# Understanding induced representations

Let $G$ be a group and $H$ be a subgroup. Let $\phi:H\rightarrow GL(V)$ be a representation of $H$. There are three constructions in Wikipedia, but I am not really convinced by these.

My question is: are there any good intuitive ways to understand induced representation $Ind_H^G(\phi)$?

• Why is $K[G] \otimes_{K[H]} V$ unintuitive to you? – user27126 Dec 28 '12 at 14:56
• It is not as clear as the restriction of representation. In my opinion, there are many "induced" representation in contrast to restriction, which is obviously unique. "The" induced representation should be canonical in some sense; I think people call $Ind_H^G(\phi)$ "the induced representation" because of the Frobenius reciprocity. But now I don't see why your definition gives rise to such a duality. This is one of the reasons why I feel uneasy with it. – M. K. Dec 29 '12 at 9:28
• It's nice that the two answers below convinced you that induced rep is a good notion, although I'm personally curious what you mean by "there are many induced rep in contrast to restriction" - in fact I can't see how you may cook up a $G$-rep naturally from a $H$-rep other than the extension of scalars above. – user27126 Dec 29 '12 at 13:06
• I think the many induced representations refer to induction vs. coinduction, which are different for infinite $G:H$. – darij grinberg Dec 30 '12 at 14:34
• @darij: It is possible the OP is thinking about the situation where two different representations of a group have the same restricted representation of a proper subgroup (picking the trivial subgroup of course makes this very obvious). If this is the case, it is noteworthy that 'induced' doesn't refer to any unique rep restricting to a given one (indeed, induction changes the degree of a rep!). – anon Dec 30 '12 at 18:38

When $[G:H]$ is finite, you can think of it as taking the direct sum of $[G:H]$ copies of $\phi$, indexed by the elements of $G/H$. (Note that if $H$ is not normal, this will not be a group, but we can still define it as the set of left cosets of $H$, and $G$ will still act on it.) An element of $G$ acts on this space by (1) permuting the summands, via its action on $G/H$, and (2) acting inside each summand, via $\phi$.

There's then the question of why you would want to do this. Well, it's a natural way of extending representations of a subgroup to representations of the whole group. In fact, it's left adjoint to the restriction functor from $G$-reps to $H$-reps, so if you agree that restriction is the right way to shrink representations (and you're categorically minded), you should agree that induction is the right way to grow them.

This is my intuition about the induced representation, but it's not quite the whole story. To actually say how to do (2) above, you need to get down to the technical details. To me the clearest way to do this (now assuming $G$ finite) is the 'Algebraic' method on the Wikipedia page you linked. That is, rather than indexing the summands by the left cosets of $H$, choose representatives $x_i$ for these cosets; any $g\in G$ will then send each $x_i$ to a unique $x_j h$ with $h \in H$, and you let $g$ act on the $x_i$ factor by sending it to the $x_j$ index and applying $\phi(h)$ to it.

Hope this helps.

• To me, the middle paragraph really drives the point home. – anon Dec 28 '12 at 20:30
• Thank you for the answer, Paul. It makes my understanding much clearer and better. – M. K. Dec 29 '12 at 9:46

Suppose you have a representation $V$ of the subgroup $H< G$ and wish to construct the 'most free' representation of $G$ that would restrict to the given representation of $H$. The way to do this is to allow formal multiplication of the vectors in $V$ by elements of $G$, subject only to the condition that expressions involving elements of $H$ can simplify by invoking its known linear action on $V$.

This idea of freeness in constructions is very general - and can also be very concrete (as opposed to involving abstract universal properties from category theory): if you have an algebraic structure and wish to extend it, you can simply append a symbol to the underlying set and then allow every expression that can be constructed from the original structure and that new symbol into the mix.

For example, if you have a ring $R$, you can create the polynomial ring $R[x]$ in the symbol $x$ by adding all of the powers of $x$ to $R$ and then allowing linear combinations. This is sufficient: by the distributive property, the product of two polynomials will again be a polynomial, so nothing beyond these polynomials is needed to make a ring with underlying ring $R$, and element $x$ satisfying no special relations (except for $x$ commuting with everything, although there is a noncommutative polynomial ring construction as well, denoted $R\langle x\rangle$).

Another example: the free group generated by a set. You allow all "words" involving the symbols from the set, as well as formal inverses, and you automatically have a group - not letting them satisfy any relations (that is, two distinct words being equal as elements in the group) means that we have a free construction. Similarly, to make the free product $G\star H$ of the groups $G$ and $H$, we allow the formal multiplication to be done by elements of $G$ and $H$ together, and the only "simplifications" that take place are when two elements of $G$ are put together or two elements in $H$, in which case we allow the multiplication to go as it would in $G$ or $H$ originally. (Also $1_G=1_H$.)

