Chateau Paradis Casseuil 2006

Château Paradis Casseuil gets its name from the combination of the registered name of the main parcel of vineyards called “Vines of Paradise” and Casseuil county. Taken under Domaines Barons de Rothschild (Lafite)’s wing in 1984, Château Paradis Casseuil then included 14 hectares of vines. In 1989, the estate grew by 9 hectares, and chais were included in the heart of the Sainte Foy la Longue vineyard. The Château Paradis Casseuil chais, located in the heart of the Sainte Foy la Longue vineyard are used for producing red wines. White wines are made at Château Rieussec, benefiting from the technical capacities of that great estate.

Lively and intense ruby color. Fine nose, red fruit and slight licorice aromas. The attack is supple (slight sensation of sweetness) with silky tannins.

2007
After a wet winter, the high temperatures in March and April helped give a good start to the vegetation. The following months were moderate until August. The fine weather settled early September encouraging the ripening of the grapes.

Beautiful crimson colour. A fresh nose with a touch of redcurrant and mint. The first impression on the palate is pleasant, frank with intense fruit. This wine can be appreciated now or kept for a few years when it will have reached its peak.

2006
The winter of 2005/2006 was dry and cold and the spring months were mild with little rain. August was disconcerting – cool and wet – and at the beginning of September ripeness levels were very low. However, the weather then became summery allowing the grapes to finish ripening well.

Beautiful straw yellow colour with hints of green. Discreet on the nose, but when swirled, the aromas are revealed with a fine lime bouquet. The attack is supple and the finish is fresh.

2005
The end of 2004 and the first few months of 2005 were dry. Moreover, maturation took place in perfect conditions.

Pale yellow colour. Notes of citrus fruit, mainly grapefruit, on the nose.

First impressions on the palate are full, lively and well rounded, leading to a silky finish marked by hints of hazelnuts.

2004
The year was marked by stormy weather until July, with no effects on the vines.
The beginning of the year was warmer than in 2003, but from March onwards the trend was reversed and an average drop of 2°C was noted. Rainfall was about the average for the past three vintages, with a dry June. July and August were very damp. Maturation was therefore slow but at harvest time the grapes for the dry whites were ripe.

Pale yellow colour. Very open and fresh on the nose: aromas of white flowers and violets.

Delicate first impressions on the palate. Notes of fresh fruit and Granny Smith apples.

Region : Sainte Foy la Longue, Médoc
Grape Varietals : Cabernet Sauvignon 50%, Merlot 45% and Cabernet franc 5%
Average wine production : 12 000 cases per year.

Use of SU(2) & SO(3)

Every Lie group homomorphism gives rise to a Lie algebra homomorphism. In the case of a simply-connected matrix Lie group G, a Lie algebra homomorphism also gives rise to a Lie group homomorphism. In fact, for a simply-connected matrix Lie group, there is a natural one-to-one correspondence between the representations of G and the representations of the Lie algebra g. Each of the representations π_m of su(2) was constructed from the corresponding representation Π_m of the group SU(2).

SU(2) is simply connected but SO(3) is not (SU(2) can be thought of (topologically) as the three-dimensional sphere S³ sitting inside R⁴. It is well known that S³ is simply connected). There exists a Lie group homomorphism Φ which maps SU(2) onto SO(3) and which is two-to-one. Therefore, SU(2) and SO(3) are almost isomorphic.

Consider the space V of all 2x2 complex matrices which are self-adjoint (i.e., A^* = A) and have trace zero. This is a three-dimensional real vector space with the following basis: A₁ = (0,1: 1,0) ; A₂ = (0,-i : i,0) ; A₃ = (1,0 : 0,-1) . Define the inner product on V by <A, B> = 1/2 trace(AB). {A₁, A₂, A₃} is an orthonormal basis for V. Next we are going to identify V with R³. Suppose U is an element of SU(2) and A is an element of V. Consider UAU^-1, trace(UAU^-1) = trace(A) = 0 and (UAU^-1)^* = UAU^-1. So UAU^-1is again in V. Because the map A --> UAU^-1 is linear. Therefore, we can define a linear map Φ_U of V to itself by Φ_U = UAU^-1. Given A, B ∈ V, <Φ_U(A), Φ_U(B)> = <A, B>. Thus, Φ_U is an orthogonal transformation of V.

Once we identify V with R³ using the above orthonormal basis, we may think of Φ_U as an element of O(3). Since Φ_U1U2 = Φ_U1Φ_U2, we see that Φ (the map U --> Φ_U) is a homomorphism of SU(2) into O(3). SU(2) is connected, Φ is continuous, and Φ_I is equal to I, which has determinant one. It follows that Φ must map SU(2) into the identity component of O(3), namely SO(3). However, the map Φ_U is not one-to-one, since for any U ∈ SU(2), Φ_U = Φ_-U. Actually Φ_U is a two-to-one map of SU(2) onto SO(3) (recall that every element of O(3) has determinant ± 1).

