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4

These algebras are not isomorphic, since $\mathfrak{so}(5,\mathbb R)$ is a compact real form of its complexification while $\mathfrak{sp}(4,\mathbb R)$ is a split real form of its complexification. The basic source of isomorphism of that type is the representation of $\mathfrak{sp}(4,\mathbb R)$ on $\Lambda^2\mathbb R^4$. This has dimension $6$, but there ...

3

Hint: One helpful observation is that for all $y$, $(\operatorname{ad}_x^3 + \operatorname{ad}_x)y = 0$. If $\operatorname{ad}_x$ is diagonalizable over the reals, its only eigenvalue can be $0$.

2

Those formulas just tell you that $\rho$ is a function that turns elements into matrices; that's what a representation is, a recipe for turning elements into matrices. The set $\rho(\mathfrak{g})$ is a set of matrices that are "structurally similar" to your algebra $\mathfrak{g}$, and it's tempting to blur the distinction between the representation (the ...

2

If you are happy with one of them being abelian, take an abelian one, and a non-abelian one. For instance, in dimension $2$, take $L_{1}$ to have a basis $a, b$, and $[a, b] = b$. If you want both of them to be non-abelian, go to dimension $3$, and take $L_{1}$ to have a basis $a, b, c$ with $$[a, b] = b, [a, c] = [b, c] = 0$$ and $L_{2}$ to have a basis ...

2

(1) We don't know that an arbitrary subspace is a Lie subalgebra, but the statement is that $C_{H_0}(S)= C_{H_0}(\mathfrak{m})$ for some Lie subalgebra $\mathfrak{m}$ (indeed, $\mathfrak{m}$ is just the Lie subalgebra generated by $S$). (2) Complete reduciblity means that $W$ decomposes into a direct sum of simple submodules. The reason this works is that ...

2

No. It's still true that there are only a finite set of possible angles between two roots in a root system. Given infinitely many vectors in $\mathbb{R}^n$ (which aren't scalar multiples of each other), the angles between them must get arbitrarily small (basically because the unit sphere $S^{n-1}$ is compact).

2

For a positive definite bilinear form, the classified finite root systems are the only ones that can exist (up to isomorphism). If we allow arbitrary, possibly degenerate, bilinear forms we can get infinite root systems that satisfy these axioms. These correspond to the root systems of infinite dimensional Kac-Moody Lie algebras. If we then drop the ...

2

One way to do this is to write down an exact sequence $$0\longrightarrow\mathfrak{sl}_2\longrightarrow\mathfrak{gl}_2\longrightarrow\mathbb{C}\longrightarrow 0$$ where the first map is the obvious embedding and the second is the trace map (of course you need to check that both are Lie algebra homomorphisms, but that is more-or-less obvious). The map ...

