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 1h comment coordinate geometry : intersection of a curve with a line Hint: The line with equation $y = mx + b$ makes angle $\theta = \arctan m$ with the rightward-pointing $x$-axis. 13h comment Trivial sections of tautological line bundle for $\mathbb{P}_F(V)$. It's not true that there only exist trivial sections of the tautological bundle; do you have another condition in mind (such as "non-vanishing", or "holomorphic" over the complex field)? 19h comment Identifying the two-hole torus with an octagon @LaurentHayez: Modulo mistakes on my part, yes, the two gluings must be topologically equivalent. :) (The gluing in your answer is the "standard" presentation of the two-holed torus, and certainly correct.) The way I read the OP's question, their instructor stated that the "antipodal" gluing gave a two-holed torus, and they wanted to confirm this assertion. 20h answered Identifying the two-hole torus with an octagon 21h comment Integrability of dirichlet function in $\mathbb{R}^3$ @RuiCarneiro: Yes, exactly. :) In other words, one should speak of a function being "Riemann integrable on $D \subset \mathbf{R}^{n}$" rather than simply being "integrable" (though in practice the domain and its dimension are implicit). 23h answered Integrability of dirichlet function in $\mathbb{R}^3$ 1d comment The curvatures of a transformed surface under a similarity transformation The point of the second comment is that $N' = A \circ N \circ f^{-1}$, so (omitting points of evaluation) $\operatorname{tr}dN' = \frac{1}{r}\operatorname{tr}(A\, dN\, A^{-1}) = \frac{1}{r} \operatorname{tr}(dN)$. 1d comment The curvatures of a transformed surface under a similarity transformation Small typographical note: The less-than/greater-than signs in (La)TeX are spaced like binary operators rather than like delimiters, which makes inner products rather difficult to parse visually. \langle and \rangle are preferred; I generally use a macro \newcommand{\Brak}[1]{\left\langle #1 \right\rangle}, and then, e.g., -\Brak{dN_{p}(v), w}. :) 1d answered The curvatures of a transformed surface under a similarity transformation Apr15 answered Why not every homogeneous manifold is parallelizable? Apr14 comment Question about parallel displacement on a surface I haven't carefully checked the calculation, but wouldn't one expect the change in $v$ to depend on $v$? (The endomorphism of the tangent space doesn't depend on $v$, of course, but that looks consistent with your last line.) Apr14 comment Circle Geometry The distance formula coming from the Pythogrean theorem is true even if one or both legs of the "triangle" with corners $(a, b)$ and $(x, y)$ have length zero. Apr14 comment Are all continuous one one functions differentiable? @MarioCarneiro: You're perfectly correct, if course; just emphasizing for posterity that the answer to the OP's "global" question is no, not even locally. :) Apr14 comment Key differences between almost complex manifolds and complex manifolds If you'll forgive a tangential "public service comment" on terminology: It would be more logical if we all spoke of "complex manifolds" (admitting an endomorphism field $J$ with $J^{2} = -I$) and "holomorphic manifolds" (complex manifolds for which $J$ is integrable), the way we already do for vector bundles. But for this changeover to happen, students (and working mathematicians) have to embrace the terminology. :) Apr14 answered Möbius strip as a non-trivial principal bundle Apr14 comment Unbounded Geodesics and Nonpositive Curvature Posted the preceding as a comment rather than an answer, since I'm guessing "completeness" was intended...? Apr14 comment Unbounded Geodesics and Nonpositive Curvature 1., 2. If $M$ is compact, no geodesic is unbounded, so curvature alone does not guarantee unbounded geodesics. 3. If you assume only simple connectedness, your metric might be incomplete, and there is no guarantee of unbounded geodesics. 4. Non-compactness is not enough for "all"; take a cylinder or similar product example. 5. Non-positive curvature, simple-connectedness, and completeness get you "all unbounded" by the Cartan-Hadamard theorem and the behavior of Jacobi fields. Apr14 comment Short differential-form free route to understanding a specific surface integral @temo: Offhand I can't think of a reference for integrating over hypersurfaces. The material in my answer is "folklore", though if memory serves, Spivak's Calculus on Manifolds has an exercise on the generalized cross product, and it's entirely possible he treats integration over hypersurfaces (perhaps in one or more exercises). Incidentally, I found Spivak to be an especially clear introduction to forms. (Do work all the exercises!) Really, differential forms largely boil down to determinants and index combinatorics; one just has to get past the (substantial!) initial intimidation barrier. Apr14 answered Are all continuous one one functions differentiable? Apr13 comment How to find $n+1$ equidistant vectors on an $n$-sphere? @Gaffney: Could you please explain? Each $\tilde{v}_{i}$ has unit length (since the $v_{i}$ have unit length), so all $(n + 1)$ vectors $(\tilde{v}_{i})_{i=1}^{n+1}$ lie on the unit sphere centered at the origin.