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\begin{document}
\centerline{\bf Math 880 \hfill{\large\bf \quad Homework 7}\hfill Fall 2025}
\vspace{1em}

\noindent  
Give credit for significant help from a reference or person. 
\\[.5em]
{\bf 1.} Consider the ring of symmetric polynomials in $n=3$ variables,
$$\Lambda^{\!(3)}=\Q[x_1,x_2,x_3]^{S_3},$$ 
with its bases $m_\lam$ (monomial), $e_\lam$ (elementary), $h_\lam$ (homogeneous),  $p_\lam$ (power), 
indexed by partitions $\lam=(\lam_1\geq\lam_2\geq\lam_3)$.  
The degree $k=3$ component $\Lambda^{\!(n)}_k = \Lambda^{\!(3)}_3$ has dimension $p(k) = 3$ 
indexed by the partitions $\lam = (3)$, $(2,1)$, $(1,1,1)$.
\\[.5em]
{\bf a.} By hand, write each basis of $\Lambda^{\!(3)}_3$ in terms of $x_1, x_2, x_3$.
\\[.5em]
{\bf b.} Write $e_\lam$, $h_\lam$, $p_\lam$ in terms of the monomial basis $m_\lam$, 
and verify the change-of-basis coefficients for the first two denoted in Stanley as 
$M_{\lam\mu}$ (0-1 matrices) and $N_{\lam\mu}$ ($\N$-matrices). 
Which of these bases has a triangular relationship to $m_\lam$,  
for an appropriate ordering of rows and columns?
\\[.5em]
{\bf c.} Recall the involution $\omega$ on $\Lambda^{\!(3)}_3$ defined by $\omega(e_\lam) = h_\lam$.
Compute $[\omega]_m^m$, the $3\times 3$ matrix of the linear mapping $\omega$ 
with respect to the monomial basis $m = \{m_\lam\}$.
Square the matrix to verify it is an involution. 
Verify that $p_\lam$ are eigenvectors with eigenvalues $\epsilon(\lam) = \sgn(w_\lam)$,
where $w_\lam$ is a permutation with cycle type $\lambda$.
\\[.5em]
{\bf d.} Inner product. Stanley defines the inner product (symmetric, bilnear, non-degenerate) 
on $\Lambda$ by $\langle h_\lam, m_\mu\rangle = \delta_{\lam\mu}$: that is, the monomial and
homogeneous bases are orthogonal by definition. Find the $3\times 3$
matrix $J$ of the inner product with respect to
the monomial basis of $\Lambda^{\!(3)}_3$, so that for $f,g\in \Lambda^{\!(3)}_3$
corresponding to $3\times 1$ coordinate vectors $[f]_m, [g]_m\in \C^3$, we have:
$$
\langle f, g\rangle = \mbox{}^{t}[f]_m\! \cdot J \cdot [g]_m\,
$$
where the right side is a matrix product $(1\times 3)\cdot (3\times 3)\cdot (3\times 1) = (1\times 1)$.
\\
Verify that $p_\lam$ is an orthogonal, but not orthonormal, basis of $\Lambda^{\!(3)}_3$\!.
(In fact, $\langle p_\lam, p_\mu\rangle = z_\lam \delta_{\lam\mu}$,
where $z_\lam=\#N_{\S_3}(w_\lam)$ the number of permutations in $\S_3$ 
which commute with a given permutation $w_\lam$ of cycle type $\lam$.)
\\[1em]
{\bf 2.} Schur polynomials. Keeping $n=3$ variables, we defined 
$s_\lam(x_1,x_2,x_3) =\sum_{T} x^T$ over $T\in \mathrm{SSYT}_3(\lam)$,
the semi-standard Young tableaux $T$ with shape $\lam$ and entries $T(i,j)\leq 3$.
We expand $s_\lam= \sum_{\mu} K_{\lam\mu} m_\mu$ with change of basis coefficients
given by the Kostka numbers $K_{\lam\mu}$, the number of SSYT of shape $\lam$ and content $\mu$
(i.e.~$T$ contains $\mu_i$ entries equal to $i$).
\\[.5em]
{\bf a.} Enumerate the SSYT for all $|\lam|=3$,
and write out the Schur polynomials $s_\lam\in \Lambda^{\!(3)}_3$ in terms of the monomial basis.
\\[.5em]
{\bf b.} For each $\lam=(\lam_1,\lam_2,\lam_3)$ above, 
construct the irreducible representation $V_\lam$ of $\GL_3(\C)$ 
inside the ring of polynomial functions on $3\times 3$ matrices $X=(x_{ij})$, 
the ring $R = \C[X] = \C[x_{ij}]$ for $i,j\in [3]$.
Define $V_\lam$ as the $\C$-span of the polynomials
$$
\Delta^{I_1}_{[\lam_1']}(X)\,\Delta^{I_2}_{[\lam_2']}(X)\,\Delta^{I_3}_{[\lam_3']}(X)\,,
$$
where $\Delta^I_J(X)$ is a minor of the matrix $X = (x_{ij})$, the determinant of the
submatrix on rows $I$ and columns $J$.
Here the row-sets $I_j$ are arbitrary, but all columns are left-justified:
$J_i = [\lam_i']=\{1,\ldots,\lam_i'\}$, where $\lam_i'$ is the length of the $i$th column. 
A matrix $A\in \GL_3(\C)$ acts via 
$$
(Af)(X) \stackrel{\text{def}}= f(\mbox{}^t\!AX),
$$
 multiplying 
the argument $X$ by the transposed matrix $\mbox{}^t\!A$.

