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## Hot answers tagged lie-algebras

36

I don't think there is just one motivation. I will mention three. First, the Jacobi identity says precisely that the bracket is a derivation with respect to itself, where a derivation of an algebra is a map $d$ with $d(a\cdot b)=d(a)\cdot b+a \cdot d(b)$. Thus, writing $\mathrm{ad}(a)$ for the map $b \mapsto [a,b]$, the Jacobi identity may be rewritten (...

24

The group axioms are intended to abstract the properties of discrete symmetries (that is, bijections from a set to itself). That is, we may define a "concrete group" to be a group of permutations of some set with composition as the group operation. An abstract group is supposed to be a version of a concrete group that does not rely on a choice of group ...

24

The idea of an exponential is the continuous compounding of small actions. Suppose you start with an object $p$, perform an action on it $v$, and then add the result back to the original object. What happens if you instead take half as much action but do it twice? What about if you take one tenth the action but do it ten times? The exponential function tries ...

23

There is a complete characterization, in a large part due to Dixmier and Saito (both independently in 1957): If $G$ is a real (finite-dimensional) Lie group with Lie algebra $\mathfrak{g}$, then the following are equivalent: $\exp$ is injective; $\exp$ is bijective; $\exp$ is a real analytic diffeomorphism; $G$ is solvable, simply connected, ...

