I am having a hard time wrapping the exponential map around my head. So if $G$ is a Lie group we have this map : $exp : T_eG \rightarrow G $ My understanding of lie groups is that they represent stuff like rotations. So if I take $G=O(n)$ the rotations, then I know that $T_{I_n}O(n) = \text{skew symmetric matrices}$. So I take one skew symmetric matrix $S$ which as I understand it somehow represents an infinitesimal rotation. Then $exp(tS)$ is a curve for which the elements are rotations of $O(n)$. But what do these element mean? How to think intuitively about what this exponential map actually is?  Edit :  Sorry, but I don't see how the linked answers tell me how to think about the curve $\gamma(t)=exp(tX)$, and neither did OP since apparently they were not accepted...
120602
I'm trying to understand an algorithm that tries to map points from a lie group to its lie algebra using the exponential map. The background is the representation of 3d coordinate transformations as a lie group.  I can imagine, how one can map points on a manifold (eg a sphere) to the tangent space (tangent plane) at some point x on the sphere. I can imagine to create the tangent space by computing derivatives at that point x. But I cannot imagine how the exponential of some matrix is giving me that tangent space?  How can I envision the exponential of a matrix? Is there some picture or analogy I can use as intuition?
75557
The group of orthogonal transformations of $\mathbb{R}^3$ is direct product of the group of rotations and the group $Z=\langle x\mapsto -x\rangle\cong \mathbb{Z}/2$. The finite subgroups of group of rotations are well known: rotational symmetries of regular polyhedrons.   These subgroups, along with their direct product with $Z$ will give some (families) of finite groups of orthogonal transformations.  It looks that some finite groups of orthogonal transformations need not be of this form (see Shafarevich- Algebra I; I couldn't understand arguments there).  How do we cover all finite subgroups of orthogonal transformations?  A more general question in this context may be: How to get subgroups of $H\times K$ knowing subgroups of $H$ and $K$?
40249
For a lie algebra $\mathbb{g}  $ we can define the adjoint representation as:  $ ad: \mathbb{g} \rightarrow End(\mathbb{g}) $ as the map such that $ad_x(y)=[x, y] $ for all $\in \mathbb{g} $  I am wondering how to we compute adjoint matrices with respect to a given basis?     For example, let the lie algebra be $ \mathbb{g}=sl_2(\mathbb{C}) $ with the basis as follows: $X=\begin{bmatrix} 0&amp;1\\0&amp; 0 \end{bmatrix}$,$Y=\begin{bmatrix} 0&amp;0\\1&amp; 0 \end{bmatrix}$ and $H=\begin{bmatrix} 1&amp;0\\0&amp; -1 \end{bmatrix}$  Let $ ad: \mathbb{g} \rightarrow End(\mathbb{g})  $ be the adjoint representation  How would I compute the matrices $ ad_X $, $ ad_Y $ and $ ad_Z$ relative to the above matrices?
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?
61327
Let $\rho: G \to GL_n(\mathbb{C})$ be an irreducible representation, and $g\in Z(G)$. Show $\rho(g)$ is a scalar multiple of the identity matrix $I$.   I think I have it, here is my solution:  Since $\rho(g) \in Hom_G(\mathbb{C}, \mathbb{C})$ and $\rho$ is irreducible, consider a nonzero eigenvalue of $\rho(g)$, say $\lambda$, we have $\rho(g) -\lambda $ is a zero map by Schur's lemma, as the map contains a non-trivial kernel (i.e. the eigenvectors associated to $\lambda$ are in the kernel).  But I didn't use the condition that  $g\in Z(G)$, so are there something wrong with my solution?
40183
If a matrix is symmetric or skew-symmetric it commutes in the obvious way with its transpose. (For symmetric: $SS^T=S^2$ and $S^TS =S^2$)  The less obvious is the case of commutativity  for orthogonal matrix but such matrix also commutes with its transpose because $RR^T=RR^{-1}= I= R^{-1}R=R^TR$.  Questions:   Are there  other cases when a matrix commutes with its transpose ?    What is the general property of such matrix which allows it to commute with its transpose ?
25520
We represent an element $(x_1,y_1,...,x_n,y_n,x_{n+1},y_{n+1})\in S^{2n+1}$ as an element $(z_1,...,z_{n+1})\in \mathbb C^{n+1}$ where $z_k = x_k+iy_k$. Now I'm considering the action of $S^1$ on $S^{2n+1}$ as $e^{iθ}·(z_1,...,z_{n+1}) = (e^{iθ}z_1,...,e^{iθ}z_{n+1})$. Under this action we identify any two elements as $(z_1,...,z_{n+1}) ∼ e^{iθ} ·(z_1,...,z_{n+1})\forall (z_1,...,z_{n+1})\in S^{2n+1}\subseteq \mathbb C^{n+1}$ and $e^{iθ}\in S^1$. For the resulting quotient space $S^{2n+1}/S^1$, I'm trying to prove that $S^{2n+1}/S^1$ is homeomorphic to $\mathbb CP^n$. How do I prove this? Any solution will be highly appreciated.
