# The Unapologetic Mathematician

## Restricting and Inducing Representations

Two of the most interesting constructions involving group representations are restriction and induction. For our discussion of both of them, we let $H\subseteq G$ be a subgroup; it doesn’t have to be normal.

Now, given a representation $\rho:G\to\mathrm{End}(V)$, it’s easy to “restrict” it to just apply to elements of $H$. In other words, we can compose the representing homomorphism $\rho$ with the inclusion $\iota:H\to G$: $\rho\circ\iota:H\to\mathrm{End}(V)$. We write this restricted representation as $\rho\!\!\downarrow^G_H$; if we are focused on the representing space $V$, we can write $V\!\!\downarrow^G_H$; if we pick a basis for $V$ to get a matrix representation $X$ we can write $X\!\!\downarrow^G_H$. Sometimes, if the original group $G$ is clear from the context we omit it. For instance, we may write $V\!\!\downarrow_H$.

It should be clear that restriction is transitive. That is, if $K\subseteq H\subseteq G$ is a chain of subgroups, then the inclusion mapping $\iota_{K,G}K\hookrightarrow G$ is the exactly composition of the inclusion arrows $\iota_{K,H}K\hookrightarrow H$ and $\iota_{H,G}H\hookrightarrow G$. And so we conclude that \displaystyle\begin{aligned}\rho\!\!\downarrow^G_K&=\rho\circ\iota_{K,G}\\&=\rho\circ\iota_{K,H}\circ\iota_{H,G}\\&=\left(\rho\circ\iota_{K,H}\right)\!\!\downarrow^G_H\\&=\left(\rho\!\!\downarrow^H_K\right)\!\!\downarrow^G_H\end{aligned}

So whether we restrict from $G$ directly to $K$, or we stop restrict from $G$ to $H$ and from there to $K$, we get the same representation in the end.

Induction is a somewhat more mysterious process. If $V$ is a left $H$-module, we want to use it to construct a left $G$-module, which we will write $V\!\!\uparrow_H^G$, or simply $V\!\!\uparrow^G$ if the first group $H$ is clear from the context. To get this representation, we will take the tensor product over $H$ with the group algebra of $G$.

To be more explicit, remember that the group algebra $\mathbb{C}[G]$ carries an action of $G$ on both the left and the right. We leave the left action alone, but we restrict the right action down to $H$. So we have a $G\times H$-module ${}_G\mathbb{C}[G]_H$, and we take the tensor product over $H$ with ${}_HV$. We get the space $V\!\!\uparrow_H^G=\mathbb{C}[G]\otimes_HV$; in the process the tensor product over $H$ “eats up” the right action of $H$ on the $\mathbb{C}[G]$ and the left action of $H$ on $V$. The extra left action of $G$ on $\mathbb{C}[G]$ leaves a residual left action on the tensor product, and this is the left action we seek.

Again, induction is transitive. If $K\subseteq H\subseteq G$ is a chain of subgroups, and if $V$ is a left $K$-module, then \displaystyle\begin{aligned}\left(V\!\!\uparrow_K^H\right)\!\!\uparrow_H^G&=\mathbb{C}[G]\otimes_H\left(V\!\!\uparrow_K^H\right)\\&=\mathbb{C}[G]\otimes_H\mathbb{C}[H]\otimes_KV\\&\cong\mathbb{C}[G]\otimes_KV\\&=V\!\!\uparrow_K^G\end{aligned}

The key step here is that $\mathbb{C}[G]\otimes_H\mathbb{C}[H]\cong\mathbb{C}[G]$. But if we have any simple tensor $g\otimes h\in\mathbb{C}[G]\otimes_H\mathbb{C}[H]$, we can use the relation that lets us pull elements of $H$ across the tensor product. We get $gh\otimes1\in\mathbb{C}[G]\otimes_H\mathbb{C}[H]$. That is, we can specify any tensor by an element in $\mathbb{C}[G]$ alone.

November 23, 2010