List 4 (Oral Exam)

1. Find the characters ${\varphi }^{\mu }$ of all permutation modules $M^{\mu }$ of $S_4$ and write the irreducible decompositions for all $M^{\mu }$ in terms of the Specht modules ${\mathcal{S}}^{\lambda }$.

Pasted image 20241201232419.png

Pasted image 20241201232429.png

Pasted image 20241201232403.png

2. Use a basis of polytabloids of the Specht module ${\mathcal{S}}^{\left(2^2\right)}$ to calculate representation matrices for representatives of all conjugacy classes in $S_4$.

Note that there are two standard tableaus, then $\dim {\mathcal{S}}^{\left(2^2\right)}=2$. Define

$$t_1=\begin{array}{cc}1& 2\\ 3& 4\end{array}\text{mbox}and{t}_2=\begin{array}{cc}1& 3\\ 2& 4\end{array},$$

then ${\mathcal{S}}^{\left(2^2\right)}=\mathrm{span}\{v_1,v_2\}$ where $v_1=e_{t_1}$ and $v_2=e_{t_2}$. With this basis, the representation matrices are

$$(1)\mapsto \left[\begin{array}{cc}1& \\ & 1\end{array}\right],(12)\mapsto \left[\begin{array}{cc}1& -1\\ & -1\end{array}\right],(123)\mapsto \left[\begin{array}{cc}0& -1\\ 1& -1\end{array}\right],(12)(34)\mapsto \left[\begin{array}{cc}1& \\ & 1\end{array}\right],(1234)\mapsto \left[\begin{array}{cc}-1& 0\\ -1& 1\end{array}\right].$$

3. For any Young diagram $\lambda =\left({\lambda }_1,\dots ,{\lambda }_l\right)$ denote by ${\lambda }_i^{\prime }$ the number of boxes in column $i$ of $\lambda$ for all $i=1,\dots ,m$, where $m={\lambda }_1$. Then ${\lambda }^{\prime }=\left({\lambda }_1^{\prime },\dots ,{\lambda }_m^{\prime }\right)$ is called the conjugate diagram to $\lambda$.Suppose that $\lambda ,\mu ⊢n$. Prove that $\lambda ⊵\mu$ if and only if ${\mu }^{\prime }⊵{\lambda }^{\prime }$.

\begin{proof} Assume that $\lambda =({\lambda }_1,\cdots ,{\lambda }_{\ell })$ and $\mu =({\mu }_1,\cdots ,{\mu }_s)$. By induction on $n$, it suffices to show ${\lambda }_1^{\prime }⩽{\mu }_1^{\prime }$. If it holds, then consider the Young diagrams ${\lambda }^{(1)}=({\lambda }_1-1,\cdots ,{\lambda }_{\ell }-1)$ and ${\mu }^{(1)}=({\mu }_1-1,\cdots ,{\mu }_s-1,1,\cdots ,1)$ with $(s-\ell )$‘s $1$ and disregarding the zeros in it. We still have ${\lambda }^{(1)}⊵{\mu }^{(1)}$ and then ${\mu }^{(1)}{}^{\prime }⊵{\lambda }^{(1)}{}^{\prime }$. Additionally by ${\lambda }_1^{\prime }⩽{\mu }_1^{\prime }$, we have ${\mu }^{\prime }⊵{\lambda }^{\prime }$.

