I am working on the following assignment in Symplectic geometry: Let $G$ be a Lie group acting on a symplectic manifold $(M,\omega)$, and let $X_M=\dfrac{d}{dt}\bigg|_{t=0}\exp(tX)\cdot x$ for $X\in\mathfrak{g}$ be the fundamental vector field. Then the Association $\mathfrak{g}\ni X\mapsto X_M\in\text{SympVec}(M,\omega)$ is an anti-Lie algebra homomorphism. We have that $$ [X,Y]_M(x)=\frac{d}{dt}\bigg|_{t=0}\exp(t[X,Y])\cdot x=\frac{d}{dt}\bigg|_{t=0}\exp(t\cdot\frac{d}{ds}\bigg|_{s=0}\text{Ad}_{\exp(s X)}Y)\cdot x, $$ How is this equal to $-[X_M,Y_M](x)$?
122993
There are two claims in McDuff-Salamon's Introduction to Symplectic Topology, 3rd edition on p. 202 that I've been trying to figure out but haven't been able to. Let $G$ be a Lie group acting symplectically on symplectic manifold $(M,\omega)$; this means that we have a smooth group morphism $G \to \text{Symp}(M,\omega), g \mapsto \psi_g$. Here, $\psi_g$ is a symplectomorphism. Then, we get a Lie algebra morphism $\mathfrak{g} \to \mathcal{X}(M,\omega)$ on which assigns to each $\xi \in \mathfrak{g}$, a symplectic vector field $X_\xi$. The prominent property of this vector field is that the contraction $\iota_{X_\xi} \omega$ is a closed 1-form. More explicitly, we may define this $X_\xi$. $$X_\xi:= \left.\frac{d}{dt}\right|_{t=0} \psi_{\exp(t\xi)}.$$ The authors claim that it is a straightforward calculation to show the following for $\xi,\eta \in \mathfrak{g}, g \in G$:  Letting $g^{-1}\xi g:= \text{Ad}(g^{-1})\xi := \frac{d}{dt}|_{t=0} \;g^{-1}\exp(t\xi)g$, we have that $X_{g^{-1}\xi g} = \psi^*_g X_\xi$. $X_{[\xi,\eta]} = [X_\xi,X_\eta]$.  I do not have much fluency with Lie groups so I barely know where to start. Any help is appreciated.
105855
After a week in my Abstract Algebra class, the professor proposed this as a problem. I'm not entirely sure where to begin. $ V = \{ e, \tau, \tau_1, \tau_2 \}$, so I'm not sure exactly what is meant by $\text{Aut} (V)$. Is it simply saying that the only way to have an automorphism on $V$ is to rearrange the order of the elements of $V$ and therefore $\text{Aut} (V) \cong S_3$?
93857
The following is a question that comes from a counterexample to a problem about exterior algebras of modules and the notes mentions somehow this all relates to the fact that symplectic geometry is done over even dimension.    Let $F$ be a free $R$-module of rank $ n\geq 3$ where $R$ is a commutative ring with identity.  Let $T(F) = \oplus_{k=0}^{\infty} T^{k}(F)$ where $T^k(F) = F \otimes F \otimes \ldots \otimes F$ is tensor product of $k$ modules.  Let $\wedge F $ denote the exterior algebra of the $R$-module $F$, that is the quotient of the tensor algebra $T(F)$ by the ideal $A(F)$ generated by all $x \otimes x$ for $x \in F$.    Suppose $1 \leq k \leq n-1$.       How do we show $ \wedge ^k F \cong \wedge^{n-k} F$?
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?
107608
Let $D$ be the distribution on $M=\{(x,y,z),x,y,z&gt;0\}$ generated by $X=y\frac{\partial}{\partial z}-z\frac{\partial}{\partial y}$ and $Y=z\frac{\partial}{\partial x}-x\frac{\partial}{\partial z}$, show $D$ is involutive and find the general integral manifolds of $D$.  I compute the flows of $X$ and $Y$, there are a lot of $\sin$ and $\cos$ in the form of flows of $X$ and $Y$ , I got $\phi_t(x,y,z)=(x,-z\sin t+y\cos t,y\sin t+z\cos t)$ and $\psi_s(x,y,z)=(z\sin s+x\cos s,y,-x\sin s+z\cos s)$, but to find the integral manifolds at last, I can not eliminate $s$ and $t$ to get a general form
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?
122488
Suppose a torus $T=(\mathbb{C}^\ast)^n$ acts on a finite dimensional vector space $W$, and define for $m \in M$ ($M$ is the character lattice of $T$) the eigenspace $W_m$ by  $$W_m = \{w \in W \mid t\cdot w = \chi^m(t)w \text{ for all }t \in T \}$$ i.e. for $w \in W_m$ is a simultaneous eigenvector for all $t \in T$, with eigenvalue $\chi^m(t)$ depending on $t \in T$. Then it is a famous fact $$W=\underset{m \in M} \bigoplus W_m$$ Can someone provide a somewhat self-contained proof of this result? I don't know much about the theory of algebraic groups.
