Representation theory of finite groups

The representation theory of groups is a part of mathematics which examines how groups act on given structures.

Here the focus is in particular on operations of groups on vector spaces. Nevertheless, groups acting on other groups or on sets are also considered. For more details, please refer to the section on permutation representations.

Other than a few marked exceptions, only finite groups will be considered in this article. We will also restrict ourselves to vector spaces over fields of characteristic zero. Because the theory of algebraically closed fields of characteristic zero is complete, a theory valid for a special algebraically closed field of characteristic zero is also valid for every other algebraically closed field of characteristic zero. Thus, without loss of generality, we can study vector spaces over ${\displaystyle \mathbb {C} .}$

Representation theory is used in many parts of mathematics, as well as in quantum chemistry and physics. Among other things it is used in algebra to examine the structure of groups. There are also applications in harmonic analysis and number theory. For example, representation theory is used in the modern approach to gain new results about automorphic forms.

Definition[edit]

Linear representations[edit]

Let ${\displaystyle V}$ be a ${\displaystyle K}$–vector space and ${\displaystyle G}$ a finite group. A linear representation of ${\displaystyle G}$ is a group homomorphism ${\displaystyle \rho :G\to {\text{GL}}(V)={\text{Aut}}(V).}$ Here ${\displaystyle {\text{GL}}(V)}$ is notation for a general linear group, and ${\displaystyle {\text{Aut}}(V)}$ for an automorphism group. This means that a linear representation is a map ${\displaystyle \rho :G\to {\text{GL}}(V)}$ which satisfies ${\displaystyle \rho (st)=\rho (s)\rho (t)}$ for all ${\displaystyle s,t\in G.}$ The vector space ${\displaystyle V}$ is called representation space of ${\displaystyle G.}$ Often the term representation of ${\displaystyle G}$ is also used for the representation space ${\displaystyle V.}$

The representation of a group in a module instead of a vector space is also called a linear representation.

We write ${\displaystyle (\rho ,V_{\rho })}$ for the representation ${\displaystyle \rho :G\to {\text{GL}}(V_{\rho })}$ of ${\displaystyle G.}$ Sometimes we use the notation ${\displaystyle (\rho ,V)}$ if it is clear to which representation the space ${\displaystyle V}$ belongs.

In this article we will restrict ourselves to the study of finite-dimensional representation spaces, except for the last chapter. As in most cases only a finite number of vectors in ${\displaystyle V}$ is of interest, it is sufficient to study the subrepresentation generated by these vectors. The representation space of this subrepresentation is then finite-dimensional.

The degree of a representation is the dimension of its representation space ${\displaystyle V.}$ The notation ${\displaystyle \dim(\rho )}$ is sometimes used to denote the degree of a representation ${\displaystyle \rho .}$

Examples[edit]

The trivial representation is given by ${\displaystyle \rho (s)={\text{Id}}}$ for all ${\displaystyle s\in G.}$

A representation of degree ${\displaystyle 1}$ of a group ${\displaystyle G}$ is a homomorphism into the multiplicative group ${\displaystyle \rho :G\to {\text{GL}}_{1}(\mathbb {C} )=\mathbb {C} ^{\times }=\mathbb {C} \setminus \{0\}.}$ As every element of ${\displaystyle G}$ is of finite order, the values of ${\displaystyle \rho (s)}$ are roots of unity. For example, let ${\displaystyle \rho :G=\mathbb {Z} /4\mathbb {Z} \to \mathbb {C} ^{\times }}$ be a nontrivial linear representation. Since ${\displaystyle \rho }$ is a group homomorphism, it has to satisfy ${\displaystyle \rho ({0})=1.}$ Because ${\displaystyle 1}$ generates ${\displaystyle G,\rho }$ is determined by its value on ${\displaystyle \rho (1).}$ And as ${\displaystyle \rho }$ is nontrivial, ${\displaystyle \rho ({1})\in \{i,-1,-i\}.}$ Thus, we achieve the result that the image of ${\displaystyle G}$ under ${\displaystyle \rho }$ has to be a nontrivial subgroup of the group which consists of the fourth roots of unity. In other words, ${\displaystyle \rho }$ has to be one of the following three maps:

${\displaystyle {\begin{cases}\rho _{1}({0})=1\\\rho _{1}({1})=i\\\rho _{1}({2})=-1\\\rho _{1}({3})=-i\end{cases}}\qquad {\begin{cases}\rho _{2}({0})=1\\\rho _{2}({1})=-1\\\rho _{2}({2})=1\\\rho _{2}({3})=-1\end{cases}}\qquad {\begin{cases}\rho _{3}({0})=1\\\rho _{3}({1})=-i\\\rho _{3}({2})=-1\\\rho _{3}({3})=i\end{cases}}}$

Let ${\displaystyle G=\mathbb {Z} /2\mathbb {Z} \times \mathbb {Z} /2\mathbb {Z} }$ and let ${\displaystyle \rho :G\to {\text{GL}}_{2}(\mathbb {C} )}$ be the group homomorphism defined by:

${\displaystyle \rho ({0},{0})={\begin{pmatrix}1&0\\0&1\end{pmatrix}},\quad \rho ({1},{0})={\begin{pmatrix}-1&0\\0&-1\end{pmatrix}},\quad \rho ({0},{1})={\begin{pmatrix}0&1\\1&0\end{pmatrix}},\quad \rho ({1},{1})={\begin{pmatrix}0&-1\\-1&0\end{pmatrix}}.}$

In this case ${\displaystyle \rho }$ is a linear representation of ${\displaystyle G}$ of degree ${\displaystyle 2.}$

Permutation representation[edit]

Let ${\displaystyle X}$ be a finite set and let ${\displaystyle G}$ be a group acting on ${\displaystyle X.}$ Denote by ${\displaystyle {\text{Aut}}(X)}$ the group of all permutations on ${\displaystyle X}$ with the composition as group multiplication.

A group acting on a finite set is sometimes considered sufficient for the definition of the permutation representation. However, since we want to construct examples for linear representations - where groups act on vector spaces instead of on arbitrary finite sets - we have to proceed in a different way. In order to construct the permutation representation, we need a vector space ${\displaystyle V}$ with ${\displaystyle \dim(V)=|X|.}$ A basis of ${\displaystyle V}$ can be indexed by the elements of ${\displaystyle X.}$ The permutation representation is the group homomorphism ${\displaystyle \rho :G\to {\text{GL}}(V)}$ given by ${\displaystyle \rho (s)e_{x}=e_{s.x}}$ for all ${\displaystyle s\in G,x\in X.}$ All linear maps ${\displaystyle \rho (s)}$ are uniquely defined by this property.

Example. Let ${\displaystyle X=\{1,2,3\}}$ and ${\displaystyle G={\text{Sym}}(3).}$ Then ${\displaystyle G}$ acts on ${\displaystyle X}$ via ${\displaystyle {\text{Aut}}(X)=G.}$ The associated linear representation is ${\displaystyle \rho :G\to {\text{GL}}(V)\cong {\text{GL}}_{3}(\mathbb {C} )}$ with ${\displaystyle \rho (\sigma )e_{x}=e_{\sigma (x)}}$ for ${\displaystyle \sigma \in G,x\in X.}$

Left- and right-regular representation[edit]

Let ${\displaystyle G}$ be a group and ${\displaystyle V}$ be a vector space of dimension ${\displaystyle |G|}$ with a basis ${\displaystyle (e_{t})_{t\in G}}$ indexed by the elements of ${\displaystyle G.}$ The left-regular representation is a special case of the permutation representation by choosing ${\displaystyle X=G.}$ This means ${\displaystyle \rho (s)e_{t}=e_{st}}$ for all ${\displaystyle s,t\in G.}$ Thus, the family ${\displaystyle (\rho (s)e_{1})_{s\in G}}$ of images of ${\displaystyle e_{1}}$ are a basis of ${\displaystyle V.}$ The degree of the left-regular representation is equal to the order of the group.

The right-regular representation is defined on the same vector space with a similar homomorphism: ${\displaystyle \rho (s)e_{t}=e_{ts^{-1}}.}$ In the same way as before ${\displaystyle (\rho (s)e_{1})_{s\in G}}$ is a basis of ${\displaystyle V.}$ Just as in the case of the left-regular representation, the degree of the right-regular representation is equal to the order of ${\displaystyle G.}$

Both representations are isomorphic via ${\displaystyle e_{s}\mapsto e_{s^{-1}}.}$ For this reason they are not always set apart, and often referred to as "the" regular representation.

A closer look provides the following result: A given linear representation ${\displaystyle \rho :G\to {\text{GL}}(W)}$ is isomorphic to the left-regular representation if and only if there exists a ${\displaystyle w\in W,}$ such that ${\displaystyle (\rho (s)w)_{s\in G}}$ is a basis of ${\displaystyle W.}$

Example. Let ${\displaystyle G=\mathbb {Z} /5\mathbb {Z} }$ and ${\displaystyle V=\mathbb {R} ^{5}}$ with the basis ${\displaystyle \{e_{0},\ldots ,e_{4}\}.}$ Then the left-regular representation ${\displaystyle L_{\rho }:G\to {\text{GL}}(V)}$ is defined by ${\displaystyle L_{\rho }(k)e_{l}=e_{l+k}}$ for ${\displaystyle k,l\in \mathbb {Z} /5\mathbb {Z} .}$ The right-regular representation is defined analogously by ${\displaystyle R_{\rho }(k)e_{l}=e_{l-k}}$ for ${\displaystyle k,l\in \mathbb {Z} /5\mathbb {Z} .}$

Representations, modules and the convolution algebra[edit]

Let ${\displaystyle G}$ be a finite group, let ${\displaystyle K}$ be a commutative ring and let ${\displaystyle K[G]}$ be the group algebra of ${\displaystyle G}$ over ${\displaystyle K.}$ This algebra is free and a basis can be indexed by the elements of ${\displaystyle G.}$ Most often the basis is identified with ${\displaystyle G}$. Every element ${\displaystyle f\in K[G]}$ can then be uniquely expressed as

${\displaystyle f=\sum _{s\in G}a_{s}s}$ with ${\displaystyle a_{s}\in K}$.

