# Tag Info

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You're completely right, maybe it's a matter of taste. One benefit of the formulation you quote is that it is very elementary: it doesn't speak about abstract representations of Lie algebras, but only about spaces of nilpotent matrices closed under commutator - pure linear algebra, if you want. However, as far as I remember, for the usual inductive proof ...

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As I mentioned in the comments, the method to find the weights of an irreducible representation with highest weight $\lambda$ is: Take the orbit of $\lambda$ under the Weyl group. Since the set of weights of any representation is preserved by the Weyl group, all these weights are in the weight space of $V_{\lambda}$. For an irreducible representation, the ...

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These identities can be found using the formulae $(i_Y\alpha)(V_1, \dots, V_k) = \alpha(Y, V_1, \dots, V_k)$ and $$(\mathcal{L}_X\alpha)(V_1, \dots, V_k) = X(\alpha(V_1, \dots, V_k)) - \sum_{i=1}^k\alpha(V_1, \dots, V_{i-1}, [X, V_i], V_{i+1}, \dots, V_k).$$ First of all, we have \begin{align*} & (\mathcal{L}_Xi_Y\alpha)(V_1, \dots, V_k)\\ =&\ ...

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1. It suffices to show that the action of ${SU}(n)$ on a single basis polynomial ${x_1}^{a_1} {x_2}^{a_2} \dots {x_n}^{a_n}$ $($where $a_1 + \dots + a_n = m$$) is continuous. That is, let A = (a_{ij}) \in {SU}(n) be a matrix, then we need to show that the change of variables$$(x_1, x_2, \dots, x_n) \mapsto \left(\sum a_{1i} x_i, \sum a_{2i} x_i, \dots, ... 2 You "forgot" a few curves and tangent vectors.$T_eS$must be a two-dimensional vector space, but$S$is not (but almost :-), We have, for example,$0\not\in S.$To remedy this, put$a = 0$in your curve$\gamma$. Please check that this is perfectly valid. We still have $$\gamma(t) \in S,\quad \gamma(0) = I_2,$$ and in particular$$\gamma'(t) = ... 1 Hint. Differentiate with respect to$t$. 1 To proceed with direct calculation, first start with the relation$AM = -MA^T$and show by induction that$A^n M = (-1)^n M(A^T)^n$for all$n \ge 1. With these identities at hand, we compute \begin{align}\exp(At)M\exp(A^T t) &= \exp(At) \cdot \sum_{m = 0}^\infty \frac{M(A^T)^m t^m}{m!} \\ &= \exp(At) \cdot \sum_{m = 0}^\infty \frac{(-1)^m A^m ... 1 It's almost the same as Jyrki's comment, but you don't need to argue with dimensions: IfV$is an${\mathfrak s}{\mathfrak l}_2({\mathbb C})$-module for which you have found some direct sum decomposition$V = \bigoplus\limits_{n\in{\mathbb Z}} V^{\prime}_n$into${\mathbb C}$-subspaces such that$V^{\prime}_n$consists of$h$-eigenvectors of eigenvalue$n$, ... 1 In the semisimple case, the argument goes as follows: the conjugation$\theta$of a complex semisimple$\mathfrak{g}^\mathbb{C}$with respect to a compact real form$\mathfrak{g}_u$is a Cartan involution, that is,$-B(X,\theta Y)$is a positive definite bilinear form on$\mathfrak{g}^\mathbb{C}$. If$\theta$commutes with the conjugation$\bar{\cdot}$of ... 1 Disclaimer: I think the below computation is correct but it's very late and I might have made a mistake out of tiredness. I will recheck my answer tomorrow to make sure I haven't misled you somewhere. Computing weights in Verma modules is actually not bad and just uses the PBW theorem. Suppose$\mathfrak{n}^{-}, \mathfrak{n}^{+}$are the subalgebra of ... 1 In his comment, user148177 already explained that the exponential function of$\text{Sp}_{2n}({\mathbb R})$is not surjective. Two factors, however, suffice: Theorem (Polar decomposition) Any$g\in\text{Sp}_{2n}({\mathbb R})$can uniquely be written as a product$g = h\cdot\text{exp}(X)$with$h\in SO_{2n}({\mathbb R})\cap\text{Sp}_{2n}({\mathbb R})$... 1 The answer is yes. This is because$G$-submodule of any$G$-module$V$are the same as$\mathfrak{g}$-submodules of$V$. Let me start with the direction I think you already know. Let$V$be a representation of$G$and let$\phi : G \rightarrow GL(V)$be the structure map. If$W$is a$G$-submodule of$V$, then by restricting the image of$\phi$to$W$, we ... 1 There is a bijective correspondence between real connected simple Lie groups and real simple Lie algebras. Since the Lie algebras${\frak{sp}}(2n,\mathbb{R})$are simple, every connected Lie group with Lie algebra${\frak{sp}}(2n,\mathbb{R})$is simple. On the other hand, there is no generally accepted definition of a simple Lie group. Sometimes it is ... 1 You are confusing the definition of$gl_n(R)$. This lie algebra consists of all matrices, not just invertible ones. In order to prove the claim, use the following general fact: Let$f : M \to N$be a smooth map between manifolds and let$p \in M$be a regular point. Then$M' = f^{-1}(f(p))$is a submanifold of$M$and$T_p M'$is the kernel of$T_p M \to ...

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It looks to me like it is true that $B_n$ is uniquely determined. Let $K_n$ be the kernel of the map $A_n\to B_n$ and let $E_n$ be the kernel from $C_n$ to $D_n$. Then we have an exact sequence $0\to E_n\to K_{n+1}\to K_n$. $E_n$ is known and $K_n$ is known inductively. That means that $K_{n+1}$ can be determined as the subspace of $A_{n+1}$ generated by all ...

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Notice that, for any form $\omega$, $i_X(F^*\omega)=F^*(\iota_Y\omega)$. Hence, using the fact that $d$ commutes to $F^*$, and the Cartan formula, \begin{eqnarray} L_X(F^*\omega)&=&\iota_X(d(F^*\omega))+d(\iota_X(F^*\omega))\\ &=&\iota_X(F^*d\omega)+d(F^*\iota_Y\omega)\\ &=&F^*\iota_Y(d\omega)+F^*d(\iota_Y\omega)\\ ...

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