Consider a group $\,G\,$, a Hilbert space $\,{\mathbb{V}}\,$ with a dot product $\,\langle~,~\rangle\,$, and a space $\,{\cal{L}}^G\,$ of functions $\varphi$ on this group: $$ {\cal{L}}^G\;=\;\left\{~\varphi~\Big{|}~~~\varphi:\,~G\longrightarrow{\mathbb{V}}\,\right\}~~. $$ Let $\,D\,$ be a representation of a subgroup $\,K\leq G\,$ in the said Hilbert space: $$ D~:\quad K~\longrightarrow~GL({\mathbb{V}})\;\;.\qquad\qquad\qquad (1) $$

On a group element $\,g\in G\,$, a function $\,\varphi\in{\cal{L}}^G\,$ assumes the value $\,\varphi(g)\in{\mathbb{V}}\,$. Since this value is a vector in the Hilbert space, we can act on it with some $\,D(k)\,$, $\,k\in K\;$: $$ k\in K~:\quad \varphi(g)\;\mapsto\;D(k)\,\varphi(g)\;\;,\qquad\varphi(g)\in{\mathbb{V}}\;\;. $$ For a fixed $\,g\,$, this is a mapping of one Hilbert-space vector to another.

However, the set of all these mappings, for all $\,g\in G\,$, generates a mapping of a function to a function: $$ k\in K~:\quad \varphi\;\mapsto\;D(k)\,\varphi\;\;,\quad\varphi\in{\cal{L}}^G\;\;. $$ $$ $$ QUESTION 1:

May I write the latter as $$ D~:\quad K~\longrightarrow~GL({\cal{L}}^G)\;\;,\qquad\qquad\qquad (2) $$ using the same notation $\,D\,$ as was used in equation (1)?

REMARK: $~$ While interconnected in an obvious way, the two $\,D$'s are two different representations, because they are acting in different spaces: one in $\,{\mathbb{V}}\,$, another in $\,{\cal{L}}^G\,$. Hence the above question. $$ $$ QUESTION 2:

Would it be possible to say that these two $\,D$'s are, in some sense, equivalent? $$ $$ QUESTION 3:

The induced representation $\,\operatorname{Ind}_K^GD\,$ is implemented with the left translations $$ U_g\varphi(x)=\varphi({g^{-1}}x)~~,\qquad g,\,x\in G\;, $$ acting in the subspace $\,\Gamma\in{\cal{L}}^G$ of the Mackey functions: $$ \Gamma\;=\;\left\{~\varphi~\Big{|}~~~\varphi:\,~G\longrightarrow{\mathbb{V}}\;,\quad \varphi(xk)=D^{-1}(k)\varphi(x)\,\right\}~~. $$ Which of the two $\,D$'s is actually being induced here? -- the $\,D\,$ given by (1) or the $\,D\,$ given by (2)?


1 Answer 1

  1. It's probably not good to use the same symbol for the action of $K$ on $V$-valued functions on $G$ as for the action of $K$ on $V$. It would be an understandable abuse of notation, but also with some risks. Literally, the action of $K$ on some space of $V$-valued functions on $G$ would be a direct sum or Hilbert direct integral of copies of $V$.

NB: for precision, you'd want to say what kind of $V$-valued functions on $G$. Some measurability conditions?

  1. The two representations are not "equivalent" in an immediate sense (though perhaps in a "Morita equivalence" sense...). Also, what is "equivalent" intended to mean?

  2. In any case, contemporary usage would say that the repn of $K$ on $V$ is what is being induced, whatever qualifications or modifiers are put on these function spaces.

EDIT: As in a comment, it is often intended that functions be "square-integrable"... so certainly measurable. When the $K$ here is compact, square-integrability on $G/K$ (or $K\backslash G$...) is equivalent to square-integrability on $G$. Also, for compact $K$, invoking uniform boundedness theorem, a $K$ repn on a Hilbert space can be assumed unitary (without changing the topology on the Hilbert space). So things are as harmonious as they could possibly be.

EDIT2: As in the comment, applying restriction from $G$ to $K$ to the induced repn does not return to the original repn of $K$ at all. In fact, (because the action of $K$ is on the opposite side of $G$, whatever one's convention), that restriction is not even any kind of sum/integral of copies of the original repn of $K$, for non-abelian $G$.

  • $\begingroup$ I have square-integrable functions in mind, if that matters. $\endgroup$ May 24, 2020 at 23:06
  • $\begingroup$ When saying "direct sum of copies of V", you probably mean that the number of these copies is equal to the number of elements in G, correct? $\endgroup$ May 24, 2020 at 23:07
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    $\begingroup$ Literally, the copies of $V$ would be in bijection with $G/K$ (ignoring measurability or other considerations...). But/and I don't think this issue is what you will care about. $\endgroup$ May 24, 2020 at 23:11
  • $\begingroup$ So, if I am inducing D in the sense of eqn (1), then I have no chance to return back to this D after performing $\operatorname{Res}_K^G\,\operatorname{Ind}_K^G D$ -- simply because this restriction will render me a representation in ${\cal{L}}^G$, not in ${\mathbb{V}}$, correct? I am referring to my previous question math.stackexchange.com/questions/3669004/… which, unfortunately, has never been answered. $\endgroup$ May 24, 2020 at 23:13
  • $\begingroup$ Regarding the copies. I understand that induction requires the use of |G/K| copies. When we are dealing with matrix representations of finite groups, then ${\mathbb{V}}$ is all we have, so we need |G/K| copies of ${\mathbb{V}}$. Thus much is clear. On this occasion, however, we are dealing not with representations by matrices, but with those on functions. So it is important for me to know that we are talking about copies of the Hilbert space, not about copies of the space of functions. This is nontrivial, because the induced rep is implemented with left-regular operators $U_g$ on functions $\endgroup$ May 24, 2020 at 23:20

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