherwise, if $v=0$, we analyze the $G$-orbit of $w$ instead. Since $$xw = (x+yx^2)u, \text{ }x^2w = (x^2+yx)u,$$ we have $$y(xw) = (yx+x^2)u = x^2w,\text{ }y(x^2w) = (yx^2+x)u = xw.$$

But $1+y$ commutes with $1+x+x^2$, so $$w+xw+x^2w = (1+x+x^2)yu = y(1+x+x^2)u = yv = 0$$ $($from $v=0$$)$. Thus $$\text{span}(\text{Orb}(w)) = \text{span}(w,xw,x^2w) = \text{span}(w,xw)$$ is $G$-invariant with dimension at most $2$, and we are fine as long as $w\ne0$.

Thus we have reduced the problem to the case in which $$(1+x+x^2)u = (1+y)u = 0$$ for all $($nonzero$)$ vectors $u\in V$. In this case, we must have $$1+\rho_y = 1+\rho_x+(\rho_x)^2 = 0$$ identically $($as operators$)$, so in particular $\rho_y$ must fix all one-dimensional eigenspaces. Now set $u$ to be an arbitrary eigenvector of $\rho_x$; then $\rho_x$, $\rho_y$ both fix the nonzero subspace $\text{span}(u)$ of dimension $1$, and we are done.

We remark that a simpler approach altogether is just to start with an eigenvector of $\rho_x$.

b.

First we classify all $1$-dimensional $($irreducible$)$ representations of $G$ $($over some $1$-dimensional $V = (v)$, $v\ne0$$)$. Write $xv = \alpha v$ and $yv = \beta v$ $($for complex $\alpha$, $\beta$$)$; then $$x^3v = y^2v = (xy)^2v = v,$$ so $$\alpha^3 = \beta^2 = (\alpha\beta)^2 = 1,$$ which forces $$\alpha^3 = \alpha^2 = 1\implies \alpha = 1,$$ and $\beta = \pm1$. Thus the only $1$-dimensional $($irreducible$)$ representations of $G$ are the trivial $($$\beta=1$$)$ and sign $($$\beta=-1$$)$ representations. $($In terms of matrices, $R_x = [\alpha]$ and $R_y = [\beta]$ for both cases.$)$

Now consider an irreducible representation of $G$ over a $2$-dimensional $V$. Take an eigenvector $v\ne0$ of $x$ with eigenvalue $\lambda$, so $xv=\lambda v$ and $\lambda^3 = 1$. Define $w = yv$, so $$yw = y^2v = v.$$ Then $$xw = xyv = yx^2v = y\lambda^2 v = \lambda^2 w.$$

Of course, $w\ne v$, or else $(v)$ would be a $1$-dimensional $G$-invariant subspace of $V$. Similarly, $\lambda\ne1$, or else we would have $xv=v$, $xw=w$, $yv+yw = w+v$, and $(v+w)$ would be a $1$-dimensional $G$-invariant subspace.

Also, without loss of generality, we may assume $\lambda = e^{2\pi i/3}$, or else we could have started with the eigenvector $w\ne0$ of $x$ $($with eigenvalue $\lambda^2$$)$ instead of $v$. It follows that for any irreducible representation of $G$ over a $2$-dimensional $V$, there exist basis vectors $v$, $w\in V$ such that the matrices $R_x$, $R_y$ of $\rho_x$, $\rho_y$ with respect to the basis $B = (v,w)$ are $$\begin{pmatrix} e^{2\pi i/3} & 0 \\ 0 & -e^{2\pi i/3} \end{pmatrix},\, \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix},$$ respectively, so we are done.

We remark that it is easy to check that the three representations presented above are in fact irreducible. The one-dimensional representations are clearly irreducible; for the two-dimensional one, we use the matrix interpretation $R_x$, $R_y$ and note that a $G$-invariant one-dimensional subspace $(v)$ $($with $v = (p,q)^\text{T}$$)$ would have to have $(q,p) = yv\in (v)$ and thus $p^2 = q^2 \ne0$. But it is easy to check that $R_x$ sends $(p,p)$ to $(\lambda p, \lambda^{-1} p)\notin (v)$ and $(p,-p)$ to $(\lambda p,-\lambda^{-1} p)\notin (v)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.