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Mar
16
comment How does a Lie derivative generate a $U(1)$ isometry?
That's correct, $V$ is tangent to the flow at any point. This is what the equation defining $\phi_t$ says.
Mar
16
comment Exterior derivatives involving representations
No problem! Notation in this area can be very confusing. I think most authors would write $\rho_2(\eta) \wedge \omega$ instead of $\rho_2(\eta) \circ \omega$.
Mar
16
comment Exterior derivatives involving representations
Let me know if my edit helps.
Mar
16
comment Associated bundles: isomorphism between spaces of differential forms.
So on $U_\alpha$ you just have a trivial vector bundle with fiber $V$ but then the transition functions tell you how to glue (i.e. identify) these trivial vector bundles together on overlaps. To get a global section you then need the sections on each piece to be compatible via the transition functions.
Mar
16
comment Associated bundles: isomorphism between spaces of differential forms.
In general, a vector bundle can be specified by the data of a vector space $V$, a cover $U_\alpha$ of your manifold, and functions $h_{\alpha\beta} : U_\alpha\cap U_\beta \to GL(V)$ satisfying the cocycle condition. Under this definition of a vector bundle, a section is a collection of functions $\zeta_\alpha : U_\alpha \to V$ such that $\zeta_\alpha(x) = h_{\alpha\beta}(x) \zeta_\beta(x)$. The associated vector bundle has the same transition functions as the principal bundle, but now acting via the rep, so the transition functions are $\rho(g_{\alpha\beta})$.
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.