Canonical decomposition

In the mathematical field of representation theory , canonical decomposition is a decomposition of representations into simpler representations.

Completely reducible representations

A representation of a group is a homomorphism of into the automorphism group of a given vector space . The group link in matches the one behind the other run of automorphisms in : . If a -dimensional vector space is over a body , then the representation accordingly consists of invertible -matrices with coefficients . ${\displaystyle \rho }$ ${\displaystyle G}$ ${\displaystyle G}$ ${\displaystyle \operatorname {Aut} (V)}$ ${\displaystyle V}$ ${\displaystyle G}$ ${\displaystyle V}$ ${\displaystyle \rho (gh)=\rho (g)\rho (h)}$ ${\displaystyle V}$ ${\displaystyle n}$ ${\displaystyle K}$ ${\displaystyle n\times n}$ ${\displaystyle K}$

The representation (or the representation space ) is called irreducible if there are only the two trivial -invariant subspaces and of . Is a direct sum of irreducible representations of G, then is called completely reducible. In particular, every irreducible representation is completely reducible. ${\displaystyle \rho }$ ${\displaystyle V}$ ${\displaystyle G}$ ${\displaystyle 0\ (=\{0\})}$ ${\displaystyle V\ (\neq 0)}$ ${\displaystyle V}$ ${\displaystyle \rho }$ ${\displaystyle \rho }$

Every representation of a finite group in a finite-dimensional complex vector space is completely reducible, see Weyl's unitary trick . More generally, the following always applies to a representation of a finite group in a vector space over a field of the characteristic : Let be an -invariant subspace of Then the complement of in exists and is also -invariant. ${\displaystyle G}$ ${\displaystyle V}$ ${\displaystyle 0}$
${\displaystyle W}$ ${\displaystyle G}$ ${\displaystyle V.}$ ${\displaystyle \textstyle W^{0}}$ ${\displaystyle W}$ ${\displaystyle V}$ ${\displaystyle \textstyle W^{0}}$ ${\displaystyle G}$

This result also applies more generally to representations of compact groups :
Every linear representation of compact groups over a field of the characteristic is a direct sum of irreducible representations. In the formulation of the modules this means: If so, the group algebra is semi-simple, ie it is the direct sum of simple algebras. This breakdown is ambiguous. However, the number of partial representations that occur, which are isomorphic to a given irreducible representation, does not depend on the decomposition chosen. ${\displaystyle 0}$
${\displaystyle K[G]}$ ${\displaystyle {\text{char}}(K)=0,}$ ${\displaystyle K[G]}$

The canonical decomposition

In order to get a clear decomposition, one summarizes all isomorphic irreducible partial representations. So the representation space is broken down into the direct sum of its isotypes. This decomposition is clear. It is called the canonical division.

${\displaystyle V=\bigoplus _{k}V_{k}}$,

where each is the sum of copies of an irreducible representation . So you have ${\displaystyle V_{k}}$ ${\displaystyle n_{k}}$ ${\displaystyle W_{k}}$

${\displaystyle V=\bigoplus _{k}W_{k}^{n_{k}}}$.

The summands are called the isotypes of the representation . ${\displaystyle V_{k}}$ ${\displaystyle V}$

properties

Let be the canonical decomposition of a representation . ${\displaystyle V=\bigoplus _{k}V_{k}}$ ${\displaystyle V}$

• Every partial representation of that is too isomorphic is contained in. ${\displaystyle W_{k}}$ ${\displaystyle V}$ ${\displaystyle V_{k}}$
• The canonical decomposition is clear; H. regardless of the original decomposition into irreducible representations.
• The endomorphism algebra is isomorphic to the matrix algebra . ${\displaystyle End^{G}(V_{k})}$ ${\displaystyle Mat(n_{k},\mathbb {C} )}$
• The endomorphism algebra is isomorphic to the direct sum , block diagonal with respect to the canonical decomposition. ${\displaystyle End^{G}(V)}$ ${\displaystyle \bigoplus _{k}Mat(n_{k},\mathbb {C} )}$

Let be canonical decomposition of two representations . Then every -equivariant forms homomorphism ${\displaystyle V=\bigoplus _{k}V_{k},V^{\prime }=\bigoplus V_{k}^{\prime }}$ ${\displaystyle V,V^{\prime }}$ ${\displaystyle G}$

