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6

This identification is called Hermite reciprocity, and I find it pretty mysterious. It's a special case of a more general problem called plethysm, which I find extremely mysterious. Here is a proof. If $V$ denotes the defining $2$-dimensional representation of $SU(2)$, let me write the desired isomorphism as $$S^n S^m V \cong S^m S^n V.$$ To show that ...


5

Morally speaking, the Lie algebra of vector fields is the Lie algebra of $\text{Diff}(M)$, the diffeomorphism group of $M$. The relationship between these is less tight than in the finite-dimensional case: for example, The exponential map can fail to be defined at any nonzero time (as mentioned by orangeskid in the comments), and Even when defined (say on ...


3

Here is a rather cheap answer to this question. Let $E_\lambda$ denote the eigenspace of $\lambda \in \mathrm{spec}(A)$. Let $F_\lambda$ denote the eigenspace of $\lambda \in \mathrm{spec}(B)$. Then $\exp(iA) =\exp(iB)$ if and only if, for every $\lambda_0 \in \mathbb{R}$, we have $$\bigoplus_{\lambda \in \mathrm{spec}(A) \cap (\lambda_0 + 2 \pi ...


2

The category $\mathcal{O}$ is closed under quotienting, submodules and finite direct sums, but not under extensions. Hence $Y$ need not be in $\mathcal{O}$ in general (see here).


2

Given $t\in[0,1]$ there exists an open nhbhd $U_t$ of $t$ in $[0,1]$ such that for every $s\in U_t$ $A_s^{-1}A_t\in V$ (since $A_t^{-1}A_t=I\in V$ and $V$ is open). The collection of $U_t$'s give you an open cover of $[0,1]$. The "standard compactness argument" referred above is that since $[0,1]$ is compact you can select a finite subcover of $[0,1]$, ...


2

Let us set $t=1$. There is a nice formal calculation of $J(1)$ which makes perfect sense if $\mathfrak g$ is nilpotent and thus $G$ sits in the enveloping algebra $U\mathfrak g$ (otherwise take it as a formal calculation that has to be made rigorous): $$J(1)=\frac{d}{d\epsilon}\exp(V+\epsilon ...


1

Just a partial answer. It is know that the preimage $ϕ^{-1}(v)$ of a regular value v is a submanifold of V. What is known about the preimage of a singular value? Is this also a manifold in this case? There is Transversality theorem, which is generalisation of known fact about regular values that you mentioned. Narasimhan in his book "Analysis on real ...


1

A free algebra $A$ on a set $S$ of generators is one with no relations among the generators. Informally, this means that the only expression of the form $\sum_{n \in \mathbb Z_{\ge 0}^S} a_n S^n = 0$ is one where all coefficients $a_n$ are $0$. Formally, it means that any set map $S \to B$, where $B$ is an algebra, extends to an algebra map $A \to B$. ...


1

The fact that you're looking at a homogeneous space is irrelevant -- what's relevant is that $M$ is compact (since you said it's homeomorphic to $\mathbb S^p\times \mathbb S^q/\mathbb Z_2$). On a compact manifold, every vector field is complete, meaning its flow exists for all time.


1

For non-commuting matrices, $e^{A}·e^B\ne e^{A+B}$. If one uses an approximation with infinitesimal time steps $dt$, then $$ γ(t+dt)\simeq γ(t)(1+β(t)·dt)\simeq γ(t)·e^{β(t)·dt} $$ so that $$ γ(t+N·dt)=γ(t)·e^{β(t)·dt}·e^{β(t+dt)·dt}·…·e^{β(t+(N-1)·dt)·dt} $$ However, since the matrices (if in a matrix representation) $β(t+k·dt)$ do not commute, you can not ...


1

Following the definition/notation here, we need to compute the matrices for $\DeclareMathOperator{\ad}{ad}$ $\DeclareMathOperator{\tr}{trace}$ $\ad(u),\ad(v),\ad(w)$ with respect to $B$. In particular, we have $$ \ad(u) = \pmatrix{ 0&0&0\\ 0&0&-1\\ 0&1&0}\\ \ad(v) = \pmatrix{ 0&0&1\\ 0&0&0\\ -1&0&0}\\ ...


1

The point is that "evaluating in $Y$" is (the restriction of) a linear map, so its derivative coincides with the map itself. This is not specific to a Lie algebra being involve. If $V$ is a vector space, $v\in V$ an element and $c:I\to GL(V)$ is a smooth curve, then the derivative of the curve $I\to V$ defined by $t\mapsto c(t)(v)$ is just $(c'(t))(v)$.


1

Here is the easiest example in terms of dimension. The $2$-dimensional Lie algebra $L=\mathfrak{r}_2(K)$ over a field $K$ is defined by the basis $(x,y)$ with Lie bracket $[x,y]=y$. It is clear that $L$ is solvable because $$ [[L,L],[L,L]]=[Ky,Ky]=0. $$ On the other hand, $L$ is not nilpotent, because it has no center.


1

The decomposition with the stronger condition is called effective. It appears in connection with pseudo-Riemannian symmetric spaces, where one deals with a special case of the symmetric decomposition $\mathfrak{g}=\mathfrak{h}\oplus \mathfrak{m}$, namely that the decomposition is effective, i.e., that it is minimal, which precisely means that ...



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