Originally based on Giuseppe Malavolta master’s thesis. It is an ongoing open project, feel free to contribute. A pdf copy of the Giuseppe’s thesis is available here
The spirit of the wiki is described in this short Introduction. Also all the Lie groups technology needed for representations is contained in Lie groups and algebras representations. It is convenient to read the Physics notation too.
In Bourbaki’s parlance, given any mathematical struture $\mathcal{S}$ its symmetries are the structure-preserving invertible self-transformations of $\mathcal{S}$. In the more modern categorical language this is expressed by saying that the symmetries of an object $X$ are the elements of its automorphism group.
Concretely, the situation one is customary faced with is the following: one has a basic strcuture $\mathcal{S}_{basic}$ which is enriched with an additional structure, becoming $\mathcal{S}_{rich}$. Symmetries of $\mathcal{S}_{rich}$ are then symmetries of $\mathcal{S}_{basic}$ preserving the additional structure.
In linear algebra one can consider a real vector space $V$ and endow it with a metric $g$; the symmetries of $(V,g)$ are then the isometries, i.e., the linear automorphisms $\varphi:V\to V$ such that $\varphi^{*}g=g$. Note that $\mathrm{Aut}(V,g)$ is the stabilizer of $g$ for the action of $\mathrm{Aut}(V)$ on the space of symmetric bilinear forms on $V$. Similarly, if $V$ is even dimensional, one can endow it with a symplectic form $\omega$, and consider the group of symplectic transformations of $(V,\omega)$. This example immediately generalizes from vector spaces to differential manifolds: one considers isometries of (pseudo-)Riemannian manifolds and symplectomorphisms of symplectic manifolds.
In classical mechanics one considers a differential manifold $M$ endowed with a Lagrangian, i.e., a smooth function $L:T M\to\mathbb{R}$. Symmetries of such a Lagrangian system are diffeomorphisms $\varphi:M\to M$ such that $(d\varphi)^{*}L=L$, where $d\varphi:T M\to T M$ is the differential of $\varphi$.
The Lagrangian example above has a natural generalization in Hamiltonian mechanics. An Hamiltonian system is a triple $(M,\omega,H)$ where $(M,\omega)$ is a symplectic manifold, and $H:M\to\mathbb{R}$ is a smooth function, called the Hamiltonian. A symmetry of an Hamiltonian system is a symplectomorphism $\varphi$ of $(M,\omega)$ such that $\varphi^{*}H=H$.
In all of the above examples, symmetries are not just a group: they are a Lie group (eventually an infinite-dimensional one). So it is meaningful to talk of infinitesimal symmetries. Mathematically speaking, these are elements of the Lie algebra of the Lie group of symmetries. It is interesting to remark that in infinite-dimensional situations, the Lie algebra structure of the vector space of infinitesimal symmetries is a perfectly well define object, even when a rigorous infinite-dimensional Lie group structure on the group od symmetries is not defined.
In example \ref{ex-hamiltonian}, infinitesimal symmetries are vector fields $X$ on $M$ such that $\mathcal{L}_{X}\omega=0$ and $\mathcal{L}_{X}H=0$, where $\mathcal{L}_{X}$ denotes the Lie derivative along $X$. Among these symmetries, of particular interest are the Hamiltonian ones, i.e., vector fields of the form $X_{f}=\{ f,-\}$, where $f$ is a smooth function on $M$ and $\{-,-\}$ is the Poisson bracket induced by the symplectic struture of $M$. Of the two conditions an Hamiltonian vector field has to satisfy in order to eb an infinitesimal symmetry of the system, the first one, $\mathcal{L}_{X_{f}}\omega=0$ is always satisfied (this is Liouville’s theorem), whereas the second one, i.e., $\mathcal{L}_{X_{f}}H=0$ is equivalent to $\{ f,H\}=0$. In other terms, the Lie algebra of infinitesimal Hamiltonian symmetries is identified with the Lie algebra of smooth functions on $M$ Poisson-commuting with $H$ (modulo the constants, i.e., the kernel of $f\mapsto X_{f}$). In terms of classical mechanics, each infinitesimal Hamiltonian symmetry $f$ is a constant of motion, i.e., if $x:\mathbb{R}\to M$ is the time evolution of the Hamiltonian system with initial datum $x(0)=x_{0}$, then $f(x(t))=f(x_{0})$ for every $t\in\mathbb{R}$.
In quantum mechanics one considers a noncommutative version of example \ref{ex-infinitesimal-hamiltonian}, where smooth functions are replaced by self-adjoint operators and Poisson brackets are replaced by commutators. More precisely, quantum states are norm 1 vectors in an Hilbert space $\mathcal{H}$ and the probability of a transition of the system from a state $\phi$ to a state $\psi$ is
Symmetries of the system will have to preserve all these transition probabilities, so they will be unitary operators $U:\mathcal{H}\to\mathcal{H}$. Moreover, the system is endowed with a distinguished self-adjoint operator, the Hamiltonian $H:\mathcal{H}\to\mathcal{H}$, and the Poisson-commutation relation $\{ f,H\}=0$ of Hamiltonain mechanics is translated in the commutation relation $[U,H]=0$. This in particular implies that any symmetry $U$ of $(\mathcal{H},H)$ preserves the $H$-eigenspaces decomposition of $\mathcal{H}$. In the quantum mechanics parlance, the scalars in the spectrum of the operator $H$ are called the energy levels of the Hamiltonian.
In the above example we restricted our attention to unitary operators by the principle of “symmetries of the rich structure are a subgroup of symmetries of the basic structure”, and assuming that the basic structure was that of an Hilbert space, so that its symmetries were the automorphisms of $\mathcal{H}$ as a Banach space. One could however consider more general symmetries, by looking at $\mathcal{H}$ just as a set endowed with the function
and then consider the set of all invertible self-transformations of the set $\mathcal{H}$ preserving $P$. But this actually brings in nothing new: a $P$-preserving self-transformation of $\mathcal{H}$ is either a unitary transformation of $\mathcal{H}$ or a unitary transformation of $\overline{\mathcal{H}}$ (Wigner’s theorem, wigner-unitary).
For a given object $X$ in some category $\mathcal{C}$, the complete description of the whole group of symmetries $\mathrm{Aut}(X)$ may be an extremely difficult problem. In concrete situations one is often more interested in distinguished (in some sense) and well-behaved subgroups of $\mathrm{Aut}(X)$, or more in general in group representation with values in $\mathrm{Aut}(X)$. For instance, if $G$ is a Lie group, a realization of $G$ as a symmetry group for a quantum mechanical system is a group homomorphism
which is smooth in the sense that for any two states $\phi$ and $\psi$ in $\mathcal{H}$, the “matrix coefficient” $\langle\phi|\rho(g)\cdot\psi\rangle$ is a smooth complex-valued function on $G$. Here we are denoting by $U(\mathcal{H})$ the group of unitary operators on $\mathcal{H}$. We will also occasionally meet linear representations
where $\mathrm{Aut}(\mathcal{H})$ is the group of automorphisms of $\mathcal{H}$ in the category of Hilbert spaces. Since the category of Hilbert spaces is a full subcategory of Banach spaces, $\mathrm{Aut}(\mathcal{H})$ is the group of automorphisms of $\mathcal{H}$ as a Banach space, i.e., the group of continuous invertible linear endomorphisms of $\mathcal{H}$ with a continuous inverse.
Let us now focus on unitary representations. If $\mathcal{H}_{0}$ is a subspace of $\mathcal{H}$ which is stable under the unitary action of the symmetry group $G$, then also $\mathcal{H}_{0}^{\perp}$ is $G$-stable. This immediately implies the following result.
All unitary representations of a group $G$ are completely reducible, i.e., are direct sums of irreducible representations.
When $G$ is compact, each linear representation $G\to\mathrm{Aut}(\mathcal{H})$ is actually an unitary representation (up to conjugation). Indeed, let $\mu_{G}$ be the Haar measure of $G$, i.e., the unique normalized biinvariant measure on $G$ and set
Then $\langle\!\langle-|-\rangle\!\rangle$ is an inner product on $\mathcal{H}$ inducing an Hilbert space structure equivalent to the original one, and the representation $\rho$ is manifestly unitary with respect to this new inner product. So, by the above proposition, we obtain that all continuous linear representations of a compact group $G$ on an Hilbert space are completely reducible.
A less trivial result is the following.
Let $G$ be a compact Lie group (or, more in general, a compact topological group). Then, irreducible representations of $G$ are finite dimensional.
This is proved by showing that if $G$ is compact then each unitary representation of $G$ is the direct sum of its finite-dimensional subrepresentations. A complete proof of this statement can be found, e.g., in brocker-tomdieck. Combining the two propositions above, one obtains the following principle: to understand compact Lie groups of symmetries in quantum mechanics, one has to study their finite dimensional irreducible representations.
Throughout this section we will assume the reader is familiar with the basics of the theory of Lie algebra and of the Lie groups/Lie algebra correspondence. So we will directly focus on the examples we will need in the sequel. Details on the general theory can be found, e.g., in fulton-harris.
