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3

You're using the wrong definition of nilradical: it's the largest nilpotent ideal in the Lie algebra. For $\mathfrak{gl}_n$, the only proper ideals are the trace-free matrices $\mathfrak{sl}_n$ and the scalar matrices $\mathbb{C}\cdot I$. The former is not nilpotent (it's simple), and the latter is. So the scalar matrices are the nilradical.


3

This is because of universal properties: for every associative algebra $X$, $$\begin{align} \hom_{\mathsf{Alg}}(U(L(A)), X) & \cong \hom_{\mathsf{Lie}}(L(A), X) \\ & \cong \hom_{\mathsf{Set}}(A, X) \\ & \cong \hom_{\mathsf{Ab}}(M_A, X) \\ & \cong \hom_{\mathsf{Alg}}(T[M_A], X) \end{align}$$ (where I didn't write the forgetful functors on ...


3

The weights are not always in a single orbit---this happens already the first time it can, for the irreducible $3$-dimensional representation of $\mathfrak{sl}_2(\mathbb{C})$ (which also happens to be the adjoint representation). The weights of this representation are $-\alpha,0$, and $\alpha$, where $\alpha$ is the unique positive root. For finite ...


3

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 ...


2

We have a map $$\phi: G \rightarrow Aut \frak{g}$$ given by the derivative (at $e$) of conjugation. Now the Lie algebra of $Aut \frak{g}$ is the space of derivations of $\frak{g}$. So we have $$\phi_\ast :\mathfrak{g} \rightarrow Der \frak{g}$$ The image of $\phi _\ast$ is the space $$ad(\mathfrak{g}):=\{ad(E): E\in \frak{g}\}$$ In Chapter 1 Section 15 of ...


2

The line that's missing is not necessarily spanned by $D$, but by some element in $D + V_1 + V_1^{\prime} + V_2$, so you have to look for some ${\mathfrak sl}_2$-invariant element in there: $D + \frac{1}{2}h$ works.


2

First of all, observe that $\textrm{Sp}_2(\mathbb{R})=\textrm{SL}(2,\mathbb{R})$. The symplectic group differs from the special linear group only in dimension $4$ and up. Similarly for the Lie algebra. This will simplify your computation enormously. Secondly, since you just have $2\times 2$ matrices of zero trace (and at most two distinct eigenvalues), I ...


2

The assumptions in the question are weaker than the ones in the title, and it depends on which one you choose as to whether the resulting statement it true or not: If we only assume $B$ negative semidefinite, the statement is not true: If ${\mathfrak g}$ is nilpotent, then its Killing form vanishes completely, but ${\mathfrak g}$ is not necessarily abelian. ...


2

Here you go :) It is also clear from what is written in the text alone: If you accept that 1) $ \exp $ is injective on the domain $ B_\epsilon $ and 2) the inverse $ \log $ is continuous, then you get $ \exp( B_{\epsilon /2} ) = \log^{-1}( B_{\epsilon /2} ) $ is open, as it is a preimage of an open set under a continuous function.


1

If $\rho: G\times M\to M$ denotes the action map, the chain rule gives you $$\left.\frac{\text{d}}{\text{d}t}\right|_{t=0}\text{exp}_{\mathfrak g}(-t\xi).m = \left.\frac{\text{d}}{\text{d}t}\right|_{t=0}\rho(\text{exp}_{\mathfrak g}(-t\xi),m)\\ = \text{d}\rho_{(e,m)}\left[\left(\left.\frac{\text{d}}{\text{d}t}\right|_{t=0}\text{exp}_{\mathfrak ...


1

So if, say, $X=\{1,2\}$, then the free Lie algebra $L(X)$, with $x\mapsto e_x$, consists of all linear combinations of the elements $e_1,e_2$, $[e_1,e_2]$, $[e_1,[e_1,e_2]]$, $[e_2,[e_1,e_2]]$, $[e_1,[e_1,[e_1,e_2]]$, $[e_1,[e_2,[e_1,e_2]]$, $[e_2,[e_1,[e_1,e_2]]$, $[e_2,[e_2,[e_1,e_2]]$, and so forth, such that skew-symmetry and the Jacobi identity holds. ...


1

Well, what about $f$ being the trace plus the determinant? Or any symmetric function applied to its two eigenvalues? Probably what's better to do is think about whether you can diagonalize the matrices by conjugating them. Remember that conjugating a matrix by another matrix is the same as writing a linear operator in a different basis.


1

My educated guess as to the meaning of this product rule is the following. The product of two monomials like $$ P=X_1^{a_1}X_2^{a_2}\cdots X_n^{a_n}\qquad\text{and}\qquad Q=X_1^{b_1}X_2^{b_2}\cdots X_n^{b_n} $$ is defined to be $$ P\cdot Q=\left(\prod_{i=1}^n\binom{a_i+b_i}{a_i}\right) X_1^{a_1+b_i}X_2^{a_2+b_2}\cdots X_n^{a_n+b_n}, $$ where the binomial ...


