
MSC Classification: 22 (Topological groups, Lie groups) 
Prerequisites: Abstract Algebra I, Abstract Algebra II
Getting Oriented
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Using mathematical formalism, a group G may be "represented'' by transformations on a vector space.
Representations
The Basics
A representation of a group $G$ is a vector space $V$ for which one can view each $g\in G$ as a transformation $g:V\to V$.
 Representation
 a pair $(\pi,V)$, where $V$ is a finite dimensional vector space over $\mathbb{C}$, and $\pi$ is a continuous homomorphism $\pi:G\to GL(V)$ into the Lie group $GL(V)$ of invertible linear transformations.
We will write $g$ for an element of $G$ and $\pi_g$ for its corresponding transformation. The dimension of the representation is the dimension of the vector space $V$. Simple examples of representations include:
 The trivial representation with $\pi_g=I$ or $g(v)=v$ for all $v\in V$.
 The standard representation for $G\subset GL(n,\mathbb{C})$ with $g(v)=g\cdot v$ for all $v\in \mathbb{C}^n$.
 A group action $G\times X\to X$ provides a representation with $V=\{f:X\to\mathbb{C}\}=X^*$ and $(\pi_g f)(x)=f(g^{1}x)$. Thus, $g$ takes the map $x\mapsto f(x)$ to the map $x\mapsto f(g^{1}x)$.
Reducible and Unitary Representations
Given an invariant subspace $U$ of a representation $(\pi,V)$, meaning $\pi_g u \in U$ for all $u\in U$, one may restrict $(\pi,V)$ to the subspace representation $(\pi',U)$ with $\pi'_g=\pi_g_U$. The existence of invariant subspaces allows one to break up a representation into smaller pieces. Those without invariant subspaces are:
 Irreducible representation
 a $(\pi,V)$ with no nontrivial invariant subspaces.
As a simple example, a 1dimensional representation must be irreducible.
Irreducible representations form the building blocks for all others, just as primes do for integers. And as we ask how to decompose a given integer into primes, it is natural to ask when and how we can write $V=V_1\oplus\cdots\oplus V_n$ where the $V_i$ are all irreducible invariant subspaces. Here are some things to consider:
 By extending a basis for an invariant subspace $U$ to a basis for $V$, we can write all $\pi_g$ in blockupper diagonal form $\begin{pmatrix}*&*\\ 0&*\end{pmatrix}$.
 If we can write $V=U_1\oplus U_2$, we can similarly write $\pi_g$ in blockdiagonal form $\begin{pmatrix}*&0\\ 0&*\end{pmatrix}$.
 Not every invariant subspace provides such a decomposition. Letting $G=(\mathbb{R},+)=\begin{pmatrix}1&a\\ 0&1\end{pmatrix}: a\in\mathbb{R}\}$ with the standard representation, we have invariant subspace $U=\begin{pmatrix}x\\0\end{pmatrix}$ but its complement $\begin{pmatrix}0\\y\end{pmatrix}$ clearly not.
It turns out that a representation $(\pi,V)$ is completely reducible, meaning it can be broken down into irreducible invariant subspace representations, when it finitedimensional and invariant under a certain inner product:
 Unitary representation
 a representation $(\pi,V)$ for which $\langle{\pi_g v,\pi_g w}\rangle=\langle{v,w}\rangle$ for some positivedefinite Hermitian inner product $\langle{\:,\:}\rangle$. To be Hermitian, it must satisfy $\langle{v,w}\rangle=\sum v_i \bar w_i$ and $\langle{v,w}\rangle=\overline{\langle{w,v}\rangle}$. This is equivalent to requiring the matrices $\pi_g$ to be unitary $\pi_g^{1}=\pi_g^*$ or $\langle{\pi_g v,w}\rangle=\langle{v,\overline{\pi_g^T}w}\rangle$.
The finitedimensional unitary representations are completely reducible for the following simple fact:
 if $U$ is an invariant subspace, then $U^\perp=\{v:\langle{u,v}\rangle \text{ for all } u\in U\}$ is also invariant.
Thus, an algorithm for finding invariant subspaces provides an algorithm for finding the irreducible ones, but only in the finitedimensional case.
Some Completely Reducible Representations
In order to show that a representation is completely reducible, it suffices to show that it is unitary. In the case of a finite group $G$, this is easy: take any positivedefinite Hermitian inner product $(\:,\:)$ on $V$ and average it over the group by defining:
(1)It is simple to show that $\langle{\:,\:}\rangle$ is again positivedefinite and Hermitian and it is obvious from the definition that it is invariant under $\pi_x$ for $x\in G$.
This averaging trick also works for any compact group $G$, although we must now integrate. Again, assuming the existence of a nondegenerate Hermitian inner product $(\:,\:)$ on $V$, we define
(2)The catch is that the measure $\mu$ must satisfy certain conditions for this new inner product to be $\pi$invariant. The appropriate notion is: :
(Right) Haar measure : a measure $\mu$ on a locally compact topological group $G$ defined on the Borel subsets $\mathcal{B}$ and satisfying (1) finiteness $\mu(K)<\infty$ for $K$ compact, (2) nondegeneracy $\mu(U)>0$ for $U$ nonempty and open, and (3) translation invariance $\mu(Eg)=\mu(E)$ for $E\in\B$ and $g\in G$.