Now we prove ${\lambda }_1^{\prime }⩽{\mu }_1^{\prime }$. Lab each position of $\lambda$ and $\mu$ from $1$ to $n$ sequentially, row by row. Then the elements of first column of $\lambda$ and $\mu$ are ${\lambda }_1+1,{\lambda }_1+{\lambda }_2+1,\cdots ,{\lambda }_1+\cdots +{\lambda }_{\ell -1}+1$ and ${\mu }_1+1,{\mu }_1+{\mu }_2+1,\cdots ,{\mu }_1+\cdots {\mu }_{s-1}+1$. As $\lambda ⊵\mu$, for any $i\in \{1,\cdots ,\ell -1\}$ we have ${\lambda }_1+\cdots {\lambda }_i+1⩾{\mu }_1+\cdots +{\mu }_i+1$. As these elements are less than or equal to $n$, we have

$$|\{{\lambda }_1+1,{\lambda }_1+{\lambda }_2+1,\cdots ,{\lambda }_1+\cdots +{\lambda }_{\ell -1}+1\}|⩽|\{{\mu }_1+1,{\mu }_1+{\mu }_2+1,\cdots ,{\mu }_1+\cdots {\mu }_{s-1}+1\}|$$

and so ${\lambda }_1^{\prime }⩽{\mu }_1^{\prime }$. \end{proof}

4. The aim of this exercise is to compute the character table of $S_5$. For a partition $\mu$ let ${\varphi }^{\mu }$ be the character of the permutation module $M^{\mu }$. Compute ${\varphi }^{\mu }$ for all partitions of 5. By taking inner products of characters find the character table of $S_5$.

Note that $5=5=4+1=3+2=3+1+1=2+2+1=2+1+1+1=1+1+1+1+1$, then there are $7$ partitions of $5$, which are denoted by ${\mu }_1,\cdots ,{\mu }_7$, respectively.

By ^i3a31d, we can compute $S^{{\mu }_i}$ as follows.

• $M^{{\mu }_1}=S^{{\mu }_1}$
• $M^{{\mu }_2}=k_1{S}^{{\mu }_1}\oplus k_2{S}^{{\mu }_2}$ and $\dim S^{{\mu }_2}=4$. Hence $k_1=k_2=1$.
• $M^{{\mu }_3}=k_1{S}^{{\mu }_1}\oplus k_2{S}^{{\mu }_2}\oplus k_3{S}^{{\mu }_3}$ and $\dim S^{{\mu }_3}=5$. We can compute that $k_1=k_2=k_3=1$.
• $M^{{\mu }_4}=k_1{S}^{{\mu }_1}\oplus k_2{S}^{{\mu }_2}\oplus k_3{S}^{{\mu }_3}\oplus k_4{S}^{{\mu }_4}$. We can compute that $k_1=1$, $k_2=2$, $k_3=1$ and then get $S^{{\mu }_4}$.

Repeat this procedure, then we get

	$(1)$	$(12)$	$(123)$	$(12)(34)$	$(1234)$	$(12)(345)$	$(12345)$
	1	10	20	15	30	20	24
$M^{{\mu }_1}$	1	1	1	1	1	1	1
$M^{{\mu }_2}$	5	3	2	1	1	0	0
$M^{{\mu }_3}$	10	4	1	2	0	1	0
$M^{{\mu }_4}$	20	6	2	0	0	0	0
$M^{{\mu }_5}$	30	6	0	2	0	0	0
$M^{{\mu }_6}$	60	6	0	0	0	0	0
$M^{{\mu }_7}$	120	0	0	0	0	0	0
$S^{{\mu }_1}=M^{{\mu }_1}$	1	1	1	1	1	1	1
$S^{{\mu }_2}=M^{{\mu }_2}-M^{{\mu }_1}$	4	2	1	0	0	-1	-1
$S^{{\mu }_3}=M^{{\mu }_3}-M^{{\mu }_2}$	5	1	-1	1	-1	1	0
$S^{{\mu }_4}$	6	0	0	-2	0	0	1
$S^{{\mu }_5}$	5	-1	-1	1	1	-1	0
$S^{{\mu }_6}$	4	-2	1	0	0	1	-1
$S^{{\mu }_7}$	1	-1	1	1	-1	-1	1

