Induced Representation #2

이번 챕터에서는 Induced Representation을 다룬다. 교수님께서 말씀하시길, 표현론에서는 induced representation이 가장 중요하다고 한다. 이후에 배울 chapter 7은 induced representation을 아래 Example 5과 같이 tensor notation을 이용해서 정의해서 Frobenius reciprocity formula, Mackey's criterion 등을 공부하게 된다. 정확히는 모르지만, 물리에서는 Wigner$($위그너$)$가 전개한 이론에서 inuduced representation이 사용되었다고 하니 잘 공부해두면 곧 도움이 되지 않을까 싶다. 

Induced representation, canonical decomposition, tensor product 같은 걸 공부하다 보니 수학과 과목은 참 어떤 큰 걸 쪼갤 수 없는 걸로 쪼개고, 작은 걸 이어 붙여서 기술하기 조차 힘든 큰 걸 만들어 내는 걸 좋아하는 것 같다. 어떻게 보면 데모크리토스의 원자론을 보는 것 같기도... 

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3. Induced Representations

Recall Undergraduate Abstract Algebra:

• $s' \equiv s$ $($mod $H$$)$ if $s^{-1}s' \in H$
• $ (G:H) = |G|/|H| = g/h $

 

Let $H \le G$,

let $ \rho: G \rightarrow GL(V)$ be a linear representation of $G$,

let $\rho_{H}$ be its restriction to $H$,

and let $W$ be a subrepresentation of $\rho_{H} \Rightarrow W$ is $H$-stable.

$^\forall g \in sH$ for any $s \in G,$ $g=sh$ for any $h \in H$

$\Rightarrow g \cdot W = (sh) \cdot W = s \cdot (hW) = s \cdot W$.

$\Rightarrow ^\forall g_{1}, g_{2} \in sH$, then $g_{1} \cdot W = g_{2} \cdot W = s \cdot W.$

$\Rightarrow ^\forall \sigma \in G/H, ^\forall g \in \sigma$, then $\sigma H = gH.$

$\Rightarrow$ We can define a subspace $W_{\sigma} = \rho_{s}W$ of $V$ for any $s \in \sigma$.

$\Rightarrow \sum_{\sigma \in G/H} W_{\sigma}$ is a subrepresentation of $V$$($It is not direct sum$)$.

 

Definition. The representation $\rho$ of $G$ in $V$ is induced by the representation $\theta$ of $H$ in $W$ if $V$ is equal to the sum of the $W_{\sigma}$ and if this sum is direct, in short, $ V = \underset{\sigma \in G/H}{\bigoplus} W_{\sigma}.$

 

Induced representation $V$ denote as $Ind^{G}_{H} (W) = Ind(W).$

Both statements are the equivalent definitions of induced representation as above:

$($i$)$ $^\forall x \in V$ can be written uniquely $\sum_{\sigma \in G/H} x_{\sigma},$ with $x_{\sigma} \in W_{\sigma}$ for each $\sigma$. 

$($ii$)$ If $R$ is a system of representatives of $G/H$, the vector space $V$ is the direct sum of the $\rho_{r} W$, with $r \in R.$

$\Rightarrow$ dim$(V) = \underset{r \in R}{\sum}$ dim$ ( \rho_{r} W) = (G:H) \cdot$ dim$(W).$

 

Example.

1. Let $V$ be the regular representation of $G$, and let $W$ be the regular representation of $H$. Then the regular representation of $G$ in $V$ is induced by the regular representation of $H$ in $W$.

$ ^\forall s \in G, s \in \sigma_{i} = g_{i} H$, then $s = g_{i} h$ where $g_{i} \in R$. 
$ \Rightarrow \sigma_{i} \cdot W = s \cdot W = (g_{i} h) \cdot W = g_{i} \cdot (hW) = g_{i} \cdot W.$
$ \Rightarrow ^\forall w \in W, \sigma_{i} \cdot w = g_{i} \cdot w = g_{i} \cdot \sum_{t \in H} a_{t} e_{t} = \sum_{t \in H} a_{t} (g_{i} \cdot e_{t}) = \sum_{t \in H} a_{t}e_{g_{i} t}$
$ \qquad \qquad \qquad \ \ \quad = \sum_{t \in g_{i}H} b_{t}e_{t} = \sum_{t \in \sigma_{i}} b_{t} e_{t}.$
On the other hand,
$ ^\forall v \in V, v = \sum_{t \in G} a_{t}e_{t} = \sum_{\sigma_{i} \in G/H} \left( \sum_{t \in \sigma_{i}} b_{t}e_{t} \right). $
$ \Rightarrow$ $V$ is equal to the sum of the $W_{\sigma}$ for each $ \sigma \in G/H$.
It is obvious that $W_{\sigma_{i}} \cap W_{\sigma_{j}} =\{0\}$ for $i \neq j$.
$ \therefore$ The regular representation of $G$ in $V$ is induced by the regular representation of $H$ in $W$. $_\Box$

 

2. Let $V$ be the permutation representation of $G$ associated with $G/H$, and let $W$ be the trivial representation of $H$. Then The permutation representation of $G$ associated with $G/H$ is induced by the trivial representation of $H$.

