Let $G$ be a Lie group which is a finite extension of an open normal subgroup $N$: $$ 1\to N\to G\to F\to 1 $$ (so $N$ and $G$ are Lie, and $F$ is finite; but I think, this is not very important, we can assume that $G$ is finite, that will be sufficient).

  1. Let me first remind the construction of the induced representation. Let $\pi:N\to {\mathcal B}(X)$ be a representation of $N$ in a Hilbert space $X$ (we can think that $\pi$ is norm-continuous, this actually is not very important). If we denote by $\varphi:G\to F$ the natural quotient map from $G$ to $F$, and choose a coretraction (not necessarily a homomorphism of groups) $\sigma:F\to G$, $$ \varphi(\sigma(t))=t,\qquad t\in F, $$ then the induced representation can be defined as the map $\pi':G\to {\mathcal B}(L_2(F,X))$, $$ \pi'(a)(\xi)(t)=\pi\Big(\sigma(t)\cdot a\cdot \sigma\big(t\cdot\varphi(a)\big)^{-1}\Big)\Big(\xi\big(t\cdot\varphi(a)\big)\Big),\qquad a\in G,\ \xi\in L_2(F,X),\ t\in F. $$

  2. Consider then a new kind of representation of $N$, namely a representation in the cartesian square ${\mathcal B}(X)^2$ with the following multiplication: $$ (A_0,A_1)\cdot(B_0,B_1)=(A_0\cdot B_0, A_0\cdot B_1+A_1\cdot B_0),\qquad A_0, A_1, B_0, B_1\in {\mathcal B}(X) $$ (${\mathcal B}(X)^2$ can be perceived as the algebra of polynomials of degree 1 with coefficients in the algebra ${\mathcal B}(X)$). Let $\pi:N\to {\mathcal B}(X)^2$ be a representation in this algebra. Then it is easy to verify that the same formula, $$ \pi'(a)_k(\xi)(t)=\pi\Big(\sigma(t)\cdot a\cdot \sigma\big(t\cdot\varphi(a)\big)^{-1}\Big)_k\Big(\xi\big(t\cdot\varphi(a)\big)\Big),\qquad a\in G,\ \xi\in L_2(F,X),\ t\in F,\ k=0,1. $$ defines a representation $\pi':G\to {\mathcal B}(L_2(F,X))^2$ that also can be called "induced" (here ${\mathcal B}(L_2(F,X))^2$ is again the algebra of polynomials of degree 1 with coefficients in ${\mathcal B}(L_2(F,X))$).

  3. Denote by $\pi_0$, $\pi_1$, $\pi'_0$, $\pi'_1$ the components of the mappings $\pi:N\to {\mathcal B}(X)^2$ and $\pi':G\to {\mathcal B}(L_2(F,X))^2$, i.e. we treat $\pi$ as two mappings, $\pi_k:N\to {\mathcal B}(X)$, and $\pi'$ also as two mappings, $\pi'_k:G\to {\mathcal B}(L_2(F,X))$, $k=0,1$, and the following identities hold: $$ \pi_0(a\cdot b)=\pi_0(a)\cdot\pi_0(b),\quad \pi_1(a\cdot b)=\pi_0(a)\cdot\pi_1(b)+\pi_1(a)\cdot\pi_0(b),\qquad a,b\in N $$ $$ \pi'_0(a\cdot b)=\pi'_0(a)\cdot\pi'_0(b),\quad \pi'_1(a\cdot b)=\pi'_0(a)\cdot\pi'_1(b)+\pi'_1(a)\cdot\pi'_0(b),\qquad a,b\in G. $$


if we require an additional condition on $\pi$, the commutativity between all operators $\pi_0(a)$ and $\pi_1(b)$, $$ \pi_0(a)\cdot\pi_1(b)=\pi_1(b)\cdot\pi_0(a), \qquad a,b\in N, $$ then this condition will not automatically hold for $\pi'$: we only obtain the identity $$ \pi'_0(a)\cdot\pi'_1(b)=R_{\varphi(a)}\cdot\pi'_1(b)\cdot R_{\varphi(a\cdot b)^{-1}}\cdot\pi'_0(a)\cdot R_{\varphi(b)}, \qquad a,b\in G, $$ where $R_s$ are shifts by elements of $F$: $$ R_s\xi(t)=\xi(t\cdot s),\qquad t,s\in F,\ \xi\in L_2(F,X). $$


is it possible to re-define the induced representation $\pi'$ in such a way that the condition $$ \pi_0(a)\cdot\pi_1(b)=\pi_1(b)\cdot\pi_0(a), \qquad a,b\in N, $$ implies the condition $$ \pi'_0(a)\cdot\pi'_1(b)=\pi'_1(b)\cdot\pi'_0(a), \qquad a,b\in G $$ ?

It seems to me that if we change something in this construction, for example, if we replace $L_2(F,X)$ by some other space, maybe a subspace in $L_2(F,X)^{\otimes 2}$ (the Hilbert tensor square), this will lead to what I need. But this escapes me, I can't find a proper construction.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.