Domenico Fiorenza Lie groups and algebras representations

Lie groups and algebras representations

Contents

Lie groups and algebras

This is meant to a be a very quick introduction to Lie groups and representation, tuned for computing the representations of the Poincaré group and of (special) unitary groups occuring in the standard model. Unless specified, we will be considering finite-dimensional representation in this section.

A Lie group is a group $G$, with a comptible differential structure, i.e., such that the multiplication

and the inverse

are differentiable mappings.

Basic examples of Lie groups are the general linear groups of order $n$: $GL_n(\mathbb{R})$ and $GL_n(\mathbb{C})$. Since they are open subsets of $\mathbb{R}^{n^2}$ and $\mathbb{R}^{2n^2}$ respectevely they have a natural differentiable structure, and both the product and the inverse are differentiable operations. Closed subgroups of $GL_n(\mathbb{R})$ and $GL_n(\mathbb{C})$ are Lie groups. Classical examples are:

We will meet a few of them as typical groups for particle physics. Notice that all the Lie groups listed above are matrix groups. This is not always the case, i.e., not every Lie group is a matrix group, but due to introductional purpose of the section, the proofs will be given only for matrix groups.

Lie algebras and their representation

We want to introduce Lie algebras as tangent spaces at the identity for Lie groups. As for any differential manifold, the tangent space at a point $p\in M$ is ralized by considering all the germs of smooth curves from $c:(-\epsilon,\epsilon)\to M$, and taking their tangent vector at $0$. Therefore, if we write

where $G$ is a Lie group and $\mathbb{I}$ is its identity element, then

Note that for a sub-Lie group $G$ of the general linear group of order $n$, the Lie algebra $\mathfrak{g}$ is a subspace of the vector space of $n\times n$ matrices. It should be remarked that $U(n)$ is a real subgroup of $GL_n(\mathbb{C})$ but not a complex subgroup; hence its Lie algebra $\mathfrak{u}_n$ is a real subspace of $M(n,n;\mathbb{C})$ but not a complex subspace.

Proposition. Let $G$ be Lie subgroup of $GL_n(\mathbb{K})$, with $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$, then $\mathfrak{g}$ is a linear subspace of $M_n(\mathbb{K})$.

Proof. A typical result borrowed by differential geometry states that the tangent space in a point for a $n$-dimensional manifold $M$, is a vector space of dimension $n$. By an easy calculation, it results that the tangent space to $GL_n(\mathbb{K})$ is $M_n(\mathbb{K})$. Then for every $G \subset GL_n(\mathbb{K})$, $T_{g}(G)$, $\forall g \in G$, is a vector space, included in $M_n(\mathbb{K})$. This completes the proof.

We are going to give $\mathfrak{g}$ a Lie algebra structure.

Definition. A Lie algebra is a vector space $V$ endowed with an antisymmetric bilinear operation $[,]: V\wedge V \to V$ such that the Jacobi identity $[[x,y],z]+[[y,z],x]+ [[z,x],y] =0$ holds, for any $x,y,z$ in $V$.

Recall that for matrix algebras the bracket has its common form:

Proposition. The vector space $\mathfrak{g}$ has a natural Lie algebra structure.

Proof. We only prove this for a matrix group $G$; hence our simplified statement will be: the subspace $\mathfrak{g}\subseteq M_n(\mathbb{K})$ is closed with respect to the Lie bracket of $M_n(\mathbb{K})$. First remark that group $G$ is closed under conjugation: if $h\in G$ then $g^{-1}h g \in G$ for every $g \in G$. Thus if $c(t)$ is a smooth curve through the identity, so is $g^{-1}c(t)g$. By differentiating, we see that $g^{-1}c'(0)g$ is in $\mathfrak{g}$. We can state that under $Ad(g)X=gXg^-1$ for $g \in G$ $X$ is sent to $\mathfrak{g}$. We obtain a smooth function in $\mathfrak{g}$: $t \to Ad(c(t))X$. By definition we have:

The right side is a multiplication for a scalar of two elements from a vector space, then is still in the vector space. Since $\mathfrak{g}$ is a closed subset in a matrix space the limit is still in $\mathfrak{g}$. Let’s compute the limit.