$ \Rightarrow V$ has basis $ \{ e_{\sigma} | \sigma \in G/H \}, W = \mathbb{C} e_{H}$ where $W$ is $H$-invariant.
$ ^\forall v \in V, v = \sum_{\sigma} c_{\sigma} e_{\sigma} = \sum_{\sigma} c_{\sigma} \left( \sigma \cdot e_{H} \right) = \sum_{\sigma} \sigma \cdot \left( c_{\sigma} e_{H} \right).$
$ \Rightarrow$ $V$ is equal to the sum of the $W_{\sigma}$ for each $ \sigma \in G/H$.
It is obvious that $W_{\sigma_{i}} \cap W_{\sigma_{j}} = \{0\}$ for $i \neq j$.
$ \therefore$ The permutation representation of $G$ associated with $G/H$ is induced by the trivial representation of $H$. $_\Box$

 

Let's consider the extreme case of $H=\{1\}$. Then the permutation representation becomes the regular representation of $G$. The unit representation of $H$ is equal to the regular representation of $H=\{1\}$. It is a re-confirmation of Example 1.

 

3. $\rho_{1}$ in $V_{1}$ is induced by $\theta_{1}$ in $W_{1}$, and $\rho_{2}$ in $V_{2}$ is induced by $\theta_{2}$ in $W_{2}$. Then $\rho_{1} \oplus \rho_{2}$ in $V_{1} \oplus V_{2}$ is induced by $\theta_{1} \oplus \theta_{2}$ in $W_{1} \oplus W_{2}$.

$ \underset{\sigma \in G/H}{\bigoplus} \left( W_{1} \oplus W_{2} \right)_{\sigma} = \underset{\sigma \in G/H}{\bigoplus} \left( \underset{i}{\bigoplus} (W_{i})_{\sigma} \right) = \underset{i}{\bigoplus} \left( \underset{\sigma \in G/H}{\bigoplus} (W_{i})_{\sigma} \right) = \underset{i}{\bigoplus} V_{i} = V_{1} \oplus V_{2}. _\Box$

 

4. $(V,\rho)$ is induced by $(W,\theta)$, and $W_{1}$ is a stable subspace of $W$. The subspace $V_{1} = \sum_{r \in R} \rho_{r} W_{1}$ of $V$ is stable under $G$, and the representation of $G$ in $V_{1}$ is induced by the representation of $H$ in $W_{1}$.

Obvious! $_\Box$

 

5. If $\rho$ is induced by $\theta$, and if $\rho'$ is a representation of $G$, and if $\rho'_{H}$ is the restriction of $\rho'$ to $H$ $(Res_{H} (\rho'))$, then $\rho \otimes \rho'$ is induced by $\theta \otimes \rho'_{H}$.

Detail Proof is shown later in Chapter 7 with tensor product notation.

 

Lemma 1. Suppose that $(V,\rho)$ is induced by $(W,\theta)$. Let $\rho':G \rightarrow GL(V')$ be a linear representation of $G$, and let $f:W \rightarrow V'$ be a linear map such that $f(\theta_{t}w) = \rho'_{t}f(w)$ for all $t \in H$ and $w \in W$. Then there exists a unique linear map $F:V \rightarrow V'$ which extends $f$ and satisfies $F \circ \rho_{s} = \rho'_{s} \circ F$ for all $s \in G$.

$($i$)$ Uniqueness
If $ x \in \rho_{s}W,$ then $\rho_{s}^{-1}x \in W$.
$ F(x) = F(\rho_{s} \rho_{s}^{-1} x) = \rho'_{s} F(\rho_{s}^{-1} x) = \rho'_{s} f(\rho_{s}^{-1} x)$ by $\rho_{s}^{-1} x \in W$ 
$\Rightarrow$ If there exist $F_{1}$ and $F_{2}$, $F_{1}(x) = F_{2}(x) = \rho'_{s} f(\rho_{s}^{-1} x) \Rightarrow ^{\exists !} F(x).$
$($ii$)$ Existence
$ \forall x \in W_{\sigma}, s \in \sigma, t \in H, \rho'_{st}f(\rho_{st}^{-1}x) = \rho'_{s}f(\rho_{s}^{-1} x).$
$\Rightarrow$ $F(x)$is well-defined function for varying in $\sigma$. 
$ \left( (\rho_{s}^{-1}) \circ F \circ \rho_{s} \right) (x) = f(x) = F(x)$ easily is obtained for $x \in W_{\sigma}. _\Box$

 

Theorem 11. Let $(W,\rho)$ be a linear representation of $H$. There exists a linear representation $(V,\rho)$ of $G$ which is induced by $(W,\theta)$, and it is unique up to isomorphism.

