nLab
induced representation

Idea

Given a group $G$ with subgroup $H\hookrightarrow G$ and a representation of $H$, there is canonically induced a representation of $G$: the induced representation.

Explanation

We start with a Lie group $G$ acting smoothly and transitively on a smooth manifold $M$. The stabilizer of your favorite point $x\in M$ will be a Lie subgroup $H\subseteq G$, and we have

$$M\cong G/H$$M \cong G/H

Starting from this, there’s a process that takes any representation $s$ of $H$ on a vector space $V$ and turns it into a vector bundle $E$ over $M$ — the so-called ‘induced bundle’. Even better, the group $G$ acts on this bundle, and the projection

$$\pi :E\to M$$\pi : E \to M

gets along with the action of $G$:

$$\pi (ge)=g\pi (e).$$\pi(g e) = g \pi(e) .

So, we say that $E$ is a $G$-equivariant vector bundle over $M$.

The ‘process’ described is actually a functor.

There’s a category

$$\mathrm{Rep}(H)$$Rep(H)

of linear representations of $H$, and a category

$$\mathrm{Vect}(M,G)$$Vect(M,G)

of $G$-equivariant vector bundles over $M$. The induced bundle construction gives a functor

$$L:\mathrm{Rep}(H)\to \mathrm{Vect}(M,G)$$L: Rep(H) \to Vect(M,G)

But, if you think about it, you’ll notice there’s also a functor going back the other way:

$$R:\mathrm{Vect}(M,G)\to \mathrm{Rep}(H)$$R: Vect(M,G) \to Rep(H)

If you give me a $G$-equivariant vector bundle $E$ over $M$, I can take its fiber over your favorite point $x$, and I get a vector space — and this becomes a representation of the stabilizer group $H$, thanks to how $G$ acts on $E$.

This functor is simpler than the induced bundle construction!

Whenever we have functors going both ways between two categories, we should suspect that they’re adjoints. The simpler functor often amounts to ‘forgetting’ something. This forgetful functor is usually the right adjoint. It’s partner going the other way, the left adjoint, usually involves ‘constructing’ something instead of ‘forgetting’ something.

And indeed, that’s what’s happening here! Technically, this is to say that

$$\mathrm{hom}(LV,F)\cong \mathrm{hom}(V,RF)$$hom(L V, F) \cong hom(V, R F)

Here $V$ is a representation of $H$ — note abuse of notation in calling it $V$, which is the name for the vector space on which $G$ acts, instead of the more pedantic full name for a representation, which is something like $s:G\to \mathrm{GL}(V)$.

Similarly, $F$ is a $G$-equivariant vector bundle over $M$ — and this should be something like $\pi :F\to M$, or something even more long-winded that gives a name to how $G$ acts on $F$ and $M$.

$LV$ is the induced bundle corresponding to $V$.

$RF$ is the fiber of $F$ over your favorite point $x$, which becomes a representation of $G$.

And this:

$$\mathrm{hom}(LV,F)\cong \mathrm{hom}(V,RF)$$hom(L V, F) \cong hom(V, R F)

says that $G$-equivariant vector bundle maps from $LV$ to $F$ are in natural 1-1 correspondence with intertwining operators from $V$ to $RF$.

Now, whenever you see any sort of ‘forgetful’ process, you should wonder if it has a left adjoint, a construction which in some loose sense is the ‘reverse’ of forgetting. Why? Because these left adjoints tend to be important.

Endowed with this heuristic, as soon as you see there’s a rather obvious ‘forgetful’ process that takes a $G$-equivariant vector bundle over $M$ and gives a representation of $H$ on the fiber over $x\in M$, you will seek the ‘reverse’ process — and then you’ll rediscover the induced bundle construction!

And why is this so great? Well, there’s also a process that takes any representation of $G$ and restricts it to a representation of $H$:

$$R\prime :\mathrm{Rep}(G)\to \mathrm{Rep}(H)$$R': Rep(G) \to Rep(H)

And this too, has a left adjoint:

$$L\prime :\mathrm{Rep}(H)\to \mathrm{Rep}(G)$$L' : Rep(H) \to Rep(G)

which is called the induced representation trick.

