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5

So let's do $f_2$. What you need to do is to first work out the derivative in $M_n$, and then project onto $SU(n)$. That is, we identify $\mathfrak{su}(n)$ with its dual via the inner product $\langle U,V \rangle = \text{Re}\bigl(\text{Tr}(U^\dagger V)\bigr)$. (You need to do this identification if you want to figure out gradient flow, because $U^{-1} ...


4

The problem is with #2. The rank of a Lie algebra is the dimension of any of its Cartan subalgebras (all of which are isomorphic). None of them need to contain diagonal matrices. You can see one such subalgebra computed for $SO(2n + 1, \mathbb{C})$ in this paper: Structure Theory of Semisimple Lie Groups - Knapp


3

The problem is that the property "diagonal matrices", or "antisymmetric matrices" depends on the basis of the underlying vector space. The standard way to represent the Lie algebra $\mathfrak{so}(n)$ faithfully by matrices, is by antisymmetric matrices. However, we may also represent $\mathfrak{so}(n)$ faithfully by matrices which are not skew-symmetric. ...


3

I. I think you are trying to show that $U$ is also a weight $L$-module. However, I'm not sure if you understand how "$h$ acts diagonally on $U$" would imply $v_{k_i}\in U$. Do you know how to prove this? (I can give you hints, but I'm not sure you need them.) You are definitely right that $U$ is the direct sum of $\mathbb{F}v_j$ whenever $v_j$ occurs as ...


3

Let $G=G_2\times G_2$ with Lie algebras $\mathfrak{g}=\mathfrak{g}_1\times \mathfrak{g}_2$. Then we have, for $x_i\in \mathfrak{g}_i$ $$ \exp_G(x_1,x_2)=(\exp_{G_1}(x_1),\exp_{G_1}(x_2)) $$ as expected, see here.


3

It follows from a much more general fact: Whenever a Lie group $G$ acts smoothly on a smooth manifold $M$, its orbits are immersed smooth manifolds. This can be derived from the following observations: For each $p\in M$, the isotropy group $G_p = \{g\in G: g\centerdot p=p\}$ is a closed subgroup of $G$. The quotient space $G/G_p$ has a unique smooth ...


3

We have $0\neq [H_i,L_\alpha] = \{\alpha(h)x\,\big|\, h\in H_i, x\in L_\alpha\}$. In particular there exists $h\in H_i$ such that $\alpha(h) \neq 0$. But this means $[H_i,L_\alpha]= L_\alpha$. Since $L_i\subseteq L$ is a simple ideal, we also have $[L_i,L_\alpha]\subseteq L_i$. Putting things together, we arrive at $$L_\alpha = [H_i,L_\alpha] \subseteq ...


3

Since $\| U^\dagger V - V^\dagger U \| \le 2$ (where $\|.\|$ is the operator norm, say), the answer is: no, not always.


3

Since $\beta$ is a positive root, it is a positive linear combination of some of the $\alpha$'s. Since $\beta \ne \alpha_i$, then $\beta$ has some component $\alpha_j$ with positive coefficient where $j \ne i$ (since higher multiples of $\alpha_i$ cannot be roots). Therefore each element of $\beta + \mathbb{Z}\alpha_i$ has a component $\alpha_j$ with ...


3

You are almost there: The group $Aut(\mathfrak{g})$ of automorphisms of a Lie algebra $\mathfrak{g}$ is closed in the group $End(\mathfrak{g})^×$ of vector space automorphisms, hence it is a Lie group. No, only one direction is true in general: If $H$ is a Lie subgroup of $G$, then $\mathfrak{h}\simeq T_eH\subset T_eG\simeq \mathfrak{g}$ is a Lie ...


2

Suppose that $K$ is the base field. Fix $h\in\mathfrak{h}$. Let $h_i\in\mathfrak{g}_i$ be such that $h=\sum_i\,h_i$ (the $h_i$'s are unique). We claim that $h_i\in\mathfrak{h}$ for all $i$. Let $t\in\mathfrak{h}$. Write $t=\sum_{i}\,t_i$ with $t_i\in\mathfrak{g}_i$. For $i\neq j$, $\left[t_i,h_j\right]\in\mathfrak{g}_i\cap\mathfrak{g}_j=\{0\}$, so ...


2

The linear map $ad(x)\colon y\mapsto [x,y]$ has a matrix associated, with respect to the basis of the Lie algebra, as any linear map has a matrix associated with respect to the basis of the vector space. So we can multiply the two matrices $ad(x)$ and $ad(y)$ and take its trace. For example, let $L=\mathfrak{sl}_2(K)$ with vector space basis $(e,f,h)$ and ...


2

The answer is affirmative. By a change of basis, we may assume that $\frac1{\|A\|}A=\operatorname{diag}(iy_1,\, iy_2,\, \ldots,\, iy_n)$, where each $y_j$ is a real number between $-1$ and $1$. Therefore each $y_j=2i\sin\theta_j$ for some real $\theta_j$. Let $W=\operatorname{diag}\left(e^{i\theta_1},\ldots,e^{i\theta_n}\right)$. Then ...


2

For every field $K$ there are just two different $2$-dimensional Lie algebras. The first one is $K^2$, which is abelian (which means the bracket is zero). The second one is solvable, non-nilpotent, hence non-abelian, and the bracket of the basis $(x,y)$ is an arbitrary nontrivial linear combination of $x$ and $y$, i.e., $[x,y]=\alpha x+\beta y$ for some ...


