Lecture 15

Solution: (A slight abuse of notation is used throughout the solution; the same notation is used for the (different) representations of $S_{4}$ and $S_{5}$). We will use Frobenius reciprocity: for an irreducible representation $\pi$ of $S_{4}$ and an irreducible representation $\sigma$ of $S_{5}$, we have

so we need the restrictions to $S_{4}$ of the irreducible characters of $S_{5}$. Recalling the character table of $S_{5}$ and then restricting to $S_{4}$ gives the following values:

The irreducible characters of $S_{4}$ are as follows:

We now compute the inner product of each irreducible character of $S_{4}$ with the restricted characters above:

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  $\mathrm{Ind}_{S_{4}}^{S_{5}}{\,\mathrm{Id}}$:

  	$\displaystyle\langle\mathrm{Ind}_{S_{4}}^{S_{5}}\chi_{{\,\mathrm{Id}}},\chi_{{\,\mathrm{Id}}}\rangle_{S_{5}}=\langle\chi_{{\,\mathrm{Id}}},\operatorname{Res}_{S_{4}}^{S_{5}}\chi_{{\,\mathrm{Id}}}\rangle_{S_{4}}=1$
  	$\displaystyle\langle\mathrm{Ind}_{S_{4}}^{S_{5}}\chi_{{\,\mathrm{Id}}},\chi_{\pi}\rangle_{S_{5}}=\langle\chi_{{\,\mathrm{Id}}},\operatorname{Res}_{S_{4}}^{S_{5}}\chi_{\pi}\rangle_{S_{4}}=1.$

  The dimension of $\mathrm{Ind}_{S_{4}}^{S_{5}}{{\,\mathrm{Id}}}$ is $[S_{5}\,|\,S_{4}]\cdot 1=5$, so no other irreducibles occur, i.e.

  $$\mathrm{Ind}_{S_{4}}^{S_{5}}({\,\mathrm{Id}},{\mathbb{C}})\cong({\,\mathrm{Id}},{\mathbb{C}})\oplus(\pi,W_{0}).$$
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  $\mathrm{Ind}_{S_{4}}^{S_{5}}\mathrm{sgn}$:

  	$\displaystyle\langle\mathrm{Ind}_{S_{4}}^{S_{5}}\chi_{\mathrm{sgn}},\chi_{\mathrm{sgn}}\rangle_{S_{5}}=\langle\chi_{\mathrm{sgn}},\operatorname{Res}_{S_{4}}^{S_{5}}\chi_{\mathrm{sgn}}\rangle_{S_{4}}=1$
  	$\displaystyle\langle\mathrm{Ind}_{S_{4}}^{S_{5}}\chi_{\mathrm{sgn}},\chi_{\mathrm{sgn}\pi}\rangle_{S_{5}}=\langle\chi_{\mathrm{sgn}},\operatorname{Res}_{S_{4}}^{S_{5}}\chi_{\mathrm{sgn}\pi}\rangle_{S_{4}}=1.$

  The dimension of $\mathrm{Ind}_{S_{4}}^{S_{5}}\mathrm{sgn}$ is $[S_{5}\,|\,S_{4}]\cdot 1=5$, so no other irreducibles occur, i.e.

  $$\mathrm{Ind}_{S_{4}}^{S_{5}}(\mathrm{sgn},{\mathbb{C}})\cong(\mathrm{sgn},{\mathbb{C}})\oplus(\mathrm{sgn}\pi,W_{0}).$$
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  $\mathrm{Ind}_{S_{4}}^{S_{5}}\pi$: Since the restriction from $S_{5}$ to $S_{4}$ of ${\,\mathrm{Id}}$ and $\mathrm{sgn}$ are ${\,\mathrm{Id}}$ and $\mathrm{sgn}$, we know that neither of these representations occur. We next compute:

