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 GG, with a comptible differential structure, i.e., such that the multiplication

G×GG G\times G \to G
(g,h)gh(g,h) \to g h

and the inverse

GG G \to G
gg 1g \to g^{-1}

are differentiable mappings.

Basic examples of Lie groups are the general linear groups of order nn: GL n()GL_n(\mathbb{R}) and GL n()GL_n(\mathbb{C}). Since they are open subsets of n 2\mathbb{R}^{n^2} and 2n 2\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()GL_n(\mathbb{R}) and GL n()GL_n(\mathbb{C}) are Lie groups. Classical examples are:

SO(n)={xGL n()|xx=1anddet(x)=1} SO(n) = \{ x \in GL_n(\mathbb{R}) | x x'=1 and det(x) =1 \}
U(n)={xGL n()|xx *=1} U(n) = \{ x \in GL_n(\mathbb{C}) | x x^*=1 \}
SU(n)={xGL n()|xx *=1anddet(x)=1} SU(n) = \{ x \in GL_n(\mathbb{C}) | x x^*=1 and det(x)=1 \}
SL n()={xGL n()|det(x)=1} SL_n(\mathbb{C}) = \{ x \in GL_n(\mathbb{C}) | det(x)=1 \}
SL n()={xGL n()|det(x)=1}. SL_n(\mathbb{R}) = \{ x \in GL_n(\mathbb{R}) | det(x)=1 \}.

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 pMp\in M is ralized by considering all the germs of smooth curves from c:(ϵ,ϵ)Mc:(-\epsilon,\epsilon)\to M, and taking their tangent vector at 00. Therefore, if we write

𝔤=T 𝕀G, \mathfrak{g}=T_\mathbb{I}G,

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

𝔤={c(0)|cisagermofsmoothcurvewithc(0)=𝕀}. \mathfrak{g}=\{ c'(0) | c is a germ of smooth curve with c(0)=\mathbb{I} \}.

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

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

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

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

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

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

[X,Y]=XYYX [X,Y] = X Y-Y X

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

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

ddtAd(c(t))X| t=0=lim t=01t[Ad(c(t))XX].\frac{d}{dt} Ad(c(t))X |_{t=0} = lim_{t=0} \frac{1}{t}[Ad(c(t))X-X].

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

ddtc(t) 1=c(t) 1c(t)c(t) 1 \frac{d}{dt} c(t)^{-1}= -c(t)^{-1}c'(t)c(t)^{-1}

Then

ddtAd(c(t))X=ddt[c(t)Xc(t) 1]=c(t)Xc(t) 1+c(t)X[ddtc(t) 1]=c(t)Xc(t) 1c(t)Xc(t) 1c(t)c(t) 1. \frac{d}{dt} Ad(c(t))X=\frac{d}{dt}[c(t)Xc(t)^{-1}]=c'(t)Xc(t)^{-1} + c(t)X[\frac{d}{dt}c(t)^{-1}]= c'(t)Xc(t)^{-1}-c(t)Xc(t)^{-1}c'(t)c(t)^{-1}.

Evaluating in t=0 t=0 results that c(0)XXc(0)c'(0)X-Xc'(0) is still in 𝔤\mathfrak{g}. Hence 𝔤\mathfrak{g} is closed respect to the Lie bracket of M n(𝕂)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 𝔤\mathfrak{g} to GG.

Definition. The exponential of a matrix AA is:

exp(A)=e A= n=0 1n!A n \exp(A)= e^A=\sum_{n=0}^{\infty}\frac{1}{n!}A^n

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

Proposition. Let AA be a squared complex matrix. then 1. e Xe Y=e X+Ye^X e^Y=e^{X+Y} whenever [X,Y]=0[X,Y]=0 1. e Xe^X is non singular 1. te tXt \to e^{tX} is a smooth curve into GL nGL_n{\mathbb{C}} 1. ddt(e tX)=Xe tX\frac{d}{dt}(e^{tX})=Xe^{tX} 1. det(e X)=e Tr(X)det(e^X)=e^{Tr(X)} 1. Xe XX \to e^X is a C C^\infty mapping from matrix space into itself.