For such a measure, we have $\int_G f(x) d\mu(x)=\int_G f(xg)d\mu(x)$, making the above inner product invariant under $\pi_x$.
Actually, a Haar measure is unique up to a constant factor, usually chosen so that $\mu(G)=1$. In the finite case, we have $\mu(E)=\frac{E}{G}$, and this process gives the same inner product as above.
Dual and Tensor Representations
An arbitrary representation $(\pi,V)$ of $G$ gives rise to a dual representation $(\check\pi,V^*)$ with $V^*$ the dual vector space to $V$ and $\check\pi$ the contragradient defined by $\langle{\check\pi_g y,x}\rangle=\langle{y,\pi_{g^{1}}x}\rangle$. Note that $y\in V^*$ and $x\in V$, so this inner product is given by $\langle{y,x}\rangle=y(x)$. In terms of matrices (using a given basis and the dual basis), we have $\check\pi_g=[\pi_g^{1}]^T$. Also, an invariant subspace $U\subset V$ corresponds to an invariant subspace $U^\perp\subset V^*$ where $U^\perp=\{y\in V^*: \langle{y,x}\rangle=0 \text{ for all } x\in U\}$.
Two representations $(\pi,V)$ of $G$ and $(\rho,W)$ of $H$ give a tensor representation $(\pi\otimes\rho,V\otimes W)$ of $G\times H$ defined in the obvious manner by $(\pi\otimes\rho)_{(g,h)}(v\otimes w)=\pi_g v\otimes\pi_h w$. In the case that $G=H$, we also have a representation on $G$ with $(\pi\otimes\rho)_g(v\otimes w)=\pi_g v\otimes\pi_g w$. A major question of interest in representation theory is:
 Given (irreducible) representations $V$ and $W$, how does one decompose $V\otimes W$ in terms of irreducible invariant subspaces?
Equivalence of Representations
A map between representations $(\pi,V)$ and $(\rho,W)$ must respect the structure of $G$, and thus is called a $G$map, meaning a linear map $A:V\to W$ such that $A(\pi_g v)=\rho_g (Av)$. These maps form the group $\Hom_G(V,W)$. This also provides the notion of equivalence:
 Equivalent Representations
 those for which there exists a $G$map $A:V\to W$ which is also a vector space isomorphism (a $G$isomorphism). In this case, the matrix of $\pi_g$ with respect to basis $\{v_i\}$ and that of $\rho_g$ with respect to basis $\{Av_i\}$ coincide. This brings up another major problem in representation theory:
 Find all irreducible representations of a given group $G$ up to equivalence.
A first step in solving this problem is the following intuitive theorem:
 Schur's Lemma
 Let a group $G$ and a $G$map $A\in\Hom_G(V,W)$ between irreducible representations $(\pi,V)$ and $(\rho,W)$ be given. We have:
 If $\pi=\rho$, then $A=\gamma I$ for some $\gamma\in\mathbb{C}$; * If $\pi\approx\rho$ (just an equivalence), then $\dim_\mathbb{C}[\Hom_G(V,W)]=1$;
 Otherwise, $\pi\not\approx\rho$, and all such $A$ vanish, so that $\Hom_G(V,W)=\{0\}$.
The proof relies on the fact that $G$maps preserve invariant subspaces, such as $\ker A$, $\mathsf{im} A$, and $\lambda$eigenspaces. This lemma is a simple but powerful tool in representation theory and will be widely used. It follows, for example, that all irreducible representations of abelian groups are 1dimensional, since all 1dimensional subspaces are invariant.
Following abelian groups, the next simplest case is that of finite groups. Representations in this case are also easily classified, and are a good model for the general theory. Before we turn to such taxonomical questions, however, we will look at classifying only pieces of representations… specifically class functions… in the PeterWeyl Theorem.
The PeterWeyl Theorem
The PeterWeyl Theorem is really just a generalization of Fourier series:
 Fourier: any function $S^1\to\mathbb{C}$ from the circle to the complex numbers can be approximated by a finite sum of exponentials.
 PeterWeyl Theorem: any class function on a compact group $G$ (those with $f(gxg^{1})=f(x)$) can be approximated by a sum of characters (a character is the trace of a representation).
There are actually many ways to write the PeterWeyl Theorem, and the classical statement says nothing about class functions and characters. Rather, this result will arise as a corollary.
Fourier Series
We can reinterpret Fourier series in the language of representation theory. First, our group is $G=\mathbb{R}/\Z=S^1$, which is abelian. Hence, all representations are 1dimensional and of the form $e_m:G\to\mathbb{C}^\times$ with $e_m(x)=e^{imx}$ for $m\in\Z$.
We can identify the representation $e_m$ with the vector space $V_m=\mathbb{C}e_m$, which is invariant under translation: $[R(g)e_m](x)=e_m(x+g)=e_m(x)e_m(g)=\lambda_g e_m(x) \in \mathbb{C}e_m$. Then, Fourier theory states that
(3)where $L^2(G)$ is the space of squareintegrable functions on $S^1$ with respect to the standard Haar measure $\int_G f(g)dg = \tfrac{1}{2\pi}\int_0^{2\pi} f(x)dx$. Moreover, $\{e_m\}$ is a complete orthonormal basis for $L^2(G)$.
