Consider a Lie group $G$, $$ L_x : G\rightarrow G, ~~~~~ L_x(y)=xy $$ is the translations from the left. A vector field $X$ on $G$ is left invariant if $$ dL_x (X)=X. $$ Let $x_t$ be the flow of $X$, then why we have $L_y\circ x_t = x_t \circ L_y$ ? There is a similar . But I can't understand it.  The answer hint   Let $\theta^{(p)}(t) = yx_t(p)$ and $\psi^{(p)}(t) = x_t(yp)$.  Compute both ${\theta^{(p)}}'(0)$ and ${\psi^{(p)}}'(0)$.  Then use uniqueness of integral curves.   But in my calculation, $$ {\theta^{(p)}}'(0) =\frac{d}{dt}|_{t=0} ~yx_t(p)=yX(p)  \\ {\psi^{(p)}}'(0)=  \frac{d}{dt}|_{t=0} x_t(yp) = X(yp) $$ Evenly, I don't know what is the $yX(p)$. Since $y$ is a element of $G$, and $X(p)$ is a  tangent vector at $p$,    how do they multiply ?
114103
Given $G$ a lie group and $X$ a left-invariant vector field. Let $\Phi_X^t$ be the local flow of $X$. Why can we conclude that $\Phi_X^t \circ L_x=L_x \circ \Phi_X^t$? Thanks!
107577
Question statement is as in the title:     $\varphi:G\to H$ is a Lie Group Homomorphism, then $\ker \varphi$ has Lie algebra $\ker D_e \varphi$   Since $\varphi$ is a Lie group homomorphism, in particular it is continuous. As $H$ must be Hausdorff, every point set is closed, and hence $\{e\}$ is closed in $H$, which implies that $\ker\varphi = \varphi^{-1}(e)$ is closed as well.  So $N=\ker\varphi$ is an embedded Lie subgroup of $G$. Therefore, the inclusion $j:N\to G$ is a Lie Group homomorphism, and induces a Lie algebra homomorphism $D_e j :T_e N \to T_e G$. So I only need to show that $D_e j$ is in fact an isomorphism between $T_e N$ and $\ker D_e \varphi$.  But I'm stuck at this point. I know that for $X\in T_e N$, we have, for $f\in C^{\infty}(H)$,  $$ D_e\varphi (D_e j(X))(f)=D_e j (X)(f\circ \varphi)=X(f\circ \varphi\circ j)=0$$  So $D_ej(T_e N) \subset \ker D_e \varphi$.  At this point, I can either show inverse inclusion: that is, for every $X\in \ker D_e \varphi$, we can find $Y\in T_e N$ such that $D_e j (Y)=X$, or for all $f\in C^{\infty}(G)$, $Y(f\circ j)=X(f)$  Or I can somehow show that dimensions of the two spaces are the same.   But I can't proceed from here.  It'd be great if someone could help me
77326
Let $ w \in \Bbb C $, and let $ \gamma, \delta : [0,1] \rightarrow \Bbb C $ be closed curves such that for all $ t \in [0,1], |\gamma(t) - \delta(t)| &lt; |\gamma(t) - w| $. By computing the winding number $n_\sigma(0)$ about the origin for the closed curve $\sigma(t) = (\delta(t) - w)/(\gamma(t) - w) $, show that $n_\gamma(w) = n_\delta(w)$.  This seems intuitively clear to me. (Informally, the inequality tells us that $\delta$ and $\gamma$ can never be on "opposite sides" of $w$, and hence their winding numbers must be equal. Making this rigorous isn't enough, however, since I need to use the winding number of $\sigma$ about 0.  I can also show that $n_\sigma(0) = n_\gamma(w) - n_\delta(w)$ (fairly elementarily), so it remains to show that $n_\sigma(0) = 0$. This is the bit that I'm stuck on.  Can someone give me a (small) hint (not a major hint)? If I still can't get anywhere, then I'll probably ask for a larger hint!  There is also a second part to this question - dependent on whether I'm able to do the first part with a hint, I may well update this question to include the second part.  (This is an example sheet question - completely non-examinable.)  Thanks! :)
105655
I saw  and it states that the Lie algebra of the special linear group $SL_n(\mathbb C)$ is the group of traceless matrices $\mathfrak{sl}_n$.  I know the definition of a Lie algebra: given a matrix group $G \subseteq GL_n(\mathbb K)$ its Lie algebra is defined to be the tangent space at $I$.  From this definition it is immediately clear what dimension the Lie algebra has to have.  What is unclear to me is:      Assuming I don't already know the answer, how do I arrive at the insight that this tangent space is traceless matrices?   I skipped through Tapp's book in search of hints and found that the exponential map is a homomorphism from the Lie algebra into the Lie group. I also found a lemma that states that   $$ \det e^A = e^{\mathrm{trace}(A)}$$  That is unfortunately only half-useful: If I already know what the Lie algebra is then this lets me prove that the image of the Lie algebra under the homomorpism defined by the exponential map is indeed in $SL_n$.   So it's kind of not so useful. Surely there must be geometric insight?