Domenico: It’s worth using the exponential to write explicitely the Lie algebras of the matrix groups mentioned above here.

A typical result from differential geometry states that the exponential map is a local homeomorphism defined between 𝔤=T aG\mathfrak{g}=T_{a}G and UGU \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 VV be a vector space over a field 𝕂\mathbb{K} A representation of 𝔤\mathfrak{g} on VV is a morphism of Lie algebras π:𝔤(End 𝕂V)\pi : \mathfrak{g} \to (End_\mathbb{K}V) such that:

π([X,Y])=[π(X),π(Y)]X,Y𝔤\pi([X,Y]) = [\pi(X),\pi(Y)] \quad \forall X,Y \in \mathfrak{g}

Domenico: What does “such that” should mean???? I guess you intended to write “a morphism of Lie algebras π:𝔤(End 𝕂V)\pi : \mathfrak{g} \to (End_\mathbb{K}V), i.e., a linear map π:𝔤(End 𝕂V)\pi : \mathfrak{g} \to (End_\mathbb{K}V) such that…”

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

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

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

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

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

g(vw)=(gv)w+v(gw). g \bullet (v \otimes w) = (g \bullet v) \otimes w + v \otimes( g \bullet w ).

If vv is an eigen vector for the action of gg, with eigenvalue α\alpha and ww is an eigenvector with eigenvalue β\beta then we obtain the following:

g(vw)=(α+β)vw. g \bullet (v \otimes w) = (\alpha + \beta) v \otimes w.

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 [𝔤,𝔤];[[𝔤,𝔤],[𝔤,𝔤]],.. [ \mathfrak{g}, \mathfrak{g}]; [[\mathfrak{g},\mathfrak{g}],[\mathfrak{g},\mathfrak{g}]], .. at a certain step is {0}\{ 0 \}. We define 𝔤\mathfrak{g} to be simple if 𝔤\mathfrak{g} is non abelian and 𝔤\mathfrak{g} has no proper nonzero ideals, and semi-simple if 𝔤\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 𝔤\mathfrak{g} be a Lie algebra, then the Killing form for 𝔤\mathfrak{g} is the bilinear form

B(X,Y)=Tr(adXadY). B(X,Y)=Tr(ad X ad Y).

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

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

Theorem. \[ Cartan's criterion \] The Lie algebra \[Cartanscriterion \[ Cartan's criterion is semi-simple if and only if the Killing form for 𝔤\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 𝔤\mathfrak{g} is semi-simple (for example using Cartan’s Criterion). Step 1. Find an abelian sub-algebra 𝔥𝔤\mathfrak{h} \subset \mathfrak{g} acting diagonally. This is the analogous of h𝔰𝔩 2()h \in \mathfrak{sl}_2(\mathbb{C}). We want such an 𝔥\mathfrak{h} that is maximal among all abelian sub-algebras acting diagonally. We call it Cartan sub-algebra. Step 2. Let 𝔥\mathfrak{h} act on 𝔤\mathfrak{g} by the adjoint representation, and decompose 𝔤\mathfrak{g} accordingly. We obtain the Cartan decomposition:

𝔤=𝔥(𝔤 α) \mathfrak{g} = \mathfrak{h}( \bigoplus \mathfrak{g}_\alpha)

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

AdH([X,Y])=[H,[X,Y]]=[[H,X],Y]+[[H,Y],X]=(α(H)+β(H))[X,Y] AdH([X,Y]) = [H,[X,Y]] = [[H,X],Y] + [[H,Y],X] = (\alpha(H)+\beta(H)) [X,Y]

i.e. X𝔤 αX \in \mathfrak{g}_\alpha acts by adjoint on 𝔤 β\mathfrak{g}_\beta carrying an element of 𝔤 β\mathfrak{g}_\beta in 𝔤 α+β\mathfrak{g}_{\alpha+\beta}. This is the analogous of ee and ff in 𝔰𝔩 2()\mathfrak{sl}_2(\mathbb{C}). One can prove that each 𝔤 α\mathfrak{g}_\alpha is one dimensional, and that if α\alpha is a root, then so is for α-\alpha. The analogous is valid for a generic representation on VV vector space. We obtain the following direct sum decomposition:

V=V α V = \bigoplus V_{\alpha}

and a computation shows that each X𝔤 αX \in \mathfrak{g}_\alpha carries vV βv \in V_\beta into X(v)V α+βX(v) \in V_{\alpha+\beta}

pics to insert.