As we have said, this example is a prototype for the PeterWeyl Theorem, the stated form of which will mimic the $L^2(G)=\bigoplus\sum_m\mathbb{C}e_m$ decomposition.
Preliminary Results
In general, we will assume that $G$ is a compact group. This compactness implies that left and right Haar measures on $G$ are equivalent; thus, all Haar measures are invariant under both left $\mu(gE)=\mu(E)$ and right $\mu(Eg)=\mu(E)$ translations.
We now give a slew of results which lead up to the PeterWeyl Theorem, and can be thought of as extensions of Schur's Lemma:
 (Lemma) For irreducible representations $(\pi,V)$ and $(\rho,W)$ of a compact group $G$, and $A\in\Hom_G(V,W)$ a $G$map, set $A^0=\int_G \rho(g^{1})A\pi(g)dg$ for a normalized Haar measure. Then, $A^0=0$ if $\pi\not\approx\rho$, but $A^0=(\frac{\tr(A)}{\deg\pi})\cdot I$ if $\pi=\rho$.
The map $A^0$ is a sort of average of $A$ between the two representations. The proof is straightforward: one must show that $A^0$ is a $G$map, so Schur's Lemma can be applied. Then one must show how the lemma's claims for the $G$map $A$ extend to $A^0$.
 (Corollary) If $v\in V, v^*\in V^*, w\in W,$ and $w^*\in W^*$, then $\int_G \langle{\pi(g)v,v^*}\rangle \langle{\rho(g^{1})w,w^*}\rangle dg$ equals 0 if $\pi\not\approx\rho$ but $\frac{1}{\deg\pi}\langle{v,w^*}\rangle \langle{w,v^*}\rangle$ if $\pi=\rho$.
This result follows from the previous lemma using the map $A(x)= \langle{x,v^*}\rangle w$. A slight change in language for unitary representations gives the next result.
 (Corollary) Likewise, for unitary representations and Hermitian inner product $(\cdot,\cdot)$, we may take $x_i\in V, y_i\in W$ and obtain $\int_G (\pi(g)x_1,x_2) \overline{(\rho(g)y_1,y_2)}dg$ equals 0 if $\pi\not\approx\rho$ but $\frac{1}{\deg\pi}(x_1,y_1)\overline{(x_2,y_2)}$ if $\pi=\rho$.
 (Schur Orthogonality Relation) Here we restrict to matrix elements: if $\pi_{ij}(g)$ and $\rho_{kl}(g)$ are matrix elements of $\pi$ and $\rho$ with respect to orthonormal bases of $V$ and $W$, then $\int_G \pi_{ij}(g)\overline{\rho_{kl}(g)}dg$ equals 0 if $\pi\not\approx\rho$ or $\frac{1}{\deg\pi}\delta_{ik}\delta_{jl}$ if $\pi=\rho$.
This follows from the above by noting that for basis elements $e_i$ and $e_j$, the matrix element is given by $\pi_{ij}(g) = (\pi(g)e_j,e_i)$.
The Theorem
We are finally ready for the main result:
 PeterWeyl Theorem
 Let $G$ be a compact group, and $\{(\pi^\lambda,V^\lambda)\}_{\lambda\in\Lambda}$ a complete set of inequivalent unitary representations of $G$. Given an orthonormal basis $\{e_i^\lambda\}$ for $V^\lambda$, define $\pi_{ij}^\lambda(g)=(\pi^\lambda(g)e_j^\lambda,e_i^\lambda)$. Then, $\{f_{ij}^\lambda\}$ where $f_{ij}^\lambda=\sqrt{\deg\pi^\lambda}\pi_{ij}^\lambda$ is a complete orthonormal set in $L^2(G)$, and $\Lambda$ must be countable.
For the proof, first note that $\{f_{ij}^\lambda\}$ is clearly orthonormal by the Schur Orthogonality Relation given above. It remains to show completeness and countability. In a separable Hilbert space, every orthonormal set is countable, and $L^2(G)$ is separable since the set of rational polynomials in $L^2(G)$ forms a countable dense subset.
Actually, all polynomials are matrix coefficients: if $p(x)$ is a polynomial of degree $d$, then $(R,V_d)$ where $R$ is right translation $R_g(q)(x)=q(xg)$ and $V_d$ is the set of polynomials of degree $\leq d$ is a representation. Moreover, selecting $e^*$ such that $\langle{e^*,q}\rangle=q(I)$ gives $p(x)=\langle{R_g(q),e^*}\rangle$ meaning $p$ is a matrix coefficient. So the matrix coefficients are dense in $L^2(G)$ since the polynomials are, and this verifies completeness.
The PeterWeyl Theorem allows us to decompose $L^2(G)$ into subspaces invariant under $G$. Specifically, we can write
(4)where $M(V^\lambda)$ is spanned by the matrix coefficients of the representation $(\pi^\lambda, V^\lambda)$. Note that $M(V^\lambda)$ is invariant under the action of $G$ by left/right translation.
Characters and Class Functions
Basics
The PeterWeyl Theorem also has consequences for characters and class functions. Characters are functions in $G$ given by the trace of a representation, while a class function is any function invariant under conjugation:
 Character
 given a finitedimensional representation $(\pi,V)$, it is the function $\chi_\pi:G\to G$ given by $g\mapsto \tr(\pi_g)$.
