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Graduate student in mathematics.


Dec
2
comment Curvature 2-form vs. Sectional Curvature
The definition of sectional curvature is in terms of the curvature 2-form (see e.g. en.wikipedia.org/wiki/Sectional_curvature)
Nov
30
comment Differential forms on $S^1$
@self-learner the cohomology groups are the kernel divided by the image (since $d^2 = 0$, the image is contained in the kernel).
Nov
23
comment Prove that any map $f\colon S^1 \to S^1$ mapping antipodal point to antipodal point has $\deg_2(f)=1$ by a direct computation.
What definition of degree are you using?
Nov
19
comment Differential forms on $S^1$
The star means the dual space, so $T^*M$ is the space of dual vectors and sections of its exterior product are differential forms. Any $n$-form on an n-dimensional manifold looks like $\mu = f dx_1 \wedge \cdots \wedge dx_n$ in coordinates. Therefore $d\mu = \sum_j \partial_j f dx_j \wedge dx_1 \wedge \cdots \wedge dx_n = 0$.
Nov
19
comment Differential forms on $S^1$
This is because $\mu$ is an $n$-form on an $n$-dimensional manifold. Thus $d\mu$ is an $n+1$-form and so must be zero since $\Lambda^{n+1} T^*M = 0$.
Nov
18
comment Differential forms on $S^1$
@orion: Not every form can be written that way.
Nov
18
comment Complex structure on a real vector bundle
$J$ is the complex structure. The $e_i$ and $\sigma_i$ are local sections of $E$ and so are acted on by $J$. Having a trivialization (i.e. iso from $E$ restricted to $U$ to $U\times \mathbb R^{2n}$) is equivalent to having $2n$ pointwise independent sections of $E$ restricted to $U$.
Nov
17
comment Complex structure on a real vector bundle
Are you familiar with how a metric on a vector bundle reduces the structure group to the orthogonal group?
Oct
27
comment Lie Bracket, Hopf Fibration, independence of choice
A horizontal lift is not unique unless you fix a connection on the fibration.
Aug
14
comment Is this a manifold?
So you're looking at the sphere with like a ring around it thats touching it? This won't be locally homeomorphic to $\mathbb R^2$.
Jul
25
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 $\mathbb C^2 - \{0\} \to \mathbb C P^1 \simeq S^2$.