${\displaystyle f\colon V\to V^{\prime }}$

${\displaystyle V_{k}}$up down. ${\displaystyle V_{k}^{\prime }}$

Projection formula

Let be the family of all irreducible representations of a group except for isomorphism. Let Let be a representation of and the set of isotypes of The projection to the canonical decomposition is given by ${\displaystyle (\tau _{j})_{j\in I}}$ ${\displaystyle G}$ ${\displaystyle g={\text{ord}}(G).}$ ${\displaystyle V}$ ${\displaystyle G}$ ${\displaystyle \{V(\tau _{j})|j\in I\}}$ ${\displaystyle V.}$ ${\displaystyle p_{j}\colon V\to V(\tau _{j})}$

${\displaystyle p_{j}={\frac {n_{j}}{g}}\sum _{t\in G}{\overline {\chi _{\tau _{j}}(t)}}\rho (t),}$

where and is the associated character . ${\displaystyle n_{j}={\text{dim}}(\tau _{j})}$ ${\displaystyle \chi _{\tau _{j}}}$ ${\displaystyle \tau _{j}}$

In the following we see how one can determine the isotype for the trivial representation.

For each representation of a group with define general is non- linear. Set Then a linear map, as for all ${\displaystyle (\rho ,V)}$ ${\displaystyle G}$ ${\displaystyle g={\text{ord}}(G)}$ ${\displaystyle V^{G}=\{v\in V\mid \rho (s)v=v\,\,\,\,\forall s\in G\}.}$
${\displaystyle \rho (s)\colon V\to V}$ ${\displaystyle G}$ ${\displaystyle \textstyle P:={\frac {1}{g}}\sum _{s\in G}\rho (s)\in {\text{End}}(V).}$ ${\displaystyle P}$ ${\displaystyle G}$ ${\displaystyle \textstyle \sum _{s\in G}\rho (s)=\sum _{s\in G}\rho (tst^{-1})}$ ${\displaystyle t\in G.}$

Proposition

The figure is a projection from to ${\displaystyle P}$ ${\displaystyle V}$ ${\displaystyle V^{G}.}$

With this proposition we can now explicitly determine the isotype for the trivial partial representation of a given representation.
The number of times the trivial representation occurs in is given by the trace of This follows, since a projection can only have the eigenvalues and and the eigenspace to the eigenvalue is the image of the projection. Since the trace is the sum of the eigenvalues, one thus obtains ${\displaystyle V}$ ${\displaystyle P.}$ ${\displaystyle 0}$ ${\displaystyle 1}$ ${\displaystyle 1}$

${\displaystyle {\text{dim}}(V(1))={\text{dim}}(V^{G})=Tr(P)={\frac {1}{g}}\sum _{s\in G}\chi _{V}(s),}$

where denotes the isotype for the trivial representation and Let be a non-trivial irreducible representation of then the isotype for the trivial representation of the null space. That is, it applies ${\displaystyle V(1)}$ ${\displaystyle g={\text{ord}}(G).}$
${\displaystyle V_{\pi }}$ ${\displaystyle G,}$ ${\displaystyle \pi }$

${\displaystyle P={\frac {1}{g}}\sum _{s\in G}\pi (s)=0.}$

Let be an orthonormal basis of Then: ${\displaystyle e_{1},\dotsc ,e_{n}}$ ${\displaystyle V_{\pi }.}$

${\displaystyle \sum _{s\in G}{\text{Tr}}(\pi (s))=\sum _{s\in G}\sum _{j=1}^{n}\langle \pi (s)e_{j},e_{j}\rangle =\sum _{j=1}^{n}\langle \sum _{s\in G}\pi (s)e_{j},e_{j}\rangle =0.}$

So it is true for a non-trivial irreducible representation ${\displaystyle V:}$

${\displaystyle \sum _{s\in G}\chi _{V}(s)=0.}$

Examples

example 1

Let be the dihedral group of the order with generators for which applies and Let be a linear representation of defined on the generators by: ${\displaystyle G=D_{6}=\{{\text{Id}},\mu ,\mu ^{2},\nu ,\mu \nu ,\mu ^{2}\nu \}}$ ${\displaystyle 6}$ ${\displaystyle \mu ,\nu ,}$ ${\displaystyle {\text{ord}}(\nu )=2,}$ ${\displaystyle {\text{ord}}(\mu )=3}$ ${\displaystyle \nu \mu \nu =\mu ^{2}.}$ ${\displaystyle \rho \colon D_{6}\to {\text{GL}}_{3}(\mathbb {C} )}$ ${\displaystyle D_{6}}$

${\displaystyle \rho (\mu )=\left({\begin{array}{ccc}{\text{cos}}({\frac {2\pi }{3}})&0&-{\text{sin}}({\frac {2\pi }{3}})\\0&1&0\\{\text{sin}}({\frac {2\pi }{3}})&0&{\text{cos}}({\frac {2\pi }{3}})\end{array}}\right),\,\,\,\,\rho (\nu )=\left({\begin{array}{ccc}-1&0&0\\0&-1&0\\0&0&1\end{array}}\right).}$