74825
Symmetry of second derivative states that for $u=u(x,y)$ if $u_x,u_y$ exists and $u_{xy},u_{yx}$ exists and continuous then $u_{xy}=u_{yx}$.  I proved that statement using the mean value theorem.   While I was looking in Wikipedia there is a section called "Sufficiency of twice-differentiability", if $u(x,y):E^{\text{open set}}\subset \Bbb R^2\to \Bbb R$ and $u_x,u_y,u_{yx}$ exists everywhere and $u_{yx}$ is continuous at a point in $E$ then $u_{xy}$ exists at that point and equal to $u_{yx}$.  My question is, while I proved Symmetry of second derivative I had to assume the continuity of $u_{xy}$ and $u_{yx}$, so how can I prove that this is true without even assuming the existence of one of the second derivative? I'm sitting on this for a long time and I couldn't think on any starting point and would love help with this.
5062
Let $M$ be a smooth ($C^\infty$) manifold. Let $\mathfrak{X}(M)$ be a set of all vector fields on $M$ and let $\mathfrak{F}(M)$ be a set of all real smooth functions on $M$. $\mathfrak{X}(M)$ is a real vector space and it is also a module over $\mathfrak{F}(M)$. We know that partial derivatives constitute a basis for tangent space at any point $p\in M$. Is there some sort of basis for $\mathfrak{X}(M)$ (as a vector space or as a module)? Do partial derivatives constitute a basis here as well? I think this question  is similar to mine, but because of the way it's written, I'm not really sure I understand the question and I'm not sure we're asking about the same thing.
79478
For a vector field $X$, the divergence in coordinates is given by $\nabla\cdot X=\sum_n\frac{X^i}{\partial x^i}$. In polar coordinates, the metric is $\begin{bmatrix}1 &amp; 0\\ 0 &amp; r^2\end{bmatrix}$, and so $\frac{1}{\sqrt{g(\frac{\partial}{\partial r},\frac{\partial}{\partial r})}}\frac{\partial}{\partial r}=\frac{\partial}{\partial r}$ and $\frac{1}{\sqrt{g(\frac{\partial}{\partial\theta},\frac{\partial}{\partial\theta})}}\frac{\partial}{\partial\theta}=\frac{1}{r}\frac{\partial}{\partial\theta}$ are unit vectors. Then for $X=X_{r}\frac{\partial}{\partial r}+X_{\theta}\frac{\partial}{r\partial\theta}$,  $\nabla\cdot X=\frac{\partial X_r}{\partial r}+\frac{\partial}{\partial\theta}\frac{X_{\theta}}{r}=\frac{\partial X_r}{\partial r}+\frac{1}{r}\frac{\partial X_{\theta}}{\partial\theta}$. But this disagrees with the usual formula given in vector calculus books. Does anyone see the error?
47843
I've been doing some more exercises in , in particular the following:  Show that $H_0(X,A) = 0$ iff $A$ meets each path-component of $X$.  "$\Leftarrow$": Let $x_i \in A \cap X_i \neq \emptyset \forall $ path-components $X_i$. Then for any $x \in X_i$ there is a path $\gamma$ from $x_i$ to $x$. They therefore differ by a boundary: $\partial \gamma = x - x_i$ and therefore $\{&#160;x \} = \{ x_i \}$ in $H_0(X,A) $. Therefore $H_0(X,A) = 0$.  "$\Rightarrow$" Now for this direction I'm not so sure and I'd be glad if you could give me a hint. Let $H_0(X,A) = 0$. By definition, the relative homology group is calculated from the following sequence of chain groups:  $$ 0 \rightarrow C_0(A) \rightarrow C_0(X) \xrightarrow{\partial_1} C_0(X) / C_0(A) \xrightarrow{\partial_0} 0$$  Then $H_0(X, A) = H_0(C_0(X) / C_0(A)) = ker \partial_0 / im \partial_0 = (C_0(X) / C_0(A) ) / im \partial_1 = 0$.  There are two cases where this equality holds:  (i) $C_0(X) / C_0(A) = 0$  (ii) $im \partial_1 = C_0(X) / C_0(A)$  In case (i), $A \cap X_i \neq \emptyset$ for all $X_i$.  In case (ii) I'm not sure how to proceed. Can you give me a hint? Many thanks for your help!  Edit Or maybe I could do the second direction like this:  $$0 = H_0(X,A) = \oplus_i H_0(X_i, A) \implies  H_0(X_i, A) = 0 \forall i$$  $$ \implies A \cap X_i \neq \emptyset \forall i$$?