If we have the basis E₁ = 1/2(i,0 : 0,-i) ; E₂ = 1/2(0,-1: 1,0) ; E₃ = (0,i : i,0) for su(2) and the basis F₁ = (0,0,0 : 0,0,1 : 0,-1,0) ; F₂ = (0,0,-1 : 0,0,0 : 1,0,0) ; F₃ = (0,1,0 : -1,0,0 : 0,0,0) for so(3). Then, we have [E₁ , E₂] = E₃ , [E₂ , E₃] = E₁ and [E₃ , E₁] = E₂ , and similarly with the E's replaced by the F's. Thus the linear map Φ : su(2) --> so(3) which takes E_i to F_i will be a linear algebra isomorphism.

Let σ_m = π_m º Φ^-1 be the irreducible complex representations of the Lie algebra so(3) (m ≥ 0). If m is even, then there is a representation ∑_m of the group SO(3) such that ∑_m(exp X) = exp(σ_m(X)) for all X in so(3). If m is odd, then there is no such representation of SO(3).

Representation of su(2) ≅ so(3) in Physics

Representation of su(2) ≅ so(3) in Physics are labeled by the parameter l = m/2. In terms of this notation, a representation of so(3) comes from a representation of SO(3) if and only if l is an integer. The representations with l an integer are called "integer spin"; the others are called the "half-integer spin."Consider the path in SO(3) consisting of rotations by angle 2πt in the (x, y)-plane, which comes back to the identity when t = 1. However, this path is not homotopic to the constant path.

If one defines ∑_m along the constant path, then one gets the value ∑_m(I) = I, as expected. If m is odd and one defines ∑_m along the path of rotations in the (x, y)-plane, then one gets the value ∑_m(I) = -I. There is no way to define ∑_m(m odd) as a "single-valued" representations of SO(3).

An electron is a "spin-1/2" particle, which means that it is described in quantum machines in a way that involves the representation σ₁ of so(3). In the quantum machines, one finds statements to the effect that performing a 360º rotation on the wave function of the electron gives back the negative of the original wave function.This reflects that if one attempts to construct the nonexistent representation ∑₁ of SO(3), then when defining ∑₁ along a path of rotations in some plane, one gets that ∑₁(I) = -I.

A Unitary Representations of SO(3)

Consider the unit sphere S² ⊂ R³, with the usual surface measure Ω. Any R ∈ SO(3) maps S² into S² . For each R, we can define Π₂(R) acting on L²(S², dΩ) by [ Π₂(R) f](x) = f(R^-1x). Then, Π₂ is a unitary representation of SO(3). Here, L²(S², dΩ) has a very nice decomposition as the orthogonal direct sum of finite-dimensional invariant subspaces. This decomposition is the theory of "spherical harmonics" in physics.

Representations of SU(2) & su(2)

su(2) ≅ so(3) and the representation of so(3) is important in the computation of angular momentum. By studying the representation theory of su(2) we can know:

(1) how to commutation relations to determine the representations of a Lie algebra.

(2) how to determine the representations of semisimple Lie algebras, e.g., su(3).

Some Representations of SU(2)

By definition, an element of U of SU(2) is a linear transformation of C². Let z denotes the (z₁, z₂) pair in C². Then, we may define a linear transformation Π_m(U) on the space V_m by the formula [Π_m(U)f] (z) = f(U^-1z). The inverse is necessary in order to make Π_m a representation.  And V_m is the space of functions of the form f(z₁, z₂) = a₀z₁^m + a₁z₁^m-1z₂  + a₂z₁^m-2z₂²  + ... + a_mz₂^m .
 

Therefore, [Π_m(U)f] (z₁, z₂) =  ∑ a_k(U₁₁^-1z₁ + U₁₂^-1z₂)^m-k (U₂₁^-1z₁ + U₂₂^-1z₂)^k for k = 0 .. m.  Π_m(U)f is a homogeneous polynomial of degree m. Thus, Π_m(U) maps V_m into V_m. Moreover, Π_m(U₁) [Π_m(U₂)f] (z) = Π_m(U₁U₂) f(z). So Π_m is a (finite-dimensional complex) representation of SU(2).

