Hot answers tagged lie-algebras
4
For a 1-form $\newcommand{\pp}[2]{\frac{\partial #1}{\partial #2}}\mathrm{d}\xi$, acting on a vector field $\displaystyle\mathbf{v} = v_1 \pp{}{x} + v_2 \pp{}{y}$ means taking the derivative along the $\mathbf{v}$:
$$
\mathrm{d}\xi(\mathbf{v}) = v_1 \pp{\xi}{x} + v_2 \pp{\xi}{y}
$$
Hence let $\omega = f(x,y)\mathrm{d}x + g(x,y)\mathrm{d}y$, then
$$
...
3
Consider the orbit of the vector $(0,0,1)$ under $SO(2,1)$; you should find that it's disconnected (note that there are elements of $SO(2,1)$ which map $(0,0,1)$ to itself, or to $(0,0,-1)$, and then show that it can not be mapped to any vector $(a,b,0)$). So this gives us a continuous map from $SO(2,1)$ to a disconnected space, which implies that ...
2
The universal enveloping algebra $U\mathfrak{g}$ is a unital associative algebra. The map $\epsilon$ you've described above would send the identity element of this algebra to 1, and the elements of $\mathfrak{g}$ (which, along with the identity, generate $U\mathfrak{g}$ as an algebra) to zero. So it's not the zero map.
2
The eigenvectors for $y$ acting on $V$ are $v_1$ with eigenvalue $-1$ and $v_1+v_2$ with eigenvalue $0$. So these have to go (up to scalars) to $x$ (eigenvalue $-1$) and $y$ in the adjoint representation. Now you just have to check what (if any) scalar multiples work to match up the $x$ action.
That is, define a linear map $\phi: V \rightarrow L$ by ...
2
As noted in the comments, the answer I give may not coincide with the definitions/assumptions of the book/lecture course you are following.
Suppose $L$ is a non-abelian semi-simple Lie algebra. Then by definition of semi-simplicity (or it is a theorem depending on the approach you take) $L=\bigoplus_{i=1}^n L_i$ for some simple Lie algebras. And because $L$ ...
1
Let $U$ be an open neighborhood of $e$ in $G$ such that the local coordinates system $x=(x^i)$ is well defined on $U\cdot U$. Given $g\in G$, denote $L_g$ the left translation by $g$, i.e.
$$L_g:G\to G,\quad h\mapsto gh.$$
When $g,h\in U$, following your notations,
...
1
Gustavo,
Are you sure this is all correct? The right approach is to use the fact that Lie derivative is a derivation and compute $L_S(X\wedge Y)$, etc. But I don't get your answers, and I think the hypotheses are mutually contradictory. In particular, the only way $[S,X]=mX$ can hold is to have $m=p$ and $X_2=0$. What is going on? :)
1
Fix a basis $x_1,\dots,x_m$ of $\frak{h}$, and consider the right invariant vector fields $X_1,\dots X_m$ extending this basis of $\frak{h} \subset \frak{g}$. Every vector field that is tangent to your distribution is a $C^{\infty}(G)$-linear combination of those vector fields. Thus, to show integrability, you only need to show that the brackets $[X_i,X_j]$ ...
1
Over $\mathbb{R}$, for a general $n$, a real matrix has a real logarithm if and only if it is nonsingular and in its (complex) Jordan normal form, every Jordan block corresponding to a negative eigenvalue occurs an even number of times. So, you may verify that $\pmatrix{-1&1\\ 0&-1}$ (as given by rschwieb's answer) and ...
1
For these sorts of questions, you want to transition between Lie algebras and Lie groups using the exponential map. The useful facts for this problem are:
$\exp(Ad_a Y) = a (\exp Y) a^{-1}$. This follows from $Ad_a$ being the derivative of conjugation and the naturality of $\exp$.
If $X \in \mathfrak g$ then $X \in \mathfrak h \subset\mathfrak g$ if and ...
1
A priori, it depends on the way you extend your vector $u,v\in \mathfrak{g}$ to vector fields in $TG$ and of the subspace $\mathfrak h$ in general.
Usually, the vector field associated to $u$ in $TG$ is the unique left-invariant vector field $X$ such that $X_e=u$.
And as you might notice, $[DR_p(e)u,DR_p(e)v]=DR_p(e)[u,v]$ so you are asking if for every ...
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