The question: Let $L$ be a three dimensional simple Lie algebra over $\mathbb{C}$. Say that $x \in L$, $x \neq 0$, and that $ad_x$ has an eigenvalue $\lambda \neq 0$. Show that $L \cong \mathfrak{sl}(2, \mathbb{C})$.  Things I've tried: We can say $[h,e] = 2e$, $[h,f] = -2f$, and $[e,f] = h$. If I say $ad_x(y) = [x,y] = \lambda y$ then maybe I can have a basis $\{x,y,z\}$. I tried to just create a map $\phi(\frac{2}{\lambda}x)=h$, $\phi(y)=e$, and $\phi(z)=f$. Then I think $x$ and $y$ work but I'm not sure how to ensure $z$ works the way I want.
121612
If we know that a 3-dimensional Lie algebra $L$ with $[L,L]=L$ is simple.   How to prove that the only (up to isomorphism) 3-dimensional complex Lie algebra $L$ with $L=[L,L]$ is $sl_2(\mathbb C)$?
9690
I would like to have a result similar to . Namely:      Let $R$ be a commutative $\mathbb{C}$-algebra, $A \in \mathcal{M}_n(R)$ such that $\mathrm{tr}(A)=\cdots=\mathrm{tr}(A^n)=0$. Does it follow that $A^n=0$?   I have no idea weather or not the analogy of eigenvalue exists for matrices over algebra so I can't simply generalize considerations from cited post. Any help will be appreciated.
92539
I would like to know whether the $GL^{+}_n(\mathbb{R})$ the set of all invertible matrices with positive determinant is simply connected or not? I guess it is not simply connected but that is just a guess only, I do not know how to prove that, i.e how to show that its fundamental group is non-trivial.  Well, I can rigorously prove that this is connected and hense path connected as it is Lie Group. Could any one rigorously tell me how to approach this kind problem and solve them from the basic knowledge of fundamental group or some other way? So basically I need some result or tools by which I can compute fundamental groups of all known classical matrix Lie groups.
122546
I thought that groups with the same Lie Algebra are automatically equivalent, but there appear to be some exceptions to this?   What sort of exceptions are there and why?
21554
I want to determine the root-system of the lie algebra $sl(3,\mathbb C)$. Does someone know a good (and complete) reference for this problem?  I know that the root-system is $A_2$ but I want to see a complete proof (calculation) for this.  Thanks in advance.
77180
I have the same question as  from a long time ago. Is there an easy way to see that the Killing form on $\mathfrak{sp}(2n)$ is $\kappa(x,y) = (4n+2) \mathrm{tr}(xy)$?  For example, the Killing form for $\mathfrak{sl}(n)$ can be found from the easier to calculate case of $\mathfrak{gl}(n)$ using the decomposition $$\mathfrak{gl}(n) = \mathfrak{sl}(n) \oplus \mathbb{C}.$$ I don't know a similar argument for $\mathfrak{sp}(2n)$ though.
76528
Let $L$ be a Lie algebra over $A$, $I$ a Lie ideal in $L$, $K$ a subalgebra of $L$. The goal is to show that $[I,K]$ (the $A$-submodule generated by $[i,k]$ for $i \in I$ and $k \in K$) is a Lie subalgebra as well. If $$ S = \{ [i,k] : i \in I, k \in K\},$$ I believe that by bilinearity of the bracket it suffices to show that $[S,S] \subseteq [I,K]$. So take two elements $i, i' \in I$, $k, k' \in K$. The goal is to then show $$ [[i,k],[i',k']] \in [I,K].$$ This is where I'm lost and would like a hint as to where to go. My initial idea is to use the Jacobi identity to get $$ [[i,k],[i',k']] = -[i',[k',[i,k]]] - [k',[[i,k],i']].$$ Since $I$ is an ideal, $[i,k] \in I$, $[[i,k],i'] \in I$, and so $-[k',[[i,k],i']] \in [I,K]$. The idea then is to show that $-[i', [k',[i,k]]] \in [I,K]$. Use the Jacobi identity again to get $$-[k', [i,k]] = [i,[k,k']] + [k,[k',i]], $$ so that $$ -[i',[k',[i,k]]] = [i',[i,[k,k']]] + [i',[k,[k',i]]].$$ Trying the Jacobi identity again gets us $$[i',[i,[k,k']]] = -[i, [[k,k'],i']] - [[k,k'],[i',i]], $$ $$[i', [k,[k',i]]] = -[k,[[k',i],i']] - [[k',i],[i',k]].  $$ It seems like this is going nowhere. If $K$ was assumed to also be a Lie ideal, then this result follows easily (in fact, $[I,K]$ would be an ideal), but otherwise it doesn't seem like there's a way to rearrange things with just the Jacobi identity so that we get a linear combination of elements in $[I,K]$. I've alternatively thought about just approaching it at the level of showing $$ [[I,K],[I,K]] \subseteq [I,K].$$ Here, we use the Jacobi identity to note $$ [[I,K],[I,K]] \subseteq [[I,[I,K]],K] + [[K,[I,K]],I].$$ Now use the fact that $I$ is an ideal to get $$ [I,[I,K]] \subseteq I,$$ so we have $$ [[I,K],[I,K]] \subseteq [I,K] + [[K,[I,K]],I].$$ The goal here is to show $$[[K,[I,K]],I] \subseteq [I,K],$$ which seems to require $$ [K,[I,K]] \subseteq K.$$ If $$ [I,K] \subseteq N_L(K) = \{x \in L : [K,x] \subseteq K\},$$ then the result follows, but I don't see why this is necessarily true either.
