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1. Introduce a new independent variable $s=bt$ to get rid of $b$, and w.l.o.g., assume that the vector $b$ points along $e_1$, $b=be_1$: $$\frac{dx}{ds} = 2x_1x-x^2e_1,$$ $$\dot x_1 = x_1^2-x_2^2-\cdots-x_n^2, \qquad \dot x_i = 2x_1x_i,$$ where, since $x_1$ is special, the index $i$ ranges over $2,\ldots,n$. 2. Introduce the change of variables $(z, w) ... 4 For a Lie group$G$, there is an exponential map$\operatorname{exp}:g\to G$(where$g$is the corresponding Lie algebra) such that$\operatorname{exp}\{tA\}$is just the one-parameter subgroup of$G$with tangent vector$A$at$e$. For matrix groups, this is just the usual matrix exponential$e^{At}$. Now note that ... 3 Howard Georgi's "Lie Algebras in Particle Physics" is good, if more intended for the physicist going towards the math than vice versa. It should provide a lot of context, though, and there's a PDF version floating around on google. I'd say similar things about these two introductions to aspects of high-energy theory [1] [2]. I'll see if I can remember some ... 3 Left and right translations are not Lie group homomorphisms; they don't even preserve the identity, and the induced map on Lie algebras is obtained by looking at derivatives at the identity. However, conjugation by a fixed element$g \in G$is, and the induced map on$\mathfrak{g}$gives a representation$G \to \text{Aut}(\mathfrak{g})$called the adjoint ... 2 The products and linear combinations of gamma matrices form a clifford algebra; the algebraic properties of elements of this algebra do not depend in any way whatsoever on the actual matrices you use to represent the basis elements. It helps, then, to think about the gamma matrices not as matrices but as basis vectors for Minkowski spacetime. In ... 2 Peter Woit, the author of the book "Not Even Wrong" and a blog by the same name, has been working on a book on quantum mechanics as described by representation theory. The latest draft may be found at the following link: Quantum Theory, Groups and Representations: An Introduction. 2 What you write probably makes sense only for matrix Lie groups and algebras: the last term$\gamma'(0) Y (-\gamma(0)^{-1} \gamma'(0) \gamma(0)^{-1})$should probably be$\gamma(0) Y (-\gamma(0)^{-1} \gamma'(0) \gamma(0)^{-1})$. Then the rest follows from the matrix identity $$\frac{d}{dt} A^{-1}(t)=-A^{-1}(t) \big(\frac{d}{dt} A(t)\big) \,A^{-1}(t)$$ ... 2 There is a purely geometrical explanation. First, let$n=2$, and consider$\mathrm{SL}(2,\mathbb{R})$. Let$X : [0,1) \to \mathrm{SL}(2,\mathbb{R})$be a smooth path, with$X(0) =E$, where$E$is the identity. $$X(t) = \left[\begin{array}{cc} a(t) & b(t) \\ c(t) & d(t) \end{array}\right]$$ where$\det[X(t)] = (ad-bc)(t) = 1$for all$t$. The ... 2 Your action 2 is not even well-defined if$\mathfrak{h}$is not commutative: Let$i<j$such that$[x_i,x_j]\neq0$. Then $$(x_ix_j)\cdot1=x_ix_j=(x_jx_i)\cdot1,$$ hence: $$0\neq[x_i,x_j]=[x_i,x_j]\cdot1=(x_ix_j-x_jx_i)\cdot1=0,$$ which is impossible. 2 If the Lie algebra derive from an associative Algebra, with the definition $$[u,v]=uv-vu,$$ then you can easily prove that $$[uv,w]=u[v,w]+[u,w]v.$$ For a situation like this: $$[x_1^ex_2^fx_3^g,x_5]$$ with$e$,$f$,$gpositive integers, you should have \begin{align} [x_1^ex_2^fx_3^g,x_5]&=\sum_{h=0}^{e-1} x_1^h ... 2 From the definition of the set\Pi$it is immediate that$\sigma(\Pi)=\Pi$for all$\sigma\in W$. By Lemma 13.2A we have$\sigma\mu\prec\mu$for all dominant weights$\mu$and all$\sigma\in W$. This allows us to redescribe the set$\Pi$as $$\Pi=\{\mu\in\Lambda\mid \sigma\mu\prec\lambda\ \text{for all \sigma\in W}\}.\qquad(*)$$ Let then$\mu\in\Pi$and ... 