This is used to compute inner products. \end{proof}

5. Consider the 5-dimensional Specht module ${\mathcal{S}}^{(2,2,1)}$ over $S_5$.

• a). Calculate the representation matrices for all four generators $s_k=(kk+1)$ of $S_5$ in the basis of standard polytabloids $e_{t_i}$ with $i=1,\dots ,5$, where

$$t_1=\begin{array}{cc}1& 2\\ 3& 4\\ 5& \end{array},t_2=\begin{array}{cc}1& 2\\ 3& 5\\ 4& \end{array},t_3=\begin{array}{cc}1& 3\\ 2& 4\\ 5& \end{array},t_4=\begin{array}{cc}1& 3\\ 2& 5\\ 4& \end{array},t_5=\begin{array}{cc}1& 4\\ 2& 5\\ 3& \end{array}.$$

• b). By the branching rule, the restriction of ${\mathcal{S}}^{(2,2,1)}$ to the subgroup $S_4$ is a direct sum of two submodules, ${\mathcal{S}}^{(2,2,1)}=V\oplus W$. Find basis vectors of each of the submodules $V$ and $W$ in the form of linear combinations of standard polytabloids.

\begin{proof} a) By ^b2c5aa, we have $e_{\sigma t}=\sigma e_t$ and it deduces that

$$\begin{array}{rlr}& \rho ((12))=\left(\begin{array}{ccccc}1& & -1& & \\ & 1& & -1& \\ & & -1& & \\ & & & -1& \\ & & & & -1\end{array}\right),& \rho ((23))=\left(\begin{array}{lllll}& & 1& 0& \\ & & 0& 1& \\ 1& 0& & & \\ 0& 1& & & \\ & & & & -1\end{array}\right),\\ & \rho ((34))=\left(\begin{array}{lllll}1& & -1& & \\ & -1& & & \\ & & -1& & \\ & & & 0& 1\\ & & & 1& 0\end{array}\right),& \rho ((45))=\left(\begin{array}{llllc}0& 1& & & \\ 1& 0& & & \\ & & 0& 1& \\ & & 1& 0& \\ & & & & -1\end{array}\right).\end{array}$$

b) By ^5voxbj,

$${\mathcal{S}}^{(2,2,1)}={\mathcal{S}}^{(2,1,1)}\oplus {\mathcal{S}}^{(2,2)}:=V\oplus W.$$

Since the set of standard polytabloids is a basis of a Specht module, there are bases

$$\{e_t:t\in T_1\}\text{mbox}and\{e_t:t\in T_2\}$$

for $V$ and $W$, respectively, where

$$\begin{array}{rl}& T_1=\left\{\begin{array}{cc}1& 2\\ 3& \\ 4& \end{array},\begin{array}{cc}1& 3\\ 2& \\ 4& \end{array},\begin{array}{cc}1& 4\\ 2& \\ 3& \end{array}\right\},\\ & T_2=\left\{\begin{array}{cc}1& 2\\ 3& 4\end{array},\begin{array}{cc}1& 3\\ 2& 4\end{array}\right\}.\end{array}$$

Now we finish the solution. \end{proof}

**6. Prove the followings.

• a). For variables $x_1,\dots ,x_n$ set $\Delta \left(x_1,\dots ,x_n\right)={\prod }_{1⩽i<j⩽n}\left(x_i-x_j\right)$ and prove the identity

$$\sum _{i=1}^n{x}_i\Delta \left(x_1,\dots ,x_i+t,\dots ,x_n\right)=\left(x_1+\cdots +x_n+\left(\genfrac{}{}{0ex}{}{n}{2}\right)t\right)\Delta \left(x_1,\dots ,x_n\right)$$

where $t$ is another variable.