If the Lie group $G$ is connected, then its group structure is entirely determined by an arbitrary small neighborhood of the identity element $e$ and so, ultimately, by the Lie algebra structure of its tangent space at the identity. The Lie algebra $\mathfrak{g}=T_{e}G$ is called the Lie algebra of the Lie group $G$; it is canonically identified with the Lie algebra of left-invariant vector fields on $G$ by the Lie algebra homomorphism
If $\{\mathbf{e}_{i}\}_{i\in I}$ is a linear basis of $\mathfrak{g}$, then the Lie algebra structure of $\mathfrak{g}$ is completely encoded in the structure constants of the Lie bracket:
The Lie algebra $\mathfrak{so}_{3}$ of the Lie group $SO(3)$ of rotations in $\mathbb{R}^{3}$ is the Lie algebra of $3\times3$ real antisymmetric matrices. Its canonical basis is given by the generators of rotations around the $x$-, $y$-, and $z$-axis, respectively:
These are colloquially called the infinitesimal rotations around the coordinate axes. It is starightforward to check the commutation relations
i.e.,
In the physics literature it is customary to consider the complexified Lie algebra $\mathfrak{so}_{3;\mathbb{C}}=\mathfrak{so}_{3}\otimes\mathbb{C}$ with basis $\{ J_{1},J_{2},J_{3}\}$, where $J_{i}=\mathbf{i}X_{i}$. The commutation relations of the $J_{i}$ are
The Lie algebra $\mathfrak{su}_{2}$ of the Lie group $SU(2)$ is the Lie algebra of $2\times2$ complex anti-Hermitean matrices. The inner product $(A,B)\mapsto\mathrm{tr}(A^{*}B)$ makes $\mathfrak{su}_{2}$ a 3-dimensional Euclidean space. The action of $SU(2)$ by conjugation on $\mathfrak{su}_{2}$ is isometric with respect to this inner product, thus giving a Lie group homomorphism $SU(2)\to O(3)$. Since $SU(2)$ is connected, the image of this homomorfism is necessarily contained in the connected component $SO(3)$. Moreover, one checks that $SU(2)\to SO(3)$ is actually surjective and that its kernel is $\{\mathrm{Id},-\mathrm{Id}\}$. Thus the homomorphism $SU(2)\to SO(3)$ is a degree 2 covering map, and since $SU(2)$ is simply connected this exhibits $SU(2)$ as the universal cover of $SO(3)$:
A covering map of Lie groups induces an isomorphism of the corresponding Lie algebras. So we have a natural isomorphism $\mathfrak{su}_{2}\cong\mathfrak{so}_{3}$. Explicitly, a basis of $\mathfrak{su}_{2}$ is given by
One has
and the isomorphism $\mathfrak{su}_{2}\xrightarrow{\sim}\mathfrak{so}_{3}$ is manifest. In physics literature, it is customary to consider the complexified Lie algebra $\mathfrak{su}_{2;\mathbb{C}}=\mathfrak{su}_{2}\otimes\mathbb{C}$ with basis $\{ J_{1},J_{2},J_{3}\}$ given by $J_{i}=\mathbf{i}Y_{i}$. The commutation relations among the $J_{i}$ are, clearly,
as in the $\mathfrak{so}_{3;\mathbb{C}}$ case. Note that
where $\{\sigma_{1},\sigma_{2},\sigma_{3}\}$ are the Pauli matrices.
The Lie group $SO(3)$ acts as a group of diffeomorphisms on $\mathbb{R}^{3}$. Hence we have an injective group homomorphisms $SO(3)\to\mathrm{Diff}(\mathbb{R}^{3})$ inducing an injective Lie algebra homomorphism
\mathfrak{so}_{3}\hookrightarrow\{\text{left $SO(3)$-invariant vector fields on $\mathbb{R}^{3}$}\}.
In particular, we can see $\mathfrak{so}_{3}\hookrightarrow\{\text{left $SO(3)$-invariant vector fields on $\mathbb{R}^{3}$ as a $SO(3)$-invariant vector field on $\mathbb{R}^{3}$. Explicitly,
which is best written in the compact form
where we have used the metric in $\mathbb{R}^{3}$ to raise the index in the derivation, i.e., $\partial^{i}=\delta^{ij}\partial_{j}$. Complexifying this construction, we can identify the basis elements $J_{i}$ of $\mathfrak{so}_{3;\mathbb{c}}$ with complex vector field on $\mathbb{R}^{3}$. It is customary to write
With this notation the commutation relations read
Any associative algebra $A$ defines a Lie algebra $A_{Lie}$ simply by taking the commutator as the Lie bracket:
This construction is a functor
Remarkably, this functor has a left adjoint
That is, given a Lie algebra $\mathfrak{g}$ there exists an associative algebra $U(\mathfrak{g})$ sucht that for any associative algebra $A$,
In other words, any Lie algebra morphism $\mathfrak{g}\to A_{Lie}$ can be uniquely extended to an associative algebra morphism $U(\mathfrak{g})\to A$. Since $U(\mathfrak{g})$ is characterized by an universal property, it is unique up to natural isomorphism. The algebra $U(\mathfrak{g})$ is called the universal envelping algebra of $\mathfrak{g}$. Note that, taking $A=U(\mathfrak{g})$ one obtatins a canonical Lie algebra morphism $\iota_{\mathfrak{g}}:\mathfrak{g}\to U(\mathfrak{g})_{Lie}$, induced by the identity of $U(\mathfrak{g})$. This morphism is actually injective and $U(\mathfrak{g})$ is generated, as an associative algebra, by the image of $\iota_{\mathfrak{g}}$ (Poincaré-Birkhoff-Witt theorem).
A typical situation in which $U(\mathfrak{g})$ occurs is the following. Consider a linear representation of a Lie group $G$, i.e., a Lie graoup homomorphism $G\to GL(V)$, where $V$ is some vector space. This induces a Lie algebra homomorphism $\mathfrak{g}\to\mathfrak{gl}(V)=\mathrm{End}(V)_{Lie}$, and so an associative algebra homomorphism
Let $M$ be a differential manifold, and let $\mathcal{X}(M)$ be the Lie algebra of vector fields on $M$. Finally, let $\mathcal{D}(M)$ be the associative algebra of differential operators on $M$. Then $\mathcal{X}(M)$ is a sub-Lie algebra of $\mathcal{D}(M)_{Lie}$. If a smooth action of a Lie group $G$ on $M$ is given, the above inclusion is refined to an inclusion $\mathcal{H}(M)^{G}\hookrightarrow\mathcal{D}(M)^{G}_{Lie}$ of left $G$-invariant vector fields on $M$ into the Lie algebra of left $G$-invariant differential operators on $M$. In particular, if $M=G$ acting on itself by left multilication, we get a Lie algebra inclusion
where we have used the identification of $\mathfrak{g}$ with left-invariant vector fields on $G$. By the universal property of $U(\mathfrak{g})$, this gives an associative algebra morphism from $U(\mathfrak{g})$ to $\mathcal{D}(G)^{G}$, which turns out to be an isomorphism:
More in general, if $M$ is a differential manifold with a smooth $G$-action, we have a natural morphism of asociative algebras
By definition, a Casimir element for a Lie algebra $\mathfrak{g}$ is an element in the center of its universal enveloping algebra, i.e. is an element $C$ in $U(\mathfrak{g})$ such that $[C,x]=0$ for any $x$ in $U(\mathfrak{g})$. Since the algebra $U(\mathfrak{g})$ is generated by the linear subspace $\mathfrak{g}$, this is equivalent to $[C,x]=0$ for any $x$ in $\mathfrak{g}$. As remarked in the previous section, if $\mathfrak{g}$ is the Lie algebra of a Lie group $G$, then $U(\mathfrak{g})$ is identified with the algebra of left-invariant differential operators on $G$. Therefore Casimir elements are identified with biinvariant differential operators on $G$. More in general, if $M$ is a differential manifold endowed with a smooth $G$-action, Casimir elements induce $G$-biinvariant differential operators on $M$. In what follows we will say $G$-invariant differential operator to mean that a differential operator is biinvariant, whereas we will always say left $G$-invariant or right $G$-invariant when the operator is invariant only with respect to left- or right- translations.
For $\mathfrak{su}_{2}\cong\mathfrak{so}_{3}$ a Casimir operator is
Indeed,
The operator $\| J\|^{2}$ is called angular momentum operator in physics. If $\rho:SU(2)\to U(\mathcal{H})$ is a unitary representation of $SU(2)$, then the image of $\mathfrak{su}_{2}$ in $\mathrm{End}(\mathcal{H})$ consists of anti-Hermitean operators. Since multiplication by $\mathbf{i}$ turns an anti-Hermitean operator into an Hermitean one, the operators (corresponding to the) $J_{i}$ are Hermitean, and so is also
since
since $\delta$ is real and symmetric. In particular $\| J\|^{2}$ is diagonalizable, its spectrum is real, and its eigenspaces are subrepresentations of $\rho$. This means that $J^{2}$ acts as a scalar on irreducible unitary representations of $SU(2)$. This scalar is a numerical invariant attached to irreducible complex representations of $\mathfrak{su}_{2}\cong\mathfrak{so}_{3}$. More precisely, we will see in Section \ref{su2} that for any nonnegative half-integer $\ell$ there exist exactly one unitary irreducible representation $\rho_{\ell}$ of dimension $2\ell+1$, and that these are all possible unitary irreducible representations of $SU(2)$. The operator $\| J\|^{2}$ acts as the multiplication by $\ell(\ell+1)$ in the representation $\rho_{\ell}$.
