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5

The Killing form can be explicitly calculated. Let $A,B\in \mathfrak{sl}_n(\mathbb{C})$, Killing form is $$\kappa(A,B) = 2n\text{tr}(AB)$$ it is straightforward to show this is non-degenerate. Briefly, if $A$ is such that $\text{tr}(AB) = 0$ for all $B\in \mathfrak{sl}_n(\mathbb{C})$, then testing with $B=e_{ij}, i\neq j$ shows $a_{ji}=0$, testing with $B=e_{...


4

It is the good idea, let $p:g\rightarrow g/I$ the quotient map $p(rad(g))$ is a solvable ideal, since $g/I$ is semi-simple, $p(rad(g))=0$ and $rad(g)\subset I$.


2

Like I mentioned in my comment, the Killing form of the quotient Lie algebra $\mathfrak{g}/\mathfrak{g}^\perp$ does not necessarily equal the reduction of the Killing form of $\mathfrak{g}$ to $\mathfrak{g}/\mathfrak{g}^\perp$. Elaborating: As mentioned in the comment of Torsten Schoeneberger here, a good example is a 3-dimensional Lie algebra so that the ...


2

As to the question "why": The Killing form of $\mathfrak{a}$ is given by $$k_1(x,y) := Tr (ad_{\color{red}{\mathfrak{a}}}(x) \circ ad_{\color{red}{\mathfrak{a}}}(y))$$ for $x,y \in \mathfrak{a}$, whereas the Killing form of $\mathfrak{g}$ restricted to $\mathfrak{a}$ is given by $$k_2(x,y) := Tr (ad_{\color{red}{\mathfrak{g}}}(x) \circ ad_{\color{red}{\...


2

It is sufficient to notice that this algebra is semisimple, which may be verified by checking that its Killing form is nondegenerate. Of course this requires some computation also. Another route is to find a Cartan subalgebra and the corresponding root system. This is done in many textbooks in Lie theory. I would be very happy to learn about a way to get the ...


2

You are representing the three algebra elements $q, p, 1\!\!1$ by hermitian operators acting on functions of q. For simplicity, non-dimensionalize $\hbar=1$--only the clueless keep it unscaled. Multiplying linear combinations of such operators by i and exponentiating yields the generic unitary group element representation for you , $$ e^{ic + ibq + a\...


2

Given a vector space basis $(e_1,\ldots ,e_n)$ and Lie brackets $$ [e_i,e_j]=\sum_{r=1}^n c_{ij}^r e_r, $$ the Jacobi identity is equivalent to the system of polynomial equations $$ \sum_{r=1}^n (c_{ij}^r c_{lr}^s+c_{jk}^r c_{ir}^s+c_{ki}^r c_{jr}^s)=0 $$ for all $1\le i<j<k\le n,\; 1\le s\le n$. In general, we have to verify all these equations. Of ...


2

Hint: We have $\mathfrak{gl}_n(\Bbb C)\cong \mathfrak{sl}_n(\Bbb C)\oplus \Bbb C$, where the first ideal is simple and the second one is abelian. In general, if we have direct sum of ideals $L=L_1\oplus L_2$, we can express ${\rm Der}(L)$ by ${\rm Der}(L_1)$, ${\rm Der}(L_2)$ and the spaces ${\rm Hom}(L_1,Z(L_2))$, ${\rm Hom}(L_2,Z(L_1))$. Since $L_1$ is ...


2

A proof can be found in Section 1.9 of James E. Humphreys’ Reflection groups and Coxeter groups. However, this proof does not use Weyl chambers but instead argues that any relation $s_{\alpha_{i_1}} \dotsm s_{\alpha_{i_n}} = 1$ in $W$ can be deduced from the given relations $(s_{\alpha_i} s_{\alpha_j})^{n_{ij}} = 1$. (Disclaimer: I’ve never read through the ...