• b). If $\lambda =\left({\lambda }_1,\dots ,{\lambda }_n\right)$ is a partition of $n$, set ${\ell }_i={\lambda }_i+n-i$ for $i=1,\dots ,n$. Use the previous part to show that the numbers

$$F\left({\ell }_1,\dots ,{\ell }_n\right)=\frac{n!}{{\ell }_1!\dots {\ell }_n!}\prod _{1⩽i<j⩽n}\left({\ell }_i-{\ell }_j\right)$$

satisfy the recurrence relation

$$nF\left({\ell }_1,\dots ,{\ell }_n\right)=\sum _{i=1}^{n}F\left({\ell }_1,\dots ,{\ell }_i-1,\dots ,{\ell }_n\right)$$

• c). Hence prove by induction that the number $f^{\lambda }$ of standard tableaux of shape $\lambda$ is given by

$$f^{\lambda }=\frac{n!}{{\ell }_1!\dots {\ell }_n!}\prod _{1⩽i<j⩽n}\left({\ell }_i-{\ell }_j\right).$$

(This also proves the hook length formula).

\begin{proof} a) Notice that the coefficient of $t$ of LHS is $c\cdot \Delta (x_1,\cdots ,x_n)$ where

$$c=\sum _{i=1}^n{x}_i\left(\sum _{k=1}^{i-1}\frac{1}{x_i-x_k}-\sum _{k=i+1}^n\frac{1}{x_k-x_i}\right)=\sum _{i=1}^n\left(\sum _{k=1}^{i-1}\frac{x_i}{x_i-x_k}-\sum _{k=i+1}^n\frac{x_i}{x_k-x_i}\right).\text{\hspace{1em}(*)}$$

For each fixed $s<k$, there are two terms with denominator $x_k-x_s$, that is, $\frac{x_k}{x_k-x_s}$ and $-\frac{x_s}{x_k-x_s}$. In $(\ast )$, there are $n(n-1)$ terms and they form $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ pairs in the form of $\left\{\frac{x_k}{x_k-x_s},\frac{-x_s}{x_k-x_s}\right\}$. It deduces that the coefficient of $t$ of LHS equals $\left(\genfrac{}{}{0ex}{}{n}{2}\right)\Delta (x_1,\cdots ,x_n)$. In addition, it is easy to show when $t=0$ the equality holds, and so the constant parts of LHS and RHS are equal. Thus, it suffices to show the coefficients of $t^m$ with $m>1$ is equal to $0$.

For any $m>1$, the coefficients of $t^m$ is the sum of the following elements

$$S:=\left\{\frac{(-1{)}^{p_i}{x}_i\Delta (x_1,\cdots ,x_n)}{(x_i-x_{k_1})(x_i-x_{k_2})\cdots (x_i-x_{k_m})}:i\ne k_s,s\in \{1,\cdots ,m\},p_i\text{mbox}isthenumberof(i,k_s)\text{mbox}suchthati>k_s\right\}$$

and the sum of the terms in $S$ containing $x_i:=x_{k_0},x_{k_1},\cdots ,x_{k_m}$ is

$$\frac{{\sum }_{i=0}^m(-1{)}^{p_{k_i}}{x}_{k_i}({\prod }_{j<\ell ,j,\ell \ne i}(x_{k_j}-x_{k_{\ell }}))}{{\prod }_{0⩽j<\ell ⩽m}(x_{k_j}-x_{k_{\ell }})}\Delta (x_1,\cdots ,x_n)$$

and we claim that it equals $0$. To show it, it suffices to show

$$\sum _{k=0}^m(-1{)}^k{x}_k\Delta (x_0,\cdots ,{\hat{x}}_k,\cdots ,x_m)=0,$$

which can be proved by

$$0=\det \left[\begin{array}{cccc}{x}_0& x_1& \cdots & x_m\\ 1& 1& \cdots & 1\\ x_0& x_1& \cdots & x_m\\ ⋮& ⋮& & ⋮\\ x_0^{m-1}& x_1^{m-1}& \cdots & x_m^{m-1}\end{array}\right]=\sum _{k=0}^m(-1{)}^k{x}_k\Delta (x_0,\cdots ,{\hat{x}}_k,\cdots ,x_m).$$

Therefore, the coefficients of $t^m$ with $m>1$ is $0$. Now we finish the proof.

