I'm pretty sure the answer is out there, see , but that is unfortunately way over my head :).    I'm interested in the case that $G$ is a Lie group (e.g. $U(1)$), but I don't need to know the answer for general manifolds -- just for boring old Euclidean space.    For example, I sort of feel like there is only one $U(1)$-bundle over $\mathbb{R}^n$ for any $n$ (up to isomorphism), and possibly only one $SU(2) \times U(1) \times SU(3)$-bundle, too. But I have no idea how to prove/disprove that.  Do I need to go ahead and try to understand these classifying spaces referenced in the link above?  Or is there a more elementary argument?    If there can be multiple different $G$-bundles over $\mathbb{R}^n$, what is the simplest Lie group $G$ and the lowest value of $n$ for which this occurs?  Is there a simple explicit construction?
21325
What's the easiest way to see this? The only thing I could think of was to try to patch together trivializations. I couldn't find a way to make that work. Thank you! edit: For the record, here's why I asked about this special case of the more general result about fiber bundles over contractible spaces. In the much beloved book by Bott and Tu, it's claimed that the Leray Hirsch theorem can be proven in the same way the Kunneth theorem is proven:  induct on the size of a finite good cover for the base space, then apply the Mayer Vietoris sequence and the Poincare lemma for the induction step. It's assumed that there exists a finite good cover for the base space, but it's not assumed that this cover is a refinement of the cover of the base space, which gives the local trivializations of the fiber bundle. Therefore, to apply the Poincare lemma in the induction step, it seems that you need to know that the result I asked about is true. Since fiber bundles had just been introduced in the text, I thought there may be a short, elementary proof that the authors had taken for granted.
26055
I'm trying to prove the following. Let $\frak{g}$ and $\frak{h}$ be (semisimple) Lie algebras. Then every representation $d$ of $\frak{g}\oplus\frak{h}$ is the tensor product of representations $d^1$ on $\frak{g}$ and $d^2$ on $\frak{h}$; that is $d=d^1\otimes d^2$.  $d^1$ and $d^2$ are cannot be defined by restriction (see comment below). I don't know how to define them so that their tensor product must give back $d$. In particular how do I know that the vector space $V$ of $d$ decomposes appropriately for this to work?  Any hints would be much appreciated!
72284
Let $G$ be a group and $H&lt;G$ such that $[G:H]&lt;\infty$. There exists a subgroup $N\triangleleft G$ such that $[G:N]&lt;\infty$.  I have to show this fact (that according to my book is due to Poincaré), but I think that the statement, written in this way, is trivial: for every group $G$, I can take $N=G$, in fact $G\triangleleft G$ and $[G:G]=1$.  Where am I wrong?  If I'm not wrong, do you know a similar statement?
11001
After googling this fact (which is used in Bott and Tu to show the existence of good covers of manifolds) I've gotten the impression this is somewhat difficult to prove.  But I also came across this homework problem (problem 0) with a hint:    which gives me the impression that it is maybe not so difficult as I thought.  If anyone has any thoughts about how to prove this result I'd be very appreciative. Thanks for your time.
113106
I was reading the first chapter of Robert Gilmore's "Lie groups, physics and geometry" and I came across a brief statement regarding the number of preimages of an element under an homomorphism which I can't seem to prove. Here's the whole paragraph:  "The group $H$ is called a homomorphic image of $G$. Several different group elements in $G$ may map to a single group element $H$. Every element $h_i \in H $ has the same number of inverse images $g_i \in G$."  Anyone knows why is it that this is the case? I've been trying to construct a proof but getting nowhere.  Thanks in advance.
11375
Inspired by the , I am particularly looking into the exact sequences of SU(N), but I will like to loosen the conditions of the previous ,     Q1. $$1 \to A \to SU(N) \to B \to 1$$   where any of $A$,$B$ can contain finite group or discrete groups.    I am particularly interested in SU(5),     Q2. $$1 \to A \to SU(5) \to B \to 1$$    I wonder whether there is any example such that both $A$ and $B$ contains the same number of generators of continuous part of Lie groups (there can be an additional discrete finite group parts such as additional $\mathbb{Z_N}$)? (for example $SU(5)$ has 24 generators. Is there any example such that $A$ and $B$ have 12 generators for continuous part of Lie groups for each $A$, $B$.) How about  $$A=U(1) \times SU(2) \times SU(3)\;\;\; \text{or} \;\;\;A \supset U(1) \times SU(2) \times SU(3)?$$  Or whether there is any example such that $B$  $$B=U(1) \times SU(2) \times SU(3)\;\;\; \text{or} \;\;\;B \supseteq U(1) \times SU(2) \times SU(3)?$$ (there can be an additional discrete finite group parts for $B$ such as additional $\mathbb{Z_N}$.)
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?