2

I assume you are talking about a split semisimple Lie algebra. This is always the case if you work over an algebraically closed field of characteristic $0$, like $\Bbb C$. Then it's part of the standard structure theory that such a Lie algebra $L$ decomposes as $$ L \simeq H \oplus \bigoplus_{\alpha \in R} L_\alpha$$ where $H$ is a Cartan subalgebra, $R$ ...


1

The definition is correct, you just have to interpret $d_0$ as the zero map. Hence $H^0(\mathfrak g) = \Bbb C$. That looks tautological but keep in mind cohomology makes sense for any $\mathfrak g$-module $M$, and in this case we have $H^0(\mathfrak g, M) = M^{\mathfrak g} := \{m \in M : xm = 0, \forall x \in \mathfrak g\}$. This is because the differential ...


1

You are right in principle. Little correction: $s_c(b+c)=b+c$. Just for completeness, in a next step you would get the new positive root $s_b(a+b+2c)=a+2b+2c$, and then we have all 9 positive roots. By the way, this is the root system $B_3$. You can make the procedure a bit more efficient by using the methods of this answer. Then e.g. in the first step, the ...


1

Yes, the direct summands are ideals by definition, since a Lie algebra direct sum $L=L_1\oplus L_2$ is defined with the Lie bracket $[L_1,L_2]=0$. Hence we have $$[L_1,L]=[L_1,L_1\oplus L_2]=[L_1,L_1]\oplus[L_1,L_2]=L_1,$$ hence $L_1$ is an ideal. The same is true for $L_2$.


1

For $n\times n$ matrices over a field $K$ of characteristic zero, it is exactly because of the trace, that $[A,B]=I$ is impossible. Actually, all matrices in $M_n(K)$ of trace zero are commutators of the form $[A,B]$. This is usually proved with Lie algebra theory, because the Lie algebra $\mathfrak{sl}_n(K)$ of trace zero matrices with Lie bracket $[A,B]=AB-...


1

As I read it, the text in the book just claims that the 1-dimensional subalgebra spanned by one nonzero $x_s$ is a non-zero subalgebra consisting of semisimple elements. (I agree the formulation is not completely clear.)


1

Post-Lie algebras and post-Lie algebra structures have two Lie algebra structures on a vector space $V$, say a Lie bracket $[x,y]$ and a Lie bracket $\{x,y\}$, together with a bilinear product $x\cdot y$ such that \begin{align} x\cdot y -y\cdot x & = [x,y]-\{x,y\} \label{post5}\\ [x,y]\cdot z & = x\cdot (y\cdot z) -y\cdot (x\cdot z) \label{post6}\\ ...


1

Let $L=\mathfrak{so}_3(\Bbb C)$ with basis $(x,y,z)$ and Lie brackets $$ [x,y]=z,\; [z,x]=y,\; [y,z]=x. $$ Then the operators $ad(x)$ are real skew-symmetric matrices in $M_3(\Bbb C)$, which are diagonalizable since they are normal. However, the Lie algebra $L$ is non-abelian. References: is a real skew-symmetric matrix diagonalizable? Skew-symmetric non-...


1

It is well known that $$\mathrm{exp}:\mathfrak{g}\to G, $$ satisfies \begin{eqnarray}\mathrm{d}\left(\mathrm{exp}\right)_X = \mathrm{d}L_{e^X}\circ T_X,\quad (*)\end{eqnarray} where $$T_X = \frac{ e^{\mathrm{ad}(X)} -1}{\mathrm{ad}(X)} =\sum_{k\geq0}\frac{1}{(k+1)!}(\mathrm{ad}(X))^k. $$ Since $G$ is compact, all eigenvalues of $\mathrm{ad}(X)$ are purely ...


1

Isomorphic Lie algebras have equivalent representations. The problem here seems to be that people are sloppy in keeping track of what is a representation of either the complexified Lie algebras or the original real Lie algebras. Here, the representations of the complex Lie algebras $\mathfrak{so}(3,1)_{\Bbb C}$ and $\mathfrak{so}(4)_{\Bbb C}$ are equivalent ...


