For a partition $\lambda$ it is very well-known operation to take its conjugate partition $\lambda'$ which is obtained by transposing the Young diagram of $\lambda$.

A partition $\lambda$ can be viewed as a dominant weight of $GL_n$ for some $n \geq l(\lambda)$.

Suppose $m \geq l(\lambda),l(\lambda')$. Is there a 'natural' operation on $\mathbb{Z}^m$, the weight lattice of $GL_m$, which takes $\lambda \mapsto \lambda'$ ?

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    $\begingroup$ For one thing, $n\geq l(\lambda)$ does not imply $n\geq l(\lambda^\top)$, so this is unlikely to be something simple. $\endgroup$ Jun 24, 2022 at 19:00

1 Answer 1


Computing some simple examples it doesn't appear to be a particularly nice operation. It certainly isn't linear in the weights or anything so simple as that. Let $\omega_1,\dots,\omega_n$ be the fundamental weights. Here are some examples of the "transpose the young diagram" operation on weights which I computed using LiE: $$\begin{array}{|c|c|} \hline \omega_1& \omega_1 \\ \hline k\omega_1& \omega_k \\ \hline \omega_1 + \omega_2 & \omega_1 + \omega_2 \\ \hline \omega_1 + \cdots + \omega_n & \omega_1 + \cdots + \omega_n \\ \hline 2\omega_1 + \omega_2& \omega_1 + \omega_3 \\ \hline \omega_1 + 2\omega_2 & \omega_2 + \omega_3 \\ \hline 2\omega_1 + 2\omega_2 & \omega_2 + \omega_4 \\ \hline k\omega_k& k\omega_k\\\hline (k-i)\omega_k & k\omega_{k-i}\\\hline k\omega_1 + k\omega_2 & \omega_k + \omega_{2k} \\ \hline k\omega_1 + k\omega_3 & 2\omega_k + \omega_{2k} \\ \hline k\omega_2 + k\omega_3 & \omega_k + 2\omega_{2k} \\ \hline \end{array}$$ Here $i<k\leq n$ (and $2k \leq n$ where it is used). Caveat: these are patterns I observed and extrapolated so are not proved true for all $k,n,i$.

As you can see there are certainly some patterns but I would be hard-pressed to give a general rule for any weight.

  • $\begingroup$ Interesting... did you put specific values of $k$ and $n$ in LiE and then extrapolate these or can you do all these formally without specifying $k$ and $n$? $\endgroup$
    – ArB
    Jun 21, 2022 at 8:13
  • $\begingroup$ LiE has functions to_part, from_part and trans_part which I joined together (note: you have to append a bunch of $0$'s as trans_part strips them off for some reason which will make from_part give the wrong answer) And then I tried that with various weights. I put in specific values of $k$ and $n$ is the length of the vector you put in (although as long as $n$ is high enough this doesn't change the answers). The patterns are just those I observed so they are not completely guaranteed/may break down for edge cases. $\endgroup$
    – Callum
    Jun 21, 2022 at 10:57
  • $\begingroup$ Another observation I made is that if you decompose $V^{\otimes k}$ into irreducibles the conjugate components seem to have the same multiplicity. Thus you can just compute p_tensor($k,[1,0,\dots,0]$) and pair off things with the same coefficient. I imagine this must be mentioned in some treatments of tensor products of representations but I don't know where off the top of my head. $\endgroup$
    – Callum
    Jun 21, 2022 at 11:04
  • $\begingroup$ Thinking about it a bit harder, my last comment is evident from the hook length formula. The multiplicity of each irreducible is given by $f_\lambda = \frac{n!}{\prod h_\lambda(i,j)}$ where $h_\lambda(i,j)$ is the hook length of the cell $(i,j)$ and then $h_\lambda(i,j) = h_{\lambda'}(j,i)$ so $f_{\lambda'} = f_\lambda$ although of course there is no guarantee that there aren't other representations with this multiplicity. $\endgroup$
    – Callum
    Jun 21, 2022 at 11:47

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