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comment Characterization of gradient vector fields
@JoshBurby unfortunately I have not thought about this anymore.
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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
answered How does a Lie derivative generate a $U(1)$ isometry?
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
revised Exterior derivatives involving representations
added 711 characters in body
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 $\rho(g_{\alpha\beta})$.
Mar
16
answered Principal bundles, connection forms and fundamental vector fields
Mar
16
answered Exterior derivatives involving representations
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).