Step 3. Find the copies of 𝔰𝔩 2()\mathfrak{sl}_2(\mathbb{C}) in 𝔤\mathfrak{g}. We already remarked in step 2 that each 𝔤 α\mathfrak{g}_\alpha is one-dimensional and for any given 𝔤 α\mathfrak{g}_\alpha a 𝔤 α\mathfrak{g}_{-\alpha} exists. The commutator [𝔤 α,𝔤 α][\mathfrak{g}_\alpha,\mathfrak{g}_{-\alpha}], due to calculation done in step 2. is in 𝔥\mathfrak{h} and has dimension at most 1, next ,by proving that the commutator is not 00, it results that:

𝔰 α=𝔤 α𝔤 α[𝔤 α,𝔤 α].\mathfrak{s}_\alpha = \mathfrak{g}_\alpha \oplus \mathfrak{g}_{-\alpha} \oplus [\mathfrak{g}_\alpha,\mathfrak{g}_{-\alpha}].

is a copy of 𝔰𝔩 2()\mathfrak{sl}_2(\mathbb{C}) in 𝔤\mathfrak{g}.

Step 4. We can now build a lattice of weight for a given representation. We have two fundamental lattices: Λ r\Lambda_r generated by roots, and Λ\Lambda the lattice generated by the weights of the standard representation. Due to the calculation above, any “legal” value for a weight α\alpha lays on Λ\Lambda and any other weight must be congruent modulo Λ r\Lambda_r to α\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 𝔐\mathfrak{M} generated by reflections of the weights α\alpha respect to the orthogonal hyper-plane. Now suppose that VV is a representation with a certain decomposition V=V βV = \bigoplus V_\beta. The weights β\beta appearing in this decomposition can then be broken up into equivalence classes mod α\alpha, and the direct sum V [β]= nV β+nαV_{[\beta]} = \bigoplus_{n \in \mathbb{Z}} V_{\beta + n\alpha} of the eigenspaces in a given equivalence class will be a subrepresentation of VV for 𝔰 α\mathfrak{s}_\alpha. It follows that the set of weights of VV congruent to any give β\beta mod α\alpha will be invariant under action of W αW_\alpha, reflection of the Weyl group respect to the hyperplane orthogonal to α\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 ll irrational respect to Λ r\Lambda_r (i.e. l(α)0αΛ rl(\alpha)\ne 0 \forall \alpha \in \Lambda_r ). This leads to a classification of weights in

R=R +R R = R^+ \cup R^-

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

Proposition.For any semisimple Lie algebra 𝔤\mathfrak{g}, any VV finite-dimensional representation of 𝔤\mathfrak{g} possesses a highest weight vector, and any WVW \subset V invariant subspace generated by successive application of 𝔤 α,αR \mathfrak{g}_\alpha, \alpha \in R^- on vv highest weight vector of VV 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 𝔤\mathfrak{g} of a Lie subgroup of GL n()GL_n(\mathbb{C}) the complexification 𝔤 \mathfrak{g}_\mathbb{C} is the tensor product

𝔤 =𝔤 . \mathfrak{g}_\mathbb{C} = \mathfrak{g} \otimes_\mathbb{R} \mathbb{C}.

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

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

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

Lie groups representations

A representation for a Lie group GG acting on a vector space is a homomorphism Π:GEnd(V)\Pi : G \to End(V) i.e. Π(g 1g 2)=Π(g 1)Π(g 2)\Pi(g_1g_2)= \Pi(g_1)\Pi(g_2). We already remarked that a Lie Algebra 𝔤\mathfrak{g} is the tangent space to the identity for a certain Lie group GG. So the first problem is associating 𝔤 \mathfrak{g} to a Lie group GG. 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 GG connected Lie group, a neighborhood of the identity UU generates GG .