The multiplication in ${\displaystyle K[G]}$ extends that in ${\displaystyle G}$ distributively.

Now let ${\displaystyle V}$ be a ${\displaystyle K}$–module and let ${\displaystyle \rho :G\to {\text{GL}}(V)}$ be a linear representation of ${\displaystyle G}$ in ${\displaystyle V.}$ We define ${\displaystyle sv=\rho (s)v}$ for all ${\displaystyle s\in G}$ and ${\displaystyle v\in V}$. By linear extension ${\displaystyle V}$ is endowed with the structure of a left-${\displaystyle K[G]}$–module. Vice versa we obtain a linear representation of ${\displaystyle G}$ starting from a ${\displaystyle K[G]}$–module ${\displaystyle V}$. Additionally, homomorphisms of representations are in bijective correspondence with group algebra homomorphisms. Therefore, these terms may be used interchangeably.^[1][2] This is an example of an isomorphism of categories.

Suppose ${\displaystyle K=\mathbb {C} .}$ In this case the left ${\displaystyle \mathbb {C} [G]}$–module given by ${\displaystyle \mathbb {C} [G]}$ itself corresponds to the left-regular representation. In the same way ${\displaystyle \mathbb {C} [G]}$ as a right ${\displaystyle \mathbb {C} [G]}$–module corresponds to the right-regular representation.

In the following we will define the convolution algebra: Let ${\displaystyle G}$ be a group, the set ${\displaystyle L^{1}(G):=\{f:G\to \mathbb {C} \}}$ is a ${\displaystyle \mathbb {C} }$–vector space with the operations addition and scalar multiplication then this vector space is isomorphic to ${\displaystyle \mathbb {C} ^{|G|}.}$ The convolution of two elements ${\displaystyle f,h\in L^{1}(G)}$ defined by

${\displaystyle f*h(s):=\sum _{t\in G}f(t)h(t^{-1}s)}$

makes ${\displaystyle L^{1}(G)}$ an algebra. The algebra ${\displaystyle L^{1}(G)}$ is called the convolution algebra.

The convolution algebra is free and has a basis indexed by the group elements: ${\displaystyle (\delta _{s})_{s\in G},}$ where

${\displaystyle \delta _{s}(t)={\begin{cases}1&t=s\\0&{\text{otherwise.}}\end{cases}}}$

Using the properties of the convolution we obtain: ${\displaystyle \delta _{s}*\delta _{t}=\delta _{st}.}$

We define a map between ${\displaystyle L^{1}(G)}$ and ${\displaystyle \mathbb {C} [G],}$ by defining ${\displaystyle \delta _{s}\mapsto e_{s}}$ on the basis ${\displaystyle (\delta _{s})_{s\in G}}$ and extending it linearly. Obviously the prior map is bijective. A closer inspection of the convolution of two basis elements as shown in the equation above reveals that the multiplication in ${\displaystyle L^{1}(G)}$ corresponds to that in ${\displaystyle \mathbb {C} [G].}$ Thus, the convolution algebra and the group algebra are isomorphic as algebras.

The involution

${\displaystyle f^{*}(s)={\overline {f(s^{-1})}}}$

turns ${\displaystyle L^{1}(G)}$ into a ${\displaystyle ^{*}}$–algebra. We have ${\displaystyle \delta _{s}^{*}=\delta _{s^{-1}}.}$

A representation ${\displaystyle (\pi ,V_{\pi })}$ of a group ${\displaystyle G}$ extends to a ${\displaystyle ^{*}}$–algebra homomorphism ${\displaystyle \pi :L^{1}(G)\to {\text{End}}(V_{\pi })}$ by ${\displaystyle \pi (\delta _{s})=\pi (s).}$ Since multiplicativity is a characteristic property of algebra homomorphisms, ${\displaystyle \pi }$ satisfies ${\displaystyle \pi (f*h)=\pi (f)\pi (h).}$ If ${\displaystyle \pi }$ is unitary, we also obtain ${\displaystyle \pi (f)^{*}=\pi (f^{*}).}$ For the definition of a unitary representation, please refer to the chapter on properties. In that chapter we will see that (without loss of generality) every linear representation can be assumed to be unitary.