\noindent
Verify that a basis for $V_\lam$ is given by 
$\Delta^{I_1}_{[\lam_1]}\Delta^{I_2}_{[\lam_2]}\Delta^{I_3}_{[\lam_3]}$
where $I_1,I_2, I_3$ run over the {\it column entries} of each semistandard tableau $T\in \mathrm{SSYT}_3(\lam)$.
Thus the character 
$$
\chi_\lam(x_1,x_2,x_3) = \mathrm{trace}(\mathrm{diag}(x_1,x_2,x_3) \mid V_\lam)
$$
is equal to the Schur polynomial $s_\lam(x_1,x_2,x_3)$.
\\[.5em]
{\bf c.} For each $V_\lam$, let $U_\lam$ be the subspace of vectors with character $x_1x_2x_3$,
corresponding to {\it standard} Young tableaux $T$.
Show that $U_\lam$ is a representation of the permutation matrix group $\S_3\subset \GL_3(\C)$,
and that this gives all the irreducible representations of $\S_3$. (If necessary, look up the requisite facts about
representations of finite groups.)
\\[.5em]
{\bf d.} Consider the vector space $V$ of matrices $X\in M_3(\C)$ with zero trace: $\mathrm{tr} (X)=0$. 
Define the {\it adjoint action} of $\mathrm{GL}_3(\C)$ by conjugation:
$$
\rho(A)X\ =\ AXA^{-1}.
$$
Find the character $\chi_V)(x_1,x_2,x_3)$ of this representation and write it in terms of Schur polynomials,
and hence identify the representation in terms of the $V_\lam$.
\\
{\it Hint:} Tensor by a power of the 1-dim determinant representation $V_{111}$.
\\[.5em]
{\bf e.} Repeat (d) for the representation of $\mathrm{GL}_n(\C)$ on traceless
$n{\times}n$ matrices, which is the {\it adjoint representation}
on the Lie algebra of $\mathrm{SL}_n(\C)$.
Write it in terms of Schur polyomials $s_\lam(x_1,\ldots,x_n)$.


\end{document}


Cauchy's original idea for Schur polynomials (long before the theory of representations)
was to produce a symmetric polynomial as a quotient of two anti-symmetric ones.
Specifically, let $a_\lam = \sum_{w\in S_3} \sgn(w) x_{w(1)}^{\lam_1} x_{w(2)}^{\lam_2}
x_{w(3)}^{\lam_3}$ be an anti-symmetrized monomial. 
The minimal non-zero $a_\lam$ is $a_\delta$ for $\delta = (2,1,0)$. 
Verify the Vandermonde determinant:
$$
a_\delta \ = \ 
\det\!\left[\begin{array}{@{\, }c@{\ }c@{\ }c@{\, }}
1 & 1& 1\\
x_1 & x_2 &x_3\\[.5em]
x_1^2&x_2^2& x_3^2
\end{array}\right]
\ =\  \prod_{ i<j} (x_i{-}x_j).
$$
It turns out every $a_\lam$ is divisible by $a_\delta$, and
Cauchy's bialternant formula is: 
$$
s_\lam \ =\ \frac{a_{\lam+\delta}}{a_\delta}.
$$
Verify this for $|\lam|= 3$.