For the defining representation of $SO(3)$ on $\mathbb{R}^{3}$, the elements $J_{i}$ in $\mathrm{End}(\mathbb{R}^{3})$ are described in Example \ref{ex-so3}. One therefore sees that
Complexifying this representation one obtains the $\ell=1$ representation of $SU(2)$.
The defining representation of $SU(2)$ on $\mathbb{C}^{2}$ is the $\ell=1/2$ representation of $SU(2)$. For this representation, the elements $J_{i}$ in $\mathrm{End}(\mathbb{C}^{2})$ are described in Example \ref{ex-su2}. One therefore sees that
in this case.
Let $\mathbb{R}^{1,3}$ be the standard Minkowski space with metric $\eta$ of signature $(+,-,-,-)$. The Lorentz group is the group $O(1,3)$ of isometries of $\eta$. In matrix form, an element $\Lambda$ of the Lorenz group is a $4\times4$ real matrix $\Lambda^{i}_{j}$ such that
The determinant map
is surjective since
is an element of $O(1,3)$. The subgroup of Lorentz transformations of determinant 1 is called the subgroup of proper Lorentz transformations, and is denoted $SO(1,3)$. In contrast with what happens with $O(4)$, where $SO(4)$ is the connected component of the identity, the group of proper Lorentz transformations is not connected. Indeed, since the columns of $\Lambda^{i}_{j}$ are an $\eta$-orthonormal basis of $\mathbb{R}^{4}$, the first column $\Lambda^{i}_{0}$ satisfies $\eta(\Lambda_{0},\Lambda_{0})=1$, i.e.,
This implies $(\Lambda^{0}_{0})^{2}\geq1$ and so there are two disjoint possibilities: $\Lambda^{0}_{0}\geq1$ or $\Lambda^{0}_{0}\leq-1$. In the first case the Lorenz transformation is called orthocronous, in the seconda case anorthocronous. An example of anorthochronous Lorenz transformation is
The map $O(1,3)\to\{\pm1\}$ given by $\Lambda\mapsto\mathrm{sgn}(\Lambda^{0}_{0})$ is actually a group homomorphism, and so orthocronous Lorentz transformations are a subgroup of $O(1,3)$. One can then show that the connected component of the identity in $O(1,3)$ is precisely the subgroup $SO^{+}(1,3)$ of proper orthocronous Lorentz transformations. In particular, $O(1,3)$ has exactly four connected components.
The group $SO^{+}(1,3)$ is a 6-dimensional connected Lie group. It is not simply connected, and its universal cover is the Lie group $SL(2;\mathbb{C})$. This is conveniently seen as follows. Let $\mathfrak{her}_{2}$ be 4-dimensional real vector space of $2\times2$ Hermitean matrices. Since the determinant of an Hermitean matrix is a real number, we have a quadratic form
whose signature turns out to be $(1,3)$. This is seen by the linear isomorphism $\mathbb{R}^{4}\to\mathfrak{her}_{2}$ given by
Indeed,
The group $SL(2;\mathbb{C})$ acts on $\mathfrak{her}_{2}$ by
and this action clearly preserves the quadratic form $\det$, so that we get a group homomorphism $SL(2;\mathbb{C})\to O(1,3)$. Since $SL(2;\mathbb{C})$ is connected, the image of this homomorphism is contained in $SO^{+}(1,3)$; moreover the morphism induced at the Lie algebra level is an isomorphism and so $SL(2;\mathbb{C})\to SO^{+}(1,3)$ is a covering. The kernel of this map is $\{\pm\mathrm{Id}\}$ so $SL(2;\mathbb{C})\to SO^{+}(1,3)$ is a two-fold covering; moreover, since $SL(2;\mathbb{C})$ is simply connected, this is the universal covering of $SO^{+}(1,3)$:
{The Lorentz Lie algebra} The Lie algebra $\mathfrak{so}_{1,3}$ of the Lorentz group is a 6-dimensional real Lie algebra. As a matrix algebra, it is the algebra of $4\times4$ real matrices $A^{i}_{j}$ such that
As above, we will be interested in the complexification $\mathfrak{so}_{1,3;\mathbb{C}}=\mathfrak{so}_{1,3}\otimes\mathbb{C}$. A linear basis of $\mathfrak{so}_{1,3;\mathbb{C}}$ is given by the six matrices
where $E^{i}_{j}$ is the elementary matrix with 1 in position $(i,j)$ and 0 elsewhere. Indeed,
where we used the identities $\eta_{ij}=\eta_{ji}$ and $\eta^{ij}+\eta_{ij}$ for ay $i,j$. The commutation relations of the basis elements $J^{ab}$ are
where we have set $J^{aa}=0$ and $J^{ba}=-J^{ab}$ if $a\lt b$. It is convenient to represent the basis elements $J^{ab}$ as $SO^{+}(1,3)$ left-invariant vector fields on $\mathbb{R}^{4}$, as we did for the Lie algebra of $SO(3)$:
where $\partial^{i}=\eta^{ij}\partial_{j}$. The group $SO(3)$ embeds in $SO^{+}(1,3)$ via
and so $\mathfrak{so}_{3}$ is a Lie subalgebra of $\mathfrak{so}_{1,3}$. The generators of the copy of $\mathfrak{so}_{3}$ inside $\mathfrak{so}_{1,3}$ are clearly $J^{12}$, $J^{23}$ and $J^{13}$. For $i,j,k\in\{1,2,3\}$ one writes
so that
In the physics lieterature, the generators $L_{1},L_{2},L_{3}$ are called the infinitesimal rotations aroud the spatial axes of $\mathbb{R}^{1,3}$. Note that these precisely corresponds to the elements od $\mathfrak{so}_{3}$ we denoted by $J_{1},J_{2},J_{3}$ in \ref{ex-so3}.
The reamining three generators of $\mathfrak{so}_{1,3}$ are the elements
These are called the boosts in the physics literature. The commutation relations of the generators $J^{ab}$ are conveniently written in terms of the $L_{i}$ and the $K_{j}$:
Therefore, if we consider the complex basis
the commutation relations become
So we have an isomorphism of complex Lie algebras
It follows that irreducible complex representations of the Lorentz Lie algebra are tensor products of an irreducible representation of the left'' $\mathfrak{su}_{2}$ subalgebra and of an irreducible representation of the
right’‘ $\mathfrak{su}_{2}$ subalgebra. In particular, as we will show in Section \ref{su2}, complex irreducible representations of $\mathfrak{so}_{1,3}$ are indexed by a pair of nonnegative half integers $(j^{+},j^{-})$, and we will write
to mean that the $\mathfrak{so}_{1,3}$ representation indexed by $(j^{+},j^{-})$ is the tensor product of the representation of the left copy of $\mathfrak{su}_{2}$ indexed by $j^{+}$ and of the representation of the right copy of $\mathfrak{su}_{2}$ indexed by $j^{-}$.
We now extend the Lorentz group by adding to it translations of $\mathbb{R}^{4}$. More precisely, we are considering the semidirect product
where the Lorentz group acts on $\mathbb{R}^{4}$ by its defining representation. In particular we have a shoert exact sequence
This semidirect product is the group of isometries of $\mathbf{R}^{1,3}$ as a Minkowskian manifold, and is called the Poincaré group. It is a (noncompact) 10-dimensional Lie group. Its Lie algebra $\mathfrak{p}$ is obtained by adding to the Lorentz Lie algebra $\mathfrak{so}_{1,3}$ four new generators, corresponding to a basis of the abelian Lie algebra $\mathbb{R}^{4}$ of infinitesimal translations of $\mathbb{R}^{1,3}$. It is customary to take as basis of $\mathbb{R}^{4}$ the infinitesimal translations along the coodinate axis. In the identification of elements of $\mathfrak{p}$ with vector fields on $\mathbb{R}^{1,3}$ these are just the coordinate vector fields $\partial_{i}$. In the complexified Poincaré Lie algebra we set
The commutation relations involving the new generators $P^{i}$ are then easily seen to be
A Casimir element for $\mathfrak{p}$ is
This is an immediate consequence of the fact that the Poincaré group acts isometrically on $\mathbb{R}^{1,4}$. Indeed, one trivially has $[\| P\|^{2},P^{a}]=0$ and, if $\Lambda=(\Lambda^{a}_{b})$ is an element in the Lorentz subgroup, then
where we have used that, since $\Lambda$ is an element of the Lorentz group,
Hence
And the invariance of $\| P\|$ under the conjugacy action of the Lorentz group immediately gives the invariance of $\| P\|$ under the adjoint action of the Lorentz Lie algebra.