 Class Function
 a function $f:G\to\mathbb{C}$ such that $f(gxg^{1})=f(x)$.
Thus, a class function is really a function on the conjugacy classes of the group. Of course, all characters are class functions since $\tr(ABA^{1})=\tr(B)$. Characters have the following properties:
 Equivalent representations have the same character: $\pi\approx\rho$ implies $\chi_\pi=\chi_\rho$. Actually, the reverse implication also holds.
 For direct sum and tensor product representations: $\chi_{\pi\oplus\rho}=\chi_\pi+\chi_\rho$ and $\chi_{\pi\otimes\rho}=\chi_\pi \chi_\rho$.
 For the dual representation: $\chi_{\check\pi}(g)=\chi_\pi(g^{1})$; for unitary representations, $\chi_{\check\pi}(g)=\chi_\pi(g)=\chi_\pi(g^{1})$.
 (Schur Orthogonality for Characters) Given irreducible representations $(\pi,V)$ and $(\rho,W)$ of $G$, the characters satisfy $(\chi_\pi,\chi_\rho) = \delta_{\pi,\rho}$: 1 if $\pi\approx\rho$ and 0 if $\pi\not\approx\rho$.
Characters for Counting
The Schur Orthogonality relation above implies that a lot of information about a representation can be read directly from its character. For example, if a representation $(\pi,V)$ has irreducible direct summands $(\pi_i,V_i)$, and $\rho$ is an irreducible character of $G$, then
(5)the number of pieces of $\pi$ equivalent to $\rho$. Thus, we could write $\pi\approx \sum_\rho m_\pi(\rho) \rho.$ Note that this implies that $\pi$ is irreducible iff $(\chi_\pi,\chi_\pi)=1$, or more generally $(\chi_\pi,\chi_\pi)=\sum_\rho m_\pi(\rho)^2$.
Characters are especially useful because they form a complete orthonormal basis for $L^2$ class functions. This is perhaps the most important consequence of the PeterWeyl Theorem. We present the justification in a number of steps:
 Given $f\in L^2(G)$, let $\pi_f:V\to V$ be given by $\pi_f=\int_G f(g) \pi_g dg$, meaning $\langle{\pi_f(v),v^*}\rangle=\int_G\langle{f(g) \pi_g(v),v^*}\rangle.$ This essentially gives a representation over $L^2(G)$.
 Given a class function $f$ and $\pi$ irreducible, then $\pi_f=\alpha\cdot I$, where $\alpha=\frac{1}{\deg\pi}(f,\bar\chi_\pi)$. This follows based on a result prior to the PeterWeyl Theorem for the matrix $A=\pi_f$, noting that $\tr(\pi_f)=(f,\bar\chi_\pi)$.
 Given an orthonormal basis $\{e_i\}$ of $V$ with $\pi_{ij}(a)=(\pi_a e_j,e_i)$, then $(f,\pi_{ij})=\frac{1}{\deg\pi}(f,\chi_\pi)\delta_{ij}$. A straightforward evaluation demonstrates this, noting that $(\check f,\chi_\pi)=(f,\chi_\pi)$.
These results give an easy demonstration of:
 Theorem
 Given a class function $f$ and an irreducible representation $(\pi,V)$ of $G$, we can write $f=\sum_{\lambda\in\Lambda} (f,\chi^\lambda)\chi^\lambda$, where $\Lambda$ is the set of all representations on $G$. Hence, $\{\chi^\lambda\}$ is a complete orthonormal basis for $L^2$ class functions.
We can also decompose a representation uniquely in terms of its $\rho$isotypic subspace, the direct summand of all irreducible components equivalent to $\rho$. Thus,
(6)Uniqueness does not hold for the decomposition into irreducible subspaces however.
Characters can be used to show that the representation $\pi\otimes\rho$ of $G\times H$ is irreducible iff $\pi$ and $\rho$ are, and a given irreducible representation $\tau$ of $G\times H$ can be decomposed $\tau=\pi\otimes\rho$ into irreducibles.
Finally, given $A\in\Hom_G(V,V)$, one can write (i) $A=\sum_{i=1}^n \alpha_i \pi(a_i)$ where $\alpha_i\in\mathbb{C}$, $a_i\in G$, (ii) $A=\pi_f$ where $f\in C(G)$ or $f$ is a matrix coefficient of $(\pi,V)$. The proof relies on the fact that $(\tau,\Hom_G(V,V))$ is irreducible, where $\tau_{a,b}T=\pi_a T\pi_{b^{1}}$. One must show that the subspaces generated by functions of the forms given in (i) and (ii) are both invariant, hence all of $\Hom_G(V,V)$.
Representations of Finite Groups
Representation theory for finite groups contains a surprisingly large amount of the theory used in the more general case. We have already seen that all finite representations are unitary using the averaging technique
(7)a fact which will bring us to several more formula allowing us to completely classify the irreducible representations of a finite group.
Representations via the Group Algebra
Classifying the Representations of Permutation Groups
The methods used to classify the representations of the symmetric groups $S_d$ are surprisingly adaptable to the case of general Lie groups. In particular, we introduce Young Tableau and the Young Symmetrizer, which will come in again with the classification of general $SL(n)$ Lie groups.