49140
See the related question . This is the second part of question 4-C in Milnor and Stasheff's book on characteristic classes. In the solution to the first part, we rely on the fact that having a field of tangent $1$-planes allows us to split the tangent bundle into a direct sum. It doesn't seem like this should be true for a field of tangent $2$-planes (otherwise the question wouldn't just be asking about $P^4$ and $P^6$), so I'm not sure exactly where to start.  edit: actually, since $P^n$ can be given a Riemannian metric, any subbundle should be a Whitney summand; this doesn't answer the question as to why they only ask for $P^4$ and $P^6$ though, so I'm still confused.
105623
I have homework in a course on Lie groups, in which I must show that $(\pi_m,\mathbb C_m[x,y])$ are the only irreducible representations of finite dimension in SL(2,ℝ). Here $ C_m[x,y])$ is the vector space of homogeneous polynomials, and $\pi_m(g)$ acts via $\pi_m(g)(f(x,y))=f((x,y)\cdot g)$. It was not too complicated to calculate the characters, taking the basis $\{p_k:0\leq k \leq n\}$ with $p_k = x^ky^{n−k}$, since $\pi_m(exp(H))p_k=exp(d\pi_m(H))p_k=exp(2k−n)p_k$, but the question is how one could define an inner product for these matrices. But, I don't have the volume form of SL(2,ℝ) to define the integral $\langle f,g \rangle=\int_Gf(z)⋅\overline{g(z)}dz$
68085
Let $G$ be a Lie group acting smoothly on a manifold $M$. Then, for each $p \in M$, it is obvious that the isotropy group $G_p = \{ g \in G : gp=p\}$ is a closed subgroup of $G$. A standard theorem () about Lie groups tells you that closed subgroups are (embedded) submanifolds, so this is true in particular of the isotropy group $G_p$. I must confess, however, that for me this theorem is really more of a "fact". The proof looks a little complicated, and I have never gone through it. So, I would like to know     Question: Is there a way to see that $G_p$ is a submanifold more directly, i.e. can this be shown with less work than it takes to prove the more general statement about closed subgroups?
21045
I try currently to revise my classic. Could you help me please to demonstrate : $$\frac{\partial T^{\prime\mu}}{\partial x^{\prime \alpha}}=\frac{\partial}{\partial x^{\prime \alpha}}\left(\frac{\partial x^{\prime \mu}}{\partial x^{v}} T^{v}\right)$$ i.e by using a changing of coordinates between $x^{\mu}$ and $x^{\prime\mu}$. $T^{\mu}$ represents the contravariant components of a &quot;vector&quot; (I mean a tensor of rank 1). This demostration would be the same than proving : $$T'^{\mu}\text{d}x^{\nu}=T^{\nu}\text{d}x^{\prime\mu}$$ How to justify this equality if I apply a 1-form $\text{d}x^{\mu}$ or another 1-form $\text{d}x^{\prime\mu}$ on a vector $T$ ?
24621
I saw this Lagrangian in notes I have printed:  $$ L\left(x,\frac{dx}{dt}\right) = \frac{m^2}{12}\left(\frac{dx}{dt}\right)^4 + m\left(\frac{dx}{dt}\right)^2\times V(x) -V^2(x). $$ (It appears in the exercises in the first chapter of Goldstein.)  What is it? Is it even physical? It seems like it doesn't have the right units of energy.
