In Diaconis’ book Group representations in Probability and Statistics, freely online there is the following formula on p. 40 labeled (D-2):
Let $\tau$ be any transposition in $S_n$, $\lambda$ any partition of $n$, and $r(\lambda)=\chi_\lambda(\tau)/d_\lambda$, where $d_\lambda$ is the degree of the representation corresponding to $\lambda$ and the numerator is the character of the representation corresponding to $\lambda$ evaluated at $\tau$. Then $$r(\lambda)=\frac{1}{{n \choose 2}} \sum_j {\lambda_j \choose 2} - {\lambda_j^t \choose 2}$$
Here $\lambda_j^t$ is the $j$-th entry of the transpose of $\lambda$. Precisely if $\lambda = (\lambda_1, \dotsc, \lambda_m)$ with $\sum_j \lambda_j =n$ with $\lambda_i$ being decreasing positive integers, then the transpose of $\lambda$ is given by reflecting the Ferrers diagram (or called Young’s table) over the diagonal, equivalently counting the number boxes in the columns rather than rows. For example $(3,1)$ has transpose $(2,1,1)$. I apologize I don’t know how to typeset Ferrers diagrams at this hour otherwise I would include more context about them.
My question How does the above summation formula for $r$ make sense if the indices range over different lengths for a partition and its transpose?
For symmetric partitions, there is no issue and the sum is zero, but the formula is stated in general. I can’t make any sense of this, and this is especially not my area of expertise, if any. It cannot be interpreted as merely summing up to the length of the original partition because its transpose may be greater in length or may be lesser in length than the original. Perhaps it is meant to sum up to the smaller of the two?
It is a standard convention that any partition is extended so that $\lambda_i = 0$ for all $i$ greater than the length of the partition. Then the sum can be interpreted as being over all integers $j \ge 0$.