Proposition. Let GG and HH be Lie groups with GG connected. A map ρ:GH\rho : G \to H is uniquely determined by its differential dρ e:T eGT eHd \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 eGGexp : T_{e}G \to G is a local homeomorphism in a neighborhood of the identity, we can write each element gGg \in G as product g=x 1x 2..x ng=x_1x_2..x_n and realize each x ix_i as exponential: x i=e A iA i𝔤x_i = e^{A_i} \quad \exists A_i \in \mathfrak{g}. So by knowing the action of dρd\rho we can reconstruct ρ\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:

{ConnectedLiegroupshomomorphisms}{Liealgebrashomomorphisms} \{Connected Lie groups homomorphisms \} \to \{Lie algebras homomorphisms \}

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 gG\forall g \in G as a product of x iUx_i \in U with UU neighborhood of the identity, it is not always true that this form is unique. The basic idea is that we can join ee and gg in GG by several paths. Each path is a different choice of x ix_i in the neighborhood to write gg. 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

{ConnectedLiegroupshomomorphisms}longleftrightarrow{Liealgebrashomomorphisms}\{ Connected Lie groups homomorphisms \} \longleftrightarrow \{Lie algebras homomorphisms \}

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

π:𝔤𝔤𝔩(V)=End(V)\pi : \mathfrak{g} \to \mathfrak{gl}(V) = End(V)

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:GU(n) 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:GU(n) U : G \to U(n) be a unitary representation of a Lie group G acting on \mathbb{H}, Hilbert space of finite dimension nn \in \mathbb{N}. Then UU is completely reducible. Moreover, if UU is irreducible and if G G si a connected, simple non compact Lie group, then UU 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 U be a unitary irreducible representation (UIR from now on) of a compact Lie group GG, acting on \mathbb{H}, Hilbert space. Then \mathbb{H} is finite dimensional. Moreover, every unitary representation of GG 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 HGH \subset G and “inducing” from it a representation for a Lie group GG. We will want to use this approach for computing irreps for a group GG that is a semi-direct product of two groups.

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

Ind H G(V)={smoothf:GVs.t.f(hg)=hf(g)hH}.Ind^G_H(V)= \{smooth f : G \to V s.t. f(hg)=hf(g) \quad \forall h \in H \}.

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

G \times_H V = G \times V / \tilde

where \tilde is the following equivalence relation:

(gh,v)(˜g,hv)gG,hH,vV.(gh,v) \tilde (g,hv) \quad \forall g \in G, \forall h \in H, \forall v \in V .

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

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

Proof. We produce the two applications explicitly. Given a section sΓ(G/H,G× HV)s \in \Gamma(G/H,G \times_H V), let f sInd H G(V)f_s \in Ind^G_H(V) be f s(g)=g 1(s(g)) f_s(g)= g^{-1} (s(g)) . Conversely, given fInd H G(V)f \in Ind^G_H(V), let s fΓ(G/H,G× HV)s_f \in \Gamma(G/H,G \times_H V) be s f(g)=(g,f(g))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}H= \{e \} and V=V=\mathbb{C}, it results ind H G(V)=C(G)ind^G_H(V) = C(G) that is an enormous one. 1. A priori, even if the representation VV of HH is irreducible, we can’t state anything on the irreducibility of ind H G(V)ind^G_H(V).

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

Theorem. Leg GG be a group and HH a closed subgroup of GG. If VV is a representation of HH and WW is a representation of GG, then as vector spaces:

Hom G(W,Ind H G(V))=Hom H(W| H,V)Hom_G(W, Ind^G_H(V)) = Hom_H(W|_H, V)

where W| HW|_H is the representation WW with the same action of GG but restricted to the elements of HGH \subset G .

to be continued

Irreducible representations for GL(V)GL(V)

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

Let S dS_d be the group of symmetries of dd elements. We know that we can write dd as sum of integer in 2 d2^d combinations. To each of 2 d2^d writing we can associate a Young diagram: say λ=(λ 1,...,λ k)\lambda = (\lambda_1, ..., \lambda_k) the partition of dd such that λ 1>λ 2>...>λ k\lambda_1 \gt \lambda_2 \gt ... \gt \lambda_k then a Young diagram for this partition is

young

with λ i\lambda_i boxes in the i-th row, lined up on the left. We define the conjugate partition λ\lambda' associated to λ\lambda, the partition of the Young diagram reflected in the 4545 degree line. For example given the Young diagram in picture, associated to λ=(3,3,1,1)\lambda=(3,3,1,1), by reflecting, we obtain the conjugate partition λ=(5,3,2)\lambda'=(5,3,2).