If one prefers a direct Lie algebra computation, then we trivially have $[\| P\|^{2},P^{a}]=0$, and
Therefore, if
is a unitary representation, the Casimir element $\| P\|^{2}$ will act as an Hermitean operator
and so $\| P\|^{2}$-eigenspaces will be subrepresentations of $\mathcal{H}$. In particular, $\| P\|^{2}$ will act as a real scalar on every irreducible unitary representation of $\mathcal{P}$. This scalar is called the mass of the irreducible unitary representation. This is the first ingredient for Wigner’s classification of irreducible unitary representations of the Poincaré group, we will describe in Section \ref{wigner}.
As remarked in Section \ref{casimir}, the Casimir operator $\| P\|^{2}$ corresponds to a (second-order) Poincaré biinvariant differential operator on $\mathbb{R}^{1,3}$. Explicitly, this second order operator is
So $\| P\|^{2}$ corresponds to the opposite of the D’Alembert operator on $\mathbb{R}^{1,3}$ and the eigenstates equation $\| P\|^{2}\phi=m^{2}\phi$ becomes the Klein-Gordon equation
We will come back to this in Section \ref{klein-gordon}
In this section we present the induced representation method. Essentially it consists in building a representation of a Lie group $G$ from a representation of a closed Lie subgroup $H \subseteq G$.
Given a representation $V$ of $H$, the induced representation of $G$, usually written $\mathrm{Ind}^{G}_{H}(V)$ is by definition the vector space
with the obvious $G$-action on it.
This construction has a nice geometric interpretation: $\mathrm{Ind}^{G}_{H}(V)$ can be naturally realized as the space of sections of a $G$-equivariant vector bundle on a $G$-homogeneous space. Recall that a $G$-equivariant vector bundle $\mathcal{V}$ on a $G$-manifold $M$ is the datum of a lifting of the $G$-action on $M$ to a $G$-action on the total space of $\mathcal{V}$ which is linear on the fibers. Given $G$, $H$ and $V$ as before, define the vector bundle $G \times_{H} V$ over $G/H$ by
where $\sim$ is the equivalence relation
The projection map $\pi: G \times_{H} V \to G/H$ is given by $\pi(g,v) = gH$ and the $G$-action is given by $g^{\prime}(g,v)= (g^{\prime}g,v)$, for $g^{\prime}\in G$. Denote by $\Gamma(G/H, G \times_{H} V )$ the vector space of sections of $G \times_{H} V$. This vector space carries a natural action of $G$: if $\sigma$ is a section of $G \times_{H} V\to G/H$ and $g$ is an element of $G$, then
There is a natural isomorphism of representations of $G$ between $\Gamma(G/H,G \times_{H} V)$ and $\mathrm{Ind}^{G}_{H}(V)$. Indeed, given a section $s \in\Gamma(G/H,G \times_{H} V)$, let $f_{s} \in\mathrm{Ind}^{G}_{H}(V)$ be the function $f_{s}:G\to V$ defined by
Conversely, given $f \in\mathrm{Ind}^{G}_{H}(V)$, let $s_{f} \in\Gamma(G/H,G \times_{H} V)$ be section
It is straightforward to check that the given construction define morphisms of $G$-representations between $\Gamma(G/H,G \times_{H} V)$ and $\mathrm{Ind}^{G}_{H}(V)$ which are inverse each other.
The induced representation contrucition also has an important functroial interpretation: it is the adjoint of the restriction functor
i.e. if $V$ is a representation of $H$ and $W$ is a representation of $G$, then there is a natural isomorphism
This is known as Frobenius reciprocity formula.
Let us remark that bigger the subgroup $H$, smaller the induced representation. For instance if $H= \{ e \}$ and $V=\mathbb{C}$, then $\mathrm{Ind}^{G}_{H}(V) = C^{\infty}(G;\mathbb{C})$, which is an enormous space.
It is also interesting to remark that in general, even if the representation $V$ of $H$ is irreducible, we can’t state anything on the irreducibility of $\mathrm{Ind}^{G}_{H}(V)$. On the other hand, if $\mathrm{Ind}^{G}_{H}(V)$ is irreducible, then clearly the $H$-representation $V$ is irreducible, since a splitting of $V$ induces a splitting of $\mathrm{Ind}^{G}_{H}(V)$.
Every unitary irreducible representation of the Poincaré group is induced by a representation of the stabiler subgroup $O(1,3)_{\vec{p}}$ of some point $\vec{p}$ for the standard action of the Lorentz group on $\mathbb{R}^{1,3}$, a result originally due to Wigner wigner-induced. Furthermore, in the next section we will classify all the possibilities for $\mathcal{P}_{\vec{p}}$.
Let
be a unitary representation of the Poicaré group on an Hilbert space $\mathcal{H}$, and let
be the induced Lie algebra represention of the complexified Poincaré Lie algebra. As in the previous sections, let $P^{\mu}$ the generators of the translations or $\mathbb{R}^{4}$, multiplied by $\mathbf{i}$. Since $\mathcal{P}$ acts on $\mathcal{H}$ by unitary operators, the real Lie algebra $\mathfrak{p}$ acts on $\mathcal{H}$ by anti-Hermitean operators, and so the $P^{\mu}$ act as Hermitean operators. In particular, they are diagonalizable, with a real spectrum. Since they commute, they are simultaneously diagonalizable, and the Hilbert space $\mathcal{H}$ can be decomposed as
with $\vec{p}=(p^{0},p^{1},p^{2},p^{3})$ ranging in $\mathbb{R}^{4}$. In the above orthogonal decomposition, $\mathcal{H}_{\vec{p}}$ denotes the $\vec{p}$-eigenspace for the $P^{\mu}$: elements of $\mathcal{H}_{\vec{p}}$ are vectors $|\vec{p}\rangle$ of $\mathcal{H}$ such that
The action of the translations subgroup of $\mathcal{P}$ is then recovered siimply by exponentiation:
where $(\vec{a}|\vec{p})=\eta(\vec{a},\vec{p})$ is the Minkowski inner product on $\mathbb{R}^{1,3}$. So in particular each subspace $\mathcal{H}_{\vec{p}}$ is stable for the action of the translation subgroup of $\mathcal{P}$, which acts by scalar multiplication on each $\mathcal{H}_{\vec{p}}$.
We are thus left with the problem of describing the action of the Lorentz group $O(3,1)$: given an eigenvector $|\vec{p}\rangle$, we want to see where it is mapped by an element $\Lambda\in O(3,1)$. The key to answer this question is to notice that the Lie subalgebra of infinitesimal translation is preserved by the conjugacy action of $P$ on $\mathfrak{p}_{\mathbb{C}}$. More precisely, recall that if $\Lambda=(\Lambda^{\mu}_{\nu})$, we have
and let us act on $\Lambda|\vec{p}\rangle$ with the operator $P^{\mu}$. We have
Therefore $\Lambda|\vec{p}\rangle$ is a $P^{\mu}$-eigenvector, with eigenvalue $(\Lambda\vec{p})^{\mu}$. We express this in compact form as
In other words, an element $\Lambda$ in the Lorentz group induces an isomorphism
Moreover, if $\pi_{\vec{p}}:\mathcal{H}\to\mathcal{H}_{\vec{p}}$ denotes the projection on the $\vec{p}$-eigenspace, then
for any $\Lambda$ in the Lorentz group. Indeed, both sides act as the zero operator on an eigenvector $|\vec{q}\rangle$ with $\vec{q}\neq\vec{p}$, whereas, acting on $|\vec{p}\rangle$ we have
since $\Lambda|\vec{p}\rangle$ is an element in the eigenspace $\mathcal{H}_{\Lambda\vec{p}}$. Summing up, for a fixed $\vec{p}_{0}$, the direct sum
where $\vec{p}$ ranges in the Lorentz orbit of $\vec{p}_{0}$, is a subrepresentation of the representation of $\mathcal{P}$ we started with. So if the original representatin was irreducible and $\vec{p}$ is in the spectrum of the $P^{\mu}$, then
Let $m^{2}=\|\vec{p}_{0}\|^{2}=p_{0\mu} {p_{0}}^{\mu}$. The Lorentz group preserves the Minkowski norm, and, if $m^{2}\neq0$, acts transitively on the set on norm $m^{2}$ vectors, so that $\vec{p}$ is in the Lorentz orbit of $\vec{p}_{0}$ if and only if $\| p\|^{2}=m^{2}$. Therefore, if $m^{2}\neq0$ we have found
Note that from this, in particular we recover that the Casimir element $\| P\|^{2}$ acts as the scalar $m^{2}$ on $\mathcal{H}$, so that $m^{2}$ is the mass of the irreducible representation $\mathcal{H}$, in the notations of Section \ref{poincare}. Indeed, by the above decomposition, for any eigenvector $|\vec{p}\rangle$ in $\mathcal{H}$ we have
Denote by $X_{m^{2}}$ the mass $m^{2}$ hyperboloid, i.e., the set
Then the collection of eigenspaces $\mathcal{H}_{\vec{p}}$ is a vector bundle $\mathcal{E}_{m^{2}}$ over $X_{m^{2}}$, and the Hilbert space $\mathcal{H}$ is naturally identified with the Hilbert space of $L^{2}$-sections of this bundle (with respect to the spectral measure on $X_{m^{2}}$). Indeed, each vector $\psi$ in $\mathcal{H}$ is identified with the section $\sigma_{\psi}$ defined by $\sigma_{\psi}:\vec{p}\mapsto\pi_{\vec{p}}(\psi)$, where $\pi_{\vec{p}}: \mathcal{H}\to\mathcal{H}_{\vec{p}}$ is the projection on the $\vec{p}$-eigenspace. Moreover the vector bundle $\mathcal{E}$ is clearly $O(1,3)$-equivariant, and the map
\sigma:\mathcal{H}\xrightarrow{\sim}\{\text{sections of $\mathcal{E}_{m^{2}}$ over $X_{m^{2}}$}\}
is an isomorphism of representations of $\sigma:\mathcal{H}\xrightarrow{\sim}\{\text{sections of $\mathcal{E}_{m^{2}}$. This last statement is nothing but the identity $\pi_{\vec{p}}=\Lambda^{-1}\pi_{\Lambda\vec{p}}\Lambda$ derived above. Therefore we see that we are precisely in the situation described in Section \ref{induced}: let $\mathcal{P}$ act on $\mathbb{R}^{1,3}$ via if we denote by $O(1,3)_{\vec{p}_{0}}$ the stabilizer of $\vec{p}_{0}$ under the Lorentz action, then the Lorentz orbit of $\vec{p}$ is
and the representation of the Lorentz subgroup of $\mathcal{P}$ on $\mathcal{H}$ is the representation induced on the space of sections of an equivariant bundle on a Lorentz-homogeneous space by a representation of the stabilizer subgroup $O(1,3)_{\vec{p}_{0}}$ on the fiber $\mathcal{H}_{\vec{p}}$. The description of the Poincaré action is completed by recalling that the subgroup of translations acts by scalar mutiplication by $e^{\mathbf{i}(\vec{a}|\vec{p})}$ on the fibers.