Detailed description

Given a group $G$ with a subgroup $H$, and a representation $s$ of $H$ on a vector space $V$, we define a left action of $H$ on the product $G\times V$ by $h\cdot (g,v)=(g{h}^{-1},s(h)v)$. We write $[(g,v)]$ for the orbit, or equivalence class, that contains $(g,v)$.

We then define $E=(G\times V)/H$ as the set of orbits of that action of $H$, $M=G/H$ as the set of left cosets of $H$, and the projection $\pi :E\to M$ by $\pi ([(g,v)])=gH$, where of course it makes no difference if we re-describe the orbit $[(g,v)]$ as $[(g{h}^{-1},s(h)v]$ for any $h\in H$ because $(g{h}^{-1})H=gH$.

For each $x\in M$, choose $g$ to be any element of $G$ such that $x=gH$. Define $E_x={\pi }^{-1}(x)$, and ${\varphi }_g:V\to E_x$, ${\varphi }_g(v)=[(g,v)]$.

The map ${\varphi }_g$ is onto: for any $[(k,w)]\in E_{(gH)}={\pi }^{-1}(gH)$, we have $k=g{h}_1^{-1}$ for some $h_1\in H$, so $k^{-1}g\in H$, $(k^{-1}g)\cdot (g,s(g^{-1}k)w)=(k,w)$, so ${\varphi }_g(s(g^{-1}k)w)=[(g,s(g^{-1}k)w)]=[(k,w)]$.

The map ${\varphi }_g$ is one-to-one: if ${\varphi }_g(v)={\varphi }_g(w)$, then $[(g,v)]=[(g,w)]$, so for some $h_1\in H$, we have $h_1\cdot (g,v)=(g,w)$, or $(g{h}_1^{-1},s(h_1)v)=(g,w)$; equating the first coordinates requires $h_1=e$, and $s$ is a representation so $s(e)=1_V$, and $v=w$.

Since ${\varphi }_g$ is a bijection between $E_x$ and the vector space $V$, we can make $E_x$ into a vector space by defining $\alpha p+\beta q\equiv {\varphi }_g(\alpha {\varphi }_g^{-1}(p)+\beta {\varphi }_g^{-1}(q))$, for all $\alpha ,\beta \in ℝ,p,q\in E_x$. But is this independent of our choice of $g$? If we chose $gh$ instead of $g$, we’d have ${\varphi }_{gh}(v)=[(gh,v)]=[(g,s(h)v)]={\varphi }_g(s(h)v)$, so ${\varphi }_{gh}={\varphi }_g\circ s(h)$, and ${\varphi }_{gh}^{-1}=s(h^{-1})\circ {\varphi }_g^{-1}$. Then:

${\varphi }_{gh}(\alpha {\varphi }_{gh}^{-1}(p)+\beta {\varphi }_{gh}^{-1}(q))=({\varphi }_g\circ s(h))(\alpha (s(h^{-1})\circ {\varphi }_g^{-1})(p)+\beta (s(h^{-1})\circ {\varphi }_g^{-1})(q))={\varphi }_g(\alpha {\varphi }_g^{-1}(p)+\beta {\varphi }_g^{-1}(q))$

in agreement with our original definition.

We define the action of $G$ on $E$ by $g_1\cdot [(g,v)]=[(g_{1}g,v)]$, or in other words $g_1\cdot {\varphi }_g(v)={\varphi }_{g_{1}g}(v)$. We then have:

$\pi (g_1\cdot [(g,v)])=\pi [(g_{1}g,v)]=(g_{1}g)H=g_1\cdot (gH)=g_1\cdot \pi ([(g,v)])$

That is, $\pi$ is a $G$-morphism. This also means that the action maps fibers to fibers, $g_1:E_{(gH)}\to E_{g_1\cdot (gH)}$. What’s more, the action of $g_1$ restricted to the fiber $E_{(gH)}$ is ${\varphi }_{g_{1}g}\circ {\varphi }_g^{-1}$, passing from $E_{(gH)}\to V\to E_{g_1\cdot (gH)}$, and this is linear simply by virtue of the way we’ve defined the vector space operations on the $E_x$.

