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The Lie algebra $\mathfrak g$ of a connected compact Lie group $G$ is unimodular: this means that for all $x\in\mathfrak g$ the map $\operatorname{ad}(x):\mathfrak g\to\mathfrak g$ has zero trace. Now let $G$ be a compact Lie group of dimension two. Its Lie algebra, by the result above, is unimodular. Now, there is exactly one Lie algebras of dimension two ...

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$U$ is a an open set containing $e$. Thus $gU$ is open and contains $g$ (note that multiplication by an element of a topological group is a homeomorphism). But $g \in \partial H$ so every neighborhood of $g$ intersects $H$ non-trivially

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$SO^+(3,1)$ is the so-called restricted Lorentz group, which is the identity component of the Lorentz group $SO(3,1)$. It is a six-dimensional real Lie group, which is not simply connected. Since $SO^+(1,3)$ is not compact, but $SU(2)\times SU(2)$ is compact, the groups cannot be isomorphic as real Lie groups. We have $SO^+(3,1)\simeq ... 3 EDIT. The argument below can be shortened, following Mariano, if we start out with the universal cover directly. Suppose$H$is a (noncommutative) connected$2$-dimensional compact Lie group. Its universal cover$\widetilde{H}$carries a Lie group structure such that the cover$p:\widetilde{H}\to H$is a morphism of groups, and its kernel is a discrete ... 3 Define$\mathfrak{t}_n$to be the set of triangular$n \times n$matrices with entries in some field$k$. To be clear, by triangular, I mean entries are allowed on the diagonal or anywhere above the diagonal, while everything below the diagonal is zero. Now define$\mathfrak{u}_n$to be the set of upper triangular$n \times n$matrices with entries in some ... 3 In short the difficulty is that although Eckmann-Shapiro tells you $$\operatorname{Ext}^n_{U(\mathfrak{h})}(k,k) \cong \operatorname{Ext}^n_{U(\mathfrak{g})}(k, k\Uparrow)$$ where$k\Uparrow$is the coinduced module, unlike in the group setting there is no canonical map of$\mathfrak{g}$-modules$k\Uparrow \to k$, and indeed in some cases there isn't any ... 3 This is false. For example,$\mathfrak{sl}_2(\mathbb{R})$is a semisimple Lie algebra, but$SL_2(\mathbb{R})$isn't compact. This is also false, and a counterexample to 1 also provides a counterexample here. For example,$U(1) \times SL_2(\mathbb{R})$isn't compact. This is also false, and a counterexample to 2 also provides a counterexample here. The ... 2 One of the axioms of a bialgebra$B$(in particular a Hopf algebra) is: (taken from Wikipedia: in their notation$K$is the base field and$B$is the bialgebra). Here$\eta$is the unit and$\Delta$is coproduct. In plain words,$\eta(\lambda \in K) = \lambda 1_B. By the commutativity of the diagram, $$\Delta(1_B) = \Delta(\eta(1_K)) = (\eta \otimes ... 2 Let A=\left(\begin{array}{ccc}a & b & c \\ d & e & f \\ g & h & i \\ \end{array}\right) Then A+A^{\mathsf{T}}=\left(\begin{array}{ccc}2 a & b+d & c+g \\ b+d & 2 e & f+h \\ c+g & f+h & 2i \\ \end{array}\right)=0 So a=e=i=0 and d=-b,g=-c, h=-f leaving three degrees of freedom. 2 A basis of M_3(\mathbb{R}) is given by the 9 matrices E_{i,j}, which have a 1 at place (i,j) and 0 everywhere else. Now the subspace of skew-symmetric matrices has dimension 3. To see this, show that S_1=E_{1,2}-E_{2,1},S_2=E_{1,3}-E_{3,1},S_3=E_{2,3}-E_{3,2} is a basis. More generally, the subspace of skew-symmetric matrices of size n ... 2 [NB: This was originally posted as a comment above, but I've been asked to repost as an answer so it can be accepted.] You might try reading the answers to this post from a year ago; it seems like the easiest way to think about the isomorphism SU(4) \cong SO(6)/\mathbb{Z}^2 is to make SU(4) act on \mathbb{C}^6 in a nice way. It's not exactly a ... 2 Yes, that's correct. {}{}{}{}{}{}{}{} 2 If I understand the notation well, the ith fundamental weight should be e_1^*+\dots+e_i^*. 2 No; even in the category of connected abelian Lie groups, tori are not injective objects. For ease of notation, we denote by \mathbb{T} the torus \mathbb{R}/\mathbb{Z}. Here is a counterexample: Fix rationally independent, irrational numbers \xi, \eta. Let f : \mathbb{R} \to \mathbb{T}^2 be given by$$ f : x \mapsto (x\xi, x\eta). $$Obviously f ... 2 You can't. You should think of \text{Der}(A) as infinitesimal automorphisms of A, or equivalently as vector fields on \text{Spec } A, and so it isn't functorial with respect to arbitrary maps in the same way that taking automorphism groups or vector fields of manifolds isn't functorial. (However, like taking automorphism groups or vector fields, it is ... 