b) By a) there is

$$\begin{array}{rl}\sum _{i=1}^{n}F({\ell }_1,\cdots ,{\ell }_i-1,\cdots ,{\ell }_n)& =\sum _{i=1}^n\frac{n!}{{\ell }_1!\cdots ({\ell }_i-1)!\cdots {\ell }_n!}\Delta ({\ell }_1,\cdots ,{\ell }_i-1,\cdots ,{\ell }_n)\\ & =\sum _{i=1}^n\frac{n!}{{\ell }_1!\cdots {\ell }_n!}{\ell }_i\Delta ({\ell }_1,\cdots ,{\ell }_i-1,\cdots ,{\ell }_n)\\ & =\frac{n!}{{\ell }_1!\cdots {\ell }_n!}\sum _{i=1}^n{\ell }_i\Delta ({\ell }_1,\cdots ,{\ell }_i-1,\cdots ,{\ell }_n)\\ & =\frac{n!}{{\ell }_1!\cdots {\ell }_n!}\left({\ell }_1+\cdots +{\ell }_n-\left(\genfrac{}{}{0ex}{}{n}{2}\right)\right)\Delta (x_1,\cdots ,x_n)\\ & =nF({\ell }_1,\cdots ,{\ell }_n)\end{array}$$

as ${\ell }_1+\cdots +{\ell }_n=n+n(n-1)/2$.

c) For a standard tableaux of shape $\lambda$, we assume that $\lambda =({\lambda }_1,\cdots ,{\lambda }_n)$ with ${\lambda }_1⩾{\lambda }_2⩾\cdots ⩾{\lambda }_n⩾0$ and $\lambda ⊢n$. Define $f({\lambda }_1,\cdots ,{\lambda }_n)=f^{\lambda }$. We aim to show

$$f({\lambda }_1,\cdots ,{\lambda }_n)=\{\begin{array}{ll}F({\ell }_1,\cdots ,{\ell }_n),& {\lambda }_1⩾{\lambda }_2⩾\cdots ⩾{\lambda }_n⩾0\\ 0,& \text{mbox}otherwise.\end{array}$$

where ${\ell }_i={\lambda }_i+n-i$. We use induction on ${\sum }_{i=1}^n{\lambda }_i$ to prove this.

If $\lambda =(1,1,\cdots ,1)$, then $f^{\lambda }=1$ and $F({\ell }_1,\cdots ,{\ell }_n)=1$ coincide. Now we assume that $({\lambda }_1,\cdots ,{\lambda }_n)\ne (1,\cdots ,1)$, so ${\lambda }_n=0$ and ${\ell }_n=0$. For a standard tableaux of shape $\lambda$, it is easy to verify the recurrence relation

$$f^{\lambda }=\sum _{i=1}^{n-1}f({\lambda }_1,\cdots ,{\lambda }_i-1,\cdots ,{\lambda }_{n-1}),$$

because ${\lambda }_n=0$ and we get a standard tableaux with $n-1$ blocks if $n$ is removed. Note that $F({\ell }_1,\cdots ,{\ell }_{n-1},0)=nF({\ell }_1-1,\cdots ,{\ell }_{n-1}-1)$. Then by induction hypothesis,

$$\begin{array}{rl}{f}^{\lambda }& =\sum _{i=1}^{n-1}f({\lambda }_1,\cdots ,{\lambda }_i-1,{\lambda }_{n-1})\\ & =\sum _{i=1}^{n-1}F({\ell }_1^{\prime },\cdots ,{\ell }_i^{\prime }-1,\cdots ,{\ell }_{n-1}^{\prime })\\ & =\frac{1}{n}\sum _{i=1}^{n}F({\ell }_1,\cdots ,{\ell }_i-1,\cdots ,{\ell }_n)\\ & =F({\ell }_1,\cdots ,{\ell }_n),\end{array}$$

where ${\ell }_i^{\prime }={\lambda }_i+(n-1)-i$. Now we finish the proof. \end{proof}

Remark. Maybe with some values, $F$ is not defined. But I believe it can be avoided by extending $F$ suitably.

