# 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 -

1. I’m enjoying every step, John. Have a happy Thanksgiving.

In my daily 2,000 words of new fiction writing (over 300,000 words since the regimen was self-imposed on the advice of Stephen King and Ray Bradbury) I found myself this morning trying to explain why it would be preposterous to discover tensor calculus computations on a 3,750-year-old Babylonian clay tablet.

In this science fiction story, tablets are unburied which have Clebsch–Gordan coefficients and Calabi-Yau manifold tables. The poor NSA officer is trying to explain to the President of the United States.

Comment by Jonathan Vos Post | November 24, 2010 | Reply

2. [...] want to work out the matrices of induced representations. Explicitly, if is a left -module of degree , where is a subgroup of , then is a left -module. [...]

Pingback by Induced Matrix Representations « The Unapologetic Mathematician | November 25, 2010 | Reply

3. [...] of Induced Representations We know how to restrict and induce representations. Now we want to see what this looks like on the level of [...]

Pingback by Characters of Induced Representations « The Unapologetic Mathematician | November 29, 2010 | Reply

4. [...] Today, we can prove the Frobenius’ reciprocity formula, which relates induced characters to restricted [...]

Pingback by (Fake) Frobenius Reciprocity « The Unapologetic Mathematician | November 30, 2010 | Reply

5. [...] Before we can prove the full version of Frobenius reciprocity, we need to see that induction and restriction are actually additive [...]

Pingback by Induction and Restriction are Additive Functors « The Unapologetic Mathematician | December 1, 2010 | Reply

6. [...] Now we come to the real version of Frobenius reciprocity. It takes the form of an adjunction between the functors of induction and restriction: [...]

Pingback by (Real) Frobenius Reciprocity « The Unapologetic Mathematician | December 3, 2010 | Reply

7. [...] really should see an example of inducing a representation. One example we’ll find extremely useful is when we start with the trivial [...]

Pingback by Inducing the Trivial Representation « The Unapologetic Mathematician | December 6, 2010 | Reply

8. [...] they’re restricted to , it also tells us how representations of decompose when they’re induced to [...]

Pingback by The Branching Rule, Part 3 « The Unapologetic Mathematician | January 31, 2011 | Reply

9. [...] we see, the right hand side is obvious, but he left takes a little more work. We consult the definition of induction to [...]

Pingback by The Branching Rule, Part 4 « The Unapologetic Mathematician | February 1, 2011 | Reply