2 The answer is no, even when the algebra is finite dimensional. Consider k = \mathbb C, and let H = \mathbb C with the usual algebra structure and the comultiplication given by \Delta(1) = 1 \otimes 1, hence \Delta(z) = z (1 \otimes 1). The counit is the identity. Then \mathbb C has no nontrivial primitive elements, because 2z = z \implies z = 0. ... 2 Let K be the kernel of \varphi. Then by the third isomorphism theorem we have$$A/K\cong (A/C)/(K/C)We use this to construct \hat{\varphi}:A/C\to B with kernel K/C, factoring through the surjection A/C\to A/K and composing with \tilde{\varphi}. 2 By definition, X=F_* \frac{\partial}{\partial x^1} and Y=F_* \frac{\partial}{\partial x^2}. If F is a diffeomorphism, then y^1:=F^1,\ldots, y^n:=F^n is a coordinate frame on \mathbb{R}^n and \frac{\partial}{\partial y^j} are commuting vector fields. But \frac{\partial}{\partial y^i}=\sum_j \frac{\partial F^j}{\partial ... 1 To me, the problem does not indicate to use the first part in order to prove the second part. For the second part, note that since X = F_{*}\frac{\partial}{\partial x^1} and Y = F_{*}\frac{\partial}{\partial x^2}, for every smooth real-valued function f on \Bbb R^n, \begin{align}[X,Y]f &= X(Yf) - Y(Xf)\\ &= \left(\frac{\partial}{\partial ... 1 The answer depends on some facts about the classifcation of complex simple Lie algebras. We can use the formula \dim (L)=\mid\Phi\mid+rank(L) $$for a simple Lie algebra L, where \mid\Phi\mid denotes the cardinality of the root system of L. Now we can list explicitly 72 roots of L=E_6, see here, with a picture of all 72 roots. Because ... 1 As Tobias already indicated in his comment, usually you consider the tensor product of two G- resp. {\mathfrak g}-modules over {\mathbb C}, but in case you mean X\otimes_G Y resp. X\otimes_{\mathfrak g} Y as (X\otimes_{\mathbb C} Y)^G resp. (X\otimes_{\mathbb C} Y)^{\mathfrak g} (that's the same as providing X with the right {\mathscr ... 1 Your computation is correct, but you were too pessimistic about the continuation: the second derivative terms in the bracket of \rho(X) and \rho(Y) cancel. It becomes less reasonable to ask for a Lie algebra repn on not-smooth functions, since natural Lie algebra repns are differential operators. Nevertheless, once you start down that path, if you go ... 1 To my knowledge, the "usual" way to embed S_{n-1} into S_n is to let S_{n-1} act on \{1,\ldots,n-1\} and fix n. We can iterate this construction to get an embedding of S_n into S_{2n}. These are the embeddings we use in the following. In the semisimple case (i.e. the characteristic of the coefficient field is greater than n), we have the ... 1 For general semisimple {\mathfrak g}, if {\mathbb Z}\Phi is the root lattice, then any {\mathfrak g}-module X in {\mathcal O} decomposes as X = \bigoplus\limits_{C\in {\mathfrak h}^{\ast}/{\mathbb Z}\Phi} X_C, where X_C := \bigoplus\limits_{\lambda\in C} X_\lambda. As vector spaces, this is ok since it is only nested way of writing the weight ... 1 Yep. Sanity checks are often a good idea. 1 Presumably$$V^e=\{v \in V \ | \ ev=0 \}$$is the Lie algebra version of fixed points---thus if V is generated by V^e and locally nilpotent for f it is a quotient of a sum of Verma modules. 1 As you wrote in your question, you may consider a k-form \omega which is of the form \omega=fd\alpha for a (k-1)-form \alpha. Step 1: Clearly, i_{[ \mathbb{X}, \mathbb{Y}]} fd\alpha= f i_{[\mathbb{X},\mathbb{Y}]} d\alpha. On the other hand,$$ L_{\mathbb{X}} i_\mathbb{Y} fd\alpha = L_{\mathbb{X}} f i_\mathbb{Y} d\alpha = (\mathbb{X}\cdot f) ... 1 Note thatX + X^T = 0$means$X^T = -X$. This means, for the entries in$X$: the entries on the diagonal are$0$. the entries above the diagonal determine those below the diagonal. This should alow you to conclude. 1 The induced module will not be irreducible in general. For example, start with a$\mathfrak g$-dominant weight$\lambda$, viewed as a weight of$L$. Then there is a finite dimensional irreducible representation$V$of$\mathfrak g$with highest weight$\lambda$. If$W$is the irreducible$L$-representation of highest weight$\lambda$, then there is a ... 1 First of all, different Lie groups may have the same Lie algebra. For example, the Lie algebras of$PSL(n,\mathbb{C})$and$SL(n,\mathbb{C})$are both isomorphic to$\mathfrak{sl}_n(\mathbb{C})$. Secondly, different real simple Lie algebras may become isomorphic over the complex numbers, e.g.,$\mathfrak{so}_3(\mathbb{R})\$ and ...

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