The Lie algebra representation of Π_m can be computed as π_m(X) = d/dt Π_m(e^tX) |_t=0. So (π_m(X)f) (z) = d/dt f(e^-tXz) |_t=0. Let z(t) be the curve e^-tXz. We have z(0) = z and (dz/dt) |_t=0 = -Xz. Since z(t) can also be written as z(t) = (z₁(t), z₂(t)), with z_i(t) ∈ C. By chain rule, π_m(X)f = ∂f / ∂z₁(dz₁/dt) |_t=0 + ∂f / ∂z₂(dz₂/dt) |_t=0 . We have

π_m(X)f = -∂f / ∂z₁(X₁₁z₁ + X₁₂z₂)  - ∂f / ∂z₂(X₂₁z₁ + X₂₂z₂). 

Because every finite-dimensional complex representation of the Lie algebra su(2) extends uniquely to a complex-linear representation of the complexification of su(2). And the complexification of su(2) is sl(2;C). Therefore, the representation π_m of su(2) given by above extends to a representation of sl(2;C). 

Consider H = (1,0 : 0,-1)

(π_m(H)f) (z) = -(∂f / ∂z₁)z₁ + (∂f / ∂z₂)z₂ .

 π_m(H) = -z₁(∂f / ∂z) + z₂(∂f / ∂z₂). Apply this to a basis element z₁^kz₂^m-k , we have

π_m(H) z₁^kz₂^m-k = -k z₁^kz₂^m-k  + (m-k)z₁^kz₂^m-k  = (m-2k) z₁^kz₂^m-k .

Thus, z₁^kz₂^m-k is an eigenvector for π_m(H) with eigenvalue (m-2k). In particular, π_m(H) is diagonalizable. 

Let X and Y be the elements, X = (0,0 : 1,0) , Y = (0,1 : 0,0) in sl(2;C). We have π_m(X) = -z₂(∂f / ∂z₁) and π_m(X) = -z₁(∂f / ∂z₂). Apply these to the basis element 

π_m(X) z₁^kz₂^m-k = -k z₁^k-1z₂^m-k+1  

π_m(Y) z₁^kz₂^m-k = -(m-k) z₁^k+1z₂^m-k-1  

It suffices to show that every nonzero invariant subspace of V_m is equal to V_m. Let W be such a space. Since W is assumed nonzero, there is at least one nonzero element w in W. Then w can be written uniquely in the form 

w =  a₀z₁^m + a₁z₁^m-1z₂  + a₂z₁^m-2z₂²  + ... + a_mz₂^m

with at least one of the a_k's nonzero. Let k₀ be the smallest value of k for which  a_k ≠ 0 and consider π_m(X)^m-k0 w. Since π_m(X) lowers the power of z₁ by 1, it will kill all the terms in w except a_k0 z₁^m-k0 z₂^k0 . So we have  π_m(X)^m-k0 z₁^m-k0 z₂^k0 = (-1)^m-k0(m-k₀)! z₂^m . Since W is assumed invariant, W must contain z₂^m . Furthermore, π_m(Y)^k z₂^m is a nonzero multiple of z₁^kz₂^m-k for all 0 ≤ k ≤ m. Because these elements form a basis for V_m . In fact W = V_m. Therefore, representation π_m is an irreducible representation of sl(2;C). 

Irreducible Representations of su(2)

Every finite-dimensional complex representation π of su(2) extends to a complex-linear representation of the complexification of su(2), namely sl(2;C). Studying the irreducible representations of su(2) is equivalent to studying the irreducible representation of sl(2;C). Passing to the complexified Lie algebra makes computations easier, in that there is a nice basis for sl(2;C) that has no counterpart among the bases of su(2).

We can use commutation relations to determine the representation of a Lie algebra. Consider the following basis for sl(2;C) and commutation relations:

H = (1,0 : 0,-1);  X = (0,0 : 1,0);  Y = (0,1 : 0,0)

[H, X] = 2X,  [H, Y] = -2Y,  [X, Y] = H.

If V is a (finite-dimensional complex) vector space and A, B, and C are operators on V satisfying

[A, B] = 2B,  [A, C] = -2C,  [B, C] = A, then

because of the skew symmetry and bilinearity of brackets, the linear map π : sl(2;C) --> gl(V) satisfying π(H) = A, π(X) = B, π(Y) = C will be a representation of sl(2;C).

We call π(X) the "raising operator", because it has the effect of raising the eigenvalue of π(H) by 2, and call π(Y) the "lowering operator". Since [π(H), π(X)] = π([H, X]) = 2π(X). Let u be an eigenvector of π(H) with eigenvalue α ∈ C. Thus,

π(H)π(X)u = π(X)π(H)u + 2π(X)u

= π(X)(αu) + 2π(X)u = (α + 2)π(X)u.