2

Let $V$ be a $\mathfrak h$-module, where $\mathfrak h$ might not be abelian. That is, an Lie algebra homomorphism $\mathfrak h \to \mathfrak{gl}(V)$. Then we have for all $v\in V$, $h_1, h_2 \in \mathfrak h$, $$[h_1, h_2]v = h_1 (h_2 v) - h_2 (h_1 v)$$ Now if we use the definition of $V_\lambda$, this means $$[h_1, h_2] v = 0\ \ \ \ \forall h_1, h_2 \in ... 1 The 4-dimensional Lie algebra \mathfrak{gl}_2(\mathbb{C}) has a basis e_1=E_{12}, e_2=E_{21}, e_3=E_{11}-E_{22} amd e_4=E_{11}+E_{22}=I_2, where E_{ij} denotes the matrix with entry 1 at position (i,j) and zero entry otherwise. The Lie bracket is given by matrix commutator. Obviously we have [e_1,e_2]=e_3, [e_1,e_3]=-2e_1 and ... 1 The Lie algebra \mathfrak{t}_n(K) of upper-triangular matrices cannot be nilpotent, because it has a non-nilpotent solvable Lie subalgebra of dimension 2, e.g., generated by E_{12} and E_{11}-E_{22}. Hence it cannot be nilpotent. 1 Using the left multiplication of G on itself one can show that T^*G\cong \mathfrak{g}^*\times G, and under that identification the lifted action is simply h\cdot (\xi,g)=(\xi,h\cdot g), and the momentum mapping is \mathfrak{g}^*\times G\ni (\xi,g)\mapsto(\mathrm{Ad}_{g^{-1}})^*(\xi)\in\mathfrak{g}^*. The preimage by the momentum mapping of a point ... 1 There is a program around this called Spetses. The aim is to find some Lie theoretic object whose `Weyl group' is a complex reflection group (which includes the non-crystallographic reflection groups above). I am not entirely clear on what has been accomplished. The notes linked allude to some results for general Coxeter groups, but I haven't looked into ... 1 The nilpotent elements e=\begin{pmatrix} 0 & 0 \cr 1 & 0\end{pmatrix} and f=\begin{pmatrix} 0 & 1 \cr 0 & 0\end{pmatrix} are conjugate (as matrices) in \mathfrak{sl}_2(K), namely we have ses^{-1}=f with s=\begin{pmatrix} 0 & 1 \cr 1 & 0\end{pmatrix}\in \mathfrak{sl}_2(K). The elements are also conjugated over the complex ... 1 The two representations have different dimensions (standard is 2, adjoint is 3), so they can't be equivalent in the normal definition of equivalence between representations. If what you mean is that they have the same commutation relations, then that's what expected from all representations of the same algebra. (PS. I wanted to add this as a comment, ... 1 I think you have to assume that H is connected for the result to be true. I'just give a hint how to prove the result you are looking for, let me know if you want me to add details. Under the assumption, that H is connected, you first prove that G_H=G_{\mathfrak h}:=\{g\in G:Ad(g)(\mathfrak h)\subset\mathfrak h\}. Second, you can use ... 1 I gather that you want to prove that L_{\tilde{X}}\tilde{Y}(e) = ad(X)(Y), where \tilde{X}, \tilde{Y} are the left invariant vector fields corresponding to X,Y respectively. First observe that from definition we have ad(X)(Y) = \lim_{t \rightarrow 0}\frac{Ad(exp(tX))_*Y-Y}{t} Now we want to compute L_{\tilde{X}}\tilde{Y} and thus we need to ... 1 Before calculating this by hand, I suggest that you should recognize this as the Cartan matrix of the simple Lie algebra of type B_2. It has two simple roots \alpha_1 and \alpha_2, and the set of positive roots is \Phi^+=\{\alpha_1,2\alpha_1+\alpha_2,\alpha_1+\alpha_2,\alpha_2\}, and \Phi=\Phi^+\sqcup(-\Phi^+). From this, the character of the ... 1 I cannot add much to David's answer except for a comment - the first thing to notice is that the given Cartan matrix has size 2, which means that we are looking for a root system in \mathbb{R}^2, i.e., for a simple Lie algebra of rank 2. So we can start with two linear independent vectors \alpha and \beta in \mathbb{R}^2, and try to obtain the ... 1 If one takes the Lie bracket of two vector fields defined as the commutator (thinking of vector fields as derivations on the commutative algebra of smooth functions), this is exactly minus the infinitesimal counterpart of the adjoint action of diffeomorphisms (change of coordinates). In complicated language. Let M be a manifold (smooth, Hausdorff, ... 1 See 4.29 (on page 54) Peter W. Michor: Topics in Differential Geometry. Graduate Studies in Mathematics, Vol. 93 American Mathematical Society, Providence, 2008.(pdf) for complete formulas involving the the general terms of the BCH formula. You can use this to derive some expressions for the formula you want. 1 It's hard to find anything as useful as the one you mention, but this might be of interest. http://mathoverflow.net/questions/115192/power-log-distance-between-matrices/115227#115227 (ps. again I would have made this a comment rather than an answer, but I still can't do that) 1 Yes, [I,J]+[I,J]\subset [I,J] by definition of [I,J]. 1 Yes, although I don't know why you'd call it the Lie group functor as opposed to the Lie algebra functor. 1 Looking at a general basis will not help much, I guess, but what you have done can be completed: Note that the subspace U = \def\<#1>{\left<#1\right>}\<A_1, A_2> of \def\sl{\mathfrak{sl}(2, \mathbf R)}\sl generated by A_1 and A_2 is a subalgebra, as [A_1, A_2] = A_2 \in U. So \sl has a two-dimensional subalgebra. We will proof ... 1 The roots of \mathfrak{g}^{(1)}\oplus \mathfrak{g}^{(2)} are \Phi^{(1)}\sqcup\Phi^{(2)}, where \Phi^{(s)} is the root system for \mathfrak{g}^{(s)}. In particular, since \alpha is a root of \mathfrak{g}^{(1)}\oplus \mathfrak{g}^{(2)}, \alpha\in\Phi^{(1)} or \alpha\in\Phi^{(2)}. In turn, this forces \mathfrak{g}_\alpha to belong to either ... 1 The Lie algebra L of the cross product is isomorphic to \mathfrak{so}_3(\mathbb{R}), and ad(L) consists of 3\times 3 real skew-symmetric matrices. Since the eigenvalues of a real nonzero skew-symmetric matrix are imaginary it is not possible to diagonalize them over \mathbb{R} - except for the zero matrix. 1 The Cartan decomposition of a simple Lie algebra L of type A_n implies that$$ |\Phi|=\dim L-\dim H=(n+1)^2-1-n=n^2+n  Here $\dim A_n=(n+1)^2-1$ and $\dim H$, the dimension of a Cartan subalgebra $H$, is equal to the rank of $A_n$, which is $n$. In fact, $H$ consists of diagonal matrices of size $n+1$, but with trace zero. So your result is correct.

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