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0

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, $H$ ...


0

Hint: the fact that $G_x(\Omega) $ preserves a bounded open $U$ set implies that the modulus of its elements is $1$ (to see this consider an open ball $B$ containing $x$ contained in $U$, if $g^n(B) \subset U$, for every integer $n$, then the modulus of the eigenvalues of $g$ is $1$.) https://mathoverflow.net/questions/241565/groups-of-matrices-in-which-all-...


1

I only know the case where $\Phi,\Phi'$ are assumed to be irreducible root systems. Recall that for irreducible systems, roots of the same length are conjugate under the Weyl group $\mathcal{W}$. (These can be found in Humphreys' Introduction to Lie Algebras and Representation Theory, Section 10.4) Since and isomorphism of roots system comes from an ...


0

There might be a language problem involved. By "generators" you mean Lie algebra elements, and not group elements that preserve the symplectic matrix, so your Gs are the exponentials of the 36 generators you are seeking. The symplectic Lie algebras are for real symmetric matrices what the orthogonal Lie algebras are for antisymmetric ones. Consider the ...


0

This answer only tried to give a picture vision of the situation, hoping that things become clear already. After the answer there is a cited reference, Free Lie Algebras, which is the better answer. (Because it is a structural answer, and the structure is beautiful, just switch to the reference and enjoy!) First, in my pictural opinion a Lie polynomial in ...


1

I think, the answer is negative. Taking $\mathfrak{sl}_3(K)$ and $(e,f,h)$ given by $$ e=\begin{pmatrix} 0 & 1 & 0 \cr 0 & 0 & 0 \cr 0 & 0 & 0\end{pmatrix},\; f=\begin{pmatrix} 0 & 0 & 0 \cr 1 & 0 & 0 \cr 0 & 0 & 0\end{pmatrix},\; h=\begin{pmatrix} 1 & 0 & 0 \cr 0 & -1 & 0 \cr 0 & 0 & 0\...


0

Let $\Phi$ be a root system with inner product $\langle, \rangle$ which w.l.o.g. is chosen as invariant under the automorphism group $A(\Phi)$. As remarked in a comment, the question is equivalent to: for a given root $\alpha$, of what type is the root system $\alpha^\perp := \lbrace \gamma \in \Phi: \gamma \perp \alpha\rbrace$? Remark first that we can ...


0

another way to write co-acobi equaton is the following: Consider $g$ as vector space ad $\Lambda g$ (the exterior algebra on $g$) as algebra. From a linear map $\delta:g\to g\wedge g$, you can extend it as a degree 1 derivation (notice $\Lambda g$ is super commutative and free as such, so any super-derivation is determined by its values on $g$. Also, since ...


0

Most of this can be found in Naive Lie Theory by John Stillwell. Proof of simple connectivity of $SU(2)$: $SU(2)$ is homeomorphic to the space $S^3$. A loop $p:[0,1]\to S^3$ can be contracted by: Ideally, picking a point not in the image of $p$, stereographically projecting from that point onto $\mathbb R^3$, and then contracting in $\mathbb R^3$ (which is ...


2

Here is an example of an indecomposable nilpotent Lie algebra $L$ of dimension $7$ with $C^5(L)=0$ but $C^4(L)\subsetneq Z(L)$. Here $C^1(L)=L$ and $[L,C^k(L)]=C^{k+1}(L)$. Consider the Lie brackets with respect to a basis $(x_1,\ldots ,x_7)$ given by $$ [x_1,x_2]=x_4,\; [x_1,x_4]=x_5,\;[x_1,x_5]=x_7,\; [x_2,x_3]=x_7,\; [x_2,x_4]=x_6. $$ Then $Z(L)=\langle ...


2

$df(1)\in i\mathbb{R}$ it is of the form $ic, c\in\mathbb{R}$ let $\alpha={c\over {2\pi}}$, $df(1)=2\pi i\alpha$ and $df(t)=df(1).t=2\pi i\alpha t$.


1

No, Jordan decomposition is not preserved in general. For example, take a one-dimensional Lie subalgebra generated by any matrix: the matrix may be semi-simple or nilpotent (or anything in between), but the abstract Lie algebra is always the same (abelian). The most general results here are the following: for $\mathfrak{g}$ a semi-simple Lie algebra over ...


1

$\newcommand{e}{\mathbf e}$ Let $\e$ be an arbitrary vector, say $(1,0)^T$. Let $f:SL(2,\mathbb R) \to \mathbb R^2 \setminus \{(0,0)\}$ be defined as $$f(M) = M\e.$$ Observe that $f$ is continuous, and never maps to the origin. Let $p: [0,1] \to SL(2, \mathbb R)$ be defined as $$p(t) = \begin{pmatrix}\cos(2\pi t) & -\sin(2\pi t) \\ \sin(2\pi t) & \...