Either π(X)u = 0 or π(X)u is an eigenvector for π(H) with eigenvalue α+2. More general, π(H)π(X)ⁿu = (α + 2n)π(X)ⁿu. Similarly, for [π(H), π(Y)] = -2π(Y), we have π(H)π(Y)u = (α - 2)π(Y)u.

An operator on a finite-dimensional space can have only finitely many distinct eigenvalues. Therefore, there is some N ≥ 0 such that π(X)^N+1u = 0.

Define u₀ = π(X)^Nu and λ = α + 2N. Then,

π(H)u₀ = λu₀, π(X)u₀ = 0. 

Define u_k = π(Y)^ku₀ , for k ≥ 0. Thus, we have π(H)u_k = (λ - 2k)u_k . Since π(H) can have only finitely many eigenvalues, the u_k's cannot be all be nonzero.

For k = 1,

π(H)u₁ = π(H)π(Y)u₀ = (α - 2)π(Y)u₀ = (α - 2)u₁.

π(X)u₁ = π(X)π(Y)u₀ = (π(Y)π(X) + π(H))u₀ = π(H)u₀ = λu₀ (as π(X)u₀ = 0)

π(Y)u₁ = π(Y)π(Y)u₀  = π(Y)²u₀ = u₂  

If π(X)u_k = [kλ - k(k-1)]u_k-1. By induction, π(X)u_k+1= π(X)π(Y)u_k

= (π(Y)π(X) + π(H))u_k = π(Y)π(X)u_k + (λ - 2k)u_k

= π(Y)[kλ - k(k-1)]u_k-1 + (λ - 2k)u_k  = [kλ - k(k-1) + (λ - 2k)]u_k

= [(k+1)λ - (k+1)k]u_k  

Because π(H) can have only finitely many eigenvalues, the u_k's cannot all be nonzero. For all k ≤ m, u_m+1 = π(Y)^m+1u₀  = 0. If u_m+1 = 0. Then π(X)u_m+1  = (m+1)(λ - m)u_m  = 0. Since m ≠ 0 and m + 1 ≠ 0. So we have λ = m, where m is a non-negative integer. 

In summary, given a finite-dimensional irreducible representation π of sl(2;C) acting on a space V and putting λ = m, there exists an integer m ≥ 0 and nonzero vectors u₀, ..., u_m such that

π(H)u_k = (m - 2k)u_k, 

π(Y)u_k = u_k+1 (k < m),

π(Y)u_m = 0, 

π(X)u_k = [km - k(k-1)]u_k-1 (k > 0),

π(X)u₀ = 0.

The vectors u₀, ..., u_m must be linearly independent, since they are eigenvectors of π(H) with distinct eigenvalues. Moreover, the (m+1)-dimensional span u₀, ..., u_m is explicitly invariant under π(H), π(X), and π(Y). Hence under π(Z) for all Z ∈ sl(2;C). Since π is irreducible, this space must be all of V. 

The (m+1)-dimensional representation Π_m described above must be equivalent to π. This can be seen explicitly by introducing the following basis for V_m :

u_k = [π_m(Y)]^k (z₂)^m = (z₂)^m (m! / (m-k)!) z₁^kz₂^m-k  (k ≤ m).

In other words, π_m have a basis of the form π(H), π(X), and π(Y). π's have the right commutation relations to forma representation of sl(2;C) and that this representation is irreducible. 

Complexification

Complexification of a Real Lie Algebra

The complexification of a finite-dimensional real vector space V, as denoted by V_c, is the space of formal linear combinations v₁+ iv₂ , with v₁, v₂ ∈ V. 

Let g be a finite-dimensional real Lie algebra and g_c its complexification (as a real vector space). Then, the bracket operation on g has a unique extension to g_c which makes g_c into a complex Lie algebra. The complex Lie algebra g_c is called the complexification of the real Lie algebra g. 

Isomorphisms of Complex Lie Algebra

The Lie algebra gl(n;C), sl(n;C), so(n;C), and sp(n;C) are complex Lie algebras. In addition, there are also following isomorphisms of complex Lie algebras:

gl(n;R)_c  ≅ gl(n;C)

u(n)_c  ≅ gl(n;C)

su(n)_c  ≅ sl(n;C)

sl(n;R)_c  ≅ sl(n;C)

so(n)_c  ≅ so(n;C)

sp(n;R)_c  ≅ sp(n;C)

sp(n)_c  ≅ sp(n;C)

(u(n) is the space of all nxn complex skew-self-adjoint matrices. If X is any nxn complex matrix, then X = (X-X^*)/2 + i(X+X^*)/2i. Thus, X can be written as a skew matrix plus i times a skew matrix. Every X in gl(n;C) can be written uniquely as X₁ + iX₂, with X₁ and X₂ in u(n). It follows that u(n)_c  ≅ gl(n;C). If X has trace zero, then so do X₁ and X₂ , which has su(n)_c  ≅ sl(n;C))

Note that u(n)_c  ≅ gl(n;R)_c  ≅ gl(n;C). However, u(n)_c is not isomorphic to gl(n;R), except when n = 1. The real algebra u(n) and gl(n;R) are called real forms of the complex Lie algebra gl(n;C). 