Using the convolution algebra we can implement a Fourier transformation on a group ${\displaystyle G.}$ In the area of harmonic analysis it is shown that the following definition is consistent with the definition of the Fourier transformation on ${\displaystyle \mathbb {R} .}$

Let ${\displaystyle \rho :G\to {\text{GL}}(V_{\rho })}$ be a representation and let ${\displaystyle f\in L^{1}(G)}$ be a ${\displaystyle \mathbb {C} }$-valued function on ${\displaystyle G}$. The Fourier transform ${\displaystyle {\hat {f}}(\rho )\in {\text{End}}(V_{\rho })}$ of ${\displaystyle f}$ is defined as

${\displaystyle {\hat {f}}(\rho )=\sum _{s\in G}f(s)\rho (s).}$

This transformation satisfies ${\displaystyle {\widehat {f*g}}(\rho )={\hat {f}}(\rho )\cdot {\hat {g}}(\rho ).}$

Maps between representations[edit]

A map between two representations ${\displaystyle (\rho ,V_{\rho }),\,(\tau ,V_{\tau })}$ of the same group ${\displaystyle G}$ is a linear map ${\displaystyle T:V_{\rho }\to V_{\tau },}$ with the property that ${\displaystyle \tau (s)\circ T=T\circ \rho (s)}$ holds for all ${\displaystyle s\in G.}$ In other words, the following diagram commutes for all ${\displaystyle s\in G}$:

Such a map is also called ${\displaystyle G}$–linear, or an equivariant map. The kernel, the image and the cokernel of ${\displaystyle T}$ are defined by default. The composition of equivariant maps is again an equivariant map. There is a category of representations with equivariant maps as its morphisms. They are again ${\displaystyle G}$–modules. Thus, they provide representations of ${\displaystyle G}$ due to the correlation described in the previous section.

Irreducible representations and Schur's lemma[edit]

Let ${\displaystyle \rho :G\to {\text{GL}}(V)}$ be a linear representation of ${\displaystyle G.}$ Let ${\displaystyle W}$ be a ${\displaystyle G}$-invariant subspace of ${\displaystyle V,}$ that is, ${\displaystyle \rho (s)w\in W}$ for all ${\displaystyle s\in G}$ and ${\displaystyle w\in W}$. The restriction ${\displaystyle \rho (s)|_{W}}$ is an isomorphism of ${\displaystyle W}$ onto itself. Because ${\displaystyle \rho (s)|_{W}\circ \rho (t)|_{W}=\rho (st)|_{W}}$ holds for all ${\displaystyle s,t\in G,}$ this construction is a representation of ${\displaystyle G}$ in ${\displaystyle W.}$ It is called subrepresentation of ${\displaystyle V.}$ Any representation V has at least two subrepresentations, namely the one consisting only of 0, and the one consisting of V itself. The representation is called an irreducible representation, if these two are the only subrepresentations. Some authors also call these representations simple, given that they are precisely the simple modules over the group algebra ${\displaystyle \mathbb {C} [G]}$.

Schur's lemma puts a strong constraint on maps between irreducible representations. If ${\displaystyle \rho _{1}:G\to {\text{GL}}(V_{1})}$ and ${\displaystyle \rho _{2}:G\to {\text{GL}}(V_{2})}$ are both irreducible, and ${\displaystyle F:V_{1}\to V_{2}}$ is a linear map such that ${\displaystyle \rho _{2}(s)\circ F=F\circ \rho _{1}(s)}$ for all ${\displaystyle s\in G.}$, there is the following dichotomy:

• If ${\displaystyle V_{1}=V_{2}}$ and ${\displaystyle \rho _{1}=\rho _{2},}$ ${\displaystyle F}$ is a homothety (i.e. ${\displaystyle F=\lambda {\text{Id}}}$ for a ${\displaystyle \lambda \in \mathbb {C} }$). More generally, if ${\displaystyle \rho _{1}}$ and ${\displaystyle \rho _{2}}$ are isomorphic, the space of G-linear maps is one-dimensional.
• Otherwise, if the two representations are not isomorphic, F must be 0.^[3]

Properties[edit]

Two representations ${\displaystyle (\rho ,V_{\rho }),(\pi ,V_{\pi })}$ are called equivalent or isomorphic, if there exists a ${\displaystyle G}$–linear vector space isomorphism between the representation spaces. In other words, they are isomorphic if there exists a bijective linear map ${\displaystyle T:V_{\rho }\to V_{\pi },}$ such that ${\displaystyle T\circ \rho (s)=\pi (s)\circ T}$ for all ${\displaystyle s\in G.}$ In particular, equivalent representations have the same degree.