1

$Ad (k)$ is an automorphism of the Lie Algebra of G. But you are confusing it with an element of (subalgebra of) Lie (G). The given condition merely means that under the automorphism $Ad(k)$ of Lie (G) a particular subalgebra is mapped into itself (stabilized).


1

Okay so $\underline{\pi}\cdot\underline{\tau}$ means $\pi_1\tau_1+\pi_2\tau_2+\pi_3\tau_3$, where $\tau_1,\tau_2,\tau_3$ are the Pauli matrices. Then $$ \Phi(\sigma,\underline{\pi})=\sigma I+ \pi_1i\tau_1+\pi_2i\tau_2+i\pi_3\tau_3$$ $$ = \begin{pmatrix} \sigma+\pi_3i & \pi_2+\pi_1i \\ -\pi_2+\pi_1i & \sigma-\pi_3i \end{pmatrix}. \tag{$\circ$}$$ ...


0

In order for $L=\bigoplus L_i$ to be $\mathbb{Z}$-graded, you need $[L_i,L_j]\subseteq L_{i+j}$, or in other words $[x,y]\in L_{i+j}$ whenever we have $x\in L_i$ and $y\in L_j$. In this case, we may define $$ L_{-1}=Ke \qquad L_0=Kh \qquad L_1=Kf. $$ (And $L_i=0$ otherwise.) To check it is $\mathbb{Z}$-graded, it suffices to check the commutator conditions ...


2

Since $SU(2)$ is simple, the Killing form of $su(2)$is negative definite, the adjoint representation preserves the Killing form, thus its image is contained in $SO(3)$, it is a local diffeomorphism and since $dim SU(2)=dim SO(3)$ the adjoint representation is a covering (Ehresmann Lemma).


1

You are missing the fact that the theorem about the existence of a root space decomposition is stated from the start as a theorem about finite-dimensional semisimple Lie algebras.


0

For one dimension, one can have generators $x$, $y$, $z$ with $[x,y]=y$, $[x,z]=[y,z]=0$. For two dimensions: $[x,y]=y$, $[x,z]=z$, $[y,z]=0$.


3

Suppose that $\dim\mathfrak h>0$. There is then some $X\in\mathfrak h\setminus\{0\}$. So, $(\forall t\in\mathbb R):\exp(tX)\in H$. So, consider the map$$\begin{array}{rccc}\gamma\colon&\mathbb R&\longrightarrow&H\\&t&\mapsto&\exp(tX).\end{array}$$Then $\gamma'(0)=X\neq0$ and so $\gamma$ is not constant near $0$. But then every ...


2

There are several ways to conclude in the last step. Maybe the most elegant here is to use that $\mathfrak{b}_2$ is two-dimensional, and that over any field, up to isomorphism there are only two such Lie algebras: The abelian one, and a solvable but non-nilpotent one. See e.g. here and here. It is immediate then that for $char(\Bbb F) \neq 2$, you are in ...


0

Thanks to the helpful comments of Ted and user10354138, I was able to puzzle this together now: First of all, for a smooth map $\phi:\mathbb R^2\to\mathbb R$, $(r,s)\mapsto\phi(r,s)$ one may define the path $\gamma:\mathbb R\to\mathbb R^2$, $t\mapsto(t,t)$ to get the derivative in question via the chain rule: $$ \frac{d}{dt}\phi(t,t)=\frac{d}{dt}(\phi\circ\...


1

My guess is that whoever wrote the second sentence meant basis of the Lie algebra of $G$ which is a subset of the Lie algebra of $M$.


1

If you allow arbitrary $\mathsf{k}$-Lie algebras, then any field extension $\mathsf{K}$ of $\mathsf{k}$ can be the endomorphism algebra of some simple representation. Let $\mathfrak{g}=\mathsf{K}$ considered as a commutative $\mathsf{k}$-Lie algebra (i.e., with zero Lie bracket). Then since $\mathfrak{g}$ is commutative as a Lie algebra and $\mathsf{K}$ is ...


1

The best way to understand these things is to cut through the fog of (mostly unnecessary) terminology by adopting a few conventions that will dramatically simplify everything and make it totally transparent. So, let's do that here, first. A Lie group G is a manifold. As such, each element g ∈ G has a tangent space TgG. As is the case for all manifolds, if t ...


2

No to both. 1: try with a choice of direct product $G=N\times H$... 2: take $G=\mathrm{SL}_4$ and take $H=\mathrm{SL}_2\times\mathrm{SL}_2$ embedded as group of block-diagonal matrices each of determinant 1. What is the normalizer then?