The situation for $m^{2}=0$ is completely similar, but we have to distinguish two cases. Indeed the Lorentz action on the set of zero-norm vectors is not transitive but there are two orbits: one consisting of the $0$ vector alone (the vacuum state of physics parlance), and the other consisting of all nonzero zero-norm vectors of $\mathbb{R}^{1,3}$ (the light cone in the physics jargon). Also, almost nothing changes if instead of the Poincaré group $\mathcal{P}$ we consider the universal cover of the connected component of the identity:
Indeed, nothing changes at the Lie algebra level, and at the group level the only difference is that the orbit space for the action of the universal cover $SL(2;\mathbb{C}$ of the proper orthochronous Lorentz group $SO^{+}(1,3)$ on $\mathbb{R}^{1,3}$ is more refined than the orbit space for the full Lorentz group.
Therefore we have finally proven the following.
[Wigner] Irreducible unitary representation of the Poincaré group $\mathcal{P}$ are classified by pairs $(\mathcal{O},s)$, where $\mathcal{O}$ is a Lorentz orbit in $\mathbb{R}^{1,3}$ and $s$ is (the isomorphism class of) an irreducible representation of the stabilizer $O(1,3)_{\vec{p}}$ of a point $\vec{p}$ in $\mathcal{O}$ under the Lorentz action. Moreover, the $(\mathcal{O},s)$-representation is induced by the $s$-representation of $O(1,3)_{\vec{p}}$. The analogous statement holds for the universal cover $\mathbb{R}^{4}\rtimes SL(2;\mathbb{C})$ of the identity element in the Poincaré group.
The stabilizer subgroups of $SL(2;\mathbb{C})$ in Wigner’s theorem are called little groups in the physics terminology.
We now exhibit the classification $SL(2;\mathbb{C})$-orbits on $\mathbb{R}^{1,3}$ and the corresponding little groups. In view of Wigner’s theorem this gives a classification of the irreducible unitary representations of $\mathcal{P}$. When the unitary irreducible representation of the little group involved is finite dimensional, the corresponding representation of $\mathbb{R}^{4}\rtimes SL(2;\mathbb{C})$ is called an elementary particle in physics.
The origin $\{0 \} \in\mathbb{R}^{4}$ is a singleton orbit, stabilized by the whole $SL(2;\mathbb{C})$. As we are going to show below, the only finite dimensional unitary irreducible representation of $SL(2;\mathbb{C})$ is the trivial one. The corresponding representation of the Poincaré group is called the vacuum state in physics.
$\{ \vec{p} \in\mathbb{R}^{1,3} | p_{\mu}p^{\mu}=0, p_{0} \gt 0 \}$. Up to rescaling, a representative for this orbit is $\vec{p}_{0}=(1,0,0,1)$. Recalling the definition of the $SL(2;\mathbb{C})$-action on $\mathbb{R}^{1,3}$, the corresponding little group is
Writing
the equation $AM_{\vec{p}_{0}}A^{*}=M_{\vec{p}_{0}}$ becomes
and so we find $|a|=1$, $b \in\mathbb{C}$ and $c=0$, i.e.,
Consider now the bijection $\mathbb{C}\times U(1)\leftrightarrow H$ given by
With this notation, the multiplication in $H$ reads
Hence, the group $H$ is identified the semidirect product $\mathbb{C}\rtimes U(1)$ with $U(1)$ acting on $\mathbb{C}$ by $(z,\zeta)\mapsto z^{2}\zeta$. This is the double cover of the semidirect product of $\mathbb{C}$ with $U(1)$ given by the standard $U(1)$ action $(z,\zeta)\mapsto z\zeta$. Since the standard $U(1)$-action on $\mathbb{C}$ is naturally identified with the $SO(2)$-action on $\mathbb{R}^{2}$, the little group $H$ is the double cover of the group $SE(2)$ of orientation preserving (affine) isometries of the Euclidean plane. Its finite dimensional unitary irreducible representations are classified by an half-integer $\varepsilon$, called the helicity of the representation. The corresponding elementary particle is called a massless helicity $\varepsilon$ particle.
$\{ p \in\mathbb{R}^{4} | p_{\mu}p^{\mu}=0, p_{0} \lt 0 \}$. The little group is the analogous of the precedent case. The corresponding elementary particle is called a massless helicity $\varepsilon$ antiparticle.
$\{ p \in\mathbb{R}^{4} | p_{\mu}p^{\mu}\gt0, p_{0}\gt0 \}$. Up to rescaling, a representative for this orbit is $\vec{p}_{0}=(1,0,0,0)$, which corresponds to the Hermitean matrix
Hence the little group is
We will show next that irreducible unitary representations of $SU(2)$ are classified by a nonnegative half-integer $\ell$ called the spin. The corresponding elementary particle is called a mass $m$ spin $\ell$ particle.
$\{ p \in\mathbb{R}^{4} | p_{\mu}p^{\mu}\gt0, p_{0}\lt 0 \}$. The little group is the analogous of the previous case. The corresponding elementary particle is called a mass $m$ spin $\ell$ antiparticle.
$\{ p \in\mathbb{R}^{4} | p_{\mu}p^{\mu}\lt 0 \}$ Up to rescaling, a representative for this orbit is $\vec{p}_{0}=(0,1,0,0)$. The little group for this orbit is then:
It is convenient to rewrite the defining relation for elements of $H$ as $A=M_{\vec{p}_{0}}(A^{*})^{-1}M_{\vec{p}_{0}}^{-1}$. The right-hand side of this expression is
so that $H=SL(2;\mathbb{R})$. Corresponding elementary particles are nonphysical since they would have negative mass-square.
Here we prove that the only finite dimensional unitary representation of $SL(2;\mathbb{C})$ is the trivial one. We borrow the following argument from knapp-trapa, where the case of $SL(2;\mathbb{R})$-representations is treated. Let
be a continuous reperesentation, and consider the subset of $SL(2;\mathbb{C})$ consisting of the matrices of the form
All these matrices have the same Jordan normal form, so they are in the same $SL(2;\mathbb{C})$-conjugacy class. Therefore their images $\rho(A(z))$ are in the same $U(n)$-conjugacy class. Let us call this conjugacy class $C$. Since $U(n)$ is compact, conjugacy classes in $U(n)$ are closed, so
is an element of $C$ (here we have used the continuity of $\rho$). But this means that $C$ is the conjugacy class of the identity in $U(n)$ and so consists of the identity alone. Since by construction $\rho(A(z))$ lies in $C$ for every $z\neq0$, we have thus shown that $\rho(A(z))=\mathrm{Id}_{U(n)}$ for every $z$ in $\mathbb{C}$. The identical argument applies to the matrices
The matrices $A(z)$ and $B(w)$ generate $SL(2;\mathbb{C})$, as is easily seen by noticing that the two matrices
generate the Lie algebra $\mathfrak{sl}_{2}(\mathbb{C})$. Therefore, $\rho$ is the trivial representation.
It is worth remarking that $SL(2;\mathbb{C})$ admits nontrivial infinite dimensional unitary representations, see, e.g. dao-nguyen.
In this section we show that irreducible unitary finite-dimensional representations of the double cover $\widetilde{SE}(2)$ of the group $SE(2)$ of affine isometries of $\mathbb{R}^{2}$ are classified by an half-integer $\varepsilon$; moreover $\varepsilon$ is an integer precisely when the given reresentation $\widetilde{SE}(2)\to U(n)$ factors through $\widetilde{SE}(2)\to SE(2)$. We thank Andrea Maffei for having shown us this proof.