7. Prove the determinant formulas connecting symmetric functions:

$$p_n=\left|\begin{array}{ccccc}{e}_1& 1& 0& \dots & 0\\ 2e_2& e_1& 1& \dots & 0\\ ⋮& ⋮& \ddots & & ⋮\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ n{e}_n& e_{n-1}& e_{n-2}& \dots & e_1\end{array}\right|,n!e_n=\left|\begin{array}{ccccc}{p}_1& 1& 0& \dots & 0\\ p_2& p_1& 2& \dots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ p_{n-1}& p_{n-2}& \dots & \dots & n-1\\ p_n& p_{n-1}& \dots & \dots & p_1\end{array}\right|$$

and

$$(-1{)}^{\mathrm{n}-1}{p}_n=\left|\begin{array}{ccccc}{h}_1& 1& 0& \dots & 0\\ 2h_2& h_1& 1& \dots & 0\\ ⋮& ⋮& \ddots & & ⋮\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ n{h}_n& h_{n-1}& h_{n-2}& \dots & h_1\end{array}\right|,n!h_{\mathrm{n}}=\left|\begin{array}{ccccc}{p}_1& -1& 0& \dots & 0\\ p_2& p_1& -2& \dots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ p_{n-1}& p_{n-2}& \dots & \dots & -n+1\\ p_n& p_{n-1}& \dots & \dots & p_1\end{array}\right|.$$

\begin{proof} i) By the Newton identity, $n{e}_n={\sum }_{r=1}^n(-1{)}^{r-1}{p}_r{e}_{n-r}$ and it deduces that

$$\begin{array}{rl}\left|\begin{array}{ccccc}{e}_1& 1& 0& \dots & 0\\ 2e_2& e_1& 1& \dots & 0\\ ⋮& ⋮& \ddots & & ⋮\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ n{e}_n& e_{n-1}& e_{n-2}& \dots & e_1\end{array}\right|& =\left|\begin{array}{ccccc}{p}_1& 1& 0& \dots & 0\\ p_1{e}_1-p_2& e_1& 1& \dots & 0\\ ⋮& ⋮& \ddots & & ⋮\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ {\sum }_{r=1}^n(-1{)}^{r-1}{p}_r{e}_{n-r}& e_{n-1}& e_{n-2}& \dots & e_1\end{array}\right|\\ & =\left|\begin{array}{ccccc}0& 1& 0& \dots & 0\\ -p_2& e_1& 1& \dots & 0\\ ⋮& ⋮& \ddots & & ⋮\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ {\sum }_{r=2}^n(-1{)}^{r-1}{p}_r{e}_{n-r}& e_{n-1}& e_{n-2}& \dots & e_1\end{array}\right|\\ & =\cdots \\ & =\left|\begin{array}{ccccc}0& 1& 0& \dots & 0\\ 0& e_1& 1& \dots & 0\\ ⋮& ⋮& ⋮& & ⋮\\ 0& e_{n-2}& e_{n-3}& \cdots & 1\\ (-1{)}^{n-1}{p}_n& e_{n-1}& e_{n-2}& \dots & e_1\end{array}\right|\\ & =(-1{)}^{n-1}(-1{)}^{n-1}{p}_n\\ & =p_n.\end{array}$$

ii) Still by the Newton identity, we have $(-1{)}^{n-1}n{e}_n=p_n-e_1{p}_{n-1}+\cdots +(-1{)}^{n-1}{p}_1{e}_{n-1}$. Define

$$M=\left[\begin{array}{ccccc}{p}_1& 1& 0& \dots & 0\\ p_2& p_1& 2& \dots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ p_{n-1}& p_{n-2}& \dots & \dots & n-1\\ p_n& p_{n-1}& \dots & \dots & p_1\end{array}\right],$$

