Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I'm reading about induced representations for research. Particularly, I'm trying to get a firm grasp on the finite group case before venturing on to the locally compact case. I've been looking at Wikipedia and more or less get the idea with one somewhat significant issue: why should we look to $\bigoplus g_iV$ (where the $g_i$ are coset representatives of for $H$) when defining the induced representation? Why shouldn't the induced representation on $G$ act on simply $V$ or even $V^{|G|}$ instead of $[G:H]$ copies of $V$?

share|improve this question
This is the direct generalization of the standard linear representation of finite groups (representation via permutation matrices). In the permutation representation you induce from the trivial group and its 1-dimensional representation. –  studiosus Jul 8 at 3:20
If it helps, there is an analogue of extending an $H$-set $X$ to a $G$-set $G\times_HX$ (which is $G\times X$ modulo the relations $(gh,x)=(g,hx)$, with $G$ acting by left-muliplication on the first coordinate). Considering $H$ itself as an $H$-set, extending it to a $G$-set yields the coset space $G/H$. –  blue Jul 8 at 4:32

5 Answers 5

up vote 4 down vote accepted

For a general pair $G, H$ and a representation $V$ of $H$ there may be no way to extend the action of $H$ to an action of $V$. By making the $[G:H]$ copies we have enough wiggle room for $G$ to act.

More or less I like to think of it as follows: Take $[G : H]$ copies of the representation indexed by the cosets of $H$. When you act by something in $H$ you act on each copy separately, and when you act by something else you permute the copies according to how that element permutes the cosets of $H$.

This isn't quite correct, as other elements of $G$ not in $H$ will act on the separate copies as well as permute them, but I think it gives good intuition for what is going on. The wikipedia article gives a more explicit formula for how general $g \in G$ acts.

share|improve this answer
I get that but I guess it just seems rather.. ad hoc I suppose. I suppose my question is more along the lines of: what would suggest to look at copies of $V$? What would possess someone to even think about that? Nothing else in representation theory to that point suggested copies of $V$ being of interest. –  Cameron Williams Jul 8 at 2:32
I personally think that group actions of a finite group on a finite set arise much more naturally than linear representations when one is first learning math. Any transitive group action is isomorphic to the action of the group on the cosets of some subgroup. So from that perspective this may be a somewhat natural way to try to extend the action from $H$ to $G$. –  Nate Jul 8 at 13:31
What do you mean by transitive group action? –  Cameron Williams Jul 8 at 14:18
I'm not sure what you are asking. Are you asking about what it means for a group action to be transitive or do you need to know what a group action is? –  Nate Jul 8 at 14:49
Oh I didn't realize it was just a group action that is transitive. I thought maybe there was some deeper meaning. I think this is actually really putting it together for me. –  Cameron Williams Jul 8 at 14:58

This is a more advanced view.

More generally, if you have a ring homomorphism $\phi:R_1\to R_2$, and $M$ is a left $R_1$-module, then there is a left $R_2$-module that is the "induced module" which is:

$$R_2\otimes_{R_1} M$$ which is an $R_2$ module.

The case of $H<G$ is then $R_1=\mathbb C[H], R_2=\mathbb C[G]$ is the case of induced group representations. In that case, $R_2$ is generated by $[G:H]$ elements when it is considered as an $R_1$-module, so this is $[G:H]$ times the 'dimension' of $M$.

If you look at the categories of $R_1\text{-Mod}$ and $R_2\text{-Mod}$, there is an obvious functor from $F_{\phi}:R_2\text{-Mod}\to R_1\text{-Mod}$. I believe, but I'm not sure, that $M\to R_2\otimes_{R_1} M$ is the adjoint of that functor, but I wouldn't swear that is true.

share|improve this answer
What are $k$ and $K$ here? –  Cameron Williams Jul 8 at 2:09
Sorry, the $K$ was a typo for $k$, meant it to be an arbitrary field, but I've changed the answer above to use $\mathbb C$ for the standard group representations. @CameronWilliams –  Thomas Andrews Jul 8 at 3:09
This is more or less the way it was presented in my course on representation theory but the module description isn't easy for me to intuit but that's probably can't be helped. –  Cameron Williams Jul 8 at 15:00
Yeah, I often find this abstract level clearer, but I can see how it seems remote from the idea of groups and representations. –  Thomas Andrews Jul 8 at 16:51

Induction of modules is in a sense related to extension by scalars. Recall that any ring morphism $f \colon R \to S$ induces a functor from left $R$-modules to left $S$-modules by sending an $R$-module $M$ to $S \otimes_R M$, where $S$ is viewed as a $(S, R)$-bimodule via $f$. So the extension by scalar functor is actually just a left tensor functor $L \otimes_R -$ for some $(S, R)$-bimodule $L$.

Now suppose that $G$ is a finite group, $H$ a subgroup of $G$, and $K$ some field. We have a $KH$-module $N$ and want to obtain a $KG$-module the most universal way. Viewing the group ring $KH$ as a subring of $KG$ and then extending by scalars suggests that $KG \otimes_{KH} N$ is our candidate. In fact, you can prove that $\text{Ind}_H^G N \cong KG \otimes_{KH} N$ as $(KG, KH)$-bimodules and more generally that the induction functor $\text{Ind}_H^G \colon KH\text{-Mod} \to KG\text{-Mod}$ between the respective module categories is naturally isomorphic to the left tensor functor $KG \otimes_{KH} -$.

