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4h
comment What is uniform continuity geometrically
Related: What's the fastest way to tell if a function is uniformly continuous or not? and Why did mathematicians introduce the concept of uniform continuity?.
16h
answered Gradient vector of parametric curve
Jul
28
comment Hodge Theory, intuition?
"Fix a Riemannian metric $g$ on $M$. A closed $p$-form is harmonic (lies in the kernel of the $g$-Laplacian) if and only if it is co-closed (with respect to $g$), if and only if it strictly minimizes the $L^{2}$-norm (with respect to $g$) in its de Rham class." Is that the type of "soft analysis" (integration by parts/$L^{2}$ geometry) intuition you're seeking?
Jul
25
comment Why do we care if a function is uniformly continuous?
Related: Why did mathematicians introduce the concept of uniform continuity?
Jul
24
answered Graphically representing relations of ordered pairs
Jul
22
comment Explicit Calabi-Yau metrics
Thank you for the reference/clarification! If you'll excuse the self promotion, A momentum construction for circle-invariant Kähler metrics may be worth a look. Particularly, the metrics Neitzke and Vafa call local $\mathbf{CP}^{2}$, local $\mathbf{CP}^{1} \times \mathbf{CP}^{1}$, and resolved conifold do have explicit constructions, in the sense that the metric can be constructed from a single explicit rational function of one variable. Are you asking just out of curiosity, or would the details be helpful to your work?
Jul
22
comment Explicit Calabi-Yau metrics
Welcome to Math.SE! I'm not familiar with your terminology ("local $\mathbf{CP}^{n}$", "conifold", etc.). Could you perhaps provide references for the metrics you mention? Also, is your definition of "Calabi-Yau metric" a complete, Ricci-flat Kähler metric? If so, the earliest reference that comes immediately to mind is Calabi's Métriques kählériennes et fibrés holomorphes.
Jul
21
answered Dimension of the $0^{\text{th}}$ de Rham cohomology group of $U$
Jul
21
answered show that $X$ is homeomorphic to the $n$ dimensional (real) projective space.
Jul
20
answered Intersection form on $S^2 \tilde \times S^2$
Jul
20
comment Why do we need to check for more than $\frac{\infty}{\infty}$ or $\frac{0}{0}$ when applying L'Hospital?
(+1): Nice example and analysis! Further, if the original limit had been$$\lim_{x \to \infty} \frac{x - \frac{1}{2}\sin x}{x + \frac{1}{2} \sin x}$$the first application of l'Hôpital would still be wrong, but for a different reason than vanishing of the denominator near $\infty$: The limit-at-infinity$$\lim_{x \to \infty} \frac{1 - \frac{1}{2}\cos x}{1 + \frac{1}{2}\cos x}$$of a non-constant periodic function doesn't exist.
Jul
17
comment Describe the Riemann surface:
Yes, this Riemann surface is complex-analytically equivalent to a sphere. :) More generally, if $p$ is a polynomial of degree $2g + 1$ or $2g + 2$, the Riemann surface $w^{2} = p(z)$ has genus $g$. (Not all Riemann surfaces have this form, however.) "Topology" is the study of "invariants" of continuous, bijective maps having continuous inverse. Loosely, it's all right to cut and deform so long as you "re-glue all the cuts the way they were". For example, it's "allowed" to cut a torus along a circle to get a cylinder, twist through a full turn like a barber pole, then re-glue the ends.
Jul
16
comment Are $X=M \times [0,T]$ and $\partial X$ smooth compact manifolds when $M$ is smooth compact Riemannian manifold?
A "manifold" is "locally modeled by an open subset of $\mathbf{R}^{n}$ at each point". :) A manifold with boundary equipped with a Riemannian metric (a notion requiring a non-trivial definition, incidentally) is never geodesically complete, essentially because "a geodesic starting at a boundary point and heading away from the interior leaves the universe immediately".
Jul
16
answered Are $X=M \times [0,T]$ and $\partial X$ smooth compact manifolds when $M$ is smooth compact Riemannian manifold?
Jul
16
comment Describe the Riemann surface:
Just saw your comments. Looks like you've worked things out, but the bottom line is, this particular Riemann surface can be represented as two copies of a slit plane, "cross-joined" along the slits (the "top" edge of a slit in one copy glued to the "bottom" edge in the other copy). There are multiple ways to achieve the construction. If you cut along the real axis from the branch points to infinity, the picture looks like this.
Jul
15
comment Describe the Riemann surface:
Often a "description" consists of fixing a branch of square root, finding a dense open set $U$ of the complex plane on which $\sqrt{1 - z^{2}}$ is analytic, taking two copies of $U$, and describing how $\sqrt{1 - z^{2}}$ on one copy of $U$ is glued to $-\sqrt{1 - z^{2}}$ on the other copy of $U$. Looking at expressions for the square roots in terms of real and imaginary parts may not be the best approach. Were you given information about what constitutes a "description" for your purposes?
Jul
15
awarded  Yearling
Jul
13
comment Prove a cycle of length l is odd if l is even?
For starters, do you understand the definitions of "length" of a cycle and "parity" of a permutation? (Separately, it would be polite to include the statement of the question in the body of the post. Titles have a way of getting separated psychologically from the rest of the post.)
Jul
12
revised Distance from curve to plane
Removed algebraic geometry tag
Jul
12
comment How would I complete my proof that $\int_a^bf(g(x))\,dx = \int_{g(a)}^{g(b)}f(x)\frac{d}{dx}(g^{-1}(x))\,dx$?
@ixsetf: Your presumed proof via the substitution rule is a formal proof. :) (You probably want to assume "$g'$ is non-vanishing"; that's not implied by "$g$ is strictly monotone".) If instead you write out a proof using Riemann sums, you can use the chain rule and the mean value theorem to establish the step you ask about, but then you're effectively re-proving the one-line substitution proof (for $f$ the identity function) interval-by-interval.