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seen Dec 13 at 17:57

Graduate student in mathematics.


Jul
24
revised Complex structures on Riemann surfaces
edited tags
Jul
24
asked Complex structures on Riemann surfaces
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).
Jul
20
awarded  Yearling
Jul
12
awarded  Good Question
Jul
8
awarded  Popular Question
Jul
2
awarded  Curious
Jul
2
awarded  Inquisitive
May
24
answered The diffential of commutator map in a Lie group
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 $t$
May
13
answered Another differential topology lemma
May
13
answered Lie group and Lie algebra automorphisms
May
4
awarded  lie-groups
May
3
awarded  Nice Answer
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.