2
$\begingroup$

It is known that the level $k$ Verlinde ring of $SU(n)$ is $R(SU(n))/I_k$, where $I_k$ is the Verlinde ideal. A set of generators of $I_k$ is given by $\{V_{(k+i)L_1}:=\text{Sym}^{k+i}V_{\text{std}}| 1\leq i\leq n-1\}$, which is a regular sequence. Thus given any $V_\mu\in I_k$, there exist $r_i\in R(SU(n)), 1\leq i\leq n-1$ such that $$V_\mu=\sum_{i=1}^{n-1}r_i\cdot V_{(k+i)L_1}$$ Is there an algorithm to find those $r_i$? I can obtain some answers in low rank case using Giambelli's formula. For example, when $n=3$, $k=3$, $$V_{4L_1+L_2}=V_{L_1}\cdot V_{4L_1}-V_{5L_1}$$ $$V_{6L_1+L_2}=-V_{4L_1}+V_{L_1+L_2}\cdot V_{5L_1}$$ However, I am not sure how to extend this technique to find $r_i$ in this example: $n=4$, $k=2$, $V_\mu=V_{2L_1+L_2}$.

Note: The list of generators is corrected. It should be $\{V_{(k+i)L_1}|1\leq i\leq n-1\}$.

Improve this question
$\endgroup$
4
  • $\begingroup$ Irrelevant aside, just out of curiosity: how does one pronounce (in a talk, say) that reflected and rotated $V$? $\endgroup$
    – Jules Lamers
    Commented Oct 2, 2021 at 21:08
  • $\begingroup$ Do you mean Lambda? $\endgroup$
    – No_way
    Commented Oct 3, 2021 at 21:36
  • $\begingroup$ Well on my phone the tip points to the right (rather than up, as for a $\Lambda$) in the two symbols with subscript $L_1+L_2$ and $2 L_1+L_2$. Maybe it's just incorrect formatting $\endgroup$
    – Jules Lamers
    Commented Oct 3, 2021 at 21:43
  • $\begingroup$ I am not sure how the symbol looks like to you, but the symbol I typed is the letter V, meaning the underlying vector space of the representations with highest weights indicated by the subscript. $\endgroup$
    – No_way
    Commented Oct 3, 2021 at 22:26

0

Reset to default

You must log in to answer this question.