Then

Evaluating in $t=0$ results that $c'(0)X-Xc'(0)$ is still in $\mathfrak{g}$. Hence $\mathfrak{g}$ is closed respect to the Lie bracket of $M_n(\mathbb{K})$, and so it has a natural induced Lie algebra structure.

The construction so far of a Lie algebra clearly shows that it is the infinitesimal part of a Lie group. Through the exponential map we can go backwards from $\mathfrak{g}$ to $G$.

Definition. The exponential of a matrix $A$ is:

The exponential has some nice properties, which we report without proof:

Proposition. Let $A$ be a squared complex matrix. then 1. $e^X e^Y=e^{X+Y}$ whenever $[X,Y]=0$ 1. $e^X$ is non singular 1. $t \to e^{tX}$ is a smooth curve into $GL_n{\mathbb{C}}$ 1. $\frac{d}{dt}(e^{tX})=Xe^{tX}$ 1. $det(e^X)=e^{Tr(X)}$ 1. $X \to e^X$ is a $C^\infty$ mapping from matrix space into itself.

A typical result from differential geometry states that the exponential map is a local homeomorphism defined between $\mathfrak{g}=T_{a}G$ and $U \subset G$ an open set in the Lie group. The exponential map plays a key role in connecting Lie algebras representation to Lie groups representation. So we want to study Lie algebras reps, first, and then furnish a way to refer them to corresponding Lie groups.

Definition. Let $V$ be a vector space over a field $\mathbb{K}$ A representation of $\mathfrak{g}$ on $V$ is a morphism of Lie algebras $\pi : \mathfrak{g} \to (End_\mathbb{K}V)$ such that:

Examples: * ad is a representation of $\mathfrak{g}$ on $\mathfrak{g}$ with $\mathbb{k}=\mathbb{K}$ * If $\mathfrak{g}$ is a sub-Lie algebra of $End(\mathbb{K}^n)=M_n(\mathbb{K})$, then the inclusion $\mathfrak{g}\hookrightarrow End(\mathbb{K}^n)$ is a representation of $\mathfrak{g}$ over $\mathbb{K}^n$, called the defining representation.

Recall that a representation is irreducible if and only if the only invariant subspace $U \in V$, such that $\pi(X)U \subseteq U$ $\forall X \in \mathcal{A}$, are $V$ and $\{ 0 \}$.

Notation. Given a representation $\pi$ as above, we will be writing $g \cdot v$ with $v \in V$ and $g \in \mathfrak{g}$ instead of $\pi(g)v$.

Recall that a vector $v \in V$ such that $g \cdot v= \alpha v$ is an eigenvector for the action of $g \in \mathfrak{g}$ with eigenvalue $\alpha$. In representation theory an eigenvalue is commonly said “weight”.

We end section by introducing the tensor product of representation. Given $V$ and $W$ representation for $\mathfrak{g}$, we define the representation $V \otimes W$, the vector space $V \otimes W$ with action:

If $v$ is an eigen vector for the action of $g$, with eigenvalue $\alpha$ and $w$ is an eigenvector with eigenvalue $\beta$ then we obtain the following:

The tensor product of two or more eigenvectors is an eigenvector with eigenvalue the sum of the eigenvalues.

Irreducible representations for a semi-simple Lie algebra

We now present a general method for studying Lie algebras representation. We will be referring to representations for sl(2,C). First we need to introduce the concept of semi-simplicity for Lie algebras. Recall that a Lie algebra is solvable if the sequence $[ \mathfrak{g}, \mathfrak{g}]; [[\mathfrak{g},\mathfrak{g}],[\mathfrak{g},\mathfrak{g}]], ..$ at a certain step is $\{ 0 \}$. We define $\mathfrak{g}$ to be simple if $\mathfrak{g}$ is non abelian and $\mathfrak{g}$ has no proper nonzero ideals, and semi-simple if $\mathfrak{g}$ has no nonzero solvable ideals. We have a criterion to recognize semi-simple Lie algebras, consisting in computing the Killing form for a Lie algebra:

Definition. Let $\mathfrak{g}$ be a Lie algebra, then the Killing form for $\mathfrak{g}$ is the bilinear form

where $ad$ is the linear application $adX(Z)=[X,Z]$

Recall that a bilinear form $C$ on a vector space $V$ is non-degenerate if and only if $rad C = \{ v \in V | C(v,u)=0 \forall u \in V \}$ is $\{ 0 \}$.