To create the formal elements $gv$ for $g\in G$ and $v\in V$, we will automatically have the vector space

$$\bigoplus_{g\in G}gV$$

where $gV$ contains every symbol $gv$ for $v\in V$ and $g$ given. It is easy to see how $G$ needs to act on this space in order for things to work out: $a(bv)=(ab)v$. Sums like $gv+gw$ will simplify as $g(v+w)$, but expressions like $av+bw$ will not simplify if $a\ne b$ in $G$. These relations are required in order for this to be a linear action of $G$. But we must impose additional relations in order to keep the original action of $H$ on $V$; to this end, let $\rho:H\to GL(V)$ be our representation, and quotient the above direct sum by the subspace spanned by vectors of the form $ghv-g(\rho(h)v)$ for all $g\in G$, $h\in H$, $v\in V$. This subspace is $G$-invariant and so the quotient makes sense as a representation.

Remark. Things get tricky when $[G:H]$ is infinite: we need to reinterpret these formal vectors as functions $G\to V$ (this is a way of making "infinite tuples" indexed by elements of $G$) and then looking at the subspace of functions satisfying certain relations. Depending on whether or not we specify these functions to have finite support, we will have done either induction or coinduction.

• Thank you for the answer, anon. I am now convinced by your "the 'most free' representation of G" argument. I really like the way you think about it. Fortunately I have two good answers but unfortunately cannot "accept" two answers. So please let me choose Paul's answer as he posts it earlier. – M. K. Dec 29 '12 at 9:49
• What do you get after quotienting? I can see why we would do such a thing (since we want to effectively set $ghv = g(\rho(h)v)$) but I can't really see what comes from this. Does this somehow give the induced representation? – Cameron Williams Jul 7 '14 at 20:40
• @CameronWilliams Since $hV=V$ for all $h\in H$, but $gV\cap V=0$ if $g\not\in H$ (basically by construction) we have $g_1V=g_2V$ (as subspaces of the quotient of $\bigoplus gV$) if and only if $g_1H=g_2H$. This is why there are $[G:H]$ copies of the vector space $H$ within ${\rm Ind}_H^GV$. When you ask "does this somehow give the induced representation," while my answer is yes, your question seems to suggest you have a particular definition of Ind(V) that you weren't mentally connecting with this one... – anon Jul 8 '14 at 18:16

To begin we note that if we have $n\in\mathbb N$ and a rings homomorphism $\alpha:A\to B$, then naturally arises homomorphism $\alpha_n:M_n(A)\to M_n(B)$ such that $(X_{i,j})_{i,j}\mapsto(\alpha(X_{i,k}))_{i,j}$ and it gives a group homomorphism $\alpha_n:GL_n(A)\to GL_n(B)$ (since the invertible elements go into invertible under the action of homomorphism).

Now let $G$ be a finite group, $H\leq G$. We can consider group algebra $F[G]$ as a right $F[H]$ module, which we denote $M$. We have a natural linear representation $r:G\to GL_{F[H]}(M)$ of $G$ over $F[H]$: $g\mapsto \varphi_g$, where $\varphi_g(x)=gx$.

After these remarks we return to induced representations. Let $\rho:H\to GL_n(F)$ be a linear representation over a field $F$. Note that if $t=(t_1,\ldots,t_m)$ - left transversal of $H$ in $G$, then $t$ is a basis of $M$, hence we can go from $r$ to its version $r_t:G\to GL_m(F[H])$ in the basis $t$. We can extend $\rho$ by linearity to a homomorphism $\tilde\rho:F[H]\to M_n(F)$ of $F$-algebras. Then $\tilde\rho_m r_t:G\to GL_{mn}(F)$ is an induced representation.

This construction gives not only an induced representation, but also a transfer map $G\to H/N$, where $H'\leq N\leq H$. Indeed let $\nu:H\to H/N$ be a natural homomorphism and $\tilde\nu:F[H]\to F[H/N]$ be its extension by linearity to a homomorphism of $F$-algebras. Then $F[H/N]$ is a commutative ring, hence we have the right to consider determinants of the matrices with entries in $F[H/N]$. Then the map $g\mapsto \det(\tilde\nu_m r_t(g))$ is the transfer map. Since it is a composition of two homomorphisms, then transfer map is a homomorphism. This construction also allows for easy to prove independence of the transfer map from the choice of transversal - it's just a consequence of the fact that the conjugate matrices have the same determinants.