By <character orthogonality>, the squared multiplicities are the <character inner product> norm of the <restriction of a character>:
$$
\sum_{i=1}^r a_i^2
=\langle\chi\mathbin{\downarrow}_H,\chi\mathbin{\downarrow}_H\rangle_H
=\frac1{|H|}\sum_{h\in H}|\chi(h)|^2.
$$
Since $\chi$ is an <irreducible character>, its norm on $G$ is one, and hence
$$
\sum_i a_i^2
\leq\frac1{|H|}\sum_{g\in G}|\chi(g)|^2
=\frac{|G|}{|H|}=[G:H].
$$
The omitted summands are nonnegative, so the <character restriction norm bound> is an equality precisely when
$$
\boxed{\chi(g)=0\quad\text{for every }g\in G\setminus H.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 2 / 19I / a / Solution

The norm
$$
\langle\chi\mathbin{\downarrow}_K,
\chi\mathbin{\downarrow}_K\rangle_K
$$
is a positive integer, because it is the sum of the squares of the multiplicities of the irreducible constituents. The <character restriction norm bound> and $[G:K]=3$ make it at most three. It is therefore \b[$1$, $2$, or $3$].

All three values occur:

* In $G=S_3$ with $K=\langle(12)\rangle$, the trivial character remains irreducible on $K$, giving norm $1$.
* For the same pair, the two-dimensional <standard representation of the symmetric group> restricts to the sum of the trivial and nontrivial linear characters of $K$, giving norm $1^2+1^2=2$.
* In $G=A_4$ with its normal <Klein four-group> $K=V_4$, the <three-dimensional irreducible representation of the alternating group on four letters> restricts to the sum of the three nontrivial linear characters of $V_4$, giving norm $1^2+1^2+1^2=3$.

Past exam of the mathematics course of the University of Cambridge / 2018 / ii / Paper 2 / 19I / c / Solution

If $\chi$ is an <irreducible character> of a finite group $G$ and $H\leq G$, then
$$
\left\langle\operatorname{Res}_H^G\chi,
\operatorname{Res}_H^G\chi\right\rangle_H\leq[G:H].
$$
Equality holds exactly when $\chi$ vanishes on $G\setminus H$.

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 Incoming links: Character restriction norm bound