Theorem. \[ Cartan's criterion \] The Lie algebra $\[ Cartan's criterion$ is semi-simple if and only if the Killing form for $\mathfrak{g}$ is nondegenerate.

There is now a “standard” way to approach representation for Lie algebras consisting of several step. Step 0. Verify that the algebra $\mathfrak{g}$ is semi-simple (for example using Cartan’s Criterion). Step 1. Find an abelian sub-algebra $\mathfrak{h} \subset \mathfrak{g}$ acting diagonally. This is the analogous of $h \in \mathfrak{sl}_2(\mathbb{C})$. We want such an $\mathfrak{h}$ that is maximal among all abelian sub-algebras acting diagonally. We call it Cartan sub-algebra. Step 2. Let $\mathfrak{h}$ act on $\mathfrak{g}$ by the adjoint representation, and decompose $\mathfrak{g}$ accordingly. We obtain the Cartan decomposition:

If $X \in \mathfrak{g}_\alpha$ then for $H \in \mathfrak{h}$, $adH(X)= \alpha(H)X$. Such an $\alpha$ is a weight for adjoint representation and is also said “root”. Let $X \in \mathfrak{g}_\alpha$ and $Y \in \mathfrak{g}_\beta$ then

i.e. $X \in \mathfrak{g}_\alpha$ acts by adjoint on $\mathfrak{g}_\beta$ carrying an element of $\mathfrak{g}_\beta$ in $\mathfrak{g}_{\alpha+\beta}$. This is the analogous of $e$ and $f$ in $\mathfrak{sl}_2(\mathbb{C})$. One can prove that each $\mathfrak{g}_\alpha$ is one dimensional, and that if $\alpha$ is a root, then so is for $-\alpha$. The analogous is valid for a generic representation on $V$ vector space. We obtain the following direct sum decomposition:

and a computation shows that each $X \in \mathfrak{g}_\alpha$ carries $v \in V_\beta$ into $X(v) \in V_{\alpha+\beta}$

pics to insert.

Step 3. Find the copies of $\mathfrak{sl}_2(\mathbb{C})$ in $\mathfrak{g}$. We already remarked in step 2 that each $\mathfrak{g}_\alpha$ is one-dimensional and for any given $\mathfrak{g}_\alpha$ a $\mathfrak{g}_{-\alpha}$ exists. The commutator $[\mathfrak{g}_\alpha,\mathfrak{g}_{-\alpha}]$, due to calculation done in step 2. is in $\mathfrak{h}$ and has dimension at most 1, next ,by proving that the commutator is not $0$, it results that:

is a copy of $\mathfrak{sl}_2(\mathbb{C})$ in $\mathfrak{g}$.

Step 4. We can now build a lattice of weight for a given representation. We have two fundamental lattices: $\Lambda_r$ generated by roots, and $\Lambda$ the lattice generated by the weights of the standard representation. Due to the calculation above, any “legal” value for a weight $\alpha$ lays on $\Lambda$ and any other weight must be congruent modulo $\Lambda_r$ to $\alpha$.

Pic to insert.

Step 5. –TOFIX – Since it now clear where a weight is located, we want to present a symmetry in the weight space. First introduce the Weyl group. It the group $\mathfrak{M}$ generated by reflections of the weights $\alpha$ respect to the orthogonal hyper-plane. Now suppose that $V$ is a representation with a certain decomposition $V = \bigoplus V_\beta$. The weights $\beta$ appearing in this decomposition can then be broken up into equivalence classes mod $\alpha$, and the direct sum $V_{[\beta]} = \bigoplus_{n \in \mathbb{Z}} V_{\beta + n\alpha}$ of the eigenspaces in a given equivalence class will be a subrepresentation of $V$ for $\mathfrak{s}_\alpha$. It follows that the set of weights of $V$ congruent to any give $\beta$ mod $\alpha$ will be invariant under action of $W_\alpha$, reflection of the Weyl group respect to the hyperplane orthogonal to $\alpha$. In concrete this is a formal proof to the fact that the weights are symmetric respect to the origin.

