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## Hot answers tagged lie-algebras

8

I) Define the Lie group $$\tag{1} O(p,q)~:=~ \{\Lambda\in {\rm Mat}_{n\times n}(\mathbb{R}) ~|~\Lambda^T\eta\Lambda= \eta \}$$ of pseudo-orthogonal matrices $\Lambda$ for the metric $$\tag{2} \eta_{\mu\nu}~=~{\rm diag} (\underbrace{1,\ldots,1}_{p~\text{times}},\underbrace{-1,\ldots -1}_{q~\text{times}}), \qquad n~=~p+q.$$ II) The groups ...

7

The matrices $M$ in $O(3,1)$ and $O(1,3)$ are defined by the condition $$M G M^T = G$$ for $$G=G_{1,3} ={\rm diag} (1,1,1,-1)\text{ and } G=G_{3,1} = {\rm diag} (1,-1,-1,-1)$$ respectively. I use the convention where the first argument counts the number of $+1$'s in the metric tensor and the second one counts the negative numbers $-1$ that follow. But ...

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No, the universal enveloping algebra can have a nontrivial center even if the Lie algebra itself has only $0$ as center. The (or at least a) concept you want to look up is "Casimir element; see for example http://en.wikipedia.org/wiki/Casimir_element .

2

If the center of $\mathfrak{g}$ is 0, this only means that $U(\mathfrak{g})$ has no central elements of degree 1, but there could be central elements of higher degree. There is a natural $\mathfrak{g}$-module map $$\mathfrak{g} \otimes \mathfrak{g} \to U(\mathfrak{g}).$$ Now if, say, $\mathfrak{g}$ is semisimple so that it has a nondegenerate Killing form ...

2

A simple Lie group is a Lie group that contains no $connected$ normal subgroups. This is not the same as being a Lie group which is simple as an abstract group. For example the real numbers under addition are a simple Lie group, but have plenty of discrete normal subgroups (the integers for example), and even dense normal subgroups (like the rational ...

2

The representation $V(\lambda)$ is finite dimensional iff $\lambda$ is dominant integral. This is the theory of highest weight representations of $L$. To compute the dimensions of the weight spaces in $V(\lambda)$, say when $\lambda$is finite dimensional, there is the Weyl character formula and the Kostant multiplicity formula.

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Just saying $\partial_y$ and $y\partial_y$ are infinitesimal generators is implicitly assuming a smooth structure on your would-be Lie group $G,$ so I don't think this is the right approach. Here's another one. First we need to notice $y\mapsto \epsilon y$ for $\epsilon=0$ gives a non-invertible transformation, so we want $G$ to have underlying smooth ...

1

I am a little bit confused with Humphreys notations, correct me if I am proving the wrong thing Let $M(\lambda)$ be the Verma modulus, $L(\lambda)$ -- irreducible factor. I am not shifting by $\rho$. Then $ch(M(\lambda))=e^\lambda\sum_{\nu \in Q_+}p(\nu)e^{-\nu}$. Weil formula can be rewritten as $ch(L(\lambda))=\sum_{w \in ... 1 This is true. Any derivation maps the solvable radical into the nilradical, hence we have $$D(rad(\mathfrak{g}))\subseteq nil(\mathfrak{g})\subseteq rad(\mathfrak{g}).$$ The proof can be found in Jacobson's book on Lie algebras, see Corollary$2$to Theorem$13$in chapter$II$, section$7$. Ideals which are invariant under all derivations are called ... 1 See my comment on your question for a simplified reformulation of the problem and the notation I am using. The main idea: consider the map$X\mapsto X(e)$that assigns to a vector field on$G$its value at the identity (en element of the Lie algebra of$G$). Restricted to l.i. (left-invariant) vector fields, this map is a$G$-equivariant isomorphism: ... 1 The key result is this: If$\mathfrak{h}_0 \subset \mathfrak{h}$is the fixed subspace of$w \in W$, then$w$(considered as a transformation on$\mathfrak{h}^*$via the Killing form) can be expressed as a product of simple reflections$w_\alpha$, where$\alpha \in \mathfrak{h}^*$is a root of$\mathfrak{g}$and$\alpha(\mathfrak{h}_0) = 0$. For a ... 1 Ok, so$exp(x)exp(y) = exp(x+y+ \frac{1}{2}[x,y] +\cdots )$and$exp(y)exp(x) = exp(y+x+ \frac{1}{2}[y,x] +\cdots )$. Therefore, using$[y,x]=-[x,y]$, $$[exp(x),exp(y)] = exp(x+y+ \frac{1}{2}[x,y] +\cdots ) - exp(x+y- \frac{1}{2}[x,y] +\cdots )$$ then, $$[exp(x),exp(y)] = [x,y] +\cdots.$$ The answer would seem to fall out of the BCH relation for higher ... 1 Since the Lie group$SO(2)$is abelian it has trivial Lie algebra, i.e., with zero Lie brackets. The "generator of rotations" is indeed$X_g$, which does not imply that the group has only one generator. Note that $$X_g=\frac{d}{d\alpha}R(\alpha)\mid_{\alpha=0},$$ where$R(\alpha)$are the rotation matrices. 1 There is a proof in Chapter 5 of Gracia-Bondia, Varilly, and Figueroa's book Elements of Noncommutative Geometry. They prove in Lemma 5.7, page 182 the result that bivectors in the Clifford algebra$Cl(V)$are closed under taking commutators, and that the adjoint action of bivectors on vectors in$V\$ induces an isomorphism of the Lie algebra of bivectors ...

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