  	$\displaystyle\langle\mathrm{Ind}_{S_{4}}^{S_{5}}\chi_{\pi},\chi_{\pi}\rangle_{S_{5}}=\langle\chi_{\mathrm{\pi}},\operatorname{Res}_{S_{4}}^{S_{5}}\chi_{\pi}\rangle_{S_{4}}=1$
  	$\displaystyle\langle\mathrm{Ind}_{S_{4}}^{S_{5}}\chi_{\pi},\chi_{\mathrm{sgn}\pi}\rangle_{S_{5}}=\langle\chi_{\mathrm{\pi}},\operatorname{Res}_{S_{4}}^{S_{5}}\chi_{\pi}\rangle_{S_{4}}=0$
  	$\displaystyle\langle\mathrm{Ind}_{S_{4}}^{S_{5}}\chi_{\pi},\chi_{\rho}\rangle_{S_{5}}=\langle\chi_{\mathrm{\pi}},\operatorname{Res}_{S_{4}}^{S_{5}}\chi_{\rho}\rangle_{S_{4}}=1$
  	$\displaystyle\langle\mathrm{Ind}_{S_{4}}^{S_{5}}\chi_{\pi},\chi_{\mathrm{sgn}\rho}\rangle_{S_{5}}=\langle\chi_{\mathrm{\pi}},\operatorname{Res}_{S_{4}}^{S_{5}}\chi_{\mathrm{sgn}\rho}\rangle_{S_{4}}=0$
  	$\displaystyle\langle\mathrm{Ind}_{S_{4}}^{S_{5}}\chi_{\pi},\chi_{\pi\wedge\pi}\rangle_{S_{5}}=\langle\chi_{\mathrm{\pi}},\operatorname{Res}_{S_{4}}^{S_{5}}\chi_{\pi\wedge\pi}\rangle_{S_{4}}=1,$

  giving

  $$\mathrm{Ind}_{S_{4}}^{S_{5}}(\pi,W_{0})\cong(\pi,W_{0})\oplus(\rho,V)\oplus(\pi\wedge\pi,\bigwedge\,\!\!\!{}^{2}W_{0}).$$
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  $\mathrm{Ind}_{S_{4}}^{S_{5}}\mathrm{sgn}\pi$: Similarly to $\mathrm{Ind}_{S_{4}}^{S_{5}}\pi$, neither ${\,\mathrm{Id}}$ and $\mathrm{sgn}$ occur in the decomposition. The remaining inner products are

  	$\displaystyle\langle\mathrm{Ind}_{S_{4}}^{S_{5}}\chi_{\pi},\chi_{\pi}\rangle_{S_{5}}=\langle\chi_{\mathrm{\pi}},\operatorname{Res}_{S_{4}}^{S_{5}}\chi_{\pi}\rangle_{S_{4}}=0$
  	$\displaystyle\langle\mathrm{Ind}_{S_{4}}^{S_{5}}\chi_{\pi},\chi_{\mathrm{sgn}\pi}\rangle_{S_{5}}=\langle\chi_{\mathrm{\pi}},\operatorname{Res}_{S_{4}}^{S_{5}}\chi_{\pi}\rangle_{S_{4}}=1$
  	$\displaystyle\langle\mathrm{Ind}_{S_{4}}^{S_{5}}\chi_{\pi},\chi_{\rho}\rangle_{S_{5}}=\langle\chi_{\mathrm{\pi}},\operatorname{Res}_{S_{4}}^{S_{5}}\chi_{\rho}\rangle_{S_{4}}=0$
  	$\displaystyle\langle\mathrm{Ind}_{S_{4}}^{S_{5}}\chi_{\pi},\chi_{\mathrm{sgn}\rho}\rangle_{S_{5}}=\langle\chi_{\mathrm{\pi}},\operatorname{Res}_{S_{4}}^{S_{5}}\chi_{\mathrm{sgn}\rho}\rangle_{S_{4}}=1$
  	$\displaystyle\langle\mathrm{Ind}_{S_{4}}^{S_{5}}\chi_{\pi},\chi_{\pi\wedge\pi}\rangle_{S_{5}}=\langle\chi_{\mathrm{\pi}},\operatorname{Res}_{S_{4}}^{S_{5}}\chi_{\pi\wedge\pi}\rangle_{S_{4}}=1,$