We get a representation $r$ of $G$ on the vector space $\Gamma (E)$ of sections of the bundle $E$ by:

$(r(g_1)f)(x)=g_1\cdot f(g_1^{-1}\cdot x)$

Unitarity

Beware! The chain of reasoning in this subsection is not complete, and I’m not confident that it’s entirely correct. I’m posting it half-finished in the hope that many hands will make lighter (and more accurate) work.

In physics, one is often concerned with unitary representations, so we should make sure that a unitary representation of $H$ will induce a unitary representation of $G$.

Let’s say $V$ has an inner product, $⟨\cdot ,\cdot ⟩$, and $s$ is a unitary representation. We can define an inner product on $E_x$ by $⟨⟨p,q⟩⟩\equiv ⟨{\varphi }_g^{-1}(p),{\varphi }_g^{-1}(q)⟩$. This definition is independent of our choice of $g$: if we chose $gh$ instead, we’d have

$$⟨⟨p,q⟩⟩=⟨{\varphi }_{gh}^{-1}(p),{\varphi }_{gh}^{-1}(q)⟩=⟨s(h^{-1})\circ {\varphi }_g^{-1}(p),s(h^{-1})\circ {\varphi }_g^{-1}(q)⟩=⟨{\varphi }_g^{-1}(p),{\varphi }_g^{-1}(q)⟩.$$\lang \lang p, q \rang \rang = \lang \phi_{g h}^{-1}(p), \phi_{g h}^{-1}(q) \rang = \lang s(h^{-1}) \circ \phi_g^{-1}(p), s(h^{-1}) \circ \phi_g^{-1}(q) \rang = \lang \phi_g^{-1}(p), \phi_g^{-1}(q) \rang.

To be really thorough, we should verify that $⟨⟨\cdot ,\cdot ⟩⟩$ is in fact an inner product, but this should follow directly from our definition of the vector space operations on $E_x$.

Now we need to show that the action of any $g_1\in G$ on the fiber $E_{(gH)}$ is unitary:

$$⟨⟨g_1\cdot p,g_1\cdot q⟩⟩=⟨⟨{\varphi }_{g_{1}g}\circ {\varphi }_g^{-1}(p),{\varphi }_{g_{1}g}\circ {\varphi }_g^{-1}(q)⟩⟩=⟨{\varphi }_{g_{1}g}^{-1}\circ {\varphi }_{g_{1}g}\circ {\varphi }_g^{-1}(p),{\varphi }_{g_{1}g}^{-1}\circ {\varphi }_{g_{1}g}\circ {\varphi }_g^{-1}(q)⟩=⟨{\varphi }_g^{-1}(p),{\varphi }_g^{-1}(q)⟩=⟨⟨p,q⟩⟩.$$\lang \lang g_1 \cdot p, g_1 \cdot q \rang \rang = \lang \lang \phi_{g_1 g} \circ \phi_g^{-1}(p), \phi_{g_1 g} \circ \phi_g^{-1}(q) \rang \rang = \lang \phi_{g_1 g}^{-1} \circ \phi_{g_1 g} \circ \phi_g^{-1}(p), \phi_{g_1 g}^{-1} \circ \phi_{g_1 g} \circ \phi_g^{-1}(q) \rang = \lang \phi_g^{-1}(p), \phi_g^{-1}(q) \rang = \lang \lang p, q \rang \rang.

Finally, we need to define an inner product on $\Gamma (E)$, and show that the representation $r$ is unitary. If we had a $G$-invariant measure $\mu$ on $G/H$, we could define the inner product of two sections of $f$ and $f\prime$ of $E$ to be

$$\int ⟨⟨f(x),f\prime (x)⟩⟩d\mu (x).$$\int \lang \lang f(x), f'(x) \rang \rang \; d\mu(x).