The Young Tableau
The irreducible representations of $S_d$ are in onetoone correspondence with the conjugacy classes of $S_d$, which may be described in terms of partitions:
 Young Diagrams/Tableaus
 A Young Diagram is a graphical representation of a partition of the integer $d$, that is a set of integers $\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_k)$ such that $\lambda_i\geq \lambda_{i+1}$ and $\sum \lambda_i=d$. One may represent this as several rows of boxes, with $\lambda_i$ boxes on the [[$i$th row. If in addition the boxes are numbered, the diagram is called a Young Tableau.
Each partition has a conjugate partition $(\mu_1,\ldots,\mu_{\lambda_1})$ where $\mu_i$ is the number of boxes in the $i$th column of the Tableau. For example, the partition [[$(3,3,2,1,1)$ of 10 is conjugate to $(5,3,2)$.
The Young Symmetrizer
The numbering in a Tableau allows the definition of%
(8)These sets will be very important for classifying the irreducible representations. In terms of the group algebra $\mathbb{C}S_d$ these give rise to the elements:%
(10)If $S_d$ acts on $\vprod d$ by permuting its tensor powers ($V$ is any vector space), then these elements can be viewed as functions, and have images:%
(11)Thus, $a_\lambda$ produces a tensor product of symmetric powers of $V$ and can be thought of as a symmetrizer function, while $b_\lambda$ produces a tensor product of alternating powers of $V$ and can be thought of as an antisymmetrizer.
Combining these functions gives:%
 Young Symmetrizer
 the group ring element (or function)%
The images of these functions give essentially all irreducible representations of $GL(V)$. Indeed, $c_\lambda^2=n_\lambda c_\lambda$ for some scalar $n_\lambda$, and if $c_\lambda$ acts on $\mathbb{C}S_d$ by right multiplication (so $V=\mathbb{C}S_d$), then the image $\mathsf{im}(c_\lambda)=V_\lambda$ is a unique irreducible representation of $S_d$. All irreps may be obtained in this way.
The element $c_\lambda'=\frac1{n_\lambda}c_\lambda$ is therefore an idempotent, meaning $c_\lambda'^2=c_\lambda'$. It is also sometimes called a projector, since it may be interpreted as a map from $\mathbb{C}S_d$ to $V_\lambda$.
Some Specific Examples
As examples, consider the first few permutation groups. $S_2$ has just two partitions: $(2)$ and $(1,1)=(2)'$ which produce representations $\mathbb{C}S_2[(e_1+e_{12})\cdot(e_1)]=\mathbb{C}[e_1+e_{12}]=V_{(2)}$, the trivial representation, and $\mathbb{C}S_2[(e_1)\cdot(e_1e_{12})]=\mathbb{C}[e_1e_{12}]=V_{(1,1)}$, the alternating representation. Here, the representations are given as $S_2$modules.
Note that the correspondence between a representation $(\rho,V)$ with $\rho:G\to GL(V)$ and a $G$module is essentially given by looking for the basis. In particular, the trivial representation is the one with $\rho(g)(v)=v$ for all $v\in G$. Because $V_{(2)}$ has basis $\{e_1+e_{12}\}$, we have $v=\alpha(e_1+e_{12})$, which is of course fixed by left multiplication of any $g\in S_2$. Similarly, $g\cdot v=g\cdot\alpha(e_1e_{12})=\sgn(g)\alpha(e_1e_{12})=\sgn(g)\cdot v$, so that the second module is just the alternating representation.
Moving to $S_3$, we have three partitions, $(3), (1,1,1),$ and $(2,1)$. Just as above, we see that $V_{(3)}$ is the trivial representation, and $V_{(1,1,1)}$ is the alternating representation. For $V_{(2,1)}$, we have%
(14)Since the permutation $(12)$ fixes $v$, we can reduce this to $\mathbb{C}\{(1),(13),(23)\}[v]=\mathbb{C}\{v,(13)v\}$, where the last step follows since $v+(13)v+(23)v=0$. Thus, $V_{(2,1)}$ is the representation with basis
(15)with $G=S_3$ acting on these basis elements in the only manner possible. This is called the standard representation, which for general $S_d$ is an irreducible quotient of $\mathbb{C}^d$ by an invariant subspace (the sum of basis elements $e_1+\cdots+e_d$), where $S_d$ acts on $\mathbb{C}^d$ by permuting coordinates.
For general $S_d$, it is clear that $(d)$ is the trivial representation and $(1,1,\ldots,1)$ is the alternating representation. It is actually also true that $(d1,1)$ is the standard representation. As above we compute $V_{(d1,1)}$:
(16)which has basis $\{v,(1d)v,(2d)v,\ldots,((d2)d)v\}$, if I had to guess.
What about the conjugate representation $V_{(2,1,\ldots,1)}$? Well, we have [[$V_{(2,1,\ldots,1)}=V_{(d1,1)} \otimes V_{(1,\ldots,1)}$, since tensoring with the alternating representation $V_{(1,\ldots,1)}$ corresponds to switching the signs of all odd permutations in the expression $c_\lambda$, so that $a_\lambda \cdot b_\lambda$ becomes $b_{\lambda'}\cdot a_{\lambda'}$, where $\lambda'$ is the partition conjugate to $\lambda$. But it is not too hard to show that $\mathbb{C}S_d[b_{\lambda'}\cdot a_{\lambda'}]\isom\mathbb{C}S_d[a_{\lambda'}\cdot b_{\lambda'}]=V_{(2,1,\ldots,1)}.$ For the exact same reason, it is true in general that $V'=V\otimes U$, where $V'$ is the conjugate representation to $V$, and $U$ is the alternating representation.
