Eric O. Korman
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 5h awarded Favorite Question Apr1 awarded Enlightened Apr1 awarded Nice Answer Mar16 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. Mar16 answered How does a Lie derivative generate a $U(1)$ isometry? Mar16 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 $\rho_2(\eta) \circ \omega$. Mar16 comment Exterior derivatives involving representations Let me know if my edit helps. Mar16 revised Exterior derivatives involving representations added 711 characters in body Mar16 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. Mar16 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 $\rho(g_{\alpha\beta})$. Mar16 answered Principal bundles, connection forms and fundamental vector fields Mar16 answered Exterior derivatives involving representations Dec2 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) Nov30 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). Nov23 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? Nov19 reviewed Approve mathematical model Nov19 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 $d\mu = \sum_j \partial_j f dx_j \wedge dx_1 \wedge \cdots \wedge dx_n = 0$. Nov19 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 $\Lambda^{n+1} T^*M = 0$. Nov18 comment Differential forms on $S^1$ @orion: Not every form can be written that way. Nov18 answered Differential forms on $S^1$