A representation ${\displaystyle (\pi ,V_{\pi })}$ is called faithful when ${\displaystyle \pi }$ is injective. In this case ${\displaystyle \pi }$ induces an isomorphism between ${\displaystyle G}$ and the image ${\displaystyle \pi (G).}$ As the latter is a subgroup of ${\displaystyle {\text{GL}}(V_{\pi }),}$ we can regard ${\displaystyle G}$ via ${\displaystyle \pi }$ as subgroup of ${\displaystyle {\text{Aut}}(V_{\pi }).}$

We can restrict the range as well as the domain:

Let ${\displaystyle H}$ be a subgroup of ${\displaystyle G.}$ Let ${\displaystyle \rho }$ be a linear representation of ${\displaystyle G.}$ We denote by ${\displaystyle {\text{Res}}_{H}(\rho )}$ the restriction of ${\displaystyle \rho }$ to the subgroup ${\displaystyle H.}$

If there is no danger of confusion, we might use only ${\displaystyle {\text{Res}}(\rho )}$ or in short ${\displaystyle {\text{Res}}\rho .}$

The notation ${\displaystyle {\text{Res}}_{H}(V)}$ or in short ${\displaystyle {\text{Res}}(V)}$ is also used to denote the restriction of the representation ${\displaystyle V}$ of ${\displaystyle G}$ onto ${\displaystyle H.}$

Let ${\displaystyle f}$ be a function on ${\displaystyle G.}$ We write ${\displaystyle {\text{Res}}_{H}(f)}$ or shortly ${\displaystyle {\text{Res}}(f)}$ for the restriction to the subgroup ${\displaystyle H.}$

It can be proven that the number of irreducible representations of a group ${\displaystyle G}$ (or correspondingly the number of simple ${\displaystyle \mathbb {C} [G]}$–modules) equals the number of conjugacy classes of ${\displaystyle G.}$

A representation is called semisimple or completely reducible if it can be written as a direct sum of irreducible representations. This is analogous to the corresponding definition for a semisimple algebra.

For the definition of the direct sum of representations please refer to the section on direct sums of representations.

A representation is called isotypic if it is a direct sum of pairwise isomorphic irreducible representations.

Let ${\displaystyle (\rho ,V_{\rho })}$ be a given representation of a group ${\displaystyle G.}$ Let ${\displaystyle \tau }$ be an irreducible representation of ${\displaystyle G.}$ The ${\displaystyle \tau }$–isotype ${\displaystyle V_{\rho }(\tau )}$ of ${\displaystyle G}$ is defined as the sum of all irreducible subrepresentations of ${\displaystyle V}$ isomorphic to ${\displaystyle \tau .}$

Every vector space over ${\displaystyle \mathbb {C} }$ can be provided with an inner product. A representation ${\displaystyle \rho }$ of a group ${\displaystyle G}$ in a vector space endowed with an inner product is called unitary if ${\displaystyle \rho (s)}$ is unitary for every ${\displaystyle s\in G.}$ This means that in particular every ${\displaystyle \rho (s)}$ is diagonalizable. For more details see the article on unitary representations.

A representation is unitary with respect to a given inner product if and only if the inner product is invariant with regard to the induced operation of ${\displaystyle G,}$ i.e. if and only if ${\displaystyle (v|u)=(\rho (s)v|\rho (s)u)}$ holds for all ${\displaystyle v,u\in V_{\rho },s\in G.}$

A given inner product ${\displaystyle (\cdot |\cdot )}$ can be replaced by an invariant inner product by exchanging ${\displaystyle (v|u)}$ with

${\displaystyle \sum _{t\in G}(\rho (t)v|\rho (t)u).}$

Thus, without loss of generality we can assume that every further considered representation is unitary.

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

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

This representation is faithful. The subspace ${\displaystyle \mathbb {C} e_{2}}$ is a ${\displaystyle D_{6}}$–invariant subspace. Thus, there exists a nontrivial subrepresentation ${\displaystyle \rho |_{\mathbb {C} e_{2}}:D_{6}\to \mathbb {C} ^{\times }}$ with ${\displaystyle \nu \mapsto -1,\mu \mapsto 1.}$ Therefore, the representation is not irreducible. The mentioned subrepresentation is of degree one and irreducible. The complementary subspace of ${\displaystyle \mathbb {C} e_{2}}$ is ${\displaystyle D_{6}}$–invariant as well. Therefore, we obtain the subrepresentation ${\displaystyle \rho |_{\mathbb {C} e_{1}\oplus \mathbb {C} e_{3}}}$ with

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

This subrepresentation is also irreducible. That means, the original representation is completely reducible:

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

Both subrepresentations are isotypic and are the two only non-zero isotypes of ${\displaystyle \rho .}$

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

Let ${\displaystyle T:\mathbb {C} ^{3}\to \mathbb {C} ^{3}}$ be any vector space isomorphism. Then ${\displaystyle \eta :D_{6}\to {\text{GL}}_{3}(\mathbb {C} ),}$ which is defined by the equation ${\displaystyle \eta (s):=T\circ \rho (s)\circ T^{-1}}$ for all ${\displaystyle s\in D_{6},}$ is a representation isomorphic to ${\displaystyle \rho .}$

By restricting the domain of the representation to a subgroup, e.g. ${\displaystyle H=\{{\text{id}},\mu ,\mu ^{2}\},}$ we obtain the representation ${\displaystyle {\text{Res}}_{H}(\rho ).}$ This representation is defined by the image ${\displaystyle \rho (\mu ),}$ whose explicit form is shown above.