93720
I have a problem deriving the conservation of energy from time translation invariance. The invariance of the Lagrangian under infinitesimal time displacements $t \rightarrow t' = t + \epsilon$ can be written as     \begin{equation} \delta L = L\left( q(t),\frac{dq(t)}{dt},t\right) -  L\left( q(t+ \epsilon),\frac{dq(t+ \epsilon)}{dt},t+\epsilon \right) = 0. \end{equation} Using Taylor series, keeping only first order terms this gives \begin{equation}\rightarrow \delta L =- \frac{\partial L }{\partial q} \frac{\partial q}{\partial t} \epsilon- \frac{\partial L }{\partial \dot{q}} \frac{\partial \dot{q}}{\partial t} \epsilon -  \frac{\partial L }{\partial t}  \epsilon = 0. \end{equation} Using the Euler-Lagrange equation and assuming that the Lagrangian does not depend explicitly on time we get \begin{equation}\rightarrow \delta L =- \frac{d}{d t} \left(\frac{\partial L(q,\dot{q},t)}{\partial \dot{q}} \right) \frac{\partial q}{\partial t} \epsilon- \frac{\partial L }{\partial \dot{q}} \frac{\partial \dot{q}}{\partial t} \epsilon  =0. \end{equation} Which we can write as  \begin{equation}\rightarrow \delta L  = - \frac{d}{d t} \left(\frac{\partial L(q,\dot{q},t)}{\partial \dot{q}}  \frac{\partial q}{\partial t} \right) \epsilon  = - \frac{d}{d t} \left(p  \frac{\partial q}{\partial t} \right) \epsilon = 0. \end{equation} But unfortunatly this  is not the Hamiltonian. This computation should yield  \begin{equation}  \rightarrow \frac{d}{dt} \left(    p \dot{q}  - L \right) = 0. \end{equation} But I can't find no reason why and how the the extra $-L$ should emerge. I can see that this term can be written at the place where it is written because we have $\delta L = - \frac{d L}{dt } \epsilon$ and therefore  \begin{equation} \rightarrow \delta L  = - \frac{d}{d t} \left(\frac{\partial L(q,\dot{q},t)}{\partial \dot{q}}  \frac{\partial q}{\partial t} \right) \epsilon  =  -  \frac{d L}{dt }  \epsilon. \end{equation} And then the desired equation would only say $0-0=0$. Any idea where i did a mistake would be much appreciated.
115656
Let $G$ be a Lie group and $i:G \rightarrow G$ denote the inversion map $i(x) = x^{-1}$. (Notation: $f_*$ is the pushforward map $F_*:T_pG \rightarrow T_{i(p)}G$ which takes $(F_{*}X)(f)=X(f\circ F)$ and $X$ is a tangent vector, $X\in T_pG$.) I wish to show that $i_{*}:T_{e}G\rightarrow T_{e}G$  is given by $i_{*}(X)=-X$ As a first step, it is trivial to prove that $i_*$ is an involution as $\mbox{Id}_{*}=(i\circ i)_{*}=i_{*}\circ i_{*}$ but I can't seem to make any further progress. Any help would be appreciated.
94111
Let $r(t)=(x(t),y(t),z(t)),t\geq0$ be a parametric curve with $r(0)$ lies on cylinder surface $x^2+2y^2=C$. Let the tangent vector of $r$ is $r'(t)=\left( 2y(t)(z(t)-1), -x(t)(z(t)-1),  x(t)y(t)\right)$. Would you help me to show that :  (a) The curve always lies on ylinder surface $x^2+2y^2=C$.  (b) The curve $r(t)$ is periodic (we can find $T_0\neq0$ such that $r(T_0)=r(0)$).If we make the C smaller then the parametric curve $r(t)$ more closer to the origin (We can make a Neighboorhood that contain this parametric curve)  My effort:  (a) Let $V(x,y,z)=x^2+2y^2$. If  $V(x,y,z)=C$ then $\frac{d}{dt} V(x,y,z)=0$. Since $\frac{d}{dt} V(x,y,z)=(2x,4y,0)\cdot (x'(t),y'(t),z'(t))=2x(2y(z-1))+4y(-x(z-1))=0$, then $r(0)$ would be parpendicular with normal of cylinder surface. Hence the tangent vector must be on the tangent plane of cylinder. So $r(t)$ must lie on cylinder surface.  (b) From $z'=xy$, I analyze the sign of $z'$ (in 1st quadrant z'>0 so the z component of $r(t)$ increasing and etc.) and conclude that if $r(t)$ never goes unbounded when move to another octan ( But I can't guarante $r(t)$ accros another octan.). I also consider the case when $(x=0, y&gt;0), (x=0, y&lt;0,z&gt;1), (x&gt;0, y=0,z&gt;1$and so on)  and draw the vector $r'(t)$.  Thank you so much of your help.
61307
Since $T_p G$ is isomorphic to $T_e G$ for all $p\in G$, it makes sense that each vector in $T_p G$ can be identified with a vector in $T_e G$. Hence, to make the map from $TG$ one to one, we must identify each vector in $T_e G$ with a point in $G$, but I'm having trouble saying this precisely.  Thanks so much!