117371
Question: Let V be an inner product space s.t. $V=W_1 \oplus  W_2$. $W_1, W_2$ with the inner products $\langle ,\rangle_{W_i}$. Prove there exists a single inner space product $\langle ,\rangle_V$ s.t :  a. $W_2=W_1^\top$  b. if $u_1,u_2 \in W_i$ then $\langle u_1,u_2\rangle_V=\langle u_1,u_2\rangle_{W_i}$  Thoughts Only thing I thought about is saying that 2 of these exist and showing they are the same one... don't really know how to do that though.. Thanx
118489
On page $10$ of Hatcher's , he gives a proof that the Whitney sum of the trivial line bundle over $\mathbb{R}P^n$ and the tangent bundle is equal to the Whitney sum of copies of the tautological line bundle.  A summary of the proof is   1) The image of the tangent bundle of the sphere under the quotient is the tangent bundle of real projective space, and the image of the normal bundle is the trivial line bundle over real projective space.  2) He says the sum of the tangent bundle and normal bundle of the sphere is trivial.  3) He then proves that the trivial line bundle over the sphere is isomorphic to the normal bundle and proves that the normal bundle under the quotient is mapped to the tautological line bundle over real projective space.  But this doesn't make any sense because he claimed earlier that the image of the normal bundle is trivial and the tautological bundle is NOT trivial. Could someone please clarify what has happened? Is the image of the normal bundle of the sphere under the quotient the tautological bundle or the trivial bundle?
1594
Let $N= GL_n(\mathbb{R})$ and $M=SL_n(\mathbb{R})$. Is $M$ normal subgroup in $N$? Why or why not?  I know how to do this with $GL_2(\mathbb{R})$ and $SL_2(\mathbb{R})$ but with $N= GL_n(\mathbb{R})$ and $M=SL_n(\mathbb{R})$ I don't even know all of their elements so I can't check the left and right cosets of $M$ in $N$.
99447
I would like to prove that $U(n)$ is a manifold, where  $$U(n) = \{A \in M_n(\mathbb C): A^*A = I\}.$$  In order to do so I thought of considering the function $M_n(\mathbb C)\to M_n(\mathbb C)$ such that $A\mapsto A^*A$.  Then if I prove that the identity is a regular value of this function I'm done.  My problem is that I am struggling in proving that.  I tried to consider that $(A^*A)_{ij}=\langle v_i,v_j\rangle $ where $v_i$ are the column vectors of $A$ but from there I don't know how to proceed.  Does anyone have any suggestion?  Thanks
77233
We know that the the general linear group is defined as the set $\{A\in M_n(R): \det A \neq 0\}$. I have a homework on how to prove that it is a smooth manifold. So far my only idea is that we can think of each matrix, say $A$, in that group as an $n^2-$dimensional vector. So i guess that every neighborhood of $A$ is homeomorphic to an open ball in $\mathbb{R}^{n^2}$ (However, i don't know how to prove this.)  Now, I'm asking for help if anyone could give me a hint on how to prove that the general linear group is a smooth manifold since I really don't have an idea on how to do this. (By the way, honestly, I don't really understand what a $C^{\infty}-$smooth structure means which is essential to the definition of a smooth manifold.) Your help will be greatly appreciated. :)
39385
Can anyone explain why a semi-simple finite dimensional Lie algebra $\mathfrak{g}$ has to be perfect ?   The natural way to prove something like that would be to look to the algebra generated by the Lie brackets, which when $\mathfrak{g}$ is not perfect would be expected to be solvable. But it doesn't seem to work.
5076
I've believed for a long time that every Lie group can be decomposed as the semidirect product of a connected Lie group and a discrete Lie group. However, in , it is mentioned that this is not true, and that there exist obstructions to this decomposition. Unfortunately, the level of complexity in the answers goes a bit over my head.  This type of decomposition is also addressed in .  With this loss of faith in my intuition, I have three questions:   Is it true that Lie groups do not necessarily decompose into the semidirect product of a connected group and a discrete group? If this decomposition doesn't always work, then is there a particularly nice argument as to why it doesn't? If this decomposition doesn't always work, then is there a nice example I can use to convince myself that my intuition is wrong?   Edit: As much as I love Qiaochu Yuan's answer, which is thorough and (as usual) a role model among Stack Exchange answers, this is not what I'm looking for. I'm aware of the general extension problem, and that not every short exact sequence splits, but my intuition has led me to think that each Lie group can be decomposed this way. I will certainly play around with his outline for a construction, but I would much prefer a more concrete answer. Once again, thank you!  Edit: As of yet, playing with the construction suggested by Qiaochu hasn't yielded anything interesting. This question is still open to answers!
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.
56252
Given an $n$-dimensional geodesically complete connected Riemannian manifold $M$, we want to prove that the dimension of its isometry group is $$\dim {\rm ISO}(M) \leq \frac{n(n+1)}2.$$  Does it suffice to say that, since Euclidean space $\mathbb{R}^n$ is expected to be maximally symmetric, and the number above is the dimension of its isometry group, namely translations plus rotations, then the bound must be true for any other manifold? Do you know a more rigorous proof?
90773
I'm doing some sheets for my Abstract Algebra class and I can't seem to remember the group defined as $\mathbb{R}^{\times}$. It's obviously some variation of $\mathbb{R}$ but I'm away from college on reading week so can't ask my tutor. If someone could clear up the confusion I'd be grateful.