$($i$)$ Existence of the induced representation $\nu$
In view of preceding Example 3, we may assume $\theta$ is irreducible. In this case, $\theta$ is isomorphic to a subrepresenation of the regular representation of $H$. which can be induced to the regular representation of G $($ Which indicates the reugular representation of $H$ not a subrepresentation of the regular representation of $H$.$)$ Applying Example 4, We conclude that $\theta$ itself can be induced.
$($ii$)$ Uniqueness of the induced representation $\rho$ up to isomorphism
Let $(V,\rho)$ and $(V',\rho')$ be two representations induced by $(W,\theta)$. Applying Lemma 1 to $f:W \rightarrow V'$, we see that $^\exists$ a linear map $F:V \rightarrow V'$ $s.t.$ $F$ is the identity on $W$ and satisfies $F \circ \rho_{s} = \rho_{s}' \circ F$ for all $s \in G$.
By second property, $ \left( F \circ \rho_{s} \right)(W) = \left( \rho'_{s} \circ F \right)(W) $.
$\Rightarrow F \left( \rho_{s} (W) \right) = \rho'_{s} \left( F(W) \right) \Rightarrow F(\rho_{s}W) = \rho'_{s}W$ $\Rightarrow F \left( \bigoplus_{s \in R} \rho_{s}W \right) = \bigoplus_{s \in R} \rho'_{s}W = V' \Rightarrow$ surjective.
On the other hand, $ ^\forall w \in W$, let $F(\rho_{s} w) = \rho'_{s}w = 0.$ By $ ^\exists (\rho'_{s})^{-1}, w = 0$. It implies that $\rho_{s} w = 0$. And, $Ker(F) = \{0\} \Rightarrow$ injective.
$\therefore F_{1} \cong F_{2}. _\Box$  

 

Theorem 12. For each $u \in G$, we have:

$$ \chi_{\rho} (u) = \underset{r^{-1}ur \in H}{\underset{r \in R}{\sum}} \chi_{\theta} (r^{-1}ur) = {1 \over h} \underset{s^{-1}us \in H}{\underset{s \in G}{\sum}} \chi_{\theta} (s^{-1}us). $$

$ ur=r_{u}t$ for $r_{u} \in R$ and $t \in H$
$\Rightarrow$ $\rho_{u}$ sends $\rho_{r}W$ into $\rho_{r_{u}}W$.
$r \neq r_{u}$ gives zero diagonal terms. The others gives the trace of $\rho_{u}$ on the $\rho_{r}W$. Then:
$$ \chi_{\rho}(u) = \underset{r \in R_{u}}{\sum} Tr_{\rho_{r}W}(\rho_{u,r}),$$
where $R_{u}$ dentes the set of $r \in R$ such that $r_{u}=r$ and $\rho_{u,r}$ is the restriction of $\rho_{u}$ to $\rho_{r}W$.
$\Rightarrow (\rho_{u,r} \circ \rho_{r})(W) = u \cdot (r \cdot W) = (ur) \cdot W = (rt) \cdot W = r \cdot (tW) = (\rho_{r} \circ \theta_{t})(W).$
$\Rightarrow \rho_{r} \circ \theta_{t} = \rho_{u,r} \circ \rho_{r}$ with $t = r^{-1}ur \in H.$
$\Rightarrow Tr(\rho_{u,r}) = Tr(\theta_{t})$, that is, $ \chi_{\theta} (t) = \chi_{\theta}(r^{-1}ur) $.
$\Rightarrow \chi_{\rho}(u) = \underset{r \in R_{u}}{\sum} \chi_{\theta}(r^{-1}ur).$
Because of $|G|/|R|=h$, second formula is obtained. $_\Box$ 

 

P.S.

If $G= g_{1}H \cup \cdots \cup g_{k}H$, then $Ind(W) = g_{1}W \oplus \cdots \oplus g_{k}W$.

$^\forall g \in G, g_{i} \in R, gg_{i} \in G$. By preceding property, $gg_{i}=g_{j}h$ for $j \in R, h \in H$.

$\therefore g$ act as $g: g_{i}W \rightarrow G_{j}W$. In addition, $h$ acts $W$. So, there are two acting on $G$ and $H$ by acting $g$. That is the former acts permutating the $\rho_{r}W$ among themselves.

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Reference.

https://www.youtube.com/watch?v=Y_QGeFWlWrY&list=PLGB0U_OssPo5BAvB6QvWiOLPJhw4_jRZG&index=9