97562
Studying for an exam, a review question...  Given $G=\langle x,y|x^4=y^4=e,xyxy^{-1}=e\rangle$.    Show $|G|\leq16$.   For this, I want to consider that $x^3=x^{-1}$ and $y^3=y^{-1}$ based on our assumptions. I am a little lost as to how to put the second part, $xyxy^{-1}=e$ to show there are no more than 16 elements.   If $|G|=16$, find the center of the group and find a group that is isomorphic to $G/\langle y^2\rangle$.   I'm pretty sure the group is centerless.
34432
I'm trying to find a smooth degree 1 map from the torus $T^2 = S^1 \times S^1$ to the $2$-sphere $S^2$.  My first thought was to use the two coordinates $(\theta_1,\theta_2)$ to map onto the usual spherical polar coordinates on the sphere - however I can't quite get that to work because it seems that one of these coordinates would cover the sphere twice and so give a degree $2$ map?  Alternatively I thought about using the winding number. So in general if we embed circles into $\mathbb{R}^3$ by the maps $f_1,f_2: S^1 \to \mathbb{R}^3$ we can then define a map $F:T^2 \to S^2$ by:  $$F(\theta_1,\theta_2) = \frac{f_1(\theta_1) - f_2(\theta_2)}{\| f_1(\theta_1) - f_2(\theta_2) \|}$$  However I'm now struggling to make suitable choices for $f_1$ and $f_2$, my guess is that I'd want to make the circles cross over each other once but I'm struggling to visualise this and write down a map so I would appreciate any help.
15705
I have just been exposed to rotation matrices, and it is showing extremely strong connection with group theory and differential geometry which I am both totally inept at.  Can someone in simple term explain what it means for a group to have a manifold structure and how this affects the mathematics of rotation group?
27021
Suppose you have a group isomorphism given by the first isomorphism theorem: $$G/\ker(\phi) \simeq \operatorname{im}(\phi)$$ What can we say about the group $\ker(\phi)\times \operatorname{im}(\phi)$? In particular, when does the following hold: $$G\simeq \ker(\phi)\times \operatorname{im}(\phi)?$$ I ask this question because i want to prove that $GL_n^+(\mathbb{R}) \simeq SL_n(\mathbb{R}) \times \mathbb{R}^*_{&gt;0}$, with $GL_n^+(\mathbb{R})$ the group of matrices with positive determinant. I proved that $SL_n(\mathbb{R})$ is a normal subgroup and that $GL_n^+(\mathbb{R})/  SL_n(\mathbb{R}) \simeq \mathbb{R}^*_{&gt;0}$, using the surjective homomorphism $\det(M)$. I tried something with semidirect products but I got stuck.
122906
Let $X = \mathbb{R}$ and $Y = \{x \in \mathbb{R} :x ≥ 1\}$, and define $G : X → Y$ by $$G(x) = e^{x^2}.$$ Prove that $G$ is onto.  Is this going along the right path and if so how do get the function to equal $y$?  $G: \mathbb{R} \to\mathbb{N}_1$. Let $y$$\in $$\mathbb{N_1}$. claim: $\sqrt{\ln y}$ maps to $y$. Does $\sqrt{\ln y}$ belong to $\mathbb{N_1}$? Yes because $y \in \mathbb{N_1}$, $G( \sqrt{\ln y})=e^{(\sqrt{\ln y})^2}$.
94101
I have read in Integrable Hamiltonian Systems on Complex Lie Groups by Velimir Jurdjevic p. 61 that the complex orthogonal Lie algebra of dimension 4, $\mathfrak o_4(\mathbb C)$ or $\mathfrak{so}_4(\mathbb C)$, (by which I mean the space of skew-symmetric matrices of size 4) is isomorphic to the Cartesian product of two copies of the special linear Lie algebra of size 2, $\mathfrak{sl}_2(\mathbb C)\times\mathfrak{sl}_2(\mathbb C)$, but I cannot come up with an explicit isomorphism. Can someone please help me find one?
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.
6822
Let $f\colon S^1\to S^1$ be any smooth map. Prove that there exists a smooth map $g\colon\mathbb{R}\to\mathbb{R}$ such that $f(\cos(t),\sin(t))=(\cos(g(t)),\sin(g(t)))$ and satisfying $g(2π)=g(0)+2\pi q$.  The book told me to show that $g(2\pi)=g(0)+2\pi q$, then extend $g(t+2\pi) =g(t)+2\pi q$.  But I don't understand what they mean. How can I get function $g$ from $f(\cos(t),\sin(t))=(\cos(g(t)),\sin(g(t)))$, and what is $g$?
47914
Let $G$ be a Lie group. Recall the standard construction of a left invariant metric on $G$: Let $\langle,\rangle_e$ be an (arbitrary) inner product on $T_eG$ and define, for $u,v  \in T_xG$, $\langle u , v\rangle_x = \langle (dL_{x^{-1}})_xu , (dL_{x^{-1}})_xv\rangle_e $ , where $L_g$:$G\rightarrow G$ denotes left multiplication by $g$. How to prove that this construction produces a smooth metric?
42966
The exponential map for the restricted Lorentz group is surjective. An outline of why is shown on the wiki page .  Is there a more general theorem that states that for some class of Lie groups or Riemannian manifolds (which includes the restricted Lorentz group), the exponential map is surjective?  There is a theorem stating that compact, connected Lie groups have surjective exponential maps. But as the restricted Lorentz group is not compact, this isn't applicable.