1

This is a purely categorical fact: 1) The free Lie algebra functor $\text{Set}\to\text{Lie}_{\mathbb k}$ is left adjoint to the forgetful functor $\text{Lie}_{\mathbb k}\to\text{Set}$. 2) The universal enveloping algebra functor $\text{Lie}_{\mathbb k}\to\text{Alg}_{\mathbb k}$ is left adjoint to the forgetful functor $\text{Alg}_{\mathbb ...


1

It looks to me like it is true that $B_n$ is uniquely determined. Let $K_n$ be the kernel of the map $A_n\to B_n$ and let $E_n$ be the kernel from $C_n$ to $D_n$. Then we have an exact sequence $0\to E_n\to K_{n+1}\to K_n$. $E_n$ is known and $K_n$ is known inductively. That means that $K_{n+1}$ can be determined as the subspace of $A_{n+1}$ generated by all ...


1

I would prove it by noting that $[a,b]=0$ if and only if $[\phi(a),\phi(b)]=0$, so $[a,b]=0$ for all $b$ if and only if $[\phi(a),c]=0$ for all $c$ since $\phi$ is surjective.


1

You've made a mistake in the definition of your curve. You should take $$ \gamma(t) = \pmatrix{1&at\\0&1} $$ You should find then that $$ T_1(S) = \left\{\pmatrix{0&a\\0&0} : a \in \Bbb R\right\} $$


1

Intriguing problem. Here's a rather ad hoc proof that they're not isomorphic. There might be an easier way to do this, but this argument has the advantage that it doesn't use any sophisticated theory of infinite-dimensional Lie algebras or Lie groups. Note that $V(\mathbb R)$ contains an element $X$ such that the map $\operatorname{ad}_X\colon V(\mathbb ...


1

For any group $G$ denote by $G=G_1\supseteq G_2\supseteq G_3\supseteq \ldots$ the descending lower central series of $G$ defined inductively by $G_1 = G$ and $G_k = [G_{k−1},G_1]$ for $k\ge 2$. Then $$ gr(G)=\bigoplus_{i\ge 1} \frac{G_i}{G_{i+1}} $$ is the associated graded Lie ring, where the Lie bracket is induced by the group commutator. Let $$ ...


1

The Lie algebra $\mathfrak{o}(n)$ is the subalgebra of $\mathfrak{su}(n)$ consisting of skew-symmetric matrices over $\mathbb{R}$, and $\mathfrak{su}(n)$ itself can be viewed as subalgebra of $\mathfrak{u}(n)$. So $\mathfrak{o}(n)$ can be viewed as a subalgebra of $\mathfrak{u}(n)$.


1

In a reflection representation, $\sigma_\alpha$ acts as the reflection at a hyperplane, which means that the hyperplane is fixed and the vectors orthogonal to it get multiplied by $-1$. Hence the eigenspace of $-1$ is indeed $1$-dimensional. However, in an arbitrary representation of the Weyl group, abstract reflections do not necessarily act as reflections ...


1

Do you ask for a proof adapted to this special case? In general, if you have a symmetrizable generalized Cartan matrix $A\in\text{Mat}_{n\times n}({\mathbb Z})$, then the Weyl group of the Kac-Moody algebra ${\mathfrak g}(A)$ associated with $A$ admits a description as a Coxeter group as follows: It has generators $s_1,...,s_n$, and relations $s_i^2=e$ and ...


1

The number of roots can be found by using the Cartan matrix to find the type of the root system. In this case the type is $B_2$ (or $C_2$), so there are $8$ roots. The Lie algebra is the direct sum of a Cartan subalgebra and the weight spaces. In this case the Cartan subalgebra is two dimensional and the 8 weight spaces are one dimensional, so the dimension ...


1

Consider the following: Let $a\in\mathbb{R}$, and define a smooth vector field $X$ on $\mathbb{R}_+$, by $X(p)=ap$. Note that $X$ is left-invariant, when thinking of $\mathbb{R}_+$ as a Lie group. Now let $\gamma$ be a trajectory of $X$, with $\gamma(0)=1$. By solving a simple ODE, one verifies that $\gamma(1)=e^a$, thus the term "exponential map" for Lie ...


1

The definition of $L'$ you have is wrong ; we usually take $$ L' \overset{def}= \langle \{ [u,v] \, | \, u,v \in L \} \rangle_k $$ where $k$ is the field of your Lie algebra $L$ (the brackets $\langle - \rangle_k$ means you take the $k$-span of the subset). Then your issue becomes trivial. Hope that helps,



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