In physics, we do not always clearly distinguish a matrix Lie algebra and its Lie algebra, or between a real Lie algebra and its complexification. For example, some references in the literature to SU(2) actually refer to the complexified Lie algebra sl(2;C). 

Representation and Complexification

Let g be a real Lie algebra and g_c its complexification. Then, every finite-dimensional complex representation π of g has a unique extension to a complex-linear representations of g_c , also denoted as π and given by π(X+iY) = π(X) + iπ(Y) for all X, Y ∈ g. Furthermore, π is irreducible as a representation of g_c  if and only if it is irreducible as a representation of g.

If π is a complex representation of the real Lie algebra g, acting on the complex vector space V. Then, saying that π is irreducible means that there is no nontrivial invariant complex subspace W ⊂ V. Even though g is a real Lie algebra, when considering complex representations of g, we are interested only in complex invariant subspaces.

Credit Default Swaps

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Chateau Sainte Barbe 2004 Merlot

Blind Test

After a blind tasting of 350 Bordeaux wines by the Belgium Wine tasting committee of the Revue Vino magazine, Merlot Sainte Barbe 2004, Chateau Sainte Barbe 2003 & 2004 were selected amongst the finalists and Chateau Sainte Barbe was awarded 2 Bacchus.

The wine's taste is fruity with harmonious aromas of ripe grapes, plums and white flowers. On the palate (roof of the mouth), it combines a well-balanced structure with a velvety (closely woven fabric of silk), fine and voluptuous flavour. Evolution: 3 to 5 years

Region: Gironde, right bank of the Garonne, Bordeaux
Grape Varieties: 100% Merlot



Reference
"In 2004, observing how some parcels were reacting to our husbandry, we decided to produce a still higher quality wine, called 'Cuvée VSP' lowering the yield to an extreme 4 bunches of grapes per vine, using picking boxes for hand harvesting and making malolactic fermentation directly in 100% new oak barrels.

The results have exceeded even our expectations."



Complex wine with an aroma of blackberries, blackcurrants, chocolate, and expresso. Medium to full bodied with rich ripe tannins.

Generating Representations

One way of generating representations is to take some representations one knows and combine them in some fashion.There are three standard methods of obtaining new representations from old, namely, direct sum of representations, tensor products of representations, and dual representations.

Direct Sum of Representations

Let G be a matrix Lie group and let Π₁, Π₂, ..., Π_m be representations of G acting on vector spaces V₁, V₂, ..., V_m. Then, the direct sum of  Π₁, Π₂, ..., Π_m is a representation Π₁ ⊕ Π₂ ⊕ ...⊕ Π_m of G acting on the space V₁ ⊕ V₂ ⊕ ... ⊕ V_m, defined by [Π₁ ⊕ Π₂ ⊕ ... ⊕ Π_m (A)] (v₁, v₂, ..., v_m) = (Π₁(A)v₁, Π₂(A)v₂, ..., Π_m(A)v_m) for all A ∈ G. 

Similarly, if g is a Lie algebra, and π₁, π₂, ..., π_m are representations of g acting on V₁, V₂, ..., V_m, then we define the direct sum of  π₁, π₂, ..., π_m , acting on  V₁ ⊕ V₂ ⊕ ... ⊕ V_m by [π₁ ⊕ π₂ ⊕ ...⊕ π_m (X)] (v₁, v₂, ..., v_m) = (π₁(X)v₁, π₂(X)v₂, ..., π_m(X)v_m) for all X ∈ g.

Tensor Products of Representations

Consider an element u of U and v of V, the "product" of these two are denoted by u⊗v. The space U⊗V is then the space of linear combination of such products. The space of elements is in the form of a₁u₁⊗v₁ + a₂u₂⊗v₂ + ... + a_nu_n⊗v_n . The product is not necessary commutative (v⊗u is in different space) but is bilinear ((u₁+ au₂)⊗v = u₁⊗ v + au₂ ⊗ v, u ⊗ (v₁+ av₂) = u ⊗ v₁ + au ⊗ v₂). 