2

For putting a manifold structure on the quotient, it's irrelevant whether $G_x$ is normal. The quotient of a Lie group modulo a closed subgroup always has a unique smooth manifold structure such that the group action is smooth -- this is the Quotient Manifold Theorem (see Theorem 21.10 in my Introduction to Smooth Manifolds, 2nd ed.). Normality of the ...


2

Hint The elements of the Lie algebra $\mathfrak{g} \cong T_{\Bbb I} G$ are the tangent vectors at the identity element $\Bbb I \in G$ to curves in $G$ through that point.


2

I think the question is probably too broad to have a very satisfying general answer. But (assuming you're looking for a smooth Lie group action) here are a couple of necessary, but certainly not sufficient, conditions: Each level set of $f$ must be a smooth submanifold, because every orbit of a smooth Lie group action is an (immersed) submanifold. Each ...


2

What if the action of $G$ and $K$ are both trivial? E.g. $k\cdot x=x=x\cdot g$ for all $k$ and $g$? Then your expression $\mathrm{exp}(tX)\,x\, \mathrm{exp}(tA)$ is identically $x$, and so has trivial derivative. Hence you can't hit $\phi'(0)$ in general.


2

The mistake is that the definition you give only works for $x\in\mathfrak{g}$. A general element is not of this form. The coproduct will be the unital algebra homomorphism having this form for each element of the Lie algebra, which is the generating set, and the coproduct of a general element is a sum of products of these.


1

Dietrich worded his answer in a way that I think is excessively confusing. What I think he means is that "diagonal" and "antisymmetric" are words that depend on a choice of basis of $\mathbb{R}^n$, not of the Lie algebra. You can get "diagonal" to mean something different by picking a basis that isn't just a differently-scaled version of the usual basis, and ...


1

Yes, it is always possible. In fact, you can always take $A=0\in\mathfrak g$. Fix $x\in M$, and consider the smooth map $F\colon K\to M$ given by $F(k) = kx$. Because $K$ acts transitively, $F$ is surjective. Moreover, $F$ is equivariant with respect to the (transitive) left actions of $K$ on $K$ and $M$: For all $k,k'\in K$, $$ k'F(k) = k'(kx) = (k'k)x = ...


1

I have no idea what you mean "without using coordinates". The best you can hope to do without a lot of work is to determine the conjugacy class of the stabiliser. For that, all you need to know is the spectrum. Clearly, if $b \in \mathfrak{su}(n)$, then $ib$ is traceless and hermitian. It can therefore be diagonalised via (special) unitary ...


1

$V$ being a faithful ${\mathfrak g}$-module means that the structure homomorphism $\rho: {\mathfrak g}\to{\mathfrak g}{\mathfrak l}(V)$ is injective. By definition, $\rho$ is a homomorphism of Lie algebras, but we may also view it as a morphism of ${\mathfrak g}$-modules if we equip ${\mathfrak g}$ with the adjoint action and ${\mathfrak g}{\mathfrak l}(V)$ ...


1

Try looking at Vilenkin's introductory 1968 book on Representation Theory and Special Functions. Then look at his 3 volumes on the same subject with a more detailed treatment.


1

If $G = \mathbb R$, then $\mathfrak g = \mathbb R$ and $\exp\colon \mathfrak g = \mathbb R \rightarrow \mathbb R = G$ is actually the identity map. Indeed, since $L_g(v) = g + v$ for $v,g\in G$, we have $dL_g(X) = X$ for $g\in G$, $X\in \mathfrak g$ and hence $\tilde X(g) \equiv \tilde X(0)$ for all $g\in G$ and all left-invariant vector fields $\tilde X\in ...


1

The coefficent is inserted at the position of the deleted simple root. An explicit example can be found at page 73 here. In addition at page 147 here Cahn writes, "... and use it in place of one of the old coefficents".


1

The continuity/smoothness assumptions to require from infinite dimensional representations are a rather subtle issue. You can see that smoothness fails for very simple and natural examples. For example, take a compact Lie group $G$ and the Banach space $V:=C(G,\mathbb C)$ of complex valued continuous functions on $G$. The left regular representation of $G$ ...


1

"Why would one define $B(X,Y)=tr(ad(X)ad(Y))$ ?" You are right, for matrix algebras I would define a bilinear form more simply, namely just by $C(X,Y)=tr(X)tr(Y)$. This is very natural, because a trace form for linear operators is the easiest thing you can imagine. If we do not have linear operators $X,Y$, then we can enforce this, by using the adjoint ...


1

The following is a faithful complex 3-dimensional representation: $h\mapsto iE_{22}$, $p\mapsto E_{12}+iE_{23}$, $q\mapsto -iE_{12}-E_{23}$, $z\mapsto -2E_{13}$ ($E_{jk}$ being the matrix with only nonzero entry $(j,k)$, equal to 1). Replacing each complex entry $a+ib$ with a matrix block $\begin{pmatrix}a & -b\\ b & a\end{pmatrix}$ yields a faithful ...


1

Since the oscillator algebra is an extension of the Heisenberg algebra, we can use the standard matrix representation of the Heisenberg algebra to find a faithful $4$-dimensional matrix representation for the oscillator algebra. A faithful linear representation of dimension $4$ is constructed in the paper Minimal Matrix Representations of Four-Dimensional ...



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