We would then have

$$\int ⟨⟨(r(g_1)f)(x),(r(g_1)f\prime )(x)⟩⟩d\mu (x)=\int ⟨⟨g_1\cdot f(g_1^{-1}\cdot x),g_1\cdot f\prime (g_1^{-1}\cdot x)⟩⟩d\mu (x)=\int ⟨⟨f(g_1^{-1}\cdot x),f\prime (g_1^{-1}\cdot x)⟩⟩d\mu (x)$$\int \lang \lang (r(g_1)f)(x), (r(g_1)f')(x) \rang \rang \; d\mu(x) = \int \lang \lang g_1 \cdot f(g_1^{-1} \cdot x), g_1 \cdot f'(g_1^{-1} \cdot x) \rang \rang \; d\mu(x) = \int \lang \lang f(g_1^{-1} \cdot x), f'(g_1^{-1} \cdot x) \rang \rang \; d\mu(x)

(because $g_1$ acts unitarily on each fiber)

$$=\int ⟨⟨f(x),f\prime (x)⟩⟩d\mu (g_1\cdot x)$$= \int \lang \lang f(x), f'(x) \rang \rang \; d\mu(g_1 \cdot x)

(because $G$ acts transitively on $G/H$)

$$=\int ⟨⟨f(x),f\prime (x)⟩⟩d\mu (x)$$= \int \lang \lang f(x), f'(x) \rang \rang \; d\mu(x)

(because $\mu$ is $G$-invariant). This shows that $r$ is unitary.

But where do we get a $G$-invariant measure on $G/H$?

Adjoint of induced bundle construction

The induced bundle construction described above is a functor that takes representations of the stabilizer subgroup $H$ to $G$-equivariant vector bundles over $M$:

$$L:\mathrm{Rep}(H)\to \mathrm{Vect}(M,G)$$L: Rep(H) \to Vect(M,G)

There is a related functor going the other way:

$$R:\mathrm{Vect}(M,G)\to \mathrm{Rep}(H)$$R: Vect(M,G) \to Rep(H)

which restricts the action of $G$ on the whole bundle to the action of the stabilizer subgroup $H$ on the fiber over the chosen point $x$. We now wish to show that $L$ and $R$ are adjoint functors.

In the diagram above, on the top left we have a generic $G$-equivariant vector bundle over $M$, $F\in \mathrm{Vect}(M,G)$, with projection ${\pi }_1:F\to M$, and a chosen point $x\in M$ whose stabilizer subgroup is $H$. The functor $R$ maps $F$ to a representation of $H$ on the fiber over $x$, ${\pi }_1^{-1}(x)$, shown on the top right.

On the bottom right, we have a generic representation of $H$ on a vector space $V$. The morphisms of $\mathrm{Rep}(H)$ are intertwiners, so we are interested in intertwiners such as $i:V\to {\pi }_1^{-1}(x)$. The functor $L$, the induced bundle construction, maps a generic representation of $H$ to a $G$-equivariant vector bundle $(G\times V)/H$, shown on the bottom left. This bundle has a projection ${\pi }_2:(G\times V)/H\to G/H$, ${\pi }_2([(g,v)])=gH$. Since $M\cong G/H$, this bundle is in $\mathrm{Vect}(M,G)$. And we are interested in the morphisms of $\mathrm{Vect}(M,G)$, such as $(f,m)$ where $f:L(V)\to F$ and $m:G/H\to M$.

In fact, we need to work with a subcategory of $\mathrm{Vect}(M,G)$ in which all morphisms preserve the point $x\in M$. When we deal with bundles over $G/H\cong M$, we will use the obvious bijection $gH\to g\cdot x$, and accordingly restrict ourselves to vector bundle morphisms that map $x$ to the coset $eH$ or vice versa.

We are assuming that $G$ acts transitively on $M$, so given any $y\in M$ there exists at least one element of $G$, say $k(y)$, such that $k(y)\cdot x=y$. We will now assume that some definite function $k:M\to G$ has been chosen with this property, and for convenience we will further assume that $k(x)=e$, the identity element in $G$. The group element $k(y)$ gives us a specific way to use the action of $G$ on $M$ to get from our chosen point $x$ to some other point $y$ — and equally, to use the action of $G$ on the whole bundle $F$ to get from the fiber over $x$ to the fiber over $y$.

Now, to show that $L$ and $R$ are adjoint functors, we need to construct a bijection between the intertwiners $i:V\to {\pi }_1^{-1}(x)$ and the $G$-equivariant vector bundle morphisms $(f,m)$, where $f:L(V)\to F$ and $m:G/H\to M$.

