I want to show that for a complex Lie algebra $\mathfrak{g}$ which is solvable that $\mathfrak{g}'$ is nilpotent. Where $\mathfrak{g}'$ is the derived algebra $[\mathfrak{g},\mathfrak{g}]$.  I have shown the converse, since $\mathfrak{g}'$ being nilpotent means that $\mathfrak{g}'$ is solvable and hence the derived series of $\mathfrak{g}'$ terminates, and hence the derived series of $\mathfrak{g}$ (which is the same except for the first step) terminates.    In regard to when $\mathfrak{g}$ is solvable, we have a terminating derived series, but I can't see how this means that the lower central series terminates.  We have $\mathfrak{g}^{(0)}=\mathfrak{g},\quad \mathfrak{g}^{(n+1)}=[\mathfrak{g}^{(n)},\mathfrak{g}^{(n)}]$ terminating, and apparently this means that:  $\mathfrak{g}^0=\mathfrak{g}^{(1)},\quad\mathfrak{g}^{n+1}=[\mathfrak{g}^{(1)},g^n]$, but I don't see it.  Feel free to use whatever notation you want, rather than mine
80031
The Wikipedia  page lists the following property as a notion equivalent to solvability: $\mathfrak{g}$ is solvable iff the first derived algebra $[\mathfrak{g},\,\mathfrak{g}]$ is nilpotent.  I can't see this at all and it does sound dodgy: solvability and nilpotency different by "only one rung" in the derived series. Is this correct (I suspect I'm missing something trivial- I can't seem to find anything relevant in Knapp "Lie Groups: Beyond an Introduction" (mine is reprint of the first edition))? If not universally correct, is it correct with further assumptions about the field which $\mathfrak{g}$ is a vector space over? (For example, a field of characteristic nought, algebraically closed, ...). Can someone either give a proof, a counterexample or a reference to one?   I'd also like to put a reference on the Wikipedia page quotation, or clearly have it stricken from the page if incorrect.
118890
Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$ and let $\exp :\mathfrak{g}\rightarrow G$ be the exponential map.  In his blog,  that if a Lie group is not simply-connected, then $\exp$ will not be injective.  Conversely, is it true that if a Lie group is simply-connected, then $\exp$ is injective?  If not, what is a counter-example?
101766
I have a simple question on characters of simple groups; stated as exercise in Isaacs' character theory book.  (2.3) Let $\chi$ be a character of $G$. If $\mathfrak{X}$ is a representation corresponding to $\chi$, define $\det\chi(x):=\det\mathfrak{X}(x)$. Show: $\det\chi$ is well-defined linear character of $G$.  This is an easy exercise; consider the next one.  (3.3) Show that no simple group can have irreducible character of degree 2. Hint: Problem 2.3 is relevant.  Question: My question is not about solution of (3.3) since I have solved it in some different way. I didn't click how (2.3) can be used for this?  Other way for (3.3): $|G|$ is divisible by $2$=degree of irreducible character. If $|G|=2.(odd)$ then it is well-known from basic group theory that $G$ can't be simple (Actually it is solvable). So $4$ should divide $|G|$. Consider subgroup $H$ of order $4$ in $G$. Let $\mathfrak{X}:G\rightarrow {\rm GL}_2(\mathbb{C})$ be irreducible with $G$ simple (non-abelian). We can show that some element $x\in H$, $x\neq 1$ goes to $\pm I$. So $Z(\chi)=\{g\in G: |\chi(g)|=\chi(1)\}$ is non-trivial normal subgroup. q.e.d.
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?
27119
I found the following statement in Humphreys' book, 'Introduction to Lie Algebras and Representation Theory', in the first chapter, section 3.1:  "Notice that for arbitrary (Lie algebra) L, $L$ / Rad$L$ is semisimple."  And as substantiation of the statement he cites the proposition "If $I$ is a solvable ideal of a Lie algebra $L$ such that $L/I$ is solvable, then $L$ itself is solvable" which was proven earlier. I could prove the proposition that he's referring to but I could not, after hours of trying, see how the statement follows from the proposition. I'd much appreciate some help on this.
