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5

Here's an algebraic approach to deformations: Definition. Let $({\mathfrak g},[-,-])$ be a finite-dimensional Lie algebra over a field ${\mathbb k}$. An infinitesimal deformation (of order $2$) of ${\mathfrak g}$ is a ${\mathbb k}[t]/(t^2)$-Lie algebra structure on ${\mathfrak g}\otimes_{\mathbb k} {\mathbb k}[t]/(t^2)$ restricting to ${\mathfrak g}$ ...

4

If a group $G$ acts on itself by left multiplication and if $p$, $q$ are arbitrary points of $G$, there is exactly one element $g$ such that $gp = L_{g}(p) = q$. Consequently, a tangent vector at $p$ corresponds to a unique tangent vector at $q$ via the tangent map $dL_{g}$. In general, however, the left $G$-action on $G/H$ has non-trivial stabilizers, and ...

3

Here are two general facts. If $G$ is a connected topological group, then any discrete normal subgroup $N$ of $G$ is in fact contained in its center. This is a standard exercise and the proof is straightforward: by hypothesis, $N$ is setwise fixed by conjugation. But since $N$ is discrete, a connected topological group can't act nontrivially on it, and so ...

3

Take $\mathcal B$ a basis wherein $x$ is diagonalizable and denote $A$ its matrix. Let $B$ the matrix of $y$. $\lambda$ is an eigenvalue of $\operatorname{ad}(x)$ if $$AB-BA=\lambda B$$ so using the components we get $$\lambda_i b_{ij}-b_{ij}\lambda_j=\lambda b_{ij}\implies \lambda=\lambda_i-\lambda_j \quad\text{if}\; b_{ij}\ne0$$ so we see that the matrix ...

3

To check that a subset $S$ of a vector space (over, say, the field $\mathbb{F}$) is a vector subspace, we need only check: That the subset is nonempty ($S \neq \varnothing$). That the subset is closed under scalar multiplication (for all $f \in \mathbb{F}$, $s \in S$, we have $fs \in S$). That the subset is closed under addition (for all $s_1, s_2 \in S$, ...

3

$\newcommand{\Span}[1]{\langle #1 \rangle}$This is a drastic simplification of my previous solution - should have thought of it before, as there is a sound theoretical reason for it. Let us rewrite $X, Y, Z$ in terms of the basis $$e_1, e_1 + e_2 - 2 e_3, e_1 + e_2 - e_3.$$ We have X : e_{1} \mapsto e_1 + e_2 - 2 e_3 \mapsto e_1 + e_2 - e_3 \mapsto 0 ... 2 The "principal" involution on \mathfrak{g} given by x\mapsto -x naturally extends to an involution of U(\mathfrak{g}), i.e., we have an anti-automorphism U(\mathfrak{g})\mapsto U(\mathfrak{g})^{op}. And an involution \psi of a ring is a bijection because of \psi^2=id. 2 The answer depends on what we understand by "just by looking at a Dynkin diagram". I guess, just by looking in the strict sense only gives the type, the rank, the dimension, and if the diagram is connected or not, i.e., if the algebra is simple or only semisimple. Maybe we also see immediately automorphisms. If we allow a second of further thought, then we ... 2 Compact Lie groups are topological groups whose topology is compact. A compact Lie algebra is the Lie algebra of a compact Lie group, therefore the name "compact". Lie algebras are vector spaces, and all finite-dimensional Lie algebras are linear, i.e., subalgebra of the Lie algebra of matrices. For Lie algebras "compact" just means that they are reductive ... 2 No, there is no quicker way. You have to check all these combinations for the Jacobi identity. Of course, j(a,b,a)=[a,[b,a]]+[b,[a,a]]+[a,[a,b]]=0 is satisfied automatically, if two elements are equal. You can take the classification of 5-dimensional Lie algebras, to have some examples. 2 Let's start with the case where A = \def\d{\mathrm{diag}}\d(\lambda_1, \ldots, \lambda_n) is diagonal. If now B\in \def\C{\mathbf C}\C^{n \times n} is given, we have \begin{align*} AB &= \begin{pmatrix} \lambda_1 b_{11} & \lambda_1 b_{12} & \cdots & \lambda_1 b_{1n}\\ \vdots & & \ddots & ... 2 In the exercises of Humphreys book on Lie algebras and Representation theory the example of \mathfrak{sl}(3) (respectively of \mathfrak{sl}(3)/Z(\mathfrak{sl}(3)) in characteristic 3 is worked out. Its Killing form is identical zero, but \mathfrak{sl}(3)/Z(\mathfrak{sl}(3)) is still a simple Lie algebra in characteristic 3. One can find all details ... 2 For the direct sum we have \begin{align*} (p_1\oplus p_2)([x,y])(v_1,v_2) & = (p_1([x,y](v_1), p_2([x,y](v_2)] \\ & = ([p_1(x),p_1(y)](v_1), [p_2(x),p_2(y)](v_2)] \\ & = [(p_1\oplus p_2)(x), (p_1\oplus p_2)(y)](v_1,v_2). \end{align*} Given two Lie algebra representations \rho_i:\mathfrak{g}\rightarrow V_i, for i=1,2, the tensor product ... 2 On page 313 Jacobson gives the following reasoning (4 lines before lemma 3): "L is simple over the algebraically closed field \Omega. Hence it has a basis over \Omega whose multiplication coefficients are in the prime field and so are in \Phi." That the coefficients are in the prime field follows from the classification. There is even a ... 1 Sure: In \mathfrak{sl}(2), the usual basis elements e, f, h satisfy \begin{align*} [e, f] &= h & [h, e] &= 2e & [h, f] &= 2f. \end{align*} Thus the k-fold nested commutator [h, [h, \dots, [h, f] \cdots ]] = 2^k f is nonzero for all k > 0. More generally, the condition that every k-fold nested commutator vanishes for some ... 1 The exponential map, although it is a local diffeomorphism on some neighborhood of 0 in the Lie algebra \mathfrak{g}, need not be a local diffeomorphism near all points of \mathfrak{g}. Take for example the Lie algebra \mathfrak{su}(2) : all points at distance π from 0 are sent to the south pole of S_3≃SU(2), so the exponential fails to be ... 1 The image is an angle preserving depiction of a certain Euclidian space together with some of its elements: The Euclidean space in question is the real part of the dual {\mathfrak h}^{\ast} of the Cartan subalgebra {\mathfrak h} of {\mathfrak s}{\mathfrak l}_3({\mathbb C}) of diagonal matrices, equipped with the restriction of the dual of the ... 1 Firstly, the commutator is linear in that [A,B+C]=[A,B]+[A,C]. Now note that [\partial_1,x^2 \partial_2]=0 since they are in different variables. Then either expand out the commutator directly, or use the identity [A,BC] = [A,B]C + B[A,C] $$to get the others. 1 A more elementary proof for the case of \mathfrak{sl}_2(\mathbb{C}) is given in [Erdmann-Wildon: Introduction to Lie algebras, Exercises 8.6 and 9.15, Solutions of which can be found in Appendix E]. 1 U mentioned that L has a \underline{basis}: {u1,...,un, v1,...vn, z}. For x\in L, you can write : x=x^zz+\sum_{i=1}^{n} x^u_iu_i +x^v_iv_i where x^z, x^u_i and x^v_i are scalars. Do the same for y, and span [x,y] using the bilinearity (and skew symmetry) of the bracket :$$[x,y]= \sum_{i,j} \alpha^z_{i}[z,u_i]+\alpha^z_{j}[z,u_j] + ...

