Conjugate of conjugate partition

Question

Let $\lambda=(\lambda_1,\dots,\lambda_r)$ is a partition of $n$ and denote $\lambda'$ the conjugate partition of $\lambda$, with $\lambda'_j=\#\{i\,|\,\lambda_i\ge j\}$.

I'm struggling to try to prove that $(\lambda')'=\lambda$ using only the definition of conjugate partition and without using Ferrers diagrams.

More specifically, I'm trying to justify why these two numbers are equal $$(\lambda')'_k=\#\{j\,|\,\lambda_j'\ge k\}=\#\{j\,|\,\#\{i\,|\,\lambda_i\ge j\}\ge k\}$$

$$\lambda_k=\#\{j\,|\,\lambda_k\ge j\}$$

but I can't figure it out with the indexes and sums appearing.

Could someone help me to develop the steps?

Answer 1

We use the convention to write a partition $\lambda=\left(\lambda_1,\dots,\lambda_r\right)$ using ordered parts: $\lambda_1\geq \lambda_2\geq\cdots\geq \lambda_r\geq 1$. We also write a partition sometimes in the form $\lambda=\left(\lambda_j\right)_{j\geq 1}$ and identify it with the $r$-tupel $\lambda=\left(\lambda_j\right)_{1\leq j\leq r}$ of its non-zero entries.

We obtain using OPs prime notation for conjugate partitions \begin{align*} \lambda&=\left(\lambda_j\right)_{j\geq 1}\\ \\ \lambda^{\prime}&=\left(\lambda_j^{\prime}\right)_{j\geq 1}=\left(\#\{q\big|\lambda_q\geq j\}\right)_{j\geq 1}\\ &=\Big(\sum_{{q\geq 1}\atop{\lambda_{q}\geq j}}1\Big)_{j\geq 1}\tag{1}\\ \\ \color{blue}{\lambda^{\prime\prime}}&=\left(\lambda_j^{\prime\prime}\right)_{j\geq 1}=\left(\#\{q\big|\lambda_q^{\prime}\geq j\}\right)_{j\geq 1}\\ &=\Big(\sum_{{q\geq 1}\atop{\lambda_{q^{\prime}}\geq j}}1\Big)_{j\geq 1}\tag{2.1}\\ &=\Big(\sum_{{q\geq 1}\atop{\#\{r|\lambda_r\geq q\}\geq j}}\Big)_{j\geq 1}\tag{2.2}\\ &=\Big(\sum_{{q=1}\atop{\#\{r|\lambda_r\geq q\}\geq j}}^{\lambda_j}\Big)_{j\geq 1}\tag{2.3}\\ &=\left(\lambda_j\right)_{j\geq 1}\tag{2.4}\\ &\,\,\color{blue}{=\lambda} \end{align*} and the claim follows.

Comment:

• In (1) we write the cardinality of the set as sum adding up $1$ whenever the condition $\lambda_{q}\geq j, q\geq 1$ for a given $j$ is true.

• In (2.1) we use the sum notation again as in (1) but now with $\lambda_{q^{\prime}}$.

• In (2.2) we write $\lambda_{q^{\prime}}$ as cardinality of the set with parts $\lambda_r$.

• In (2.3) we note that since $\lambda_j$ is at the $j$-th position of the partition the index $q$ is bounded by the upper limit $\lambda_j$. Indeed, if $q>\lambda_j$ the parts with $\lambda_r \geq q$ are left to the position of $\lambda_j$ and so this number of parts is less than $j$.

• In (2.4) we respect that in (2.3) whenever $q\leq \lambda_j$ all parts to the left of $\lambda_j$ fulfil the condition $\#\{r|\lambda_r\geq q\}\geq j$.