46687
I am trying to understand the proof of the following proposition:     A finite group $G$ is soluble $\iff G^{(k)}=\{e\}$ for some $k$, where $G^{(0)}=G$, $G^{(i+1)}=[G^i G^i]$ is the derived series.   Proof: Since $G$ is finite there is some $i$ such that $G^{(i)}=G^{(i+1)}$, and clearly $G^{(i)}=G^{(k)}\ \forall k\ge i$. So $\{e\}$ appears in the derived series $\iff G^{(i)}=\{e\}$.  Suppose that $G^{(i)}\ne\{e\}$ and let $G^{(i)}=H_0&gt;H_1&gt;...&gt;H_k=\{e\}$ be a composition series for $G^{(i)}$. Since $\color{red}{G^{(i+1)}\nleq H_1}$ and since $G^{(i+1)}$ is the derived subgroup of $G^{(i)}$, it follows that $H_0/H_1$ is non-abelian. So $G^{(i)}$ is not soluble and hence $G$ is not soluble (since any subgroup of a finite soluble group must be soluble).  Now suppose that $G^{(i)}=\{e\}$. We know that if $X$ is a finite group and $N\lhd X$, then $X$ has a composition series which includes $N$.  $\color{red}{\text{Using this fact repeatedly we get a composition}}$ $\color{red}{\text{series for } G \text{ which includes } G^{(j)} \text{ for all } j\le i}$. It is clear that the composition factors are abelian since $G^{(j)}/G^{(j+1)}$ is abelian for all $j$. So $G$ is soluble. $\quad\quad\square$  I highlighted the parts which I didn't understand. Would somebody be so kind and explain the details. :/
118697
I read that a direct product of a finite number of nilpotent groups is nilpotent. Here the definition of a nilpotent group is one that has a central series. A comment in my book following this claim says      If $G_{ij}$ is the $i^{th}$ term of a central series of the $j^{th}$ factor $H_j$, with $G_{ij}=G$ if the series has already terminated at $G$, then $\prod_j G_{ij}$ will be the $i^{th}$ term of a central series for $\prod_j H_j$.    My guess is that the central series for $\prod_j H_j$ is something like $$ 1\unlhd \prod_j G_{1j}\unlhd\prod_j G_{2j}\unlhd\cdots\unlhd\prod_j G_{rj}=\prod_j H_j $$  and additionally $$ \prod_j G_{i+1,j}/\prod_j G_{ij}\subseteq Z(\prod_j H_j/\prod_j G_{ij}). $$ I'm struggling to understand why the containment above is true. I think I need to show $$ \begin{align*} \prod_j g_{i+1,j}\prod_j G_{ij}\cdot\prod_j h_j\prod_j G_{ij} &amp;= \prod_j g_{i+1,j}\prod_j h_j\prod_j G_{ij} \\ &amp;= \prod_j h_j\prod_j g_{i+1,j}\prod_j G_{ij} \\ &amp;= \prod_j h_j\prod_j G_{ij}\cdot\prod_j g_{i+1,j}\prod_j G_{ij} \end{align*} $$ but I just don't see why the second equality would be true. I'm sure there's a nice simple explanation, and I'd be glad to see it. Thanks.
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.
112265
I would like to do  (i) but before I can even attempt it I need to be clear about his terminology.  Exercise 6  Show that the Lie algebra $gl_n(\mathbb{C})$ of the general linear group $GL_n(\mathbb{C})$ can be identified with the space $M_n(\mathbb{C})$ of $n \times n$ complex matrices, with the Lie bracket $[A,B] := AB - BA$.  When he writes "identified", does he just mean there is an isomorphism? So the exercise would be to find an explicit isomorphism from $M_n(\mathbb{C})$ to $gl_n(\mathbb{C})$? Is that correct?  Many thanks for your help.
27432
I have to prove the following: If $R$ is a finite-dimensional algebra over a field $F$, then $J(R)$ is nilpotent.  I thought about this, but there are some gaps:    Because $R$ is in particular a finite dimensional vector space, there is a composition series $0=R_0\subset R_1\subset\dots\subset R_k=R$ of VECTOR SPACES. Do I know that this is a composition series of $R$-modules? If yes, why? Or how can I find some?  If that is the case, I would proceed as follows: $J(R)R_n/R_{n-1}\subseteq R_n/R_{n-1}$. Now, I would like to use, that the quotient is simple, so if the above is a strict inclusion (how to show that?), then $J(R)R_n/R_{n-1}=0$. Now, by induction it is easy to see that $J(R)^k=\{0\}$.
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.
