# Understanding solution that $\sum_i d_i^2 \leq |G:H|$ for an induced character

Question

Let $H$ be a subgroup of $G$ and $\chi_1, ... \chi_r$ irreducible characters of G. Let $\psi$ be an irreducible character of $H$ and $d_1, ... , d_s$ integers such that $\psi \uparrow G = d_1\chi_1 + ... + d_k\chi_k$. Show that $$\sum_i d_i^2 \leq |G:H|$$

Solution

We have $|G:H|\psi(1) = d_1\chi_1(1) + ... + d_k\chi_k(1)$, where $d_i = \langle \psi \uparrow G, \chi_i \rangle_G = \langle \psi, \chi_i \downarrow H \rangle_H$ by Frobenius Reciprocity Theorem.

Hence, since $\psi$ is irreducible, $\chi_i \downarrow H = d_i \psi + \beta$ where either $\beta$ is a character of $H$ or $\beta=0$.

Thus $\chi_i(1) \geq d_i \psi(1)$ and therefore $$|G:H|\psi(1) \geq (d_1^2 + ... + d_k^2)\psi(1)$$

What I'm stuck on

I don't understand the bold italic line in the proof, I get that since in H $\psi$ is irreducible, the $\chi_i \downarrow H$ must include it some number of times each, adding up to 1 (I think), but how do you know it is $d_i$ each time? How do you know $\beta$ is non-negative?

• The italic part is just a restatement of the preceding part, using Frobenius reciprocity. – Tobias Kildetoft Nov 25 '16 at 12:00
• I can almost see that, but what does $\psi$ being irreducible tell us? – veiph Nov 25 '16 at 13:30
• $\psi$ being irreducible tells us that taking the inner product with $\psi$ actually gives the number of times $\psi$ occurs, rather than something more complicated. – Tobias Kildetoft Nov 25 '16 at 14:30

## 1 Answer

Let $$\psi \in Irr(H)$$ and assume that $$\psi^G=\sum_{i=1}^sd_i\chi_i$$, where $$\chi_i \in Irr(G)$$ and $$d_i \in \mathbb{Z}_{\geq 0}$$. Since the irreducible characters form an orthonormal basis, it follows that the inner product $$[\psi^G,\psi^G]=\sum_{i=1}^sd_i^2:=m$$. By Frobenius Reciprocity, $$[\psi^G,\psi^G]=[(\psi^G)_H,\psi]$$. So $$(\psi^G)_H=m\psi + \Delta$$, where $$\Delta$$ is a character of $$H$$ or $$\Delta \equiv 0$$. Hence $$(\psi^G)_H(1)=\psi(1)|G:H| \geq m\psi(1)$$, and so $$m \leq |G:H|$$.

Conversely, assume $$\chi_H=\sum_{j=1}^te_j\psi_j$$, where $$\psi_j \in Irr(H)$$ and $$e_j \in \mathbb{Z}_{\geq 0}$$. It follows that $$[\chi_H,\chi_H]=\sum_{j=1}^te_j^2:=n$$. But $$[\chi_H,\chi_H]=[(\chi_H)^G,\chi]$$, so $$(\chi_H)^G=n\chi + \Gamma$$, where $$\Gamma$$ is a character of $$G$$ or $$\Gamma \equiv 0$$. Hence $$(\chi_H)^G(1)=\chi(1)|G:H| \geq n\chi(1)$$, and so $$n \leq |G:H|$$.