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location New Jersey
age 28
visits member for 4 years
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Graduate student in mathematics.


2d
comment Homeomorphism between SU(4) and SO(6)
The maps $SU(2) \to SO(3)$ and $SU(4) \to SO(6)$ are not homeomorphisms but are 2-to-1 Lie group covering maps (so local homeomorphisms).
Jul
24
comment Submersions and induced homomorphism on fundamental groups
I think you need a little more justification for the existence of the $\alpha_i$. You could use for example the regular value theorem (which maybe you implicitly are).
May
14
comment Another differential topology lemma
not sure... he refers to a work done before so maybe he proved there that if $v_t$ is a 1-parameter family of vector fields, then $index(v_t)$ is constant in $t$
Apr
28
comment Three linearly independent vector fields
There is not an isomorphism-- I have just proved that there isn't.
Apr
26
comment show that the compactly supported De Rham cohomology groups $H^{p}_{DR}(\mathbb{R^n})$ are all zero for $0\leq p<n$
I meant the long exact sequence associated to a pair (see e.g. the l.e.s. at en.wikipedia.org/wiki/Cohomology_with_compact_support with $X, U$ and $Z$). Because you should be able to use that with $X = S^n, Z = pt$ and $U = X - Z \simeq \mathbb R^n$ to get your desired result.
Apr
26
comment show that the compactly supported De Rham cohomology groups $H^{p}_{DR}(\mathbb{R^n})$ are all zero for $0\leq p<n$
Does Lee talk about the long exact sequence in compactly supported cohomology associated to an open (or closed) subset?
Apr
26
comment show that the compactly supported De Rham cohomology groups $H^{p}_{DR}(\mathbb{R^n})$ are all zero for $0\leq p<n$
You can't use the fact that $\mathbb R^n$ is contractible since compactly supported cohomology is only a homotopy invariant for homotopies that are proper maps.
Apr
26
comment The kernel of a differential one-form
@Ilcapitano yes, that is correct (assuming of course $\theta$ vanishes only at $m$).
Apr
21
comment Three linearly independent vector fields
$S^1\times S^2$ is not isomorphic to $SO(3)$. One way to see this is the fundamental group of the former is $\mathbb Z$ while the fundamental group of $SO(3)$ is $\mathbb Z_2$. Maybe you thought this because there is a fibration $S^1 \to SO(3) \to S^2$. Nonetheless, $S^1\times S^2$ is parallelizable since all orientable 3-manifolds are.
Apr
19
comment The magnetic monopole and the Hopf bundle
If you think of $\mathbb R^4 - \{0\}$ as $\mathbb C^2 - \{0\}$ then it seems likely (I haven't done any computations) that $\omega$ is a connection form on the principal $\mathbb C^\times$ bundle $\mathbb C^2 - \{0\} \to \mathbb C P^1 \simeq S^2$.
Apr
16
comment Rigidity for Lie Groups
why does every auto of $\tilde G$ restrict to an auto of $\Gamma$? Did you mean $G$ instead of $\tilde G$ there?
Apr
14
comment Books in spectral theory for finite dimensional spaces
I'm not an expert but I don't think spectral theory for finite dimensional spaces extends to anything beyond the basics of eigenspaces and eigenvectors.
Apr
14
comment Action of $SO_n$ on $\mathbb{S}^{n-1}$ induces fibre bundle.
A very general theorem that tells you this is a locally trivial fibration is Ehresmann's theorem: en.wikipedia.org/wiki/Ehresmann's_theorem
Apr
12
comment Notation for the Covariant Derivative of a smooth section in the direction of a tangent vector X
I think you should answer your own question and then accept it.
Apr
11
comment What is the kernel of a Maurer-Cartan form?
Oh ok. It looks like they're using a much more general definition of Maurer-Cartan form (for example, their M.C. forms are not even on a Lie group necessarily). It seems like the prototypical example of such a form is a flat connection on a principal $G$-bundle. Note that the kernel of a connection defines a horizontal distribution and this distribution is integrable (and therefore gives rise to a foliation on the total space of the principal bundle) if and only if the connection is flat.
Apr
11
comment What is the kernel of a Maurer-Cartan form?
But do you have an example of where this appears in the foliation literature you mention?
Apr
11
comment What is the kernel of a Maurer-Cartan form?
For 1) It is a vector space valued 1-form so it gives a linear map from vector fields to the Lie algebra. Do you have more context for 2)? I don't see it can have a kernel (viewing it as a map as in 1)) since it takes a tangent vector $v$ at a point $g$ to ${L_{g^{-1}}}_* v$ which is non-zero if $v$ is non-zero since left multiplicaiton by $g$ is a diffeomorphism.
Apr
10
comment Isogenies and dimensions
Or is your question if $g = g'$?
Apr
10
comment Isogenies and dimensions
Do you mean the linear map $\mathbb C^g \to \mathbb C^{g'}$ is surjective?
Apr
10
comment Can a ring isomorphism change the structure of a module?
Yes that should be true. Though the Clifford algebra is a simpler object then the group algebra of $SO(2n)$.