It should also be noted that the other basic representation theory operations of restriction, inflation, and deflation can similarly be defined as tensor functors.

share|improve this answer

Induced representations are a special case of extension of scalars. Say you're doing linear algebra over a vector space $V$ defined over a field $k$. Often it's useful to have an algebraically closed field of scalars, since then we can decompose a space into generalized eigenspaces of a given linear map, so what do we do if our scalars aren't algebraically closed? We extend them! To this end, let $L/K$ be any extension of fields: we want the extension-of-scalars-from-$K$-to-$L$ to take the free $K$-vector space on a set $X$, and return the free $L$-vector space on the set $X$. Thus, if our space is comprised of elements which are $K$-linear combinations of a given set of basis vectors, then after extension of scalars the same is true only we have $L$-linear combinations. There is a way to do this independent of choosing a basis, which is to tensor against $L$ over $K$, i.e. $V_L:=L\otimes_KV$. The tensor product is essentially a way to "pretend multiply" vectors in $V$ by scalars in $L$ and allow ourselves linear combinations of these "pretend products." One easily checks this makes $V_L$ an $L$-vector space.

The same can be done with $R$-modules. (Vector spaces are modules over fields.) If $S$ is a ring which contains the ring $R$, and $M$ is any $R$-module, then $S\otimes_RM$ is formed by "pretending" to multiply elements of $M$ by scalars in $S$ (subject to the proviso that scalars in $R$ act the same way they did originally on $M$) and then adding these pretend products, and this turns $S\otimes_RM$ into a module over the bigger ring $S$.

Linear representations of a group $H$ over a field $k$ are basically $k[H]$-modules. If we want to extend the action of $H$ on a $k$-space $V$ to an action of $G$ on it, we need to extend the scalars of $k[H]$ to the scalars of $k[G]$. This is achieved by $k[G]\otimes_{k[H]}V$. It is comprised of linear combinations of vectors $gv$ for $g\in G$, $v\in V$, subject to $(ab)v=a(bv)$, $(a+b)v=av+bv$, $a(v+w)=av+aw$, and the rule that $hv$ for $h\in H$ is exactly as it's defined when $V$ is a $k[H]$-module.

To a category theorist, we notice that "restriction of the action of $G$ to the action of $H$" is basically a forgetful functor from the category of representations of $G$ to the category of representations of $H$. The notion of "free" or "universal" constructions is captured in categorical language as "adjoints to forgetful functors," so if the opposite of restricting actions is inducing them upwards, then we can define the representations of $G$ induced from a representation of $H$ via applying such an adjoint. Frobenius reciprocity (the $\hom$ version) essentially states that $k[G]\otimes_{k[H]}V$ is the left adjoint applied to a representation $V$ of $H$ to induce a representation of $G$.

share|improve this answer

Whenever there is some definition (or in this case, more precisely a construction), it is indeed a good idea to ask yourself "why does it have to be exactly this way".

To answer such a question, one should start by figuring out what the constructed object should satisfy. In this case, we have a group $G$, a subgroup $H$ and a representation $V$ of $H$. What we want is a representation of $G$, and we would of course like it to have some sort of relation to $V$. Since we have a very nice way to get a representation of $H$ from a representation of $G$ (restriction), this should probably somehow play into this relation.

I will discuss three possible relations one might wish for: The overly optimistic, the overly pessimistic, and the "just right".

The overly optimistic:
If we were to wish for the very best relation we could get, we would like our induced representation $W$ to be such that restricting $W$ to $H$ gives back $V$. This would obviously give us the very best possible relation we could ever hope for, but unfortunately, $G$ need not have any representation with this property (in fact, determining when this is the case is a very interesting topic in the representation theory of finite groups).

The overly pessimistic:
A relation that we should at least require is that if $V$ is simple, then when we restrict the induced module to $H$, then $V$ is either a submodule or a quotient (note that I have not specified anything about the groups or the field we work over, so these need not be equivalent). However, as we will shortly see, we can get something even better than this, and we might as well get all that we can.

The "just right":
In the overly pessimistic version, we were interested in submodules or quotients. But often a more useful thing to consider will be $\operatorname{Hom}$-spaces between modules. In this context, having a certain simple submodule is the same as having a non-zero homomorphism from said simple module (and having it as a quotient is then the same with the arrow in the other direction). So in this way, the overly pessimistic version becomes a statement about certain $\operatorname{Hom}$-spaces being non-zero. To make this more general, we have an $H$-module $V$ and we want a $G$-module $W$ such that whenever we have a $G$-module $M$, we have some sort of comparison between the spaces $\operatorname{Hom}_H(V,M)$ and $\operatorname{Hom}_G(W,M)$ (or with the entries switched).
So what sort of comparison do we want? Well, these are vectorspaces, so how about asking that they have the same dimension?
This turns out to be just the right thing to ask, since it is on the one hand a very strong condition, but on the other, we can actually construct such a $W$ (that the induced module satisfies this is known as Frobenius reciprocity).

Note that to recover the overly pessimistic version, we can take $M = W$ in the above.

As a final note, I would like to add a bit of notation. If we denote restriction from $G$ to $H$ by $\operatorname{res}_H^G$ and induction from $H$ to $G$ by $\operatorname{ind}_H^G$ then the above condition becomes $\operatorname{Hom}_H(V,\operatorname{res}_H^G M)\cong\operatorname{Hom}_G(\operatorname{ind}_H^G V,M)$, or in other word that induction is left adjoint to restriction (or right adjoint if we switch the entries), at least when everything behaves nicely as functors, which indeed it tends to do.

Also worth noting is that the above gives two possibilities for what we might require of our "induction". A "slightly more optimistic" version could be to require both to hold. And indeed, for finite groups over $\mathbb{C}$, this is what we get, but in general the two need not give the same (in fact, we might not always be sure they both exist). For example, when dealing with algebraic groups, we will usually be more interested in the version mentioned in paranthesis, since this is somewhat better behaved.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.