20

Fundamental weights correspond to fundamental roots (i.e. simple roots). Each choice of simple roots leads to a different choice of fundamental weights. There aren't really any fundamental weights associated with other (non-simple) roots (or at least this terminology isn't standard to my knowledge). [Note: The rank of $\mathfrak{sl}_N$ (or equivalently $SU(... 19 Here's a non-compact example of the non-surjectivity of$\exp$. Take$\mathfrak g=\mathfrak{sl}_2(\Bbb C)$: the matrix $$T=\left(\begin{array}{rr} -1&1\\0&-1\end{array}\right)$$ is not in$\exp(\mathfrak g)$. For if$x\in\mathfrak g$, we can find a basis in which it is triangular, say $$uxu^{-1}=\left(\begin{array}{rr} a&b\\0&-a\end{array}\... 18 Even in the compact case, \exp is not open. Look at the Lie group SU(2), whose Lie Algebra is the skew-Hermitian 2 \times 2 matrices. Look at the point x_0 = \left( \begin{smallmatrix} \pi i & 0 \\ 0 & - \pi i \end{smallmatrix} \right) in the Lie algebra. We can find an open neighborhood U of x_0 where the eigenvalues are distinct, with ... 17 Here's a slightly different way of seeing that \DeclareMathOperator{\SL}{SL(2,\mathbb{R})}\DeclareMathOperator{\USL}{\widetilde{SL}(2,\mathbb{R})}\DeclareMathOperator{\SLC}{SL(2,\mathbb{C})}\DeclareMathOperator{\sl}{\mathfrak{sl}(2,\mathbb{R})}\DeclareMathOperator{\slc}{\mathfrak{sl}(2,\mathbb{C})}\USL has no faithful finite-dimensional representations: ... 16 I think about that this way. In some sence geometry is "difficult" and algebra is "easy". So you want to obtain as much information as possible from studying Lie algebras instead of Lie groups, and then transering your results from algebras back to groups. So your bracket is the most natural operation on the tangent space that sort of allows you to do that.... 16 Lie theory connects to almost every other branch of mathematics! It's almost absurdly well connected! Just off the top of my head: analysis (e.g. harmonic analysis and the Peter-Weyl theorem), algebraic topology (e.g. principal bundles and characteristic classes), algebraic geometry (e.g. algebraic groups and flag varieties), combinatorics (e.g. root ... 15 I found myself working on this same problem (for homework), and I think I've written a fairly detailed solution. So I will post it here, in case it is helpful to anyone else. Let \mathfrak{g} be a 1-dimensional Lie algebra, and let \{E_1\} be a basis for \mathfrak{g}. Then for any two vector fields X,Y\in\mathfrak{g}, we have X=aE_1 and Y=bE_1, ... 15 The proof below is based in the following statement which is valid (at least) for linear spaces V over fields of characteristic 0: If (\cdot,\cdot)_1 and (\cdot,\cdot)_2 are non-degenerated bilinear forms on V then there is a linear autormorphism P\colon V\to V such that$$(v,w)_1=(Pv,w)_2,$$for all v,w\in V. Also, we will also use (a ... 15 The below is a short exposition I wrote a few months ago about the proof of the classification, attempting to give a "big-picture" view of it. I have copied it directly from the .tex file I wrote (and then attempted to fix all those things that fails due to that), so if anyone sees any weird things, let me know. If anyone wants the original .pdf instead, ... 15 Define the Lie group$$\tag{1} O(p,q)~:=~ \{\Lambda\in {\rm Mat}_{n\times n}(\mathbb{R}) ~|~\Lambda^T\eta\Lambda= \eta \} $$of pseudo-orthogonal real matrices \Lambda for the metric$$\tag{2} \eta_{\mu\nu}~=~{\rm diag} (\underbrace{+1,\ldots,+1}_{p~\text{times}},\underbrace{-1,\ldots, -1}_{q~\text{times}}), \qquad n~=~p+q.$$The groups O(p,q)=O(q,p) ... 15 Yes, and connectedness is not necessary. I know three proofs: Proof 1 When G is abelian, the inverse map$$i:G\to G,\quad g\mapsto g^{-1}$$is a group homomorphism. Hence, its differential at 1\in G$$di_1:\mathfrak{g}\to\mathfrak{g},\quad X\mapsto -X$$is a Lie algebra homomorphism. But then$$-[X,Y]=di_1([X,Y])=[di_1(X),di_1(Y)]=[-X,-Y]=[X,Y],$$so [... 14 It depends on the Lie algebra. We have that -w_0 induces some permutation of the simple roots since it sends the positive Weyl chamber to itself. Since -w_0 respects inner products, it must induce an automorphism of the Dynkin diagram of \mathfrak g. This immediately tell us that -w_0 is the identity for the simple Lie algebras B_n, C_n, E_7, ... 13 It's not true. Start with the Jacobi identity: [a,[b,c]] + [b,[c,a]] + [c,[a,b]] = 0. If [,] is associative then [a,[b,c]] = [[a,b],c] = -[c,[a,b]], so [b,[c,a]] = 0 for all a,b,c. In other words, all brackets must lie in the center of the Lie algebra. Conversely, if we define a bilinear skew-symmetric [,] for which [b,[c,a]]=0 holds ... 13 Yes; one can prove that the Lie bracket in the Lie algebra of the general linear group is the matrix commutator. Any matrix group is a subgroup of this with corresponding subalgebra and the bracket of the subalgebra is just the restriction of the original bracket. 13 As a take-home message, I would say that the Lie bracket encodes the structure of group multiplication. The linear structure alone is very uninformative; it really just tells you about the dimension of the manifold. More precisely, the Baker-Campbell-Hausdorff formula tells you that the group multiplication (at least for elements near the identity, and ... 12 (Let the usual symbols stand for the usual objects: M a smooth manifold, \gamma a smooth curve, etc.) (1) One reflexively uses open sets because it's easiest to think of the tangent space at x\in M as the set of velocities of curves passing through x. For this, you need to be able to differentiate a curve at x, and to differentiate, you need an ... 12 First, you are confusing Lie groups with Lie algebras. Casimir elements are objects that can be attached to certain Lie algebras. Second, Casimir elements do not always exist. For any Lie algebra \mathfrak{g}, there is a canonical bilinear form, the Killing form$$B(x, y) = \text{tr}(\text{ad}_x \text{ad}_y)$$where \text{ad}_x(y) = [x, y] is the ... 12 If you've studied the moment map for coadjoint actions, the answer is easy to see but takes a few words to describe. If you are not familiar with this, see the end of this answer for some references. The Lie algebra$$\mathfrak{so}(3) = \{A \in M_{3 \times 3}(\mathbb{R}) : A = -A^T\}$$can be identified with \mathbb{R}^3 via the identification$$\mathbb{R}... 12 It is true that matrix exponential is not an open map. For example, look at$\exp$in a neighborhood of$\left( \begin{smallmatrix} 2 \pi i & 0 \\ 0 & - 2 \pi i \end{smallmatrix} \right)$. All matrices near$\left( \begin{smallmatrix} 2 \pi i & 0 \\ 0 & -2 \pi i \end{smallmatrix} \right)$are diagonalizable (over$\mathbb{C}$) so their ... 12 In each name, the word "semisimple" means "a direct sum of simple objects" in the appropriate sense. It can be shown that a complex Lie algebra is semisimple (has radical zero) if and only if it is a direct sum of simple Lie algebras. On the other hand, saying that$\operatorname{ad} x$is diagonalizable is the same as saying that$\mathfrak{g}$decomposes ... 12 A$\frak g$-valued differential form is , as far as I know, just a section$\alpha$of the tensor product of the exterior power of the cotangent bundle$\Lambda^{\bullet}T^*M$of some manifold$M$with the trivial vector bundle$M\times\frak{g}$. As such, locally over some chart domain$U$,$\alpha$can be cast in the follwing form $$\alpha\equiv\alpha_1\... 12 As long as S is symmetric, the group of linear maps preserving the inner product induced by S will always be isomorphic to O(n) (and so in particular will always have the same Lie algebra). This is because given any inner product you can find an orthornormal basis and with respect to this basis S is just the identity matrix. The reason I'm familiar ... 12$$\theta \to \begin{pmatrix} \cos(2 \pi \theta) & \sin(2 \pi \theta) \\ -\sin(2 \pi \theta) & \cos(2 \pi \theta) \end{pmatrix}$$12 Morally speaking, the Lie algebra of vector fields is the Lie algebra of \text{Diff}(M), the diffeomorphism group of M. The relationship between these is less tight than in the finite-dimensional case: for example, The exponential map can fail to be defined at any nonzero time (as mentioned by orangeskid in the comments), and Even when defined (say on a ... 12 Using the standard basis of \mathfrak{so}(3), given by$$ e_1=\begin{pmatrix} 0 & 1 & 0 \cr -1 & 0 & 0 \cr 0 & 0 & 0 \end{pmatrix},\; e_2=\begin{pmatrix} 0 & 0 & 1 \cr 0 & 0 & 0 \cr -1 & 0 & 0 \end{pmatrix},\; e_3=\begin{pmatrix} 0 & 0 & 0 \cr 0 & 0 & 1 \cr 0 & -1 & 0 \end{pmatrix},$\$ ...

11

Dear user, the roots are a "special kind" of weights - corresponding to the adjoint representation. The fundamental representation is the "seed" of all other representations - all others may be obtained as terms in the decomposition of tensor powers of the fundamental ones. It follows that the root lattice is a subset of the weight lattice. This statement ...

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