By choosing a numeration from 11 to dd of the boxes of a Young diagram, we obtain a Young tableau:

tableau

The tableau defines two subgroup of S DS_D

P λ={gS d|gpreserveseachrow} P_{\lambda} = \{ g \in S_d | g preserves each row \}
Q λ={gS d|gpreserveseachcolumn}. Q_{\lambda} = \{ g \in S_d | g preserves each column \}.

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

Definition. Given a finite group GG, we define the group algebra 𝕂G\mathbb{K}G over a field GG, the vector space with basis e ge_g one for each gGg \in G and with product:

e ge h=e gh e_g e_h = e_{gh}

By PP and QQ we define we define two elements in the group algebra :

a λ= gP λe g a_{\lambda} = \sum_{g \in P_\lambda} e_g
b λ= gQ λsign(g)e g b_{\lambda} = \sum_{g \in Q_\lambda} sign(g) e_g

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

Imm(a λ)=Sym λ 1V...Sym λ kV Imm(a_{\lambda}) = Sym^{\lambda_1}V\otimes ... \otimes Sym^{\lambda_k}V
Imm(b λ)=Λ ν 1V...Λ ν kV, Imm(b_{\lambda}) = \Lambda^{\nu_1}V \otimes ... \otimes \Lambda^{\nu_k}V,

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

c λ=a λb λS d. c_{\lambda} = a_{\lambda} b_{\lambda} \in \mathbb{C}S_d.

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

𝕊 λV=Imm(c λ| V d). \mathbb{S}_{\lambda} V = Imm(c_{\lambda}|_{V^{\otimes d}}).

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

Theorem.. Let k=dim(V)k=dim(V). Then 𝕊 λ(V)\mathbb{S}_\lambda(V) is zero if λ k+10\lambda_{k+1} \ne 0 , then

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

The theorem furnish us many of the irreps for GL(V)GL(V). We cannot obtain all the finite irreducible representation for GL(V)GL(V) since the dual of each 𝕊 λ(V)\mathbb{S}_{\lambda}(V) is excluded. The way to the complete classification is to introduce the k-th power of the determinant D k=(Λ n(V)) kD_k = (\Lambda^n (V))^{\otimes k} , with D k=D k *D_{-k} = D_{k}^* . By tensoring with D kD_k we can define any 𝕊 λ(V)\mathbb{S}_{\lambda} (V) even if λ i0\lambda_i \le 0 for some ii. By a simple tensor product property we have:

𝕊 λ(V)D k=𝕊 λ 1+k,..,λ n+k(V), \mathbb{S}_{\lambda} (V) \otimes D_k = \mathbb{S}_{\lambda_{1+k},.., \lambda_{n + k}} (V),

then if some λ i0\lambda_i \le 0 we define:

𝕊 λ 1+k,..,λ n+k(V)D k=𝕊 λ(V). \mathbb{S}_{\lambda_{1+k},..,\lambda_{n+k} } (V) \otimes D_{-k} = \mathbb{S}_{\lambda} (V).

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

Theorem. Every finite dimensional, irreducible and complex representation of GL(V)GL(V) is isomorphic to 𝕊 λ(V)\mathbb{S}_{\lambda} (V) for a unique index λ=(λ 1,...,λ n)\lambda= (\lambda_1,...,\lambda_n) with λ 1λ 2..λ n \lambda_1 \ge \lambda_2 \ge .. \ge \lambda_n .

Revised on October 20, 2010 at 21:17:51 by giuseppe_malavolta