Given an intertwiner $i:V\to {\pi }_1^{-1}(x)$, we start by defining $m:G/H\to M$ by:

$$m(gH)=g\cdot x$$m(g H)=g\cdot x

which is independent of $i$, and is just the obvious bijection between $G/H$ and $M$. Next, we define $f:L(V)\to F$ by:

$$f([(g,v)])=g\cdot i(v)$$f([(g,v)]) = g\cdot i(v)

In other words, given the equivalence class $[(g,v)]$ we use the intertwiner $i$ to take $v\in V$ to ${\pi }_1^{-1}(x)$, and then the action of $G$ on $F$ to take the result to the fiber ${\pi }_1^{-1}(g\cdot x)$. This satisfies the compatibility condition on the projections:

$${\pi }_1(f([(g,v)]))=g\cdot x=m(gH)=m({\pi }_2([(g,v)]))$$\pi_1(f([(g,v)])) = g\cdot x = m(g H) = m(\pi_2([(g,v)]))

We also need to check that $f$ commutes with the actions of $G$ on the respective bundles:

$$f(g_1\cdot [(g,v)])=f([(g_{1}g,v)])=(g_{1}g)\cdot i(v)=g_1\cdot f([(g,v)])$$f(g_1\cdot [(g,v)]) = f([(g_1 g,v)]) = (g_1 g)\cdot i(v) = g_1\cdot f([(g,v)])

Next, given a $G$-equivariant vector bundle morphism $(f,m)$, where $f:L(V)\to F$ and $m:G/H\to M$ with $m(eH)=x$, we define an intertwiner $i:V\to {\pi }_1^{-1}(x)$ by:

$$i(v)=f([(e,v)])$$i(v)=f([(e,v)])

We know $i$ will map to ${\pi }_1^{-1}(x)$ because $f$ must map $[(e,v)]$ to a point in the fiber over $m({\pi }_2([(e,v)]))=m(eH)=x$.

We check that this is an intertwiner for the representations of $H$ on the respective vector spaces:

$$i(s(h)v)=f([(e,s(h)v)])=f([(h,v)])=f(h\cdot [(e,v)])=h\cdot i(v)$$i(s(h)v)=f([(e,s(h)v)])=f([(h,v)])=f(h\cdot[(e,v)])=h\cdot i(v)

We can also demonstrate a bijection between intertwiners and $G$-equivariant vector bundle morphisms in the other direction: intertwiners $i^{*}:{\pi }_1^{-1}(x)\to V$ and vector bundle morphisms $(f^{*},m^{*})$, where $f^{*}:F\to L(V)$ and $m^{*}:M\to G/H$.

Given an intertwiner $i^{*}:{\pi }_1^{-1}(x)\to V$, we define $m^{*}:M\to G/H$ as:

$$m^{*}(y)=k(y)H$$m^*(y) = k(y) H

We define the map $f^{*}:F\to L(V)$ by:

$$f^{*}(w)=[(k({\pi }_1(w)),i^{*}(k({\pi }_1(w){)}^{-1}\cdot w))]$$f^*(w) = [(k(\pi_1(w)), i^*(k(\pi_1(w))^{-1}\cdot w) )]

for each $w\in F$. Because $k({\pi }_1(w))\cdot x={\pi }_1(w)$, $k({\pi }_1(w){)}^{-1}$ will map the entire fiber to which $w$ belongs to ${\pi }_1^{-1}(x)$, the domain of the intertwiner $i^{*}$. And we have:

$${\pi }_2(f^{*}(w))=k({\pi }_1(w))H=m^{*}({\pi }_1(w))$$\pi_2(f^*(w)) = k(\pi_1(w)) H = m^*(\pi_1(w))

The map $f^{*}$ is a linear map between the fibers ${\pi }_1^{-1}(y)$ and ${\pi }_2^{-1}(m^{*}(y))$, because, along with the linearity of $i^{*}$, the vector space structure on the fibers of $L(V)$ is defined so all maps of the form $v\to [(g,v)]$ are linear. So, $m^{*}$ and $f^{*}$ together give us a vector bundle morphism from $F$ to $L(V)$.