Constructions[edit]

The dual representation[edit]

Let ${\displaystyle \rho :G\to {\text{GL}}(V)}$ be a given representation. The dual representation or contragredient representation ${\displaystyle \rho ^{*}:G\to {\text{GL}}(V^{*})}$ is a representation of ${\displaystyle G}$ in the dual vector space of ${\displaystyle V.}$ It is defined by the property

${\displaystyle \forall s\in G,v\in V,\alpha \in V^{*}:\qquad \left(\rho ^{*}(s)\alpha \right)(v)=\alpha \left(\rho \left(s^{-1}\right)v\right).}$

With regard to the natural pairing ${\displaystyle \langle \alpha ,v\rangle :=\alpha (v)}$ between ${\displaystyle V^{*}}$ and ${\displaystyle V}$ the definition above provides the equation:

${\displaystyle \forall s\in G,v\in V,\alpha \in V^{*}:\qquad \langle \rho ^{*}(s)(\alpha ),\rho (s)(v)\rangle =\langle \alpha ,v\rangle .}$

For an example, see the main page on this topic: Dual representation.

Direct sum of representations[edit]

Let ${\displaystyle (\rho _{1},V_{1})}$ and ${\displaystyle (\rho _{2},V_{2})}$ be a representation of ${\displaystyle G_{1}}$ and ${\displaystyle G_{2},}$ respectively. The direct sum of these representations is a linear representation and is defined as

${\displaystyle \forall s_{1}\in G_{1},s_{2}\in G_{2},v_{1}\in V_{1},v_{2}\in V_{2}:\qquad {\begin{cases}\rho _{1}\oplus \rho _{2}:G_{1}\times G_{2}\to {\text{GL}}(V_{1}\oplus V_{2})\\[4pt](\rho _{1}\oplus \rho _{2})(s_{1},s_{2})(v_{1},v_{2}):=\rho _{1}(s_{1})v_{1}\oplus \rho _{2}(s_{2})v_{2}\end{cases}}}$

Let ${\displaystyle \rho _{1},\rho _{2}}$ be representations of the same group ${\displaystyle G.}$ For the sake of simplicity, the direct sum of these representations is defined as a representation of ${\displaystyle G,}$ i.e. it is given as ${\displaystyle \rho _{1}\oplus \rho _{2}:G\to {\text{GL}}(V_{1}\oplus V_{2}),}$ by viewing ${\displaystyle G}$ as the diagonal subgroup of ${\displaystyle G\times G.}$

Example. Let (here ${\displaystyle i}$ and ${\displaystyle \omega }$ are the imaginary unit and the primitive cube root of unity respectively):

${\displaystyle {\begin{cases}\rho _{1}:\mathbb {Z} /2\mathbb {Z} \to {\text{GL}}_{2}(\mathbb {C} )\\[4pt]\rho _{1}(1)={\begin{pmatrix}0&-i\\i&0\end{pmatrix}}\end{cases}}\qquad \qquad {\begin{cases}\rho _{2}:\mathbb {Z} /3\mathbb {Z} \to {\text{GL}}_{3}(\mathbb {C} )\\[6pt]\rho _{2}(1)={\begin{pmatrix}1&0&\omega \\0&\omega &0\\0&0&\omega ^{2}\end{pmatrix}}\end{cases}}}$

Then

${\displaystyle {\begin{cases}\rho _{1}\oplus \rho _{2}:\mathbb {Z} /2\mathbb {Z} \times \mathbb {Z} /3\mathbb {Z} \to {\text{GL}}\left(\mathbb {C} ^{2}\oplus \mathbb {C} ^{3}\right)\\[6pt]\left(\rho _{1}\oplus \rho _{2}\right)(k,l)={\begin{pmatrix}\rho _{1}(k)&0\\0&\rho _{2}(l)\end{pmatrix}}&k\in \mathbb {Z} /2\mathbb {Z} ,l\in \mathbb {Z} /3\mathbb {Z} \end{cases}}}$

As it is sufficient to consider the image of the generating element, we find that

${\displaystyle (\rho _{1}\oplus \rho _{2})(1,1)={\begin{pmatrix}0&-i&0&0&0\\i&0&0&0&0\\0&0&1&0&\omega \\0&0&0&\omega &0\\0&0&0&0&\omega ^{2}\end{pmatrix}}}$

Tensor product of representations[edit]

Let ${\displaystyle \rho _{1}:G_{1}\to {\text{GL}}(V_{1}),\rho _{2}:G_{2}\to {\text{GL}}(V_{2})}$ be linear representations. We define the linear representation ${\displaystyle \rho _{1}\otimes \rho _{2}:G_{1}\times G_{2}\to {\text{GL}}(V_{1}\otimes V_{2})}$ into the tensor product of ${\displaystyle V_{1}}$ and ${\displaystyle V_{2}}$ by ${\displaystyle \rho _{1}\otimes \rho _{2}(s_{1},s_{2})=\rho _{1}(s_{1})\otimes \rho _{2}(s_{2}),}$ in which ${\displaystyle s_{1}\in G_{1},s_{2}\in G_{2}.}$ This representation is called outer tensor product of the representations ${\displaystyle \rho _{1}}$ and ${\displaystyle \rho _{2}.}$ The existence and uniqueness is a consequence of the properties of the tensor product.