1

Edit: I've restructured this quite a bit, but I think it gets at the core issue more effectively now. Given a differential map between manifolds $f:M \to N$, at each $p \in M$ we have a push forward map $f_*:T_p M \to T_{f(p)}N$ which is defined by the following: if $v \in T_p M$ and the path $\sigma: (-\epsilon,\epsilon) \to M$ satisfies $\sigma(0)=p$, $\...


0

Although the expression $f(t) = \text{det}(\gamma(t))$ is a good old 1-variable calculus function, and so its derivative is indeed obtained from the formula $$f'(0) = \lim_{h \to 0} \frac{1}{h}(f(0+h) - f(0)) $$ the problem is that the composition $f = \text{det} \circ \gamma$ is not a good old 1-variable composition. Instead, $\gamma(t)$ is a function of ...


0

The map $t\rightarrow \exp(tY)$ is a differentiable map defined on $\mathbb{R}$ and takes its values in $\alpha(H)$. This implies that its differential at $t$ is a map $T_t\mathbb{R}=\mathbb{R}\rightarrow \alpha(H)_{\exp(tY)}$ in particular if $t=0$ we have: ${d\over{dt}}_{t=0}\exp(tY)=Y$ is an element of $T_{Id}\alpha(H)=\alpha_*(h)$.


1

Since $z\in [L,L]$ is a linear combination of elements $[x,y]$ and $$ {\rm tr} ({\rm ad}([x,y]))={\rm tr} ([{\rm ad}(x),{\rm ad}(y)])=0, $$ the claim follows.


3

If $z\in L$, then $\textrm{ad}\,z$ is a linear map from $L$ to $L$. A linear map from a finite-dimensional vector space to itself has a well-defined trace. Just take the trace of the matrix representing the linear map with respect to any basis of the vector space.


0

My adviser, peer, and I figured it out, I'll let the reader verify. $M = span\{E_{11}-E_{22}, E_{12}, E_{21}\}$ $M' = span\{E_{23}, E_{13}\}$ $M'' = span\{E_{31}, E_{32}\}$ $M''' = span\{\frac{1}{2}E_{11}+\frac{1}{2}E_{22}-E_{33}\}$ $L=M\bigoplus M'\bigoplus M'' \bigoplus M'''$


1

It looks fine, but there is a simpler way. Since $\mathfrak g$ is semisimple, its ajoint representation is simisimple too and therefore there is an ideal $\mathfrak j$ of $\mathfrak g$ such that $\mathfrak g=\ker f\oplus\mathfrak j$. But then $\mathfrak g/\mathfrak\ker f\simeq\mathfrak j$, which is semisimple.


0

Coderivation The differential $d_{\mathrm{CE}}$ is of degree $-1$ so we need to show that \begin{equation} \Delta( d_{\mathrm{CE}}(c) ) = \sum_{(c)} d_{\mathrm{CE}}( c_{(1)} ) \otimes c_{(2)} + (-1)^{|c_{(1)}|} c_{(1)} \otimes d_{\mathrm{CE}}(c_{(2)}) \tag{1} \end{equation} for all $c = x_1 \wedge \dotsb \wedge x_n$ with $x_i \in \mathfrak{g}$ ...


2

It means that there is only one Lie algebra structure $[\cdot,\cdot]$ on $L$ for which it is true that$$(\forall i,j\in\{1,2,\ldots,n\}):[x_i,x_j]=\sum_{k=1}^na_{ij}^kx_k.$$This follows from the fact that $\{x_1,\ldots,x_n\}$ is a basis an that the Lie bracket is bilinear.


3

A typical element of $L$ is $u=\sum_i c_i x_i$, and I suppose another typical element of $L$ is $v=\sum_j d_j x_j$. Then $$[u,v]=\sum_{i,j}c_id_j[x_i,x_j] =\sum_k\left(\sum_{i,j}a_{i,j}^kc_id_j\right)x_k.$$ This is nothing more than the bi-linearity of the Lie bracket.


1

Denote $\mu: \mathcal{U}(\mathfrak{g})\otimes \mathcal{U}(\mathfrak{g})\rightarrow \mathcal{U}(\mathfrak{g})$ is the multiplication map $a\otimes b\mapsto ab$. For each $x\in \mathfrak{g}$ one has $\mu\Delta(x)=2x$, and more generally $\Delta$ acts as multiplication by $2^n$ on $\mathcal{U}(\mathfrak{g})^{(n)}$ by the PBW theorem. Since for a primitive ...


2

Here is another answer which may be slightly more in line with first principles and/or what is learned at introductory level. Consider the path in $G$ given by $$\gamma(t) = \left(\begin{array}{cc} a(t) & b(t) \\ 0 & a(t)^2 \end{array} \right),$$ which is differentiable, with $\gamma(0) = I$ (which implies $a(0)=1$ and $b(0)=0$). Since $\gamma$ ...