If U and V are finite-dimensional real or complex vector spaces, then tensor product (W, ϕ) of U and V is a vector space W, together with a bilinear map ϕ : U x V --> W with the following property: If ψ is any bilinear map of UxV into a vector space X, then there exists a unique linear map ψ of W into X. Bilinear maps on UxV turn into linear maps on W. Suppose e₁, e₂ , e_n is a basis for U and f₁, f₂ , f_m  is a basis for V. Then {ϕ(e_i, f_j) | 1 ≤ i ≤ n, 1≤ j ≤ m} is a basis for W. In this case, {e_i⊗f_j | 1 ≤ i ≤ n, 1≤ j ≤ m} is a basis for U⊗V and dim (U⊗V) = (dim U)(dim V). 

The tensor product (W, ϕ)  is unique up to canonical isomorphism. That is, if (W₁, ϕ₁) and (W₂, ϕ₂) are two tensor products, then there exists a unique vector space isomorphism Φ : W₁ --> W₂ . 

The defining property of U⊗V is called the universal property of tensor products. Suppose that ψ(u,v) is some bilinear expression in (u,v). Then , the universal property says precisely that there is a unique linear map T(= ψ ) such that T(u⊗v) = ψ(u,v). Let A : U --> U and B : V --> V be linear operators. Then, there exists a unique linear operator from U⊗V to U⊗V, denoted by A⊗B, such that (A⊗B)(u⊗v) = (Au)⊗(Bv) for all u ∈ U and v ∈ V. Moreover, (A₁⊗B₁)(A₂⊗B₂) = (A₁A₂)⊗(B₁B₂). 

There are two approaches to define tensor of representations. 

(A) Starts with a representation of a group G acting on a space V and a representation of another group H acting on space U and produces a representation of the product group GxH acting on space U⊗V.

(B) Starts with two different representations of the same group G, acting on spaces U and V, and produces a representation of G acting on U⊗V.

(A) Let G and H be matrix Lie groups. Let Π₁ be a representation of G acting on a space U and let Π₂ be a representation of G acting on a space V. Then, the tensor product of Π₁ and Π₂ is a representation Π₁⊗Π₂ of GxH acting on U⊗V defined by  Π₁⊗Π₂ (A,B) =  Π₁(A)⊗Π₂(B) for all A ∈ G and B ∈ H. Let  π₁⊗ π₂ denote the associated representation of the Lie algebra of GxH, namely g⊕h. Then, for all X ∈ g and Y ∈ h, π₁⊗ π₂(X,Y) = π₁(X) ⊗ I + I ⊗ π₂(Y).  

(B) Let G be a matrix Lie groups and let Π₁ and Π₂ be representation of G acting on a space V₁ and V₂. Then, the tensor product of Π₁ and Π₂ is a representation of G acting on V₁ ⊗ V₂ defined by Π₁⊗Π₂ (A) =  Π₁(A)⊗Π₂(A) for all A ∈ G. The associated Lie algebra g satisfies π₁⊗ π₂(X) = π₁(X) ⊗ I  + I ⊗ π₂(X) all X ∈ g. 

Suppose Π₁ and Π₂ are irreducible representations of a group G. If regard Π₁⊗Π₂ as representation of G, it may no longer be irreducible. If it is not irreducible, one can attempt to decompose it as a direct sum of irreducible representations. This process is called the Clebsch-Gordan theory. In the physics, the problem of analyzing tensor products of representations of SU(2) is called "addition of angular momentum."

Dual Representations

A linear functional on a vector space V is a linear map of V into C. If v₁, v₂, ..., v_n is a basis for V, then for each set of constants a₁, a₂, ..., a_n, there is a unique linear functional ϕ such that ϕ(v_k) = a_k. If V is a finite-dimensional complex vector space, then the dual space to V, denoted by V^*, is the set of all linear functionals on V. This is also a vector space and its dimension is the same as that of V.  If A is a linear operator on V, let A^tr denote the dual or transpose operator on V^*, (A^trϕ)(v) = ϕ(Av) for all ϕ ∈ V^*, v  ∈ V. Note that the matrix of A^tr is the transpose of the matrix A and not the conjugate transpose. If A and B are linear operators on V, then (AB)^tr = B^tr A^tr . 

Suppose G is a matrix Lie group and Π is a representation of G acting on a finite-dimensional vector space V. Then, the dual representation Π^* to Π is the representation of G acting on V^* given by Π^*(g) = [Π(g^-1)]^tr . Similarly, if π is a representation of a Lie algebra g acting on a finite-dimensional vector space V, then π^* is the representation of g acting on V^* given by  π^*(g) = -π(X)^tr. The dual representation is also called contragredient representation. Note that the transpose is an order-reversing operation, we cannot simply define Π^*(g) = Π(g)^tr. This would not be a representation. 