Example. We reexamine the example provided for the direct sum:

${\displaystyle {\begin{cases}\rho _{1}:\mathbb {Z} /2\mathbb {Z} \to {\text{GL}}_{2}(\mathbb {C} )\\[4pt]\rho _{1}(1)={\begin{pmatrix}0&-i\\i&0\end{pmatrix}}\end{cases}}\qquad \qquad {\begin{cases}\rho _{2}:\mathbb {Z} /3\mathbb {Z} \to {\text{GL}}_{3}(\mathbb {C} )\\[6pt]\rho _{2}(1)={\begin{pmatrix}1&0&\omega \\0&\omega &0\\0&0&\omega ^{2}\end{pmatrix}}\end{cases}}}$

The outer tensor product

${\displaystyle {\begin{cases}\rho _{1}\otimes \rho _{2}:\mathbb {Z} /2\mathbb {Z} \times \mathbb {Z} /3\mathbb {Z} \to {\text{GL}}(\mathbb {C} ^{2}\otimes \mathbb {C} ^{3})\\(\rho _{1}\otimes \rho _{2})(k,l)=\rho _{1}(k)\otimes \rho _{2}(l)&k\in \mathbb {Z} /2\mathbb {Z} ,l\in \mathbb {Z} /3\mathbb {Z} \end{cases}}}$

Using the standard basis of ${\displaystyle \mathbb {C} ^{2}\otimes \mathbb {C} ^{3}\cong \mathbb {C} ^{6}}$ we have the following for the generating element:

${\displaystyle \rho _{1}\otimes \rho _{2}(1,1)=\rho _{1}(1)\otimes \rho _{2}(1)={\begin{pmatrix}0&0&0&-i&0&-i\omega \\0&0&0&0&-i\omega &0\\0&0&0&0&0&-i\omega ^{2}\\i&0&i\omega &0&0&0\\0&i\omega &0&0&0&0\\0&0&i\omega ^{2}&0&0&0\end{pmatrix}}}$

Remark. Note that the direct sum and the tensor products have different degrees and hence are different representations.

Let ${\displaystyle \rho _{1}:G\to {\text{GL}}(V_{1}),\rho _{2}:G\to {\text{GL}}(V_{2})}$ be two linear representations of the same group. Let ${\displaystyle s}$ be an element of ${\displaystyle G.}$ Then ${\displaystyle \rho (s)\in {\text{GL}}(V_{1}\otimes V_{2})}$ is defined by ${\displaystyle \rho (s)(v_{1}\otimes v_{2})=\rho _{1}(s)v_{1}\otimes \rho _{2}(s)v_{2},}$ for ${\displaystyle v_{1}\in V_{1},v_{2}\in V_{2},}$ and we write ${\displaystyle \rho (s)=\rho _{1}(s)\otimes \rho _{2}(s).}$ Then the map ${\displaystyle s\mapsto \rho (s)}$ defines a linear representation of ${\displaystyle G,}$ which is also called tensor product of the given representations.

These two cases have to be strictly distinguished. The first case is a representation of the group product into the tensor product of the corresponding representation spaces. The second case is a representation of the group ${\displaystyle G}$ into the tensor product of two representation spaces of this one group. But this last case can be viewed as a special case of the first one by focusing on the diagonal subgroup ${\displaystyle G\times G.}$ This definition can be iterated a finite number of times.

Let ${\displaystyle V}$ and ${\displaystyle W}$ be representations of the group ${\displaystyle G.}$ Then ${\displaystyle {\text{Hom}}(V,W)}$ is a representation by virtue of the following identity: ${\displaystyle {\text{Hom}}(V,W)=V^{*}\otimes W}$. Let ${\displaystyle B\in {\text{Hom}}(V,W)}$ and let ${\displaystyle \rho }$ be the representation on ${\displaystyle {\text{Hom}}(V,W).}$ Let ${\displaystyle \rho _{V}}$ be the representation on ${\displaystyle V}$ and ${\displaystyle \rho _{W}}$ the representation on ${\displaystyle W.}$ Then the identity above leads to the following result:

${\displaystyle \rho (s)(B)v=\rho _{W}(s)\circ B\circ \rho _{V}(s^{-1})v}$ for all ${\displaystyle s\in G,v\in V.}$

Theorem. The irreducible representations of ${\displaystyle G_{1}\times G_{2}}$ up to isomorphism are exactly the representations ${\displaystyle \rho _{1}\otimes \rho _{2}}$ in which ${\displaystyle \rho _{1}}$ and ${\displaystyle \rho _{2}}$ are irreducible representations of ${\displaystyle G_{1}}$ and ${\displaystyle G_{2},}$ respectively.