Step 6. In the end we have to “choose a direction” in the weight space. This is done by evaluating roots with a linear real functional $l$ irrational respect to $\Lambda_r$ (i.e. $l(\alpha)\ne 0 \forall \alpha \in \Lambda_r$). This leads to a classification of weights in

$R^+$ are the positive ones, and $R^-$ the negative. We can now give in general the definition for highest weight and highest weight vector. Let $V$ be a representation of $\mathfrak{g}$, a non-zero vector $v \in V$ that is both eigenvector for the action of $\mathfrak{h}$ and in the kernel of $\mathfrak{g}_\alpha \forall \alpha \in R^+$ is called a highest weight vector of $V$.

Proposition.For any semisimple Lie algebra $\mathfrak{g}$, any $V$ finite-dimensional representation of $\mathfrak{g}$ possesses a highest weight vector, and any $W \subset V$ invariant subspace generated by successive application of $\mathfrak{g}_\alpha, \alpha \in R^-$ on $v$ highest weight vector of $V$ is an irreducible representation. Every irreducible representation possesses a unique highest weight vector up to scalar.

Complexification of a Lie Algebra.

Given a Lie algebra $\mathfrak{g}$ of a Lie subgroup of $GL_n(\mathbb{C})$ the complexification $\mathfrak{g}_\mathbb{C}$ is the tensor product

Sometimes it is convenient to consider the complexification instead of the Lie algebra. Moreover the following result for representation is valid:

Proposition. Let $\mathfrak{g}$ be the Lie algebra of a Lie subgroup of $GL_n(\mathbb{C})$ and let $V$ be a representation for $\mathfrak{g}$. Then $V$ is irreducible for $\mathfrak{g}$ if and only if it is irreducible for $\mathfrak{g}_\mathbb{C}$.

Proof. Given a $W \subset V$, it is invariant under the action of $\mathfrak{g}$ if and only if it is invariant under the action of $\mathfrak{g}_\mathbb{C}$: the complex coefficents of $\mathfrak{g}_\mathbb{C}$ are absorbed by the complex structure of the vector subspace $W$.

Lie groups representations

A representation for a Lie group $G$ acting on a vector space is a homomorphism $\Pi : G \to End(V)$ i.e. $\Pi(g_1g_2)= \Pi(g_1)\Pi(g_2)$. We already remarked that a Lie Algebra $\mathfrak{g}$ is the tangent space to the identity for a certain Lie group $G$. So the first problem is associating $\mathfrak{g}$ to a Lie group $G$. Remark that this operation is ambiguous: different Lie groups may have the same Lie algebra modulo isomorphism.

Transporting a Lie algebra representation to a Lie group representation it’s the same than, given an homomorphism between Lie algebras, obtaining a homomorphism between group and vice-versa. Recall that for a given $G$ connected Lie group, a neighborhood of the identity $U$ generates $G$ .

Proposition. Let $G$ and $H$ be Lie groups with $G$ connected. A map $\rho : G \to H$ is uniquely determined by its differential $d \rho_e : T_{e}G \to T_{e}H$ at the identity.

This is easily achieved by utilizing the exponential map: the basic idea is that since $exp : T_{e}G \to G$ is a local homeomorphism in a neighborhood of the identity, we can write each element $g \in G$ as product $g=x_1x_2..x_n$ and realize each $x_i$ as exponential: $x_i = e^{A_i} \quad \exists A_i \in \mathfrak{g}$. So by knowing the action of $d\rho$ we can reconstruct $\rho$ in a neighborhood of the identity, through the exponential map, and then reconstruct the action on the whole group by the precedent proposition.