Los Vascos 2006 Cabernet Sauvignon

Under the direct control of Domaines Barons de Rothschild (Lafite), Los Vascos, one of the oldest wine estates in Chile, is located in the Colchagua valley. The pre-phylloxra Bordeaux rootstock, Cabernet Sauvignon is the grape that made the estate famous.

Cabernet Sauvignon 2006 has a very red fruit nose and chocolate and bay leaf touches. It is fresh in the mouth, light, very fruity and balanced with good persistent tannins well blended into the wine. Highly concentrated with strawberry & cherries fruit notes and marked spices. International Wine Cellar 2006 Cebernet Sauvignon 88 point rating. Other years: 2004 (better), 2005.

Region: VALLE CENTRAL, Chile
Sub-Region: RAPEL
Grape Varieties: CABERNET SAUVIGNON

Representation Theory

Let G be a matrix Lie group. Then, a finite-dimensional complex representation of G is a Lie group homomorphism: Π : G --> GL(V), where V is a finite-dimensional complex vector space (with dim(V) ≥ 1). We may think of a representation as a linear action of a group on a vector space, i.e. to every g ∈ G, there is an operator Π(g) acts on the vector space. 

If g is a real or complex Lie algebra, then a finite-dimensional complex representation of g is a Lie algebra homomorphism π of g into gl(V). It is a unique representation π of g acting on the same space V such that Π(e^X) = e^π(X) for all X ∈ g. The representation π can be computed as π(X) = d/dt  Π(e^tX) |_t=0 and satisfies π(AXA^-1) =  Π(A) π(X) Π(A)^-1 for all X ∈ g  and all A ∈ G.  

If Π or π is a one-to-one homomorphism, then the representation is called faithful.

Let Π be a finite-dimensional real or complex representation of a matrix Lie group G, acting on a space V. A subspace W of V is called invariant if  Π(A)w ∈ W for all w ∈ W and all A ∈ G. An invariant subspace W is called nontrivial if W ≠ {0} and W ≠ V. A representation with no nontrivial invariant subspaces is called irreducible.

Let G be a connected matrix Lie group with Lie algebra g. Let Π be a representation of G and π the associated representation of g. Then, Π is irreducible if and only if π is irreducible.

Let G be a matrix Lie group, let H be a Hilbert space, and let U(H) denote the group of unitary operators on H. Then, a homomorphism Π : G --> U(H) is called a unitary representation of G if Π satisfies the following continuity condition (strong) condition: If A_n, A ∈ G and A --> A_n, then Π(A_n)v --> Π(A)v for all v ∈ H. A unitary representation with no nontrivial closed invariant subspace is called irreducible.  

The terms invariant, nontrivial, and irreducible are defined analogously for representations of Lie algebra.

Equivalence

Let G be a matrix Lie group, let  Π be a representation of G acting on the space V, and let ∑ be a representation of G acting on the space W. A linear map ϕ : V --> W is called an intertwining map of representation if ϕ(Π(A)v) = ∑(A)ϕ(v) for all A ∈ G and v ∈ V. The analogous property defines intertwining maps of representations of a Lie algebra. 

If ϕ is an intertwining map of representations and, in addition, ϕ is invertible, then ϕ is said to be an equivalence of representations. If there exists an isomorphism between V and W, then the representations are said to be equivalent. 

Let G be a connected matrix Lie group, let Π₁ and Π₂ be represenation of G, and let π₁and π₂ be the associated Lie algebra representations. Then, π₁and π₂ is equivalent if and only if Π₁ and Π₂ are equivalent.

Schur's Lemma

1. Let V and W be irreducible real or complex representations of a group or Lie algebra and let  ϕ : V --> W be an intertwining map. Then either ϕ = 0 or ϕ is an isomorphism.

2. Let V be an irreducible complex representation of a group or Lie algebra and let ϕ : V --> V be an intertwining map of V with itself. Then ϕ = λI, for some λ ∈ C.

3. Let V and W be irreducible complex representations of a group or Lie algebra and let ϕ₁ , ϕ₂  : V --> W , be nonzero intertwining maps. Then, ϕ₁ = λϕ₂, for some λ ∈ C. 

Applications of Representation Theory

Studying the representations of a group G (or of a Lie algebra) can give information about the group (or Lie algebra) itself. For example, if G is a finite group, then associated to G is something called the group algebra. The structure of this group algebra can be described very nicely in terms of the irreducible representations of G. 

One of the chief applications of representation theory is to exploit symmetry. If a system has symmetry, then the set of symmetries will form a group, and understanding the representations of the symmetry group allows one to use symmetry to simplify the problem. For example, if the equation has rotational symmetry, then the space of solutions will be invariant under rotations. Thus, the space of solutions will constitute a representation of the rotation group SO(3). If one knows what all of the representation of SO(3) are, this can help in narrowing down what the space of solutions can be. 