In order to be a morphism in the category of $G$-equivariant vector bundles, $f^{*}$ should also commute with the action of $G$. We have:

$$f^{*}(g\cdot w)=[(k({\pi }_1(g\cdot w)),i^{*}(k({\pi }_1(g\cdot w){)}^{-1}g\cdot w))]=[(k(g\cdot {\pi }_1(w)),i^{*}(k(g\cdot {\pi }_1(w){)}^{-1}g\cdot w))]$$f^*(g\cdot w) = [(k(\pi_1(g\cdot w)), i^*(k(\pi_1(g\cdot w))^{-1} g\cdot w) )] = [(k(g\cdot \pi_1(w)), i^*(k(g\cdot \pi_1(w))^{-1} g\cdot w) )]

Let’s abbreviate ${\pi }_1(w)$ as $y$ and define $h=k(g\cdot y{)}^{-1}gk(y)$, which takes $x$ to $x$ and so must lie in $H$. Then we have:

$$f^{*}(g\cdot w)=[(k(g\cdot y),i^{*}(hk(y{)}^{-1}\cdot w))]=[(k(g\cdot y),s(h)i^{*}(k(y{)}^{-1}\cdot w))]=[(k(g\cdot y)h,i^{*}(k(y{)}^{-1}\cdot w))]$$f^*(g\cdot w) = [(k(g\cdot y), i^*(h k(y)^{-1}\cdot w) )] = [(k(g\cdot y), s(h) i^*(k(y)^{-1}\cdot w) )] = [(k(g\cdot y) h, i^*(k(y)^{-1}\cdot w) )]

$$=[(gk(y),i^{*}(k(y{)}^{-1}\cdot w))]=g\cdot [(k(y),i^{*}(k(y{)}^{-1}\cdot w))]=g\cdot f^{*}(w)$$= [(g k(y) , i^*(k(y)^{-1}\cdot w) )] = g\cdot [(k(y) , i^*(k(y)^{-1}\cdot w) )] = g\cdot f^*(w)

Suppose we’re given a $G$-invariant vector bundle morphism $(f^{*},m^{*})$, where $f^{*}:F\to L(V)$ and $m^{*}:M\to G/H$, with $m^{*}(x)=eH$.

We make use of the linear bijection ${\varphi }_e:V\to E_{eH}$, defined by ${\varphi }_e(v)=[(e,v)]$. We introduced these linear bijections ${\varphi }_g$ when initially describing the induced bundle construction. We define $i^{*}:{\pi }_1^{-1}(x)\to V$ by:

$$i^{*}(w)={\varphi }_e^{-1}(f^{*}(w))$$i^*(w) = \phi_e^{-1}(f^*(w))

We check that this is an intertwiner between the relevant representations of $H$:

$$i^{*}(h\cdot w)={\varphi }_e^{-1}(f^{*}(h\cdot w))={\varphi }_e^{-1}(h\cdot f^{*}(w))$$i^*(h\cdot w) = \phi_e^{-1}(f^*(h\cdot w))= \phi_e^{-1}(h\cdot f^*(w))

Suppose $f^{*}(w)=[(e,v)]$ for some $v\in V$. Then $i^{*}(w)=v$, and :

$$h\cdot f^{*}(w)=[(h,v)]=[(e,s(h)v)]$$h\cdot f^*(w) = [(h,v)] = [(e,s(h)v)]

$$i^{*}(h\cdot w)={\varphi }_e^{-1}([(e,s(h)v)])=s(h)v=s(h)i^{*}(w)$$i^*(h\cdot w) = \phi_e^{-1}([(e,s(h)v)]) = s(h)v = s(h) i^*(w)

Related issues

Bill Lawvere noted a structural similarity between induced representations and quantification. See blog discussion.

If the modules over a group are considered as comodules over the function Hopf algebra over the group, then one can instead consider the induction for comodules. See cotensor product.

References

This blog entry contains related discussion:

• John Baez, Unitary Representations of the Poincare Group

The above text is taken from these comments

• Greg Egan Induced representations

• John Baez Reply