The above discussion shows that the irreps of $S_4$ are $U,U',V,V'$ (the trivial, alternating, standard, and standard conjugate representations), plus $W=V_{(2,2)}$. This last is 2dimensional, with basis computed to be $\{v,(13)v\}$ (wild guess) with
(17)As a last example, $S_5$ has irreps $U,U',V,V',W,W',\bigwedge^2 V$. How do we show that $V_{(3,1,1)}=\bigwedge^2 V$?]] Actually, it is generally true that [[$V_{(ds,1,\ldots,1)}=\bigwedge^s V$, although I'm not sure at this point why.
Projectors and the Character Table
There is a nice correspondence between the character table of a given representation and the projectors $c_\lambda$. The permutations in $S_n$ may be partitioned via their conjugacy class. The character table can be used to write down the projector $c_\lambda$: its coefficient for a given permutation is just the entry in the character table for that permutation. These projectors can then be written in terms of symmetrizers and antisymmetrizers.
For example, the standard representation $V_{21}$ of $S_3$ has character values 2 on the identity element, 0 on swaps, and $1$ on the $(123)$ conjugacy class. Thus,
(18)where the numbers in parentheses are permutations. Incredibly, this also works for any representation $V_\lambda$ of any permutation group $P_n$. Diagrams can be used to greatly simplify the representations of these projectors.
Diagrams for Class Theory
In the general formula for the projector $c_\lambda$, there are 2 important components: the symmetrizer pieces $a_\lambda$, and the antisymmetrizer pieces $b_\lambda$. We now show how to write these two pieces down in terms of diagrams.
Computational Examples
We now turn to representations of nonfinite groups, and illustrate a few cases which will be especially demonstrative of the general classification theory.
Representations of $sl_2\mathbb{C}$
The representations of $sl_2\mathbb{C}$ are a very important case because they indicate the fundamental structure behind every representation of a Lie algebra. The eigenvalues of any irreducible representation form a neat structure. In the case of $sl_2\mathbb{C}$, these eigenvalues are colinear and occur on the nice integer lattice, while in the more general case they also have a very nice structure on a (higherdimensional) lattice.
The best part about the irreducible representations of $sl_2\mathbb{C}$ is that there is one, and only one, for each integer $n\geq 0$, and we can describe how to write down every one of them. Our basic strategy involves three steps:
 Find a suitable abelian subspace $\mathsf{h}$ of the Lie algebra. Break the Lie algebra into pieces corresponding to this $\mathsf{h}$. Check that all these pieces preserve the given decomposition.
 Decompose $V$ in terms of the eigenspaces of $\mathsf{h}$. Check that this decomposition is preserved by the other pieces of the Lie algebra.
 Determine the possible configuration of the eigenspaces by examining their eigenvalues.
Actually, these are the three steps used to write down the representations of any Lie algebra, although it certainly becomes more complicated. This is especially true since $W$ will be cyclic for the case at hand, but not in general.
Step 1:
For our first task, we write down the following basis of $sl_2\mathbb{C}$:
Our abelian subspace $\mathsf{h}$ consists of the diagonal elements and is generated by $H$. Note that we have $[H,X]=2X, [H,Y]=2Y$, and $[X,Y]=H$.
Step 2:
The action of $H$ on our representation $V$ is diagonalizable (a Jordan decomposition argument), so we may decompose $V=\oplus V_\alpha$ where $H(v)=\alpha\cdot v$ for all $v\in V_\alpha$.
Step 3:
Now, the question is how the rest of $sl_2\mathbb{C}$ acts on $V_\alpha$; specifically, we must consider $X$ and $Y$. We have for $v\in V$
and so $X:V_\alpha\to V_{\alpha+2}$. Likewise, for $Y$ we have $Y:V_\alpha\to V_{\alpha2}$.
Finitedimensionality and irreducibility combine to give the existence of a vector $v$ such that $\{v,Y(v),Y^2(v),\ldots\}$ spans $V$. Here, we let $v\in V_\beta$, the largest nonzero eigenspace. This is a finitedimensional vector space, and invariant under the actions of $X$, $Y$, and $H$, hence under all of $sl_2\mathbb{C}$. Therefore, it must be all of $V$.