25436
When is the Killing form $\kappa$ of a Lie algebra $\mathfrak g$ null, i.e. $\kappa(\cdot,\cdot)=0$? Surely this is true for any Lie algebra with trivial bracket, but is this the only case? I can't seem to find any nontrivial examples.
105623
I have homework in a course on Lie groups, in which I must show that $(\pi_m,\mathbb C_m[x,y])$ are the only irreducible representations of finite dimension in SL(2,ℝ). Here $ C_m[x,y])$ is the vector space of homogeneous polynomials, and $\pi_m(g)$ acts via $\pi_m(g)(f(x,y))=f((x,y)\cdot g)$. It was not too complicated to calculate the characters, taking the basis $\{p_k:0\leq k \leq n\}$ with $p_k = x^ky^{n−k}$, since $\pi_m(exp(H))p_k=exp(d\pi_m(H))p_k=exp(2k−n)p_k$, but the question is how one could define an inner product for these matrices. But, I don't have the volume form of SL(2,ℝ) to define the integral $\langle f,g \rangle=\int_Gf(z)⋅\overline{g(z)}dz$
85854
Let $G\subseteq GL_n(\mathbb R)$ and let $\mathfrak g$ denote its Lie algebra.  Let $e: \mathfrak g \to G$ be the map $X \mapsto e^X$.     Does there exist an example of $G$ and $\mathfrak g$ such that $e$ is   not injective?   Of course I think the answer is no, there is no such example because $e: \mathbb R \to \mathbb R$ is injective.      What about $G\subseteq GL_n(\mathbb C)$?   Since the map is the same I again think there should not exist such an example.
41091
Let $L$ be a semisimple complex Lie algebra and $H$ a Cartan subalgebra. Suppose that $L=L_{1}\oplus ...\oplus L_{k}$ with each $L_{i}$ a simple ideal of $L$. Show that for $1\leq i \leq k$, $H\cap L_{i}$ is a Cartan subalgebra of $L_{i}$.  I have had no thoughts on this problem and no idea how to solve it hence I cannot make an attempt. I have seriously severe difficulties.
68692
Let $V$ be a complex-vector space and $x,y\in gl(V)$ such that $[x y]=xy-yx$ commutes with $x$ and $y$. Let $[x y]$. Prove that   (1)$tr(z^{m})=0$ , for every $m\geq 1$.  (2)Deduce that z is a nilpotent map.   I couldn't do the first one. For the second one if we can prove that the possible eigen values of $z$ is only $0$, we are done. From the first one , we have if $\lambda_{1},... \lambda_{n}$ are eigen values of $z$, then $\sum_{i=1}^{n} \lambda_{i}^{k} =0$, for every $k$. From there how can we comclude that $\lambda_{1}=\lambda_{2}=...= \lambda_{n}=0$.  Thanks in advance for any help.
113362
I need some help to show that:  If  $| \langle u,v \rangle | = \|u\| \|v\|$ then $u=\lambda v$ for some scalar $\lambda$.  We have to consider this over an arbitrary field $\mathbb{F}$.  I appreciate your suggestions.  Thanks.
27264
I came across the claim  which states that there are no complex semisimple Lie algebras of dimension $4$, $5$, or $7$. As the problem suggests, we can take a Cartan subalgebra $H$ and root system $\Phi$ so that $L=H\oplus(\oplus_{\alpha\in\Phi}L_\alpha)$ where $$L_\alpha=\{x\in L:(\forall h\in H)\ [h,x]=\alpha(h)x\}$$  As shown in Humphreys' Introduction to Lie Algebras, every $L_\alpha$ has dimension $1$, so $\dim(L)=\dim(H)+\vert\Phi\vert$. I know $\Phi$ spans $H^*$, so $\dim(H)\leq\vert\Phi\vert$. Furthermore, I know that for any $\alpha\in\Phi$, the only multiples of $\alpha$ in $\Phi$ are $\alpha$ and $-\alpha$. Since $\mathbb{C}$ has characteristic $0$, this implies $\vert\Phi\vert$ is even.  Finally, another fact which may be useful is that given nonzero vectors $x_\alpha\in L_\alpha$ and $y_\alpha\in L_{-\alpha}$, if we take $h_\alpha=[x_\alpha,y_\alpha]\in H$, then $x_\alpha$, $y_\alpha$, and $h_\alpha$ span a three-dimensional subspace of $L$ isomorphic to $\mathfrak{sl}(2,\mathbb{C})$. However, I am not sure how to connect these pieces to give the desired conclusion.  As a follow-up question, would the claim still be true if I replaced the field with an arbitrary field of characteristic $0$, or is it also necessary for the field to be algebraically closed?