Symmetric and alternating square[edit]

Let ${\displaystyle \rho :G\to V\otimes V}$ be a linear representation of ${\displaystyle G.}$ Let ${\displaystyle (e_{k})}$ be a basis of ${\displaystyle V.}$ Define ${\displaystyle \vartheta :V\otimes V\to V\otimes V}$ by extending ${\displaystyle \vartheta (e_{k}\otimes e_{j})=e_{j}\otimes e_{k}}$ linearly. It then holds that ${\displaystyle \vartheta ^{2}=1}$ and therefore ${\displaystyle V\otimes V}$ splits up into ${\displaystyle V\otimes V={\text{Sym}}^{2}(V)\oplus {\text{Alt}}^{2}(V),}$ in which

${\displaystyle {\text{Sym}}^{2}(V)=\{z\in V\otimes V:\vartheta (z)=z\}}$

${\displaystyle {\text{Alt}}^{2}(V)=\bigwedge ^{2}V=\{z\in V\otimes V:\vartheta (z)=-z\}.}$

These subspaces are ${\displaystyle G}$–invariant and by this define subrepresentations which are called the symmetric square and the alternating square, respectively. These subrepresentations are also defined in ${\displaystyle V^{\otimes m},}$ although in this case they are denoted wedge product ${\displaystyle \bigwedge ^{m}V}$ and symmetric product ${\displaystyle {\text{Sym}}^{m}(V).}$ In case that ${\displaystyle m>2,}$ the vector space ${\displaystyle V^{\otimes m}}$ is in general not equal to the direct sum of these two products.

Decompositions[edit]

In order to understand representations more easily, a decomposition of the representation space into the direct sum of simpler subrepresentations would be desirable. This can be achieved for finite groups as we will see in the following results. More detailed explanations and proofs may be found in [1] and [2].

Theorem. (Maschke) Let ${\displaystyle \rho :G\to {\text{GL}}(V)}$ be a linear representation where ${\displaystyle V}$ is a vector space over a field of characteristic zero. Let ${\displaystyle W}$ be a ${\displaystyle G}$-invariant subspace of ${\displaystyle V.}$ Then the complement ${\displaystyle W^{0}}$ of ${\displaystyle W}$ exists in ${\displaystyle V}$ and is ${\displaystyle G}$-invariant.

A subrepresentation and its complement determine a representation uniquely.

The following theorem will be presented in a more general way, as it provides a very beautiful result about representations of compact – and therefore also of finite – groups:

Theorem. Every linear representation of a compact group over a field of characteristic zero is a direct sum of irreducible representations.

Or in the language of ${\displaystyle K[G]}$-modules: If ${\displaystyle {\text{char}}(K)=0,}$ the group algebra ${\displaystyle K[G]}$ is semisimple, i.e. it is the direct sum of simple algebras.

Note that this decomposition is not unique. However, the number of how many times a subrepresentation isomorphic to a given irreducible representation is occurring in this decomposition is independent of the choice of decomposition.

The canonical decomposition

To achieve a unique decomposition, one has to combine all the irreducible subrepresentations that are isomorphic to each other. That means, the representation space is decomposed into a direct sum of its isotypes. This decomposition is uniquely determined. It is called the canonical decomposition.

Let ${\displaystyle (\tau _{j})_{j\in I}}$ be the set of all irreducible representations of a group ${\displaystyle G}$ up to isomorphism. Let ${\displaystyle V}$ be a representation of ${\displaystyle G}$ and let ${\displaystyle \{V(\tau _{j})|j\in I\}}$ be the set of all isotypes of ${\displaystyle V.}$ The projection ${\displaystyle p_{j}:V\to V(\tau _{j})}$ corresponding to the canonical decomposition is given by

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

where ${\displaystyle n_{j}=\dim(\tau _{j}),}$ ${\displaystyle g={\text{ord}}(G)}$ and ${\displaystyle \chi _{\tau _{j}}}$ is the character belonging to ${\displaystyle \tau _{j}.}$

In the following, we show how to determine the isotype to the trivial representation:

Definition (Projection formula). For every representation ${\displaystyle (\rho ,V)}$ of a group ${\displaystyle G}$ we define

${\displaystyle V^{G}:=\{v\in V:\rho (s)v=v\,\,\,\,\forall \,s\in G\}.}$

In general, ${\displaystyle \rho (s):V\to V}$ is not ${\displaystyle G}$-linear. We define

${\displaystyle P:={\frac {1}{|G|}}\sum _{s\in G}\rho (s)\in {\text{End}}(V).}$

Then ${\displaystyle P}$ is a ${\displaystyle G}$-linear map, because

${\displaystyle \forall t\in G:\qquad \sum _{s\in G}\rho (s)=\sum _{s\in G}\rho (tst^{-1}).}$

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