This representation is true. The subspace is an -invariant subspace. This means that the representation is not irreducible and there is a partial representation with This partial representation has degree 1 and is irreducible. ${\displaystyle \mathbb {C} e_{2}}$ ${\displaystyle D_{6}}$ ${\displaystyle \textstyle \rho |_{\mathbb {C} e_{2}}\colon D_{6}\to \mathbb {C} ^{\times }}$ ${\displaystyle \nu \mapsto -1,\mu \mapsto 1.}$

The complement to also -invariant, and provides the partial representation with ${\displaystyle \textstyle \mathbb {C} e_{2}}$ ${\displaystyle D_{6}}$ ${\displaystyle \textstyle \rho |_{\mathbb {C} e_{1}\oplus \mathbb {C} e_{3}}}$

${\displaystyle \nu \mapsto {\begin{pmatrix}-1&0\\0&1\end{pmatrix}},\,\,\,\,\mu \mapsto {\begin{pmatrix}{\text{cos}}({\frac {2\pi }{3}})&-{\text{sin}}({\frac {2\pi }{3}})\\{\text{sin}}({\frac {2\pi }{3}})&{\text{cos}}({\frac {2\pi }{3}})\end{pmatrix}}.}$

This partial representation is also irreducible. Our original representation is therefore completely reducible:

${\displaystyle \textstyle \rho =\rho |_{\mathbb {C} e_{2}}\oplus \rho |_{\mathbb {C} e_{1}\oplus \mathbb {C} e_{3}}.}$

Both partial representations are isotypic and they are the only non-zero isotypes of ${\displaystyle \rho .}$

The representation is unitary with respect to the standard scalar product on da and are unitary. ${\displaystyle \rho }$ ${\displaystyle \textstyle \mathbb {C} ^{3},}$ ${\displaystyle \rho (\mu )}$ ${\displaystyle \rho (\nu )}$

By taking any vector space isomorphism, a representation that is too isomorphic can be defined: Let be defined by for all ${\displaystyle \textstyle T\colon \mathbb {C} ^{3}\to \mathbb {C} ^{3}}$ ${\displaystyle \rho }$ ${\displaystyle \textstyle \eta \colon D_{6}\to {\text{GL}}_{3}(\mathbb {C} )}$ ${\displaystyle \textstyle \eta (s):=T\circ \rho (s)\circ T^{-1}}$ ${\displaystyle s\in D_{6}.}$

You can now assign the definition area of ​​the representation to a subgroup, e.g. B. limit and thus obtains This representation is defined by the image as stated above. ${\displaystyle H=\{{\text{Id}},\mu ,\mu ^{2}\},}$ ${\displaystyle {\text{Res}}_{H}(\rho ).}$ ${\displaystyle \rho (\mu )}$

Example 2

Let be the permutation group in elements. Let be a linear representation of on the generators defined by: ${\displaystyle G={\text{Per}}(3)}$ ${\displaystyle 3}$ ${\displaystyle \rho \colon {\text{Per}}(3)\to {\text{GL}}_{5}(\mathbb {C} )}$ ${\displaystyle {\text{Per}}(3)}$

${\displaystyle \rho (1,2)=\left({\begin{array}{ccccc}-1&2&0&0&0\\0&1&0&0&0\\0&0&0&1&0\\0&0&1&0&0\\0&0&0&0&1\end{array}}\right),\,\,\rho (1,3)=\left({\begin{array}{ccccc}1&0&0&0&0\\{\frac {1}{2}}&-1&0&0&0\\0&0&0&0&1\\0&0&0&1&0\\0&0&1&0&0\end{array}}\right),\,\,\rho (2,3)=\left({\begin{array}{ccccc}0&-2&0&0&0\\-{\frac {1}{2}}&0&0&0&0\\0&0&1&0&0\\0&0&0&0&1\\0&0&0&1&0\end{array}}\right).}$

This representation can then be broken down at first glance into the left-regular representation denoted here with and the representation with ${\displaystyle {\text{Per}}(3),}$ ${\displaystyle \pi ,}$ ${\displaystyle \eta \colon {\text{Per}}(3)\to {\text{GL}}_{2}(\mathbb {C} )}$

${\displaystyle \eta (1,2)={\begin{pmatrix}-1&2\\0&1\end{pmatrix}},\,\,\eta (1,3)={\begin{pmatrix}1&0\\{\frac {1}{2}}&-1\end{pmatrix}},\,\,\,\eta (2,3)={\begin{pmatrix}0&-2\\-{\frac {1}{2}}&0\end{pmatrix}}.}$