2 When you ask that$G \times G \to G$is a smooth map, it means "with respect to the smooth structure you put on$G$and the product differentiable structure on$G \times G$", so that asking this map to be smooth makes sense since it becomes a map between smooth manifolds and you can ask yourself this question. So somewhere you already assumed there was a ... 2 What you observed here (for SO(3), adjoint rep. = defining rep.) is indeed a very fine and remarkable property of SO(3). It explains, for example, the vector cross product in Lie-algebraic terms: the cross product R^3x R^3 --> R^3 is precisely the commutator of the Lie algebra, [,]: so(3)x so(3) --> so(3), i.e. the differential of the adjoint rep. of its Lie ... 2 This can be done in Sage yes! By default, Sage represents Weyl groups as matrix groups, but if you give it a prefix then it will give you the elements as products of simple reflections. sage: W = WeylGroup(["A", 3], prefix = "s") sage: list(W) [1, s1*s2*s3*s2*s1, s1*s2*s3*s1, s1*s2*s3*s2, s1*s2*s1, s3*s1*s2*s1, s3, s3*s1*s2, s1*s2, s2*s3*s1*s2*s1, s2*s3, ... 2 One counterexample is the semidirect product$\mathfrak{sl}_n(\mathbb{C}) \ltimes \mathbb{C}^n$, where$\mathfrak{sl}_n(\mathbb{C})$acts naturally on$\mathbb{C}^n$. 1 And this is how to do this in GAP: gap> W:=WeylGroup(RootSystem(SimpleLieAlgebra("A",3,Rationals))); <matrix group with 3 generators> gap> it:= WeylOrbitIterator( W, [1,1,1] );; gap> elms:= [ ];; gap> while not IsDoneIterator( it ) do > wt:= NextIterator(it); Add( elms, ConjugateDominantWeightWithWord(W,wt)[2]); > od; gap> elms; [ ... 1$\def\su{\mathfrak{su}} \def\g{\mathfrak{g}}$Suppose there exists a non-trivial$n$-dimensional unitary representation of$G$i.e. a homomorphism of Lie groups from$G$to$SU(n)$. Then, since$\su(n)$is a simple Lie algebra, the induced map from$\g \rightarrow \su(n) $is an isomorphism. Hence, the map from$G$to$SU(n)$has discrete kernel and is ... 1 Take a look in the abstracts of the 16 workshop in Bedlew (non coummutativ harmonic..) 6-12 of 7 1 That be a tough book, friend. I'd recommend going straight at it, but if you really start getting slayed, start off with Symmetry Methods for Differential Equations. 1 A useful identity is$ad_{D(x)}=[D,ad_x]$for every derivation$D. Furthermore, the Baker-Campell-Hausdorff formula is given by \begin{align} Z(x,y) & =\log \exp(x)\exp(y)\\ & =x+y+\frac{1}{2}ad_x(y)+\frac{1}{12}ad_x^2(y)+\frac{1}{12}ad_y^2(x)\\ & -\frac{1}{24}ad_y ad_x^2(y)+\frac{1}{720}ad_y^4 (x)\pm \cdots \end{align} 1 You can find this in the classical books on Lie algebras, e.g. N. Jacobson's book. Since\mathfrak{gl}_n(K\simeq \mathfrak{sl}_n(K)\oplus K$is a reductive Lie algebra, and$\mathfrak{sl}_n(K)$is simple (if the characteristic of$K$is zero, or not dividing$n$), we know all the ideals. Recall that a simple Lie algebra has only the trivial ideals. The ... 1 The$M$s are in the bispinor representation of the Lorentz group. The indices are not spacetime indices, they are spinor indices. The matrices should satisfy something of the form $$M \gamma^\mu M^{-1} = \Lambda_\nu^{\;\mu} \gamma^\nu.$$ 1 Here is a proof, assuming that you know what bilinear forms are. I will consider the case when the derived subalgebra is central (this is not be the case in general). We have the following situation:$h$is a lie algebra (over a field$k$) with the 1-dimensional center$c$(the derived subalgebra) and the abelian quotient$g=h/c$. Then for every$a, b\in g$, ... 1 Rough outline for an elementary approach. Let$L$be a finite-dimensional Lie algebra with$\dim L' = 1$. If$L'$is not contained in$Z(L)$, show that in this case$L$is the direct sum of an abelian Lie algebra and an nonabelian Lie algebra of dimension$2$. Assume then that$L'$is contained in$Z(L)$. If$L'$is properly contained in$Z(L)\$, prove ...