We begin by recalling that $\widetilde{SE}(2)$ is the semidirect product $\mathbb{C}\rtimes U(1)$, with $U(1)$ acting on $\mathbb{C}$ by $(z,\zeta)\mapsto z^{2}\zeta$. Let now
be a finite dimensional representation. The restriction of $\rho$ to the normal subgroup $\mathbb{C}$ gives a representation
of the real Lie group $\mathbb{C}$ on the complex vector space $\mathbb{C}^{n}$. This gives a splitting
where $I$ is a finite subset of $\mathrm{Hom}_{\mathbb{R}}(\mathbb{C},\mathbb{R})$ and $V_{\varphi}$ is the subspace of $\mathbb{C}^{n}$ where $\mathbb{C}$ acts via $\rho$ as
We now look at the $U(1)$-action. Since $(1,z^{2}\zeta)\cdot(z,0)=(z,0)\cdot(1,\zeta)$ we have
for any $\varphi\in I$, where $({z^{-1}}^{*}\varphi)(\zeta)=\varphi(z^{-2}\zeta)$. Indeed, if $\vec{v}$ is an element of $V_{\varphi}$, then
Therefore $U(1)$ acts as a permutation group on the elemnets of the finite set $I$. Since $U(1)$ is connected, this permutation action is trivial, so ${z^{-1}}^{*}\varphi=\varphi$ for every $\varphi\in I$. But the only $U(1)$-invariant element in $\mathrm{Hom}_{\mathbb{R}}(\mathbb{C},\mathbb{R})$ is the zero morphism, hence the normal subgroup $\mathbb{C}$ of $\mathbb{C}\rtimes U(1)$ acts trivially on $\mathbb{C}^{n}$ and the representation $\rho$ factors through $\mathbb{C}\rtimes U(1)\to U(1)$. We are therefore reduced to classify irreducible unitary representations of $U(1)$. Since $U(1)$ is an abelian Lie group, these are all 1-dimensional, and so everything boils down to the problem of describing Lie group homomorphisms from $U(1)$ to itself. And it is well known (and easy to show) that these are all of the form $z\mapsto z^{k}$ for some integer $k$. Writing $k=2\varepsilon$ for an half-integer $\varepsilon$ we finally find that irreducible unitary representations of the double cover $\widetilde{SE}(2)$ of the group $SE(2)$ of affine isometries of $\mathbb{R}^{2}$ are classified by an half-integer $\varepsilon$, and that $\varepsilon$ is an integer precisely when the given representation factors through $\widetilde{SE}(2)\to SE(2)$. In the language of particle physics, the half-integer $\varepsilon$ is called the helicity of the massless particle corresponding to the given irreducible representation of $\widetilde{SE}(2)$.
Since the Lie group $SU(2)$ is compact, all its irreducible representations are finite dimensional. Moreover, since $SU(2)$ is simply connected, its representation theory is equivalent to the representation theory of its Lie algebra. A set of generators for $\mathfrak{su}(2)$ is given by Pauli matrices
By complexifying the Lie algebra $\mathfrak{su}_{2}$ we obtain
The latter is the Lie algebra of the complex Lie group $SL(2;\mathbb{C})$. Therefore, we are reduced to studying complex finite dimensional irreducible representations of $\mathfrak{sl}_{2}(\mathbb{C})$. We ar going to show that there is exactly one such representation (up to isomorphism), for any dimension, which can be explicitly described. This is conveniently done by fixing the following set of generators for $\mathfrak{sl}_{2}(\mathbb{C})$:
with commutation rules
Having fixed this notation, we have
For each nonnegative half-integer $l$ there exists a unique (up to isomorphism) irreducible complex linear representation of $\mathfrak{sl}_{2}(\mathbb{C})$ on a complex vector space of dimension $2\ell+1$. Moreover, there is a basis
of $V$ such that:
$hv_{2(\ell-j)} = 2(\ell-j) v_{i}$,
$e v_{2(\ell-j)} = j(2\ell-j+1)v_{2(\ell-j+1)}$, with $e v_{2\ell}=0$,
$f v_{2(\ell-j)}=v_{2(\ell-j-1)}$, with $f v_{-2\ell}=0$.
Let $V$ be a complex linear irreducible representation of $\mathfrak{sl}_{2}(\mathbb{C})$ of dimension $2\ell+1$, and denote by $V_{\lambda}$ the $\lambda$-eigenspace of $V$ with respect to the operator $h$. We begin by showing that
Indeed,
and similarly for $f$. Assume $V_{\lambda}\neq\{0\}$. By finite dimensionality of $V$, only finitely many eigenspaces $V_{\lambda+2k}$ can be nonzero, so there exists $k_{0}\in\mathbb{N}$ with $V_{\lambda+2k_{0}}\neq{0}$ and $V_{\lambda+2k_{0}+2}=0$. Pick a nonzero vector $v_{2\ell}$ in $V_{\lambda+2k_{0}}$ and for any $j\geq0$, set $v_{2(\ell-j)}=f^{j}v_{2\ell}$. Since $v_{2(\ell-j)}$ is an element of $V_{\lambda+2(k_{0}-j)}$ there exist a minimum $k$ such that $v_{2(\ell-(k+1))}=0$. Then the vectors $v_{2\ell},v_{2(\ell-1)},\dots,v_{2(\ell-k)}$ are linearly independent (since they are nonzero and belong to $h$-eigenspaces for distinct eigenvalues). Now we show that $V = \mathrm{span}\{ v_{2(\ell-k)}, \dots, v_{2\ell}\}$. Since by hypothesis $V$ is irreducible, it is sufficient to show that $\mathrm{span}\{ v_{2(\ell-k)}, \dots, v_{2\ell}\}$ is stable under $e,f,h$. The $f$- and $h$-stability is obvious, so we needonly to check the $e$-stability. This results from
with $v_{2(\ell-j)} = 0$ for $j\lt 0$ and $j\gt k$, which we prove inductively. Assume we have proved the statement for $j$. Then for $j+1$ we have:
So $V=\mathrm{span}\{ v_{2(\ell-k)}, \dots, v_{2\ell}\}$. In particular $\dim V=k+1$ and so $k=2\ell$. This means that our basis for $V$ is actually
It is now easy to show that $\lambda+2k_{0}=2\ell$. Indeed, since $h=[e,f]$ is a commutator, $h$ is traceless in any representation. Computing the trace of $h$ on the vector space $\mathrm{span}\{ v_{2(\ell-j)}\}_{j=0,\dots,2\ell}$ we therefore find
Therefore, our basis vectors satisfy:
$hv_{2(\ell-j)} = 2(\ell-j) v_{i}$,
$e v_{2(\ell-j)} = j(2\ell-j+1)v_{2(\ell-j+1)}$, with $e v_{2\ell}=0$,
$f v_{2(\ell-j)}=v_{2(\ell-j-1)}$, with $f v_{-2\ell}=0$. This shows that up to isomorphisms, a $2\ell+1$-dimensional irreducible representation od $\mathfrak{sl}_{2}(\mathbb{C})$ is completely determined by its dimension. To prove the existence of an irreducible $2\ell+1$-dimensional representation of $\mathfrak{sl}_{2}(\mathbb{C})$, consider the free vector space over the set $\{ v_{-2\ell},v_{-2\ell+2}, \dots, v_{2\ell-2}, v_{2\ell}\}$ and define an $\mathfrak{sl}_{2}(\mathbb{C})$-action by defining the action of the generators $e,f,h$ on the basis elements $v_{2(\ell-j)}$ by the above formulas. To see that the representation defined this way is irreducible, let $U$ be a nonzero invariant subspace. Since $U$ is invariant under $h \cdot$, $U$ is spanned by a nonempty subset of the basis $\{ v_{2(\ell-j)}\}$. By repeatedly applying $e$ and $f$ to any vector in the basis $\{ v_{2(\ell-j)}\}$ one obtains all the others (up to nonzero scalar multiples), and so $U = V$.