and denote column vectors of $M$ as $C_1,\cdots ,C_n$. Then there is

$$C_1-e_1{C}_2+e_2{C}_3+\cdots +(-1{)}^{n-1}{e}_{n-1}{C}_n=\left[\begin{array}{c}0\\ 0\\ ⋮\\ 0\\ (-1{)}^{n-1}n{e}_n\end{array}\right]$$

and so

$$\det M=\left|\begin{array}{ccccc}0& 1& 0& \dots & 0\\ 0& p_1& 2& \dots & 0\\ ⋮& ⋮& ⋮& \ddots & ⋮\\ 0& p_{n-2}& \dots & \dots & n-1\\ (-1{)}^{n-1}n{e}_n& p_{n-1}& \dots & \dots & p_1\end{array}\right|=(-1{)}^{n-1}(-1{)}^{n-1}n{e}_n(n-1)!=n!e_n.$$

iii) & iv) Similarly, there is $n{h}_n={\sum }_{r=1}^{\infty }{p}_r{h}_{n-r}$. Use the same method we can finish the proof. \end{proof}

8. a) Show that for a partition $\lambda =\left({\lambda }_1,\dots ,{\lambda }_m\right)$

$$s_{\lambda }\left(1,x,x^2,\dots ,x^{m-1}\right)=x^r\prod _{1⩽i<j⩽m}\frac{x^{{\lambda }_i-{\lambda }_j+j-i}-1}{x^{j-i}-1},$$

where $r={\sum }_{i=1}^m(i-1){\lambda }_i$.

b) Deduce that

$$s_{\lambda }(1,1,\dots ,1)=\prod _{1⩽i<j⩽m}\frac{{\lambda }_i-{\lambda }_j+j-i}{j-i}$$

(This number coincides with the dimension of the irreducible representation of the group ${\mathrm{GL}}_m(\mathrm{C})$ associated with $\lambda$.)

\begin{proof} a) Define $y_1=x^{{\lambda }_1+m-1},y_2=x^{{\lambda }_2+m-2},\cdots ,y_m=x^{{\lambda }_m}$, then

$$s_{\lambda }=\frac{\left|\begin{array}{cccc}1& y_1& \cdots & y_1^{m-1}\\ 1& y_2& \cdots & y_2^{m-1}\\ ⋮& ⋮& & ⋮\\ 1& y_m& \cdots & y_m^{m-1}\end{array}\right|}{\left|\begin{array}{ccc}{x}_1^{m-1}& \cdots & x_m^{m-1}\\ x_1^{m-2}& \cdots & x_m^{m-2}\\ ⋮& & ⋮\\ 1& \cdots & 1\end{array}\right|},$$

where $x_i=x^{i-1}$ for $i=1,\cdots ,m$. Then by Vandermonde matrix, we have

$$\begin{array}{rl}{s}_{\lambda }& =(-1{)}^{m(m-1)/2}\frac{{\prod }_{i<j}(y_j-y_i)}{{\prod }_{i<j}(x_j-x_i)}\\ & =(-1{)}^{m(m-1)/2}\frac{{\prod }_{i<j}{x}^{{\lambda }_j+m-j}(1-x^{{\lambda }_i-{\lambda }_j+j-i})}{{\prod }_{i<j}{x}^{i-1}(x^{j-i}-1)}\\ & =\prod _{i<j}{x}^{{\lambda }_j+m-j-i+1}\frac{x^{{\lambda }_i-{\lambda }_j+j-i}-1}{x^{j-i}-1}\\ & =x^r\prod _{i<j}\frac{x^{{\lambda }_i-{\lambda }_j+j-i}-1}{x^{j-i}-1}\end{array}$$