Universal Cover

Connectedness

A matrix Lie group G is said to be path-connected (in topology) if given any two matrices A and B in G, there exists a continuous path A(t), a ≤ t ≤ b, lying in G with A(a) = A and A(b) = B. A matrix Lie group is connected if and only if it is path-connected. 

A matrix Lie group G which is not connected can be decomposed (uniquely) as a union of several pieces, called components, such that two elements of the same component can be joined by a continuous path, but two elements of different components cannot.

Examples:
Group      Connected   Components
GL(n;C)      yes             1
SL(n;C)      yes             1
GL(n;R)      no              2
SL(n;R)      yes             1
O(n)           no               2
SO(n)         yes             1
U(n)           yes             1
SU(n)         yes             1
O(n;1)        no               4
SO(n;1)      no               2
Heisenberg yes            1
E(n)           no               2
P(n;1)        no               4


Simple Connectedness

A matrix Lie group G is said to be simply connected if it is connected and, in addition, every loop in G can be shrunk continuously to a point in G. More precisely, assume that G is connected. Then, G is simply connected if given any continuous path A(t), 0 ≤ t ≤ 1, lying in G with A(0) = A(1), there exists a continuous function A(s,t), 0 ≤ s,t ≤ 1, taking values in G and having the following properties:

(1) A(s,0) = A(s,1) for all s, is a loop condition;
(2) A(0,t) = A(t), when s=0 the loop is the specified loop A(t);
(3) A(1,t) = A(1,0) for all t, when s=1 the loop is a point.


Group versus Lie Algebra Homomorphisms

(A) Lie algebra homomorphism

Every Lie group homomorphism give rise to a Lie algebra homomorphism. Let G and H be matrix Lie groups, with Lie algebras g and h, respectively. Suppose that Φ : G --> H is a Lie group homomorphism. Then, there exists a unique real linear map ϕ :   g --> h such that  for all X ∈ g. The map ϕ has the following additional properties:

(1) ϕ (AXA^-1) = Φ(A) ϕ(X) Φ(A)^-1, for all X ∈ g, A ∈ G.

(2) ϕ([X,Y]) = [ ϕ(X),  ϕ(Y)], for all  X, Y ∈ g (Lie algebra homomorphism)

(3) ϕ(X) = d/dt Φ(e^tX) |_t=0, for all X ∈ g

(B) Lie group homomorphism

Let G and H be matrix Lie groups with Lie algebras  g and h. Let ϕ :   g --> h be a Lie algebra homomorphism. If G is simply connected , then there exists a unique Lie group homomorphism Φ : G --> H such that  Φ(e^X) = e^ϕ(X) for all X ∈ g. However, if G is not simply connected, this will not be true. We should look for Ğ that has the same Lie algebra as G but such that Ğ is simply connected. 

Baker-Campbell-Hausdorff formula says that if X and Y are sufficiently small, then 

log(e^Xe^Y) = X + Y + 1/2[X,Y] + 1/12[X,[X,Y]] - 1/12[Y,[X,Y]] + ...

Because ϕ is a Lie algebra homomorphism, 

ϕ(log(e^Xe^Y)) = ϕ(X) + ϕ(Y) + 1/2[ϕ(X),ϕ(Y)] + 1/12[ϕ(X),[ϕ(X),ϕ(Y)]] - 1/12[ϕ(Y),[ϕ(X),ϕ(Y)]] + ... = log(e^ϕ(X)e^ϕ(Y))

For group homomorphism, Φ(e^Xe^Y) = exp(ϕ(e^Xe^Y) = exp(ϕ(exp(log(e^Xe^Y) ))). From above, Φ(e^Xe^Y) = exp(log(e^ϕ(X)e^ϕ(Y))) = e^ϕ(X)e^ϕ(Y) = Φ(e^X)Φ(e^Y). Therefore, by Baker-Campbell-Hausdorff formula, if X is small, Φ is a group homomorphism. 

Covering Groups

Let G be a connected Lie group. Then, a universal covering group (or universal cover) of G is a simply-connected Lie group H together with with a Lie group homomorphism Φ : H --> G such that the associated Lie algebra homomorphism ϕ :   h --> g is a Lie algebra homomorphism. The homomorphism Φ is called the covering homomorphism (or projection map). 

For any connected Lie group, a universal cover exists. If G is a connected Lie group and (H₁, Φ₁) and (H₂, Φ₂) are universal covers of G, then there exists a Lie group isomorphism Ψ : H₁ --> H₂ such that Φ₂ o Ψ =  Φ₁. 

The universal cover of a matrix Lie group may not be a matrix Lie group.