In Section \ref{casimir} we anticipated that the momentum operator $\| J\|^{2}$ acts as the scalar $\ell(\ell+1)$ on the $2\ell+1$ irreducible representation $V$ of $SU(2)$. We can now prove this statement. Let $\{ v_{2(\ell-j)}\}_{j=0,\dots,2\ell}$ be a distinguished basis of $V$ as described above. Since $\| J\|^{2}$ acts as a scalar on $V$, we can compute this scalar simply by looking at the action of $\| J\|^{2}$ on the vector $v_{2\ell}$. We have
and so
As an illustrative example of the above construction, we investigate the solutions of the Klein-Gordon equation on $\mathbb{R}^{1,3}$. We begin by considering the defining action of the Poincaré group on $\mathbb{R}^{1,3}$, lifted to an action of $\mathbb{R}^{4}\rtimes SL(2;\mathbb{C})$. Since the standard Lesbegue measure on $\mathbb{R}^{1,3}$ is translation- and Lorentz-invariant, we have an induced unitary representation on the Hilbert space $\mathcal{H}=L^{2}(\mathbb{R}^{1,3})$ of square-integrabel functions on $\mathbb{R}^{1,3}$. Passing to Lie algebras, the (universal cover of the) Poincaré group acts by vector fields, i.e., by derivations. A major technical point to be stressed is that the first-order differential operator one gets this way, i.e., corresponding to elements in $\mathfrak{p}_{\mathbb{C}}$ are only densely defined on $\mathcal{H}$. For instance, the element $P^{\mu}$ corresponds to the derivation $\mathbf{i}\partial^{\mu}$. This means that $\mathfrak{p}_{\mathbb{C}}$ can not be handled as a Lie algebra of operators on $\mathcal{H}$ within the framework of classical Hilbert spaces. A rigorous treatment can be given within the framework of rigged Hilbert spaces/Gelfan’d triples, see gelfand-vilenkin. Yet, such a rigorous treatment is not the aim of this note, so we will kindly ask the reader to pretend that $\vec{x}\mapsto e^{-\mathbf{i}(\vec{p}|\vec{x})}$ is a square-integrable function on $\mathbb{R}^{1,3}$.
Let now $m^{2}\gt 0$ and consider the $m^{2}$-eigenspace equation
for the Casimir element $\| P\|^{2}$. Since $P^{\mu}$ acts on $\mathcal{H}$ as $\mathbf{i}\partial^{\mu}$, this equation is the Klein-Gordon equation on $\mathbb{R}^{1,3}$:
where $\square=\partial_{\mu}\partial^{\mu}$ is the D’Alembert operator on $\mathbb{R}^{1,3}$. Therefore, we see that the space of solutions of the Klein-Gordon equation is a representation of the Poincaré group. Obviously, this could be directly seen by the manifest Poincaré-invariance of the D’Alembert operator. What we have gained by deriving this from the abstract nonsense of Casimir operators is that we now see the Klein-Gordon equation as a stand-alone equation, but as a piece of the larger picture of Wigner’s investigation of unitary representations of the Poincaré group. And in the larger picture we know that the representation of the Poincaré group given by solutions of the Klein-Gordon equation is induced by a representation of the little group at $\vec{p}$ on $\mathcal{H}_{\vec{p}}$, where $\vec{p}$ is a point in $\mathbb{R}^{1,3}$ with $\|\vec{p}\|^{2}=m^{2}$. By definition $\mathcal{H}_{\vec{p}}$ is the $\vec{p}$-eigenspace for the action of the infinitesimal translations $P^{\mu}$, hence it is defined as the space of the joint solutions of the first-order differential equations
Therefore we find that $\mathcal{H}_{\vec{p}}$ consists of the functions
In particular it is a 1-dimensional space, and so for $m^{2}\gt 0$ the Klein-Gordon equation describes a spin 0 mass $m$ elementary particle. We will come back to this when discussing the massive scalar field.
Since $\mathcal{H}_{\vec{p}}$ is 1-dimensional, the general discussion in Section \ref{wigner} tells us that the space of solutions of the Klein-Gordon equation is naturally identified with the space of sections of a line bundle over the mass $m^{2}$ hyperboloid. This can be nicely interpreted as the Fourier transform of the Klein-Gordon equation,
telling us that $\hat{\phi}$ is a distribution on $\mathbb{R}^{1,3}$ which is supported on the mass $m^{2}$ hyperboloid $\| p\|^{2}-m^{2}$. If $\vec{p}_{0}$ is a vector in $\mathbb{R}^{1,3}$, the $\vec{p}_{0}$-eigenspace equations become, via the Fourier transform,
which show that $\vec{p}_{0}$-eigenstates are distributions supported at the point $\vec{p}$. Since a distribution supported on a point is a finite linear combination of derivatives of $\delta$-functions at that point, we have
for suitable coefficients $a_{I}$ in $\mathbb{C}$, see, e.g., vladimirov. The condition that $\hat{\phi}$ is annihilated by the ideal $(p^{\mu}-{p_{0}}^{\mu})_{\mu=0,\dots,3}$ then implies that all the coefficients $a_{I}$ with $|I|\gt 0$ vanish, i.e. $\hat{\phi}$ is a scalar multiple of the Dirac’s $\delta$ at $\vec{p}_{0}$:
and $\sigma$ is naturally interpreted as a section of the trivial line bundle over the mass $m^{2}$ hyperboloid.
BtD85 Theodor Bröcker; Tammo tom Dieck, Representations of compact Lie groups. Graduate Texts in Mathematics, 98. Springer-Verlag, New York, 1985. x+313 pp.
DaN67 Dao Vong Duc; Nguyen Van Hieu, On the theory of unitary representations of the $SL(2,\mathbb{C})$ group, Annales de l’I.H.P., section A, tome 6, n. 1 (1967), 17-37.
FuH91 William Fulton; Joe Harris. Representation theory. A first course. Graduate Texts in Mathematics, 129. Readings in Mathematics. Springer-Verlag, New York, 1991. xvi+551 pp.
GeV Israel M. Gel’fand; Naum Ya. Vilenkin, Generalized functions. Vol. 4. Applications of harmonic analysis. Academic Press, 1964, xiv+384 pp.
KnT00 Anthony W. Knapp; Peter E.Trapa, Representations of semisimple Lie groups.Representation theory of Lie groups (Park City, UT, 1998), 7–87, IAS/Park City Math. Ser., 8, Amer. Math. Soc., Providence, RI, 2000.
Vla79 Vassilij S. Vladimirov, Distributions en physique mathématique. Mir, Moscow, 1979. 280 pp.
Wig59 Eugene P. Wigner, Group theory: And its application to the quantum mechanics of atomic spectra. Pure and Applied Physics. Vol. 5 Academic Press, New York-London 1959 xi+372 pp.
Wig39 Eugene P. Wigner, On unitary representations of the inhomogeneous Lorentz group. Ann. of Math. (2) 40 (1939), no. 1, 149–204.
Tim van Beek: …(obsolete comment deleted)…
The construction of a free scalar real field is now an example on the page “Wightman axioms” on the nLab.
Giuseppe: Thank you a lot for your comment, I just corrected. I wrote a statement too careless. The costruction of yours for a quantum field is the so-said “second quantization” isn’t it? I know it’s an inevitable approach for quantum field theory, so I planned to introduce it for the interaction part of the thesis. By the way I’m consulting about this with my advisor. Thank you a lot for your contribution.
Tim van Beek: You ‘re welcome! Feel free to discuss any topic in the corresponding thread on the nForum. “Second Quantization”: Quantum mechanics (QM) is not relativistic, that is it does not take into account special relativity. One core feature of special relativity (SR) is the famous equation $E = mc^2$, which says that matter is a manifestation of energy. Quantum field theory (QFT) combines QM and SR, so one effect QFT has to incorporate is particle creation and annihilation (energy transforms into matter, matter transforms into energy). QM does not have that. QFT needs a state space that contains states with any prescribed number of particles, that is what second quantization builds out of the state space with a fixed number of particles of quantum mechanics.
First quantization is “classical one particle state space” -> “quantum one particle state space”.
Second quantization is “quantum one particle state space” -> “multiple particle state space” (multiple means any positive integer, not one in particular).
The “second quantization” step is the part where I wrote “construct the (bosonic) Fock space”. You start with a one particle state space, that is a Hilbert space H. Two non interacting particles would live in $H \otimes H$. Three non interacting particles in $H \otimes H \otimes H$. So the general Fock space is
Now there are Bosons and Fermions in nature, Bosons live in the subspace of $F(H)$ of all symmetric states (interchanging two particles does not change anything), Fermions in the subspace of all asymmetric subspaces (interchanging two particles produces a -1 factor).
Tim van Beek:
A book that I recommend to every mathematician who would like to learn QFT for the first time is this:
We exhibit irreps for $SU(N)$ and $U(N)$ for every $N \in \mathbb{N}$. First we compute irreps for $U(1)$ and $SU(2)$ directly.
The group $U(1)$ is also called the circle group since it is the set of the complex number of norm 1:
The group $U(1)$ is not simply connected, but it is compact and connected. Hence we can’t rely on infinitesimal method for calculating irreps. The fundamental tool in this case is the Schur’s Lemma.
Lemma. Let $V$ and $W$ be two complex irreducible representations for a group $G$ then
Corollary 1. When $V=W$, $Hom_G(V,W)=\mathbb{C}I$
Corollary 2. All complex irreducible representations $V$ of an abelian group are one dimensional
Proof of the Corollary 2. Every representation of a $g \in G$, as an element of $GL(V)$ is in $Hom_G(V,V)$ and commute with the representation. Then by Corollary 1 it is a multiple of the identity. The invariant subspaces of the identity are all one dimensional. Finally the only allowed irreducible representation must be one dimensional.
This result can be applied to $U(1)$, since it is a commutative group. Then any $V$ complex representation is one dimensional. Moreover the compactness of $U(1)$ implies that all irreps are unitary. Then a representation for $U(1)$ is a map:
The problem is now to classify all possible continuous group automorphism of $U(1)$.