23572
Let $G$ be a lie group with lie algebra $\frak{g}$. Let $Aut(\frak g)$ be the automorphism group of $\frak{g}$.   Its clear to me that $Aut(\frak{g})$ $\subset GL(\frak{g})$ since any automorphism of $\frak{g}$ is also a linear transformation. Now i have several questions:   How do i prove that $Aut(\frak{g})$ is actually a lie subgroup of $GL(\frak{g})$? I've tried playing with actions to realize $Aut(\frak g)$ as an orbit or an isotropy subgroup but nothing worked out... Say $H \subset G$ is an abstract subgroup (not necessarily lie). Are the following statements equivalent?   $$\{T_e H \text{  is closed under the lie bracket of } \frak{g} \}$$  $$ \{H \text{is a lie subgroup of } G\}$$   How does the lie algebra of $Aut(\frak g)$ look like? When does $Aut(\frak g)$ $= G$? what does $Aut(\frak g)$ tell me about $G$?
107360
Consider the Lie algebra $\text{sl}(2,\mathbb{C})$ (commutator as Lie bracket) with its standard basis consisting of $$e=\begin{pmatrix}0&amp;1\\0&amp;0\end{pmatrix},\quad f=\begin{pmatrix}0&amp;0\\1&amp;0\end{pmatrix},\quad h=\begin{pmatrix}1&amp;0\\0&amp;-1\end{pmatrix}.$$  Define the map $R: \text{sl}(2,\mathbb{C})\rightarrow \text{End}(\mathbb{C}[x,y])$ on the basis as $h\mapsto x\frac{\partial}{\partial x}-y\frac{\partial}{\partial y},~e\mapsto x\frac{\partial}{\partial y},~f\mapsto y\frac{\partial}{\partial x}.$  I want to show that this is a Lie algebra homomorphism where $\text{End}(\mathbb{C}[x,y])$ is equipped with the commutator as Lie bracket. For that I will use the relations $[e,f]=h,~[h,e]=2e,~[h,f]=-2f$.  $R$ is of course $\mathbb{C}$-linear since it is defined as a linear extension.  In order to show $R([A,B])=[R(A),R(B)]$ for alle $A,B\in\text{sl}(2,\mathbb{C})$, it is enough to show this on the basis.  But here is my issue: For example $R([e,f])=R(h)=x\frac{\partial}{\partial x}-y\frac{\partial}{\partial y}\neq 0=x\frac{\partial}{\partial y}y\frac{\partial}{\partial x}-y\frac{\partial}{\partial x}x\frac{\partial}{\partial y}=[R(e),R(f)]$.  What am I doing wrong?
92158
I am considering a finitely generated, solvable group $G$ which has subexponential growth. Let $G=G_0\vartriangleright G_1\vartriangleright....\vartriangleright G_d\vartriangleright G_{d+1}=1$ denote the derived series of G.  It should be the case that $G_d$ is normal in $G$. Does anyone know why this is true? I don't know whether all of the assumptions above are necessary for the result.
123948
This time I want to show that $H(R)$ ( the heisenberg group attached to a conmutative ring $R$ is nilpotent). But I dont know how to proceed, I was thinking in first serch the normal subgroups if the Heisenberg group and then try to figure out the quotient group of that, from there get the centre, etc, but I dont think it would be a good idea (In fact I thought to use Gap, but I dont know how to ask this questions in Sage), Is there a better way to do this? Thanks a lot.   For me   $H(R) = \left \{ \begin{pmatrix} 1 &amp; a &amp; c \\ 0 &amp; 1 &amp; b \\ 0 &amp; 0 &amp; 1 \end{pmatrix}: \ \ a,b,c \in R \right \}$  For a conmutative ring $R$
106633
Exercise 14.33 in Fulton and Harris's Representation Theory claims that  If the roots of a semisimple Lie algebra lie in a collection of mutually orthogonal subsets, one sees that the Lie algebra decomposes into a direct sum accordingly.  Although intuitively I can see why this is true, could anybody direct me to a proof of this assertion, or outline how I should go about doing it?
106253
I have just started learning about Lie algebra in the context of quantum mechanics and got confused with this: Some sources say the generators are $J_0,J_1$ and $J_2$ and some use $J_0,J_+$ and $J_−$. Which set is correct? Or if both are correct what key concept am I missing here? My understanding is that if have certain commutation relations then we know that Lie Algebra is such and such. But if we have two such choices then this understanding falls apart? How do I figure out what is $\mathfrak{su}(2)$ Lie Algebra in general?