  hence

  $$\mathrm{Ind}_{S_{4}}^{S_{5}}(\mathrm{sgn}\pi,W_{0})\cong(\mathrm{sgn}\pi,W_{0})\oplus(\mathrm{sgn}\rho,V)\oplus(\pi\wedge\pi,\bigwedge\,\!\!\!{}^{2}W_{0}).$$
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  $\mathrm{Ind}_{S_{4}}^{S_{5}}\rho$: The dimension of this representation is $10$, so we start by looking at the higher-dimensional irreducibles:

  	$\displaystyle\langle\mathrm{Ind}_{S_{4}}^{S_{5}}\rho,\chi_{\pi\wedge\pi}\rangle_{S_{5}}=\langle\chi_{\mathrm{\rho}},\operatorname{Res}_{S_{4}}^{S_{5}}\chi_{\pi\wedge\pi}\rangle_{S_{4}}=0$
  	$\displaystyle\langle\mathrm{Ind}_{S_{4}}^{S_{5}}\rho,\chi_{\rho}\rangle_{S_{5}}=\langle\chi_{\mathrm{\rho}},\operatorname{Res}_{S_{4}}^{S_{5}}\chi_{\rho}\rangle_{S_{4}}=1$
  	$\displaystyle\langle\mathrm{Ind}_{S_{4}}^{S_{5}}\rho,\chi_{\mathrm{sgn}\rho}\rangle_{S_{5}}=\langle\chi_{\mathrm{\rho}},\operatorname{Res}_{S_{4}}^{S_{5}}\chi_{\mathrm{sgn}\rho}\rangle_{S_{4}}=1,$

  so

  $$\mathrm{Ind}_{S_{4}}^{S_{5}}(\mathrm{sgn},{\mathbb{C}})\cong(\rho,V)\oplus(\mathrm{sgn}\rho,V).$$

Solution: Assume that the irreducible representation $(\sigma,W)$ of $H$ occurs in the decomposition of $(\pi|_{H},V)$ into irreducibles; hence $\langle\chi_{\pi|_{H}},\chi_{\sigma}\rangle_{H}=\langle\mathrm{Res}_{H}^{G}\chi_{\pi},\chi_{\sigma}\rangle_{H}>0$. Frobenius reciprocity gives

so $(\pi,V)$ occurs in the decomposition of $\mathrm{Ind}_{H}^{G}\chi_{\sigma}$ into irreducibles, i.e. it is isomorphic to a subrepresentation of $\mathrm{Ind}_{H}^{G}\chi_{\sigma}$.

Solution: Since the irreducible characters of $G$ form an orthonormal basis of the space of class functions of $G$, we have

We now use Frobenius reciprocity:

which equals the number of copies of $\pi$ that occur in the decomposition of $\operatorname{Res}^{G}_{H}\mathrm{Ind}^{G}_{H}{\pi}$ into irreducible representations of $H$. Noting that

we thus see that $\pi$ can occur at most $[G\,|\,H]$ times in the decomposition of $\operatorname{Res}^{G}_{H}\mathrm{Ind}^{G}_{H}{\pi}$, giving

Solution: We use Problem 49 Lecture 10 and Frobenius reciprocity: Let $(\sigma,Z)$ be any irreducible representation of $G$. Then for (a), the inner product of the character of the representation in the left-hand side of the identity with $\chi_{\sigma}$ equals

proving (a). Similarly, for (b), looking at the left-hand side, we have

whereas the right-hand side gives

since $\operatorname{Res}_{K}^{H}\operatorname{Res}_{H}^{G}=\operatorname{Res}_{K}^{G}$. Finally, for c), in the left-hand side we have

and for the right-hand side,