The Unapologetic Mathematician

Mathematics for the interested outsider

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.

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November 23, 2010 - Posted by | Algebra, Group theory, Representation Theory

9 Comments »

  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


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