# Tag Info

## Hot answers tagged lie-algebras

6

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 ...

3

There are many connections to other branches of mathematics. Many interesting answers have been given to the MO-question "Why study Lie algebras". It is also worth to have a look at the prerequisites for the Langlands program, where Lie groups and Lie algebras are on top of the list.

3

$SO(3)$ consists of matrices $A\in GL(3,\mathbb{R})$ with ${}^tA\cdot A = I$. The standard representation of $SO(3)$ is just the inclusion map $SO(3) \subset GL(3,\mathbb{R})$. On the level of Lie algebras we have ${\mathfrak{gl}}(3;\mathbb{R}) = M_3(\mathbb{R})$, and ${\frak{so}}(3)$ consists of matrices $B\in M_3(\mathbb{R})$ with ${}^tB = -B$. Again the ...

3

Let $\mathfrak{g}$ be semisimple, then $\mathfrak{g}=\oplus_{i=1}^n\mathfrak{g}_i,$ where each $\mathfrak{g}_i$ is simple. The dimension of every $\mathfrak{g}_i$ is $\geq3$. If the dimension of $\mathfrak{g}$ is $\leq5$, how many $\mathfrak{g}_i$'s can there be?

2

You do need to keep careful track of the terms for the induction to go smoothly. In addition to the listed steps my calculation needed the formula (proof is straightforward) $$k\binom n k=(n-k+1)\binom n {k-1}.\qquad(*)$$ I abbreviate $$y^{[k]}=\frac{y^k}{k!}$$ and similarly with $x^{[k]}$. Things are a bit simpler if we declare $x^{[-1]}=y^{[-1]}=0$. ...

2

I'm interested in the same fields. I liked very much the approach to the Lie theory of Anthony Knapp - Lie Groups Beyond an Introduction. Here the author gives a very beautifull and complete overview on the world of Lie Groups and Lie Alebras. Starting from the topological description, looking at a lot of concrete examples, he introduces lots of deep ...

2

The polynomial function $f=(h^*)^2+x^* y^*$ takes value $$f(\ell)=h^*(\ell)^2+x^*(\ell) y^*(\ell)$$ on any $\ell \in L$. Evaluating it on $c h$ for a number $c$ therefore gives $$f(ch)=h^*(ch)^2+x^*(h) y^*(h)=c^2+0 \cdot 0=c^2.$$ This is the same as the value of $\lambda^2$ on $ch$ (by definition $\lambda$ is the dual basis element to the positive coroot ...

2

The induction hypothesis is applied for the vector space $V=\mathfrak{g}/\mathfrak{h}$, which is of lower dimension, since $\mathfrak{h}$ is a proper subalgebra. So you find a vector $x+\mathfrak{h}\neq \mathfrak{h}$ in $V$ killed by the image of $\mathfrak{h}$ in $\mathfrak{gl}(V)$. It follows that $\mathfrak{h}$ is properly contained in its normalizer ...

2

Since $SU(8)$ is a maximal regular subalgebra of $E_7$, you can identify it by looking at the extended Dynkin diagram. This is the diagram you get by appending the lowest root to the simple roots that you have. The $E_7$ Dynkin diagram looks like this: The extended diagram you get by appending the lowest root has an extra circle on the left side of the ...

2

If we extend the simple roots into a basis $B=(b_i)$ of $H^{\ast}$ (possibly with additional vectors), then we can choose a dual basis $B'=(b_i')$ of $H$ such that $$b_i(b_j')=\delta_{ij}$$ If we pick $i$ so that $b_i$ is not one of the simple roots, then $\alpha(b_i')=0$ for all roots $\alpha$.

2

I had found an argument, but actually your question is answered in Bourbaki, Groupes et algèbres de Lie, Ch. VI, 1.7, Cor. 1. My argument was as follows. One definition of positive root system is: given a hyperplane which does not contain any root, take the intersection of the root system with either half-space delimited by the hyperplane. If we can show ...

2

First of all a fundamental vector field is a vector field on $P$ not on $M$ as you wrote in your question. As you say the action of $G$ on $P$ preserves the fibers i.e. $p(u) = p(u g)$ for all $u \in P , g \in G$. If $A$ is in the Lie algebra of $G$ then $exp(tA)$ is a monoparametric group. For $u \in P$ we have a curve $\gamma(t):= u.exp(tA)$ in $P$. By the ...

2

We have $(I+J)^{2m}\subseteq I^m+J^m$ for all $m\ge 1$, hence $(I+J)^{2m}=0$ for $m$ large enough. I find this proof very natural, but indeed there is another proof using Engel's theorem and another two lemmas from representation theory; so it is perhaps more elegant, but it is also more complicated. This depends on your taste, too. For the alternative ...

1

The proof is given in Humphreys book on Lie algebras, in section $19.2$, "the classical algebras". He writes: "Therefore, to prove that $\mathfrak{so}_n$ acts irreducibly in its natural representation (i.e., $v\mapsto xv$), it will suffice to prove that all endomorphisms of $V$ are obtainable from $1$ and $\mathfrak{so}_n$ using addition, scalar ...

1

The line lnR=(theta/2*sin(theta))*(R-R') should be changed to lnR=(theta/(2*sin(theta)))*(R-R')

1

By definition of Killing form $$\kappa(t_{\lambda},t_{\mu})= \operatorname{Tr}(\operatorname{ad}t_{\lambda}\operatorname{ad}t_{\mu})$$ where $\operatorname{ad}:L\to\mathfrak{gl}(L)$, so, being $\operatorname{ad}t_{\lambda}\operatorname{ad}t_{\mu}\in\mathfrak{gl}(L)$, you can split the trace over the well known decomposition ...