With the help of the character irreducibility criterion , we can see that is irreducible and not irreducible. Because it applies to the scalar product of characters . The subspace of is invariant under the left regular representation. The trivial representation results, restricted to this subspace. The orthogonal complement to is restricted to this subspace, which is also -invariant according to the above results , results in the representation given by ${\displaystyle \eta }$ ${\displaystyle \pi }$ ${\displaystyle (\eta |\eta )=1,(\pi |\pi )=2}$
${\displaystyle \mathbb {C} (e_{1}+e_{2}+e_{3})}$ ${\displaystyle \mathbb {C} ^{3}}$
${\displaystyle \mathbb {C} (e_{1}+e_{2}+e_{3})}$ ${\displaystyle \mathbb {C} (e_{2}-e_{1})\oplus \mathbb {C} (e_{1}+e_{2}-2e_{3}).}$ ${\displaystyle G}$ ${\displaystyle \tau ,}$

${\displaystyle \tau (1,2)={\begin{pmatrix}-1&0\\0&1\end{pmatrix}},\,\,\,\tau (1,3)={\begin{pmatrix}{\frac {1}{2}}&{\frac {3}{2}}\\{\frac {1}{2}}&-{\frac {1}{2}}\end{pmatrix}},\,\,\,\,\tau (2,3)={\begin{pmatrix}{\frac {1}{2}}&-{\frac {3}{2}}\\-{\frac {1}{2}}&-{\frac {1}{2}}\end{pmatrix}}.}$

As above, the irreducibility criterion for characters is used to check that is irreducible. But and are isomorphic, since it holds for all , where is given by the matrix ${\displaystyle \tau }$ ${\displaystyle \eta }$ ${\displaystyle \tau }$ ${\displaystyle \eta (s)=B\circ \tau (s)\circ B^{-1}}$ ${\displaystyle s\in {\text{Per}}(3)}$ ${\displaystyle B\colon \mathbb {C} ^{2}\to \mathbb {C} ^{2}}$

${\displaystyle M_{B}={\begin{pmatrix}2&2\\0&2\end{pmatrix}}.}$

We designate the trivial representation temporarily with A decomposition of into irreducible sub-representations is then: with the representation space The canonical decomposition is obtained by combining all isomorphic irreducible sub- representations: is the isotype of and the canonical decomposition is given by ${\displaystyle 1.}$ ${\displaystyle (\rho ,\mathbb {C} ^{5})}$ ${\displaystyle \rho =\tau \oplus \eta \oplus 1}$ ${\displaystyle \mathbb {C} ^{5}=\mathbb {C} (e_{1},e_{2})\oplus \mathbb {C} (e_{4}-e_{3},e_{3}+e_{4}-2e_{5})\oplus \mathbb {C} (e_{3}+e_{4}+e_{5}).}$
${\displaystyle \rho _{1}:=\eta \oplus \tau }$ ${\displaystyle \tau }$ ${\displaystyle \rho }$

${\displaystyle \rho =\rho _{1}\oplus 1,\,\mathbb {C} ^{5}=\mathbb {C} (e_{1},e_{2},e_{4}-e_{3},e_{3}+e_{4}-2e_{5})\oplus \mathbb {C} (e_{3}+e_{4}+e_{5}).}$

Infinite or non-compact groups

That the above theorems on the decomposition for infinite groups are no longer necessarily valid, should be illustrated here with an example: Let Then with the matrix multiplication, a group of infinite thickness is not compact. The group operates on by matrix-vector multiplication. We consider the representation for all. The subspace is an -invariant subspace. But there is no -invariant complement for this subspace . The assumption that such a complement exists leads to the contradictory result that every matrix can be diagonalized over . That is, if we consider infinite groups, the case may arise that a representation is not irreducible, but nevertheless does not break down into the direct sum of irreducible partial representations. ${\displaystyle \textstyle G=\{A\in {\text{GL}}_{2}(\mathbb {C} )|\,A\,\,{\text{ ist}}{\text{ obere}}{\text{ Dreiecksmatrix}}\}.}$ ${\displaystyle G}$ ${\displaystyle G}$ ${\displaystyle \mathbb {C} ^{2}}$ ${\displaystyle \rho (A)=A}$ ${\displaystyle A\in G.}$ ${\displaystyle \mathbb {C} e_{1}}$ ${\displaystyle G}$
${\displaystyle G}$ ${\displaystyle \mathbb {C} }$

literature

• Siegfried Echterhoff, Anton Deitmar: Principles of harmonic analysis. Springer-Verlag, 2009

Web links

• Teleman: Representation Theory