1

As Qiaochu points out, this only holds in the finite dimensional case and the proof is somewhat involved. I think it's better to give a reference here than try and repeat a long argument you could find in any text on the subject: See Chapter II.6 of Humphrey's book Introduction to Lie algebras and representation theory. In particular, Theorem 6.3 of that ...

1

To complement the excellent suggestions for computing an explicit formula made by @JyrkiLahtonen in the comments, here is a somewhat different strategy: show that the element $1-e^{-\alpha}$ generates the ideal of functions in $\mathbf{Z}[P]$ vanishing on the fixed space of $s_\alpha$ (acting on the Cartan), and then observe that $u-s_\alpha u$ is zero on ...

1

The answer to your doubt can be found in chapter $8$ of Lie algebras and particle physics by Georgi. In particular, a simple root is a root that cannot be written as a sum of positive roots. Given two simple roots $\vec{a}, \vec{b}$ then $\vec{a}- \vec{b}$ is not a root, otherwise - say $\vec{b}$ is larger - $\vec{b} = \vec{a} + (\vec{b}-\vec{a})$ and ...

1

The Lie algebra of a compact, connected Lie group $G$ is abelian if and only if $G$ is a torus (that is, if and only if $G\cong \mathbb{S}^1\times\cdots\times \mathbb{S}^1$ is a finite product of circle groups); to prove this, note that the exponential map is the universal cover. So, abelian Lie subalgebras of the Lie algebra of a Lie group give rise to ...

1

Without more context I cannot be certain, but I don't think your interpretation is the one your author has in mind. Given an associative algebra (like a Banach algebra), say $\mathcal{A}$, one gets a Lie algebra using the commutator bracket: $[x,y]=xy-yx$ where $x,y \in \mathcal{A}$. Here "$xy$" denotes $\mathcal{A}$'s (associative) multiplication. So ...

1

Lie algebras are vector spaces, in this case subspaces of the vector space of matrices. Over the real and complex numbers, vector spaces are never compact unless they are trivial.

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The formal answer is that $d$ is the differential of the left translation map $L_g:G\to G$ given by the rule $L_g(h)=gh$ for $h\in G$. The map is smooth by the definition of "Lie group" and so there is a notion of the differential of this map. In particular, $d(L_g)_h$ denotes the differential of $L_g$ at $h\in G$; it is a map of tangent spaces ...

1

Since the given Lie algebra is perfect, i.e., satisfies $$L=L^{(1)}=[L,L],$$ the derived series and the lower central series are constant to $L$. This is consistent with the definition of a solvable respectively nilpotent Lie algebra. A perfect Lie algebra is never solvable or nilpotent. Indeed, $L^{(1)}$ contains $x=[y,z]$, $y=[z,x]$ and $z=[x,y]$, ...

1

Edit: Answer may be incomplete. I will revisit when I have time. Note that this isn't a very elegant answer. Perhaps there is a better way. Taking a cue from Jean-Claude Arbaut's comment, I've used octave to verify that $Y^2 = 0$ and $Z^2 = 0$. I've also verified that $X^2 Y = 0$, $X^2 Z = 0$ and $X Y Z = 0$. Note also that $X^3 = Y^3 = Z^3 = 0$, since ...

1

When defining the symmetric and antisymmetric tensor representations of the Lie algebra, is the action of the Lie algebra on the symmetric and antisymmetric subspaces defined the same way as above? Yes. If so, are the symmetric and antrisymmetric subspaces separate invariant subspaces...meaning that every tensor product representation is ...

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