This means that all eigenspaces are 1dimensional, and that $V$ is the direct sum of $V_\beta, V_{\beta2}, \ldots, V_{\beta2k}$. Now, if $w\in V_{\beta}$, then $Y^{k+1}(w)=0$ but $Y^k(w)\neq 0$, and so $X(Y^{k+1}(v))=0=k(\betak+1)Y^k(v)$. Thus, $\beta=k1$ is an integer, and an irreducible representation $V^{(n)}$ looks like:
(21)We can actually show that $V^{(n)}$ is isomorphic to the representation $\mathsf{Sym}^nV$ by analyzing eigenvalues. It is clear that a representation is uniquely defined by its set of eigenvalues. This fact allows us to solve the ClebschGordon Problem: write down formulae for various products of representations in terms of the irreducibles. In particular, since the eigenvalues of $W\otimes W'$ are just the sum of those in $W$ and those in $W'$, we have:
(22)We also have $\mathsf{Sym}^n(\mathsf{Sym}^2V)=\bigoplus_{a=0}^{[n/2]} \mathsf{Sym}^{2n4a}V.$ One could also obtain these formulae through algebraic geometry…
Representations of $sl_3\mathbb{C}$
As might be expected, raising the dimension greatly complicates the problem of writing down the irreducible representations. It is still quite possible, however, and still involves the same three basic steps. The approach must be generalized somewhat, however, and also streamlined. We will introduce a few new terms here which occur in the general theory.
Step 1:
The first difficulty to overcome is that the abelian subspace $\mathsf{h}$ from above is no longer generated by a single element. Thus, we must redefine our notions of eigenvector, eigenvalue, and eigenspace to accommodate several generators. Given an abelian subspace $\mathsf{h}$, an eigenvector will satisfy the equation $H(v)=\alpha_H\cdot v$ for all $H\in \mathsf{h}$. Of course, now $\alpha_H$ depends on $H$, and thus can be viewed as a element $\alpha \in \mathsf{h}^*$; so the eigenvalues are now elements of the dual space rather than scalars.
Remarkably, for a suitable choice of $h$ in a general Lie algebra, the action of $\mathsf{h}$ on the Lie algebra $\mathsf{g}$ and all its representations is simultaneously diagonalizable. What this means is that we can decompose $sl_3\mathbb{C}=\mathsf{h}\oplus(\oplus \mathsf{g}_\alpha)$, where $g_\alpha$ is an eigenspace for the adjoint action of $h$: whenever $H\in \mathsf{h}$, $Y\in \mathsf{g}_\alpha$ we have $\ad(H)(Y)=[H,Y]=\alpha_H\cdot Y$.
Our choice of $\mathsf{h}$ will (again) be the set of diagonal matrices. Finding the $\mathsf{g}_\alpha$ is not difficult; the only elements $M$ with $[D,M]=\lambda M$ for all diagonal matrices $D$ are those only one nonzero entry. Hence, the subspaces $\mathsf{g}_\alpha$ are generated by the matrix $E_{ij}$ whose only nonzero entry is $e_{ij}=1$. Letting $D=(a_1,0,0;0,a_2,0;0,0,a_3)$, the eigenvectors are elements of the space $h^*$ generated by $L_i:D\mapsto a_i$. Since $[D,E_{ij}]=(a_ia_j)E_{ij}$, we see that $E_{ij}$ generates the space $\mathsf{g}_{L_iL_j}$. In particular, we have:
(23)Clearly, $h$ preserves this eigenspace decomposition, but it is also true that each $\mathsf{g}_\alpha$ preserves it. In fact, by considering elements of the given subspaces, one can show that $\ad(\mathsf{g}_\alpha):\mathsf{g}_\beta\to \mathsf{g}_{\alpha+\beta}$.
Step 2:
Now, we consider an arbitrary representation $V$ of $sl_3\mathbb{C}$, and break it up into eigenspaces $V=\oplus V_\alpha$, so that whenever $v\in V_\alpha$ and $H\in h$ we have $Hv = \alpha_H \cdot v$. When $X\in \mathsf{g}_\beta$ is outside $h$, we have $H(X(v))=(\alpha_H+\beta_H)\cdot X(v)$, and so the action of $\mathsf{g}_\beta$ on the representation carries $V_\alpha$ to $V_{\alpha+\beta}$. In particular, we see that the eigenvalues of the representation are all in some translate of the lattice $\Lambda_R$ in $h^*$ generated by $\{L_iL_j\}$.
Now for some terminology. The eigenvalues $L_iL_j$ of the adjoint action on $sl_3\mathbb{C}$ are called roots, with the corresponding $\mathsf{g}_{L_iL_j}$ root spaces. The lattice formed by the roots is called the weight lattice, and denote $\Lambda_R$. For an arbitrary representation, the eigenvalues $\alpha \in \mathsf{h}^*$ are called weights, and the $V_\alpha$ weight spaces. So, the roots determine how we decompose the Lie algebra, while the weights determine how we decompose the representation.
Step 3:
To finish the proof requires a generalization of an ‘extremal’ vector. In one dimension, the notion is trivial, but here we must choose a linear functional which is positive on half the roots and negative on the other half. The mechanism used to define an extreme vector is not so important, however. What is important is that there is a vector $v\in V_\alpha$ which is killed by each of $E_{12}, E_{13}$, and $E_{23}$. This corresponds to the case for $sl_2(\mathbb{C})$ when $v$ was the extremal vector killed by $X$.
Given this, we see as before that $V$ is generated by the images of $v$ under the opposite elements $E_{21}, E_{31}$, and $E_{32}$. This restricts us to a sort of ‘corner region’ of $h^*$. There are a few more key observations to make:
 The eigenspaces on the boundaries of the region correspond to irreducible representations on $sl_2\mathbb{C}$.
 There is a symmetry about the three lines $\langle{[E_{ij},E_{ji}],L}\rangle=0$.