1

For any nilpotent Lie subalgebra $\mathfrak{h}\subseteq L$ we have $$\kappa (h,h')=\sum_{\alpha\in \mathfrak{h}^{\ast}}\dim (L_{\alpha})\alpha (h)\alpha(h'),$$ which follows from Lie's theorem, giving a basis such that all $ad(h)$ on $L_{\alpha}$ are simultaneously upper triangular matrices with diagonal elements $\alpha(h)$. For $L$ semisimple, ...

1

I think you can do it in the following way: you know that $L$ decomposes as $L=H\oplus\bigoplus_{\alpha\in\Phi}L_\alpha$. The Killing form is $\kappa(t_\lambda,t_\mu)=\operatorname{tr}(\operatorname{ad} t_\lambda\operatorname{ad}t_\mu)$. But $H$ is abelian and nilpotent, so $t_\mu$ and $t_\lambda$ commute and are ad-nilpotent, so on $H$, ...

1

Since $\{\omega:|\omega|=1\}\cong S^1$ is one-dimensional, your one-dimensional subspace contained in the one-dimensional identity tangent space must be the whole tangent space.

1

Two ingredients here: (1) The Lie bracket is actually the Lie derivative $\mathscr L_X Y$ of the vector fields. By definition, this is $$\mathscr L_X Y(p) = \frac d{dt}\Big|_{t=0} \phi_{-t*}Y_{\phi_t(p)},$$ where $\phi_t$ is the flow of $X$. (2) The flow of the right-invariant vector field $X$ is given by left multiplication by $\exp(tX_e)$. (Solve ...

1

In general, if you have a right action $R : (g,x) \mapsto g.x$, then you can define a left action by $L : (g,x) \mapsto g^{-1}.x$ (and reciprocally). This comes from the fact that the inverse is an anti-automorphism of G : $(gh)^{-1}=h^{-1}g^{-1}$.

1

If you only want to take tensor products of non-algebraic representations here is an answer: Non-polynomial algebraic representations always become polynomial by tensoring with a large enough power of the determinant representation $D = (1,1,...,1)$. Tensoring with a one dimensional representation always preserves irreducibility, and in this case ...

1

For a redcutive Lie algebra $\mathfrak{g}$ the adjoint representation is semisimple. Hence also its restriction to $[\mathfrak{g},\mathfrak{g}]$ is semisimple, so that there is a complement $U$ to $[\mathfrak{g},\mathfrak{g}]\cap Z(\mathfrak{g})$, i.e., we have $U\oplus ([\mathfrak{g},\mathfrak{g}]\cap Z(\mathfrak{g}))=[\mathfrak{g},\mathfrak{g}]$. But we ...

1

First, the formula in 3. does not define a bilinear form: if you multiply $(X,z)$ by $\lambda$, the RHS is multiplied by $\lambda^2$. You want a sum, not a product. In Bourbaki, one of the axioms of root system is that the roots should generate the whole space. This reflects only semisimple groups. If you want to describe reductive groups, the good notion ...

1

Continuing with @TobiasKildetoft's comment: I really think one should think that the Jacobi identity is exactly the assertion that "ad" is a Lie algebra homomorphism. So, for example, it is not a good idea to try to present the literal Jacobi-identity formula in a "symmetric" form (as some sources do), which would necessarily obscure the assertion that "ad" ...

1

As indicated in the question and the other answer, this is really a fact about $\mathfrak{sl}_2$ representations and doesn't depend at all upon the adjoint representation, so perhaps it's worth observing that it follows from results found in standard references: First, for a finite dimensional irreducible $\mathfrak{sl}_2$-module $V$ with highest weight ...

1

You should complain to them, since this result is simply wrong. Consider how $\mathbb{R}^2$ acts on itself by translating with the usual symplectic form. The functions $x$ and $y$ are Hamiltonian functions for generators of the Lie algebra (you can work this out; I'll just get bogged down into notation), and since the adjoint action is trivial, you should ...

1

Yes, all irreducible representations of a nilpotent Lie algebra are $1$-dimensional by Lie's theorem (over the complex numbers). For representation theory in general of nilpotent Lie algebras see also our article concerning Faithful Lie algebra representations of nilpotent Lie algebras. For $2$-step nilpotent Lie algebras of dimension $n$ we can prove that ...

1

I guess this should work. Take $X \in \text{Der}(A)$ and consider $\gamma_t = \exp(tX)$. $\gamma$ is a curve on $GL(A)$ such that $\gamma_0=\text{Id}$ and $\gamma'_0=X$. We want to show that $\gamma$ is actually $G$-valued. Remark that $\gamma'_t=X \circ \gamma_t=\gamma_t \circ X$. Let's fix $a,b \in A$ and consider $\phi : t \mapsto \gamma_t(ab)$ and $\psi ... 1 I still don´t know exactly where the$-1$comes from, but I succeeded in solving the problem: Look at the formula before applying$r=m+n-2i$. The first summand is$1$exactly when$r\leq m+n$, the second when$r\leq m-n-2 \iff r-m+2\leq -n$, the third when$r\leq -m+n-2 \iff r-n+2 \leq -m$and the forth when$r\leq -m-n-4\$. If only the first inequation is ...

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