2

At the identity $a=1$ and $b=0$. Near the identity, the group is parameterised as $$\pmatrix{1+t&u\\0&(1+t)^2}=I+t\pmatrix{1&0\\0&2}+u\pmatrix{0&1\\0&0}+O(\text{higher powers of $t$ and $u$}).$$ The tangent space at $I$ is spanned by $$A=\pmatrix{1&0\\0&2}\qquad\text{and}\qquad B=\pmatrix{0&1\\0&0}$$ and these matrices ...


1

We know that the set of left-invariant vector fields on a Lie group under Lie bracket is isomorphic to the Lie algebra of the group. On the other hand, the bracket operation in the Lie algebra of an Abelian group vanishes. Here's why: If $G$ is Abelian, then $\mathrm{inv}: G\to G$ given by $g \mapsto g^{-1}$ is a homomorphism. Therefore, its differential at ...


1

This is true in any finite-dimensional Lie algebra over any field. Recall that a Cartan subalgebra is a nilpotent subalgebra equal to its normalizer. In particular it's maximal nilpotent. If we have $\mathfrak{g}=\prod_{i=1}^n\mathfrak{g}_i$ a direct product of such Lie algebras and $\mathfrak{h}$ is nilpotent and $\mathfrak{h}_i$ is its projection to $\...


1

That's the adjoint representation. The standard representation is the action of $\mathfrak{sl}(2,\mathbb C)$ in $\mathbb C^2$ defined by:$$(M,v)\mapsto M\cdot v.$$


4

No. This would say that vectors in a decomposition $V=\bigoplus_\lambda V_\lambda$ with a particular dominant integral weight are unique, and appear only as the highest weight vectors of the irreducible components. This is not the case. For a quick example, let $\mathfrak{g}=\mathfrak{sl}_2$, and consider the representation $V=V_5 \otimes V_3$. By the ...


2

There are exactly two different Lie algebras of dimension $2$ over an arbitrary field. Let $(x,y)$ be a basis. One Lie algebra is the abelian Lie algebra $K^2$, with $[x,y]=0$. The second Lie algebra is the non-abelian solvable Lie algebra $\mathfrak{r}_2(K)$, with Lie bracket $[x,y]=x$. It is easy to see that every $2$-dimensional Lie algebra with bracket $[...


0

There are two and (up to isomorphism) only two real $2$-dimensinal Lie algebras: The abelian $2$-dimensional Lie algebra $\mathfrak g$. In this case, $(\forall X,Y\in\mathfrak g):[X,Y]=0$. It is the Lie algebra of three connected Lie groups. $(\mathbb R^2,+)$, $(\mathbb R,+)\times(S^1,\cdot)$ and $(S^1,\cdot)\times(S^1,\cdot)$; The $2$-dimensional Lie ...


3

Hint: $gl(2, \Bbb C)$ here is really just a confusing way to write the $2\times 2$-matrices with entries in $\Bbb C$. Now a standard simple module on which $sl(2, \Bbb C)$ acts by multiplication from the left is the column vectors $\pmatrix{x\\y}$. Can you find such columns in the $2\times2$-matrices? Added: To spell it out completely, look at $$V_1 := \...


1

Let $G$ be a finite cyclic subgroup of $S^1 \subset GL(1,\mathbb{C})$ of 12 elements, for example. Clearly its Lie algebra $\mathfrak{g}$ is zero, but there is an element $A\in G$ with $0\neq \Vert A-I \Vert = |A-1| < 1$. This $A$ satisfies $\log A \neq 0$ so we are done.


0

The calculations were wrong as I forgot the transpose. The final solution is 3 and M looks like: $M=\begin{pmatrix} 0 & b &c\\ -b & 0 &f \\ c&f&0 \end{pmatrix}$


1

Any nonzero ideal of $\mathfrak{sl}_2$ contains one of $H$, $X$ and $Y$ As $[H,aH+bX+cY]=2bX-2cY$ and $[H,[H,aH+bX+cY]]=4bX+4cY$ then if $I$ is an ideal of $\mathfrak{sl}_2$ and $aH+bX+cY\in I$ then $aH\in I$, $bX\in I$ and $cY\in I$. If $H\in I$ then $[H,X]=2X\in I$ and $[H,Y]=-2Y\in I$ so that $I=\mathfrak{sl}_2$. If $X\in I$ or $Y\in I$ then $H=[X,Y]\...


2

Just to give a definite answer : Q1 : No. For example, any simple modules can be expressed as some quotient of $U(\mathfrak g)$ (look for "Verma modules"), and there is so such module in $Rep(\mathfrak g)$. Q2 : No. For example, the natural sequence $$ 0 \to M(-2) \to M(0) \to L(0) \to 0 $$ does not split (here $M(-2), M(0)$ are two Verma modules for $\...


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