# On the permanent dominance conjecture for symmetric group

The Lieb's permanent dominance conjecture states that the expression $$\frac{d_{\chi}^HA}{\chi(e)}\le per(A)$$ holds for all positive semidefinite matrices $$A$$, where $$d_{\chi}^HA=\sum_\limits{\sigma\in H}\chi(\sigma)\prod_\limits{i=1}^na_{i\sigma(i)}\ \ ,\{a_{ij}\}= A\ \ ,H\le S_n$$ is the immanent function of the matrix $$A$$ and $$per(A)$$ is the permanent of $$A$$.

Now, I think for $$H=S_n$$, the symmetric group of order $$n$$, it is sufficient to show that $$\chi(e)$$, the irreducible character corresponding to the trivial conjugacy class, or the dimension of the representation, dominates all the other irreducible characters; i.e $$\chi(e)\ge \chi(\sigma)$$ for any $$\sigma\in S_n$$. Now, $$\chi(e)$$ is straightly given by hook length formula. The other irreducible characters can be found by the Murnaghan-Nakayama rule, or the determinantal formula. Since the Murnaghan-Nakayama formula involves the recurrence on the number of rim hooks(Border-strip Tableau) and the length of each of which do not exceed the proper hook in the hook's formula, therefore, it seems intuitive to think the conjecture could be proven in the case $$H=S_n$$. Any way to rigorously look at the difference between the counting of characters in the Murnagahan-Nakayama rule and the hook length formula? Thanks beforehand.

• For any finite group and any character $\chi$ it is clear that $\chi(e)\geq \chi(\sigma)$, since $\chi(\sigma)$ is a sum of $\chi(e)$ roots of unity. I don't see how this implies Lieb's conjecture since we could have $\prod_{i=1}^n a_{i\sigma(i)}<0$. Jan 12, 2020 at 2:26
• @RichardStanley ok, so then the conjecture is trivial for matrices with positive entries, right? Jan 12, 2020 at 16:17
• That is correct. Jan 13, 2020 at 15:39