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15

Here's some intuition. First suppose that $x_0$ is a local minimizer for the problem \begin{align} \tag{$\spadesuit$} \text{minimize} & \quad f(x) \\ \text{subject to} &\quad Ax = b. \end{align} If $Au = 0$, then the directional derivative $D_u f(x_0) = \langle \nabla f(x_0),u \rangle$ is nonnegative. Otherwise, it would be possible to improve ...


5

The cuspidal cubic $C := \{(x, y) : y^2 = x^3\} \subset \Bbb R^2$ certainly admits a smooth structure: The projection $C \to \Bbb R$ onto the $y$-axis is