7446
I am trying to show that the Lie algebra of strictly upper triangular matrices $\mathfrak{u}(n,\mathbb{C})$ is soluble for all $n\geq 2$. This is not an assessed question, but is an exercise in the notes. I am having real difficulty visualising the derived series. I know how to calculate it, and have done some test cases for $n=2$, $n=3$ and $n=4$. But I just can't quite visualise what is going on to generalise it. I know the formula for the bracket of the basis vectors $e_{ij}$ and $e_{jk}$: $$[e_{ij},e_{kl}] = \delta_{jk}e_{il}-\delta_{il}e_{kj}$$  and have obtained the following  for $L = \mathfrak{u}(2,\mathbb{C})$  $$ L^{(1)} = \left\{0\right\} $$  for $L = \mathfrak{u}(3,\mathbb{C})$  $$ L^{(1)} = \left&lt;e_{13}\right&gt;_\mathbb{C} $$  $$ L^{(2)} = \left\{0\right\} $$  and for $L = \mathfrak{u}(3,\mathbb{C})$  $$ L^{(1)} = \left&lt;e_{13}, e_{14}, e_{24}\right&gt;_\mathbb{C} $$  $$ L^{(2)} = \left\{0\right\} $$  but I can't visualise what is actually going on here and explain it, and that is preventing me from going any further.  How should I visualise this? I do notice that $L^{(1)}$ always removes the entries on the next diagonal, but I don't understand why this is the case.
108165
I am trying to prove the following.  Proposition Let $g$ be a Lie algebra such that the first derived algebra $Dg$ is a proper ideal of $g$. Consider the quotient $g/D^1 g$, and show that the preimage of any codimension 1 subspace is   an ideal, and has codimension 1.   I could show that the preimage is indeed an ideal, but I need some help with the part about the dimensions:  So suppose that $\phi: x \mapsto [x]$ is the quotient projection. And write $[h] + Dg$  for a codimension 1 subspace of $g$ (here $h\subset g$). Then the preimage of this subspace is just the linear span $h+Dg\leq g$. Let $y+z \in h+Dg$ with $y\in h$ and $z\in Dg$. Let $x\in g$. We want to show that $[x, y+z]\in h$, ie that $h + Dg \lhd g$. We can write $x$ as $x_0+x_1$ for $x_0 \in Dg$, $x_1\not\in Dg$. Then  $$[x,y+z] = [x_0, y+z] + [x_1, y] + [x_1,z] \in h + Dg$$  The inclusion follows since $Dg$ is an ideal so the first and third terms belong $Dg$; and $[x_1, y] \in D_g$, as $g/Dg$ is abelian.
121835
I believe I am very close to finishing this proof, but I cannot figure out the last part. If anybody could check my work and maybe give me a little hint, it would be greatly appreciated!  Let $V$ be a finite-dimensional vector space and $T \in \mathcal{L}(V)$, and let $\mathbf{u,v} \in V$ be eigenvectors of $T$.  Claim:  If $\mathbf{u} + \mathbf{v}$ is an eigenvector of $T$, then $\mathbf{u}, \mathbf{v}$, and $\mathbf{u+v}$ all correspond to the same eigenvalue.  Proof (So far!): Suppose $T(\mathbf{u}) = \lambda_1 \mathbf{u}$ and $T(\mathbf{v}) = \lambda_2 \mathbf{v}$ with $\mathbf{u}, \mathbf{v} \neq \mathbf{0}$.   Now suppose $T(\mathbf{u+v}) = \lambda_3(\mathbf{u+v})$ with $\mathbf{u+v}\neq \mathbf{0}$.  Then, $$ T(\mathbf{u}) + T(\mathbf{v}) = \lambda_3 \mathbf{u} + \lambda_3 \mathbf{v}\\ \lambda_1\mathbf{u} + \lambda_2\mathbf{v} =  \lambda_3 \mathbf{u} + \lambda_3\mathbf{v}\\ \lambda_1\mathbf{u} + \lambda_2\mathbf{v} - \lambda_3 \mathbf{u} -\lambda_3\mathbf{v} = \mathbf{0}\\ (\lambda_1 - \lambda_3) \mathbf{u} + (\lambda_2 - \lambda_3)\mathbf{v} = \mathbf{0} $$  Now I know in order to show that $\lambda_1 = \lambda_2 = \lambda_3$, I must show that the only solution to the last line is the trivial one. This would imply that $\mathbf{u}$ and $\mathbf{v}$ are linearly independent which I am unconvinced of!  The only information I have to my advantage I haven't used yet is the fact that $\mathbf{u}, \mathbf{v}, \mathbf{u+v} \neq \mathbf{0}$. I really cannot see how this information can help me though.  Perhaps I am going about this wrong, but that's why I want to ask! Thanks for your help.