The following construction is achieved:

The opposite is not always possible by considering Lie groups that are just connected. We need to know which homomorphisms between vector space are differential of a homomorphism between Lie groups. The main issue about this problem is that, while it is always possible writing $\forall g \in G$ as a product of $x_i \in U$ with $U$ neighborhood of the identity, it is not always true that this form is unique. The basic idea is that we can join $e$ and $g$ in $G$ by several paths. Each path is a different choice of $x_i$ in the neighborhood to write $g$. If two paths can be made the same by an homotopy, then the two different “writings” can be made the same continuosly, else there is a discontinuity. In the second case, attempting a pullback from a vector space homomorphism results in an ambiguity of the definition. The obvious request to do for a Lie group to avoid this ambiguity, is to be simple-connected.

Example todo

The results so far can be applied to representation theory by remarking that a representation $\pi$ of $\mathfrak{g}$ on a vector space $V$ an application:

linear, that preserves brackets. We have that every representation of a connected and simply connected Lie group are in one-to-one correspondence with representation of its Lie algebra.

Unitary representation for compact and non-compact Lie groups

An important type of representation is the unitary one.

Definition. A representation $U : G \to U(n)$ is said unitary.

Their importance resides in being the principal type of representation we want to consider in physics. First we have a result for non-compact Lie group:

Theorem. Let $U : G \to U(n)$ be a unitary representation of a Lie group G acting on $\mathbb{H}$, Hilbert space of finite dimension $n \in \mathbb{N}$. Then $U$ is completely reducible. Moreover, if $U$ is irreducible and if $G$ si a connected, simple non compact Lie group, then $U$ is the trivial representation.

In concrete, the only interesting unitary representation for a non-compact Lie group are infinite dimensional. On the other hand for a compact Lie group we have a nicer result:

Theorem. Let $U$ be a unitary irreducible representation (UIR from now on) of a compact Lie group $G$, acting on $\mathbb{H}$, Hilbert space. Then $\mathbb{H}$ is finite dimensional. Moreover, every unitary representation of $G$ is a direct sum of UIRs (the sum may be infinite).

Induced Representations

This is a way of a approaching representation by considering a representation from a Lie subgroup $H \subset G$ and “inducing” from it a representation for a Lie group $G$. We will want to use this approach for computing irreps for a group $G$ that is a semi-direct product of two groups.

Given a representation $V$ for $H$, the induced representation of $H$ over $G$, usually written $ind^G_H(V)$ is by definition:

A more geometrical approach to induced representations is the following. Taken $G$, $H$ and $V$ as before, define the homogeneous vector bundle $G \times_H V$ over $G/H$ by :

where \tilde is the following equivalence relation:

The two definitions are the same when we identify the functions in $Ind^G_H(V)$ as sections of the homogeneous vector bundle.

Theorem Let $G$ be a Lie group, $H$ a closed subgroup of $G$, and $V$ a representation of $H$. There is a linear bijection between $\Gamma(G/H,G \times_H V)$ and $Ind^G_H(V)$ .

Proof. We produce the two applications explicitly. Given a section $s \in \Gamma(G/H,G \times_H V)$, let $f_s \in Ind^G_H(V)$ be $f_s(g)= g^{-1} (s(g))$. Conversely, given $f \in Ind^G_H(V)$, let $s_f \in \Gamma(G/H,G \times_H V)$ be $s_f(g)= (g,f(g))$. These maps are well defined and one is the inverse of the other.

Remark two fundamental facts by definition: 1. the “bigger” the group, the “smallest” is the dimensionality of the induced representation. For example if $H= \{e \}$ and $V=\mathbb{C}$, it results $ind^G_H(V) = C(G)$ that is an enormous one. 1. A priori, even if the representation $V$ of $H$ is irreducible, we can’t state anything on the irreducibility of $ind^G_H(V)$.

A partial solution to point 2. is given by the following:

Theorem. Leg $G$ be a group and $H$ a closed subgroup of $G$. If $V$ is a representation of $H$ and $W$ is a representation of $G$, then as vector spaces:

where $W|_H$ is the representation $W$ with the same action of $G$ but restricted to the elements of $H \subset G$ .

to be continued

Irreducible representations for $GL(V)$

We conclude this overview on Lie groups representation by describing the irreps for the general linear group of a vector space $V$. We need to introduce the concept of Young diagram and the Weyl construction.