where $r={\sum }_{i<j}({\lambda }_j-j-i+m+1)$. Note that

$$\begin{array}{rl}r& =\sum _{i<j}({\lambda }_j-j-i+m+1)\\ & =(m-1)m(m+1)/2+\sum _{i=1}^{m-1}\sum _{j=i+1}^m({\lambda }_j-j-i)\\ & =(m-1)m(m+1)/2+\sum _{j=2}^m\sum _{i=1}^{j-1}({\lambda }_j-j-i)\\ & =(m-1)m(m+1)/2+\sum _{j=2}^m(j-1){\lambda }_j-\frac{3}{2}\sum _{j=2}^{m}j(j-1)\\ & =\sum _{j=2}^m(j-1){\lambda }_j=\sum _{j=1}^m(j-1){\lambda }_j\end{array}$$

as ${\sum }_{j=2}^{m}j(j-1)=(m-1)m(m+1)/3$.

b) Note that

$$\underset{x\to 1}{\lim }\frac{x^{{\lambda }_i-{\lambda }_j+j-i}-1}{x^{j-i}-1}=\frac{{\lambda }_i-{\lambda }_j+j-i}{j-i},$$

then by a) we get $s_{\lambda }(1,1,\dots ,1)={\prod }_{1⩽i<j⩽m}\frac{{\lambda }_i-{\lambda }_j+j-i}{j-i}$. \end{proof}

9. Since the generators $h_k$ of the algebra of symmetric functions $\Lambda$ are algebraically independent, we can specialize them in an arbitrary way and forget about the original variables $x_i$. In other words, we can take the generating function $H(t)$ to be an arbitrary power series in $t$ with the constant term 1 for the relations between symmetric functions to remain valid.

• a) Show that

$$H(t)=\prod _{k=1}^{\infty }{\left(1-t^k\right)}^{-1}$$

is the generating function for the sequence $p(n)$, where $p(n)$ is the number of partitions of $n$.

• b) By Euler’s pentagonal number theorem, for the generating function $E(-t)$ we have

$$E(-t)=\prod _{k=1}^{\infty }\left(1-t^k\right)=1+\sum _{m=1}^{\infty }(-1{)}^m\left(t^{\left(3m^2-m\right)/2}+t^{\left(3m^2+m\right)/2}\right)$$

Deduce the recurrence relation

$$p(n)=\sum _{\mathrm{m}=1}^{\infty }(-1{)}^{m-1}\left(p\left(n-\frac{3m^2-m}{2}\right)+p\left(n-\frac{3m^2+m}{2}\right)\right)$$

where $n⩾1$.

\begin{proof} a) Assume that ${\prod }_{k=1}^{\infty }(1-t^k{)}^{-1}={\prod }_{k=1}^{\infty }(1+t^k+t^{2k}+\cdots +t^{mk}+\cdots )={\sum }_{n=1}^{\infty }{a}_n{t}^n$. Each element in LHS with degree $n$ can be written as $t^{m_1\cdot 1}{t}^{m_2\cdot 2}\cdots t^{m_n\cdot n}=t^n$, and it corresponds to a partition where $n$ is written as a sum of integers $1+1+\cdots +1+2+\cdots +2+\cdots +n$, where the number of $1’$s is $m_1$, the number of $2$’s is $m_2$, and so on. Note that it is a 1-1 corresponding to elements in LHS and partitions, so $H(t)$ is the generating function for the sequence $p(n)$, where $p(n)$ is the number of partitions of $n$.

b) Note that $H(t)E(-t)=1$. The coefficient of $t^n$ in $H(t)E(-t)$ is

$$p(n)+\sum _{m=1}^{\infty }\left((-1{)}^{m}p\left(n-\frac{3m^2-m}{2}\right)+(-1{)}^{m}p\left(n-\frac{3m^2+m}{2}\right)\right)=0$$

and so $p(n)={\sum }_{\mathrm{m}=1}^{\infty }(-1{)}^{m-1}\left(p\left(n-\frac{3m^2-m}{2}\right)+p\left(n-\frac{3m^2+m}{2}\right)\right)$. \end{proof}