This implies that the weights (eigenvalues) of the representations are contained in either a hexagon or a triangle (with $\Z_3$ symmetry). It is clear that the interior must be filled out, and the interior weights must have multiplicity 1. This completes the $sl_3\mathbb{C}$ case.
The General Case
In order to classify all (irreducible/finitedimensional) representations for an arbitrary (semisimple) Lie algebra, one must take steps analogous to those above for $sl_(2\mathbb{C})$ and $sl_3(\mathbb{C})$. The theory is extremely practical, and provides a mechanical calculation of possibilities. We will go through the steps in detail here.
The Cartan Decomposition (Step 1)
We need to find a suitable abelian subspace, analogous to the set of diagonal matrices for $sl_n(\mathbb{C})$. Our choice is restricted to maximal abelian subalgebras which act diagonally on the adjoint representation. This is called a Cartan subalgebra, and is a welldefined piece of every semisimple Lie algebra.
The Cartan subalgebra $\mathsf{h}$ is used to define the Cartan decomposition of the Lie algebra: $\mathsf{g}=\mathsf{h}\oplus (\oplus \mathsf{g}_\alpha)$. There will be a finite number of eigenvalues $\alpha$ called roots, with corresponding subspaces $\mathsf{g}_\alpha$ called root spaces. As we saw before, the roots comprise a subset $R\subset \mathsf{h}^*$ of the dual space and their configuration therein is a useful depiction of the structure of $g$. Note that the adjoint action of $\mathsf{g}_\alpha$ takes $\mathsf{g}_\beta$ to $\mathsf{g}_{\alpha+\beta}$, so that the decomposition is preserved by the action of $\mathsf{h}$.
We note a few facts regarding this decomposition. First, each root space $\mathsf{g}_\alpha$ is onedimensional. Second, $R$ generates a lattice $\Lambda_R \subset \mathsf{h}^*$ with rank equal to $\dim(\mathsf{h})$, meaning the roots span $\mathsf{h}^*$. Last, $\alpha$ is a root iff $\alpha$ is also a root.
Decomposing the Representation (Step 2)
We now turn our attention to an arbitrary representation $V$, and note that the action of $\mathsf{h}$ on $V$ gives the decomposition $V=\oplus V_\alpha$. The eigenvalues are now called weights and the $V_\alpha$ are called weight spaces. The dimension of $V_\alpha$ is called the multiplicity of the weight. A simple diagram in $\mathsf{h}^*$ suffices to depict the weights and their dimensions.
Note that the weights are all congruent modulo the root lattice, meaning they differ by linear combinations of the roots of the Lie algebra. This is related to the fact that a piece $\mathsf{g}_\beta$ of the Cartan decomposition takes $V_\alpha$ into $V_{\alpha+\beta}$.
Analyzing the Configuration of Eigenvalues/Weights (Step 3)
This third step is by far the most detailed, although it is really not much harder than the previous two. It basically amounts to finding an extremal piece of the representation (which turns out to be an $sl_2(\mathbb{C})$ representation, and using it to fill in the eigenvalues for the rest of the representation.
First, note that for any root space $g_\alpha$, we have a corresponding subalgebra $s_\alpha=\mathsf{g}_\alpha \oplus \mathsf{g}_{\alpha} \oplus [\mathsf{g}_\alpha,\mathsf{g}_{\alpha}]$ which is, in fact, isomorphic to $sl_2(\mathbb{C})$. To see this, one can choose elements $X_\alpha\in \mathsf{g}_\alpha$, $Y_\alpha\in \mathsf{g}_{\alpha}$, and $H_\alpha\in [\mathsf{g}_\alpha,\mathsf{g}_{\alpha}]$ which will act exactly like $X,Y,H\in sl_2(\mathbb{C})$. Thus, when we restrict to $\mathsf{s}_\alpha$ we obtain a representation of $sl_2(\mathbb{C})$. In particular, all eigenvalues $\beta\in \mathsf{h}^*$ of the representation must be $\mathbb{Z}$valued on all elements $H_\alpha$.
We can say more than that about the eigenvalues of $H_\alpha$, however: they are symmetric about the hyperplane $\{\langle{H_\alpha,\beta}\rangle=0\}$. Letting $W_\alpha(\beta)= \beta\frac{2\beta(H_\alpha)}{\alpha(H_\alpha)}\alpha= \beta\beta(H_\alpha)\alpha$ be the reflection in this hyperplane, we see that $\alpha$equivalence classes of $V$ such as $V_{[\beta]}=\oplus V_{\beta+n\alpha}$ are all invariant under the action of $\mathsf{s}_\alpha$, and so fixed by the $W_\alpha$ involution. Moreover, from the structure of $sl_2(\mathbb{C})$ representations, one sees that each equivalence class can be written $V_{[\beta]}=V_{\betam}\oplus V_{\betam+2} \oplus \cdots \oplus V_{\beta+m}$.
The set of such hyperplane reflections $W_\alpha$ forms a group called the Weyl Group. In general, the set of weights, and the multiplicities of the weights, of any representation of $\mathsf{g}$ is invariant under the action of the Weyl group. One can see this by noting that it is true for each equivalence class $V_{[\beta]}$. Here is another fact: every element of the Weyl group is induced by an automorphism of $\mathsf{g}$ which fixes $\mathsf{h}$.