1

Given a vector space $V$ of dimension $n$, I can form the module $$ S^{(1^{nd})}(V)$$ where $(1^{nd})$ denotes the partition $$ (1,1,\dots,1)\vdash nd$$ Now for $d=1$, I know that this module is in fact $\bigwedge^n(V)$. However I don't understand what it is for $d>1$. Is it still irreducible as a $GL(V)$-module? Does it equal $(\bigwedge^n(V))^{\otimes d}$?

Thanks in advance.

Levent
  • 4,804
  • Do you mean $(1,\dots, 1) \ (n \text{ times})$ and $(d, \dots, d) \ (n \text{ times})$? – Parthiv Basu Aug 04 '19 at 15:36
  • Also, are you aware of Shur modules and Young diagrams? – Parthiv Basu Aug 04 '19 at 15:39
  • @ParthivBasu $1^{(nd)}$ denotes the partition with $nd$ rows and $1$ column. I know Young diagrams and I have partial knowledge on Schur modules. – Levent Aug 04 '19 at 15:40
  • If that's what you mean, then for any $R$-module, the Schur module corresponding to $(1, \dots, 1) \ (k \text{ times})$ is always the $k$-th exterior power. And it has to be zero if $k > \mathrm{dim}(V)$. On the other hand, the case $(d, \dots, d) \ (n \text{ times})$ is more complicated. – Parthiv Basu Aug 04 '19 at 15:48
  • @ParthivBasu Oh, I see, thank you very much. Then can I ask what is the $GL(V)$-module corresponding to the group character $\det^d$? I thought it should be $S^{(1^{nd})}(V)$ but as you said it is $0$. – Levent Aug 04 '19 at 15:53
  • I am afraid I cannot help you here. My knowledge is very limited to only the construction of Shur modules because I once needed them to understand a certain statement. – Parthiv Basu Aug 04 '19 at 15:59

0 Answers0