1

"What are some enlightening examples of nilpotent Lie algebras? " The answer depends on what you will find enlightening, and in which context. There are several examples arising from geometry. To give an example, there is a paper by Yves Benoist "Une nilvariété non affine" which corresponds to a "very interesting" $11$-dimensional nilpotent Lie algebra ...


1

Given two elements $X,Y$, there is not always some $A$ with $X=[A,Y]$. For example, consider the case $X=Y$, and for convenience, $L=\mathfrak{su}(2)$, with a basis $(e_1,e_2,e_3)$ and Lie brackets $[e_1,e_2]=e_3, [e_3,e_1]=e_2,[e_2,e_3]=e_1$. Then $[A,e_3]=e_3$ for $X=Y=e_3$ is impossible, since $[L,e_3]=\langle e_1,e_2\rangle$.


1

With which definitions do you want to work? We can define the Lie algebra $\mathfrak{g}$ associated to $G$ as left invariant vector fields on (connected component of) $G$, then $\mathfrak{g}$ has a bracket, induced by one on the vector fields. Equivalently, such vector fields are in bijection with tangent vectors at any given point on $G$, we choose $e$ just ...


1

We cannot use Engel's theorem (which holds in arbitrary characteristic). Consider the following counterexample for the Heisenberg Lie algebra in characteristic $p>2$. It is has an irreducible representation of dimension $p$, see here. The assumption of Engel's theorem is that any element of a Lie algebra acts by a nilpotent operator. However, even for an ...


1

The question 'what happens' is best answered by looking at an example, so the first question is: where do we find examples of subalgebras that are not ideals? Here is a class of examples. consider the real Lie algebra $\mathfrak{g} = \mathfrak{sl}_n$ of traceless $n$-by-$n$ matrices. It has a Cartan decomposition $\mathfrak{g} = \mathfrak{k} \oplus \...


1

Yes, it is true. A semisimple Lie algebre is a direct sum $\bigoplus_{j=1}^n\mathfrak g_j$ of simple Lie algebras. And $[\mathfrak g_j,\mathfrak g_j]$ is an ideal of $\mathfrak g_j$. Since $\mathfrak g_j$ is simple, this ideal can only be $\mathfrak g_j$ itself. So$$[\mathfrak g,\mathfrak g]=\bigoplus_{j=1}^n[\mathfrak g_j,\mathfrak g_j]=\bigoplus_{j=1}^n\...


1

Perhaps the commutator in question is the graded commutator, i.e., $[A,B]=AB-(-1)^{pq}BA$. Because in this case, the formula simplifies to $$d[A,B]=[dA,B]+(-1)^p[A,dB],$$ which seems nicer than anything you get with the (non-graded) commutator.


1

Since the center $Z(L)$ of $L=\mathfrak{gl}_n(\Bbb C)\cong \mathfrak{sl}_n(\Bbb C)\oplus Z(L)$ is a characteristic ideal of $L$, it is preserved by all derivations of $L$.


1

The one information you have about your $\mathfrak{g}$ is that it is contained in some $M_n(\Bbb C)$, so we should use that somehow. Well, $M_n(\Bbb C)$ is an associative algebra with unit, and the inclusion can be written fancier as an injective (!) map $$j: \mathfrak{g} \hookrightarrow \mathcal A := M_n(\Bbb C).$$ Now use property 3 to conclude that $i$ ...


1

the exponential map $\exp\colon\mathfrak{g}\to G$ works for all Lie group $G$. just write down the Poisson bracket and see. If $G$ is not connected, $\psi$ at the nonidentity components of $G$ can pick out symplectomorphisms that we don't know by just looking at the identity component $G_0$. The equivariance of $\mu$ with respect to $\psi_g$ and $\...


1

Let $A\subset R^k$ be a compact convex subset with nonempty interior invariant under a closed subgroup $H< GL(k,R)$. Since $A$ is $H$-invariant then so is the symmetrization $B=SA$ of $A$ (which is the convex hull of $A\cup -A$). Now, define the norm $||\cdot||$ on $R^k$ for which $B$ is the unit ball. This norm then is preserved by $H$. Therefore, $H$ ...


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