Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $G$ be a finite group, and let $d_1,d_2,\dots,d_n$ be the dimensions of the irreducible representations. It is well-known that $\sum_{i=1}^n d_i^2=|G|$. If I am not mistaken, one has the following inequality $$ \sum_{i=1}^n d_i^{3/2}\geq \sqrt{|G|}. $$ Is this known or obvious (or false)? For abelian groups the inequality is far from sharp, because we can replace the right hand side with $|G|$. But for the symmetric group it seems pretty sharp. For example, for $S_6$, the left-hand side is 27.2688 and the right-hand side is $26.8328$.

share|improve this question
1  
Hi Harm, welkom bij mathoverflow :) Did you check small examples, say, with GAP? GAP has lots of character tables stored (or it can compute them, too). –  Dima Pasechnik Nov 1 '12 at 17:06
1  
The character degrees of S_6 are {1, 5, 9, 5, 10, 16, 5, 10, 9, 5, 1}. How do you get 27.2688? I get 227.96. –  John Wiltshire-Gordon Nov 1 '12 at 17:33
    
it was a typo - see Harm's own answer below... –  Dima Pasechnik Nov 1 '12 at 17:37
add comment

2 Answers

Never mind. The inequality that I wrote down is obvious, because $$ (\sum_{i=1}^n d_i^{3/2})^2\geq \sum_{i=1}^n d_i^3\geq \sum_{i=1}^n d_i^2=|G|. $$ I accidentally calculated $\sum_{i=1}^n \sqrt{d_i}$ on the left-hand side. So it might be an interesting question whether $$ \sum_{i=1}^n \sqrt{d}_i\geq \sqrt{|G|}. $$ I'm pretty sure this fails for large symmetric groups, because the left-hand side is less or equal than $p(n)|S_n|^{1/4}$ where $p(n)$ is the number of partitions. $p(n)$ grows subexponentially, whereas $|S_n|$ grows superexponentially. So we have $p(n)<|S_n|^{1/4}$ for large $n$. So the left-hand side will eventually be smaller than $\sqrt{|S_n|}$.

share|improve this answer
2  
the 2nd displayed inequality already fails for $S_7$. One gets $55.1\geq 70$... –  Dima Pasechnik Nov 1 '12 at 17:30
add comment

Let $G$ have $k$ conjugacy classes. Then Cauchy Schwarz seems to give $\sum_{i=1}^{k} \sqrt{d_i} \leq \sqrt{k}\sqrt{\sum_{i=1}^{k} d_i }$ and this (using C-S) again is at most $k^{\frac{3}{4}} |G|^{\frac{1}{4}}$. So as long as $k^{3} < |G|,$ the inequality you state is violated. There are many groups $G$ for which $k^{3} < |G|.$ One example is the alternating group $A_{6}$ which has $7$ conjugacy classes and order $360 >7^{3}.$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.