Proposition. Every continous automorphism of $U(1)$ can be written as:
for a choice of $n \in \mathbb{Z}.$
Proof.Obviously $f_n$ is a continuous automorphism for $S^1$. The viceversa can be proven by remarking that $\{ f_n \}_{n \in \mathbb{Z}}$ is a basis for $L^2[S^1]$, then any continuous function $S^1$-valued can be written as linear combination of $f_n$. Remark that the sum may be infinite and the $S^1$-valued continuous function are contained in $L^2[S^1]$. Now the condition
compels the sum to be over a single element $f_n$. Otherwise let $f$ be the sum of at least two terms $f_m$ and $f_n$:
Then we can prove that $f$ does not satisfy the (1)
This completes the proof. We have a one on one correspondance between $\mathbb{Z}$ and the irreducible representations of $U(1)$.
The group $SU(2)$ is the closed sub-group of $U(2)$ with determinant one. Since $U(2)$ is compact, so it is $SU(2)$. Moreover, as a Lie group it is three dimensional and the generators are the followings:
We don’t want to work with $\mathfrak{su}(2)$: by the general theory it is equivalent to consider the complexification $\mathfrak{sl}(2)$ generated by:
with commutation rules:
Finally the group $SU(2)$ is topologically equivalent to a sphere, then it is simply connected. Utilizing the infinitesimal method for Lie group representation, we can study the representations of its algebra complexified $\mathfrak{sl}(2)$ generated by $H,E,F$ and extend the results directly to $SU(2)$. Remark that we have already given a concrete realization for the irreducible representation of $\mathfrak{sl}(2)$ in Lie groups and algebras representations, so that the majority of the following computations could result in a repetition. We want to proceed anyway for emphasizing the connection between group and algebra representation for compact and simply connected groups. First we introduce a typical $SU(2)$ representation. Define
The set $V_n(\mathbb{C}^2)$ is actually a vector space with base $\{ z_1^k z_2^{n-k} | o \le k \le n \}$ with $Dim(V_n(\mathbb{C}^2)) = n+1$.
Let $z$ be $z=(z_1,z_2)$. The $SU(2)$ action is the following:
If
then
The action (7) defines a representation since with an easy calculation the following relation can be proven:
Naturally the algebra too acts over $V_n(\mathbb{C}^2)$. Given an element
the action is explicitly written:
This is easily calculated by doing the exponential of the matrix $tX$, $t \in \mathbb{C}$, then deriving in $t$ and finally evaluating the expression in $t=0$.
Proposition. The representation $V_n(\mathbb{C}^2)$ for $\mathfrak{su}(2)$, with action (9), is irreducible.
Proof. Through a substitution we compute the action of $H,E$ and $F$:
By these three relations, the representation $V_n(\mathbb{C}^2)$ results irreducible by construction: with the action of $E$ and $F$ we obtain all $n+1$ possible elements of the base, so that the only allowed invariant sub-space are $\{0\}$ and $V_n(\mathbb{C}^2)$.
Now given any irreducible representation $V$ for $\mathfrak{su}(2)$, we want identify $V$ with $V_n(\mathbb{C}^2)$, $\exist n \in \mathbb{N}$. The vector space $V_n(\mathbb{C}^2)$ is a good candidate since by varying $n$ it can cover any finite dimension. We remark again that we can work with a representation for $\mathfrak{su}(2)$ without losing any generality, since $SU(2)$ is connected and simply connected (see Lie groups and algebras representations).
Let $V$ be a representation for $\mathfrak{su}(2)$ then $\exist v_0 \in V$ such that $Hv_0=\lambda v_0$, $\lambda \in \mathbb{C}$ and $Ev_0 = 0$. Remark that we are not writing explicitly the action of $\mathfrak{su}(2)$ over $V$. Since $E$ is nihilpotent, so it is the action over $V$. Then $\exist k$ such that $E^k v =0$ and $E^{k-1}v \ne 0$, for a given $v \in V$. Define $E^{k-1} v = v_0$. Now using commutation rules we show that $v_0$ is an eigenvector for the action of $H$:
Hence $Hv_0 = \lambda v_0$, $\lambda \in \mathbb{C}$.
Define $v_i = F^i v_0$. We want to compute $Hv_i$. First remark that
Then by iterating
An analogous computation can be made for $E$. It results:
Call $n$ the smallest natural number satisfying $v_{n+1} = 0$. The set $\{v_i\}_{i=0..n}$ is a basis for $V$. If $W= span(\{v_i\}_{i=0..n})$ is such that $W \subsetneq V$ then $W$ is a sub-representation for $V$, but $V$ is irreducible, then a contraddiction occurs. Finally, we know that the trace of the action of $H$ is zero, also by making the sum over the eigen-value of $v_i$ we obtain the following expression for the trace:
Then $\lambda=n$. We can conclude that $V \cong V_n(\mathbb{C}^2)$ as representations by mapping each $v_i$ with $z_i = z_1^{i}z_2^{m-i}$ and remarking that such a isomorphism between vector space commute with the action of $\mathfrak{su}(2)$ due to relations (10). It results that every irrep $V$ is of the type $V_n(\mathbb{C}^2)$, for a given $n$. Then we classified any irrep for $\mathfrak{su}(2)$ and, consequently, for $SU(2)$.
The group $SU(N)$ is connected, simply connected and compact for every $N$. Hence we can study the representation of the algebra $\mathfrak{su}(n)$. As for $\mathfrak{su}(2)$, the algebra $\mathfrak{su}(n)$ is:
The complexified algebra $\mathfrak{su}(n)_\mathbb{C} = \mathfrak{su}(n) \otimes \mathbb{C}$ is $\mathfrak{sl}(n) = \{ A \in GL_n(\mathbb{C}) | Tr(A) = 0 \},$, as already seen for $SU(2)$. The problem is now classifying the irreducible representations for $\mathfrak{sl}(n)$. Since it is a semi-simple Lie algebra we can follow the general theory (see Lie groups and algebras representations).
First we find a maximal Cartan’s sub algebra $\mathfrak{h} \in \mathfrak{sl}(n)$, i.e. the biggest commutative subalgebra. This is easily achieved by considering the algebra $\mathfrak{h}$ generated by the $n-1$ traceless diagonal matrices: they commute each other, form a vector space and it is maximal (any try of adding another element of $\mathfrak{su}(n)$ to $\mathfrak{h}$ results in losing the commutative property).
Given a representation $V$ for $\mathfrak{sl}(n)$, as usual we decompose $V$ in eigenspaces $V_\alpha$ such that $h \bullet v = \alpha v$ for $v \in V_\alpha$ and $h \in \mathfrak{h}$. When an highest weight is found between the eigenvalues, say $\beta$, we can generate an irreducible representation $W \subset V$ by the action of $g \in \mathfrak{sl}(n)$ such that $g \bullet v \ne 0$ for $v \in V_\beta$. This representation is irreducible by construction.
We present now a more specific characterization of such an irreducible representation. We introduce a more fitting notation: let $L_i : \mathfrak{sl}(n) \to \mathbb{C}$ be the linear functional such that
Then
Let $V$ be the standard representation. Given the standard basis of $V$, $e_1, \dots , e_n$ this is a basis of eigenvectors with weights $L_i$. Choose an $L_i$, say $L_1$, then the representation $V$ has highest weight $L_1$. Following the general theory, the set $\{L_i\}_{i=1,..,n}$ generates a lattice in $\mathbb{R}^n$, so that every possible weight for any representation lays in this lattice. According to the algebra representation theory, a typical highest weight could be written as: $\alpha = (a_1+ .. +a_{n-1})( L_1 ) + .. + a_{n-1} L_{n-1}$. Such an $\alpha$ could be realized as eigenvalue of an eigenvector in the vector space $X = Sym^{a_1}(V)\otimes .. \otimes Sym^{a_{n-1}}(\Lambda^{n-1}V)$ (since the eigenvalue of a tensor product of eigenvectors is the sum of each eigenvalue). Remark that the representation $X$ is not irreducible in general, but it contains an irrep $W$, generated through the action of $\mathfrak{sl}(n)$ over $v \in X_{\alpha}$, as previously stated.
Referring to Lie groups and algebras representations, we know that the Schur functor applied to the standard representation $V$ of $GL_n(\mathbb{C})$ furnishes all finite dimensional irreducible representations $\mathbb{S}_{\lambda}(V)$ for each $\lambda= (\lambda_1,..., \lambda_n)$, such that $\lambda_1 \ge ... \ge \lambda_n$. It follows that $\mathbb{S}_{\lambda}(V)$ is irreducible for $SL_n(\mathbb{C})$ too, since it is a subgroup of $GL_n(\mathbb{C})$. In particular it is an irrep for the Lie algebra $\mathfrak{sl}(n)$. Now we have to check that every possible highest weight value for a representation of $\mathfrak{sl}(n)$ is assumed in $\mathbb{S}_{\lambda}(V)$. The answer is the following proposition:
Proposition. The representation $\mathbb{S}_{\lambda}(V)$, with $V$ the standard representation, is the irreducible representation of $\mathfrak{sl}(n)$ with highest weight $\lambda_1 L_1 + ... + \lambda_n L_n$.