Exercise 12.1.8

Question

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

Let $t_1,t_2,t_3$ defined as in (12.15).
\be
\item[(a)]
Lagrange noted that any transposition fixes exactly one of $t_1,t_2,t_3$ and interchanges the other two, possibly changing the sign of both. Prove this and use it to show that $t_1t_2t_3$ is fixed by all elements of $S_4$.
\item[(b)]
Use the methods of Chapter 2 to express $t_1t_2t_3$ in terms of the $\sigma_i$. The result should be the identity (12.16).
\ee

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}% intervalles d'entiers 
\def\dcro{\mbox{]\hspace{-.15em}]}}

ewcommand{\be} {\begin{enumerate}}

ewcommand{\ee} {\end{enumerate}}

ewcommand{\deb}{\begin{eqnarray*}}

ewcommand{\fin}{\end{eqnarray*}}

ewcommand{\ssi} {si et seulement si }

ewcommand{\D}{\mathrm{d}}

ewcommand{\Q}{\mathbb{Q}}

ewcommand{\Z}{\mathbb{Z}}

ewcommand{\N}{\mathbb{N}}

ewcommand{\R}{\mathbb{R}}

ewcommand{\C}{\mathbb{C}}

ewcommand{\F}{\mathbb{F}}

ewcommand{\U}{\mathbb{U}}

ewcommand{e}{\,\mathrm{Re}\,}

ewcommand{\im}{\,\mathrm{Im}\,}

ewcommand{\ord}{\mathrm{ord}}

ewcommand{\Gal}{\mathrm{Gal}}

ewcommand{\legendre}[2]{\genfrac{(}{)}{}{}{#1}{#2}}

\begin{proof}
\be
\item[(a)] By (12.15),
\begin{align*}
t_1 &= x_1+x_2-x_3-x_4,\
t_2 &= x_1-x_2+x_3-x_4,\
t_3 &= x_1-x_2-x_3+x_4.
\end{align*}

Since $H(t_1) = \langle (1\,2),(3\,4) angle$ has order 4, the orbit ${\cal O}_{t_1}$ of $t_1$ under $S_n$ has $4!/4 = 6$ elements, so $${\cal O}_{t_1} =  \{t_1,t_2,t_3,-t_1,-t_2,-t_3\}.$$

$$(1 \,2) \cdot t_1 = t_1, \quad (1 \,2) \cdot t_2 = -t_3, \quad (1 \,2) \cdot t_3 = -t_2,$$
therefore 
$$(1 \,2) \cdot (t_1t_2t_3) = t_1(-t_3)(-t_2) = t_1t_2t_3.$$
\begin{align*}
(1 \,2 \, 3 \, 4) \cdot t_1 &= x_2+x_3-x_4-x_1\
&=-x_1+x_2+x_3-x_4\
&=-(x_1-x_2-x_3+x_4)\
&=-t_3
\end{align*}
With similar computations, we obtain
$$(1 \,2 \, 3 \, 4) \cdot t_1 = -t_3, \quad (1 \,2 \, 3 \, 4)\cdot t_2 = -t_2, \quad (1 \,2 \, 3 \, 4)\cdot t_3 = t_1,$$
thus
$$ (1 \,2 \, 3 \, 4) \cdot (t_1t_2t_3) = (-t_3)(-t_2)t_1 = t_1t_2t_3.$$
Since $(1 \,2) \cdot (t_1t_2t_3)  = t_1t_2t_3, (1 \,2 \, 3 \, 4) \cdot (t_1t_2t_3) = t_1t_2t_3$, and $S_4 = \langle (1 \,2), (1 \,2 \, 3 \, 4) angle$, then $t_1t_2t_3$ is fixed by all elements of $S_4$, and so is in $F(\sigma_1,\sigma_2,\sigma_3,\sigma_4)$.

\item[(b)] With the methods of Chapter 2, the following Sage instructions
\begin{verbatim}
e = SymmetricFunctions(QQ).e()
e1,e2,e3,e4 = e([1]).expand(4),e([2]).expand(4),
        e([3]).expand(4),e([4]).expand(4)
R.<x0,x1,x2,x3,y1,y2,y3,y4> = PolynomialRing(QQ, order = 'lex')
J = R.ideal(e1-y1,e2-y2,e3-y3,e4-y4)
G = J.groebner_basis()
t1= x0+x1-x2-x3; t2 = x0-x1+x2-x3; t3 = x0-x1-x2+x3
u = t1*t2*t3
var('sigma_1,sigma_2,sigma_3,sigma_4')
v = u.reduce(G).subs(y1=sigma_1, y2 = sigma_2,y3=sigma_3,y4=sigma_4);v
\end{verbatim}
give
$$\sigma_{1}^{3} - 4 \, \sigma_{1} \sigma_{2} + 8 \, \sigma_{3}.
$$
So
\begin{align*}
t_1t_2t_3 &= (x_1+x_2-x_3-x_4)(x_1-x_2+x_3-x_4)( x_1-x_2-x_3+x_4)\
& = \sigma_{1}^{3} - 4 \, \sigma_{1} \sigma_{2} + 8 \, \sigma_{3}.
\end{align*}
\ee
\end{proof}

Exercise 12.1.8

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Exercise 12.1.8

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