Let $S_d$ be the group of symmetries of $d$ elements. We know that we can write $d$ as sum of integer in $2^d$ combinations. To each of $2^d$ writing we can associate a Young diagram: say $\lambda = (\lambda_1, ..., \lambda_k)$ the partition of $d$ such that $\lambda_1 \gt \lambda_2 \gt ... \gt \lambda_k$ then a Young diagram for this partition is

with $\lambda_i$ boxes in the i-th row, lined up on the left. We define the conjugate partition $\lambda'$ associated to $\lambda$, the partition of the Young diagram reflected in the $45$ degree line. For example given the Young diagram in picture, associated to $\lambda=(3,3,1,1)$, by reflecting, we obtain the conjugate partition $\lambda'=(5,3,2)$.

By choosing a numeration from $1$ to $d$ of the boxes of a Young diagram, we obtain a Young tableau:

The tableau defines two subgroup of $S_D$

Recall that an algebra could be associated to a finite group.

Definition. Given a finite group $G$, we define the group algebra $\mathbb{K}G$ over a field $G$, the vector space with basis $e_g$ one for each $g \in G$ and with product:

By $P$ and $Q$ we define we define two elements in the group algebra :

Given a vector space $V$, the group algebra $\mathbb{C}S_d$ acts on $V^{\otimes d}$, by permuting the factors of the tensor product. In particular we can calculate the images of $a_{\lambda}$ and $b_{\lambda}$ as endomorphisms of $V^{\otimes d}$:

where $\nu$ is the conjugate partition of $\lambda$. We set

This is called the Young symmetrizer. For example given $\lambda=(d)$, $c_{\lambda} = a_{\lambda}$ and $Imm(c_{\lambda}) = Sym^d(V)$; given $\lambda=(1,1,..,1)$, $c_{\lambda}= b_{\lambda}$ and $Imm(c_{\lambda}) = \Lambda^d(V)$. In general, given $V$ a vector space, and $GL(V)$ its general linear group, we define:

The functor $\mathbb{S}$ is called the Schur functor or simply Weyl construction. Each $\mathbb{S}_\lambda(V)$ is a representation for $GL(V)$, moreover one can prove that it is irreducible. This is summarized in the following theorem.

Theorem.. Let $k=dim(V)$. Then $\mathbb{S}_\lambda(V)$ is zero if $\lambda_{k+1} \ne 0$, then

1. $$dim($$
2. Let $m_{\lambda}$ be the dimension of the irredicible representation $V_{\lambda}$ of $S_d$ corresponding to $\lambda$. Then $$V^{\otimes d } \cong \bigoplus_{\lambda} \mathbb{S}_{\lambda}(V^{\otimes m_{\lambda}})$$
3. Each $\mathbb{S}_{\lambda}(V)$ is an irreducible representation of $GL(V)$.

The theorem furnish us many of the irreps for $GL(V)$. We cannot obtain all the finite irreducible representation for $GL(V)$ since the dual of each $\mathbb{S}_{\lambda}(V)$ is excluded. The way to the complete classification is to introduce the k-th power of the determinant $D_k = (\Lambda^n (V))^{\otimes k}$, with $D_{-k} = D_{k}^*$ . By tensoring with $D_k$ we can define any $\mathbb{S}_{\lambda} (V)$ even if $\lambda_i \le 0$ for some $i$. By a simple tensor product property we have:

then if some $\lambda_i \le 0$ we define:

The following theorem guarantees that we exhaust all possibilities for $GL(V)$ finite dimensional irreducible representation.

Theorem. Every finite dimensional, irreducible and complex representation of $GL(V)$ is isomorphic to $\mathbb{S}_{\lambda} (V)$ for a unique index $\lambda= (\lambda_1,...,\lambda_n)$ with $\lambda_1 \ge \lambda_2 \ge .. \ge \lambda_n$.