15573
For a free particle of mass $m$, with Hamiltonian  $$\hat{H} = \frac {\hat{P}^2} {2m},$$  where $$\hat{P} = -i \hbar \frac{\partial} {\partial x}.$$  The commutative relation is given by  $$[\hat{X}, \hat{H}] = \frac {i\hbar} {m} \hat{P}\tag{1}$$  In the common eigenstate of $\hat{H}$ and $\hat{P}$, $|e, p\rangle$, can we do the following?  $$\langle e, p| [\hat{X}, \,\hat{H}] |e, p\rangle = \langle e, p|\hat{X} (\hat{H}|e, p\rangle) -  (\langle e, p|\hat{H}) \hat{X}|e, p\rangle  \\ = \langle e, p|\hat{X} (e|e, p\rangle) -  (\langle e, p|e) \hat{X}|e, p\rangle \\ = e( \langle e, p|\hat{X}|e, p\rangle -  \langle e, p|\hat{X}|e, p\rangle ) \\ = 0 $$  Since the $\hat{H}$ is Hermitian, the above derivation doesn't seem to show any flaw. Given the commutative relation, Eq (1), we know the result is wrong. What's wrong with the above derivation?  [EDIT]  Following the comment by Luboš Motl, I have worked out the solution and would like to share it here. The link provided by Qmechanic had the solution closely related to this question.  $$ \langle e', p'| [\hat{X}, \,\hat{H}] |e, p\rangle \\ = \langle e', p'|\hat{X} (\hat{H}|e, p\rangle) -  (\langle e', p'|\hat{H}) \hat{X}|e, p\rangle  \\ = (e - e') \langle e', p'|\hat{X}|e, p\rangle $$  Note that:   $$ e - e' = \frac{p^2}{2m} - \frac{p'^2}{2m} = \frac{(p+p')(p-p')}{2m} $$  $$ \langle e', p'|\hat{X}|e, p\rangle = -i\hbar \delta'(p - p') $$  where $\delta'(\cdot)$ is the derivative of the Dirac function, with respect to $p$.  Then we get  $$ (e - e') \langle e', p'|\hat{X}|e, p\rangle \\ = -i\hbar \frac{(p+p')}{2m} \cdot (p - p')\delta'(p - p') \\ = - \frac{i\hbar (p+p')}{2m} \cdot (-\delta(p - p')) \\ = \frac{i\hbar (p+p')}{2m} \delta(p - p') $$  As we take the limit $p \rightarrow p'$:  $$ lim_{p \rightarrow p'}    \frac{i\hbar(p+p')}{2m} \delta(p - p') \\ \rightarrow \frac{i\hbar}{m} p \delta(p - p') $$
10343
I'm currently working at the start of Alexei Tsvelik's book Quantum Field Theory in Condensed Matter Physics. I'm kinda stumped on a few essential steps.  Starting with the action:  $$S = \int dt \int d^3 x \mathcal{L}_{\mathrm{EM}} = \int dt \int d^3 x \left( -\frac{1}{4} F_{\mu \nu} F^{\mu \nu} + j_\mu A^{\mu}\right).$$  In the absence of any sources $j_\mu = 0$ and $c=1$, which after pulling in the negative sign implies  $$S = \frac{1}{2}\int dt \int d^3 x  \,\,\left(E^2 - B^2\right)$$ The next step is to take $\vec{E} = -\vec{\nabla} \phi + \partial_t \vec{A}$ and $\vec{B} = \vec{\nabla} \times \vec{A}$. Letting $\phi = 0$ and directly inserting these into the above yields:  $$S = \frac{1}{2} \int dt \int d^3 x  \,\left( (\partial_t \vec{A})^2 - (\vec{\nabla} \times \vec{A})^2 \right) $$  Now considering $\vec{A} = \vec{A}_0 + \delta \vec{A}$ as some minimal change in our vector potential, with $\vec{A}_0$ gives a minimum for the action and putting this into the above expression we have: $$\delta S = \int dt \int d^3 x \left(\partial_0 \vec{A}_0 \partial_0 \delta \vec{A} + \left(\vec{\nabla} \times \vec{A}_0\right)\cdot\left(\vec{\nabla} \times \delta \vec{A} \right) \right)+ \mathcal{O}(A^2)\,\,\,\,\,\,\, (1)$$ where I have the term  $$\mathcal{O}(A^2) = \int dt \int d^3 x \left( (\partial_0 \vec{A}_0)^2 + (\partial_0 \delta \vec{A})^2 - (\vec{\nabla}\times\vec{A}_0)^2-(\vec{\nabla}\times \delta\vec{A})^2\right)$$  The next step is where I lose our author and go from "Oh ok I get this" to "completely clueless". He rewrites $\delta S$ as:  $$\delta S =  \int dt \int d^3 (\delta x) \vec{A} \mathcal{F(\vec{A}_0)}+ \mathcal{O}(A^2)$$  where $\mathcal{F} = \frac{\delta S}{\delta A}$. What is the purpose of $x\to \delta x$ in the above expression? Is the functional derivative $\mathcal{F}$ a vector quantity? He has it bolded with a bolded $\vec{A}$ in the above expression. One last question takes the above expression further. We assume $\delta \vec{A}$ vanishes at infinity, as it should, and integration by parts yields:  $$\delta S =  -\int dt \int d^3 x \left( \partial^2_0 \vec{A}_0 - (\vec{\nabla} \times \vec{\nabla})\times \vec{A}_0  \right)\delta \vec{A}\,\,\,\,\,\,\,\, (2)$$ How exactly did we go from (1) to (2)? Thanks I appreciate your time!
30824
The Jacobi identity of a Lie algebra says that $ad: \mathfrak g \to End(\mathfrak g)$ is a derivation. I am a bit emberassed but what is the easieast way to see that for every $X \in \mathfrak g$, $ad_X: \mathfrak g \to \mathfrak g, Y \mapsto [X,Y]$ is a Lie homomorphism, i.e., $[ad_X Y, ad_X Z] = ad_X([Y,Z])$ ?  Edit: The identity in the above line is wrong. It is $ad_X$ a derivation (endomorphism satisfying the product rule), not a Lie homomorphism (respecting the Lie bracket). I was a bit confused here, sorry.
28584
I am currently learning group theory and I learnt that the fundamental representation and the anti-fundamental representation of $\text{SL}(2,\mathbb{C})$, $2 \times 2$ matrix with determinant of $1$, are not equivalent.  This means that no similarity transformation can map one of them to the other.  My professor gave an explanation (on the 2nd last paragraph on page 75 of the following document ) but I don't see how the difference in the signs in the exponent imply that the representations are inequivalent.  Can anyone please explain the explanation of my professor, or perhaps give another explanation?
98987
We have some representation $\rho:L\longrightarrow End(V)$ of a $d$-dimensional complex Lie algebra $L$. Let $\{X_1,...,X_d\}$ be any basis of $L$, and let $\{\xi_1,...,\xi_d\}$ be another basis such that $\kappa_\rho(X_i,\xi_j):=tr(\rho(X_i)\circ\rho(\xi_j))=\delta_{ij}$ where $\kappa_\rho$ is the Killing form. Define the Casimir invariant by $$\Omega_{\rho}:=\sum\limits_{i=1}^{d}\rho(X_i)\circ\rho(\xi_i).$$ I want to prove that $\forall a\in L:$ $$[\Omega_{\rho},\rho(a)]=0.$$ By taking [] to be the commutator and thinking of $\circ$ as a multiplication (matrix multiplication? I think this is justified because a basis has been chosen for $L$), I've reached $$[\Omega_{\rho},\rho(a)]=\sum\limits_{i=1}^{d}\rho(X_i)\rho(\xi_i)\rho(a)-\rho(a)\rho(X_i)\rho(\xi_i)=...$$ $$...=\sum\limits_{i=1}^{d}\rho(X_i)[\rho(\xi_i),\rho(a)]- [\rho(a),\rho(X_i)]\rho(\xi_i)$$ but I don't know where to go from here. Any push in the right direction would be much appreciated.
26131
Let $(G,\cdot)$ be a group and  let $\mathfrak{S}(G)$ be the set of all bijective mappings from $G$ to $G$.   Show that: If $G$ is non-abelian, then $Inn(G):=\{\kappa_a  \vert a\in G\}$ is not a normal subgroup of $(\mathfrak{S}(G),\circ)$  First of all, this is not homework, but we had it in university as an exercise and didn't get a solution from our lecturer. So I am just looking for a correct proof, since we have easter holidays and it is not possible to ask him.  Thanks in advance!
3917
I'd like to know a reference for a simple proof that $\{c_\mu\mid \mu\vdash n\}$ is a basis for the centre of the symmetric group algebra $\mathbb{C}\mathfrak{S}_n$, where $c_\mu$ is the sum of all elements of a given cycle type (i.e. the sum of every element in the conjugacy class corresponding to $\mu$).
56066
I am reading Lee's book on Differential geometry. In Chapter 4, Lee has this exercise.  Let $M$, $N$ be smooth manifolds, and let $f:M\to N$ be a smooth map. Define $F : M\to M \times N$ by $F(x) = (x, f(x) )$. Show that for every smooth vector field $V$ of $M$ , there is a smooth vector field on $W$ on $M \times N$ that is $F$-related to $V$.  I can understand that we must have $W_{(x,f(x))}=V_x\oplus Df(x)V_p$ for all $x\in M.$ The set $\{(x,f(x)):x\in M\}$ is a closed set in $M\times N.$ Therefore, if we can show that for all $(p,f(p))\in M\times N$ there is a neighbourhood $U_p$ and a smooth vector field on $U_p$ extending $V_x\oplus Df(x)V_x$ we are done by partition of unity. But I cannot show that. Can someone help me out?
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?
114007
I have already know that at a point $p\in S^2$ with $u,v$ the tangent vectors at $p$. Then the standard symplectic form is $\omega_p(u,v) :=\langle p,\,u\times v\rangle$ where $u\times v$ is the outer product. My question is: How can one deduce from this that the symplectic form is $dz\wedge d\theta$ in the cylindrical polar coordinates $(z,\theta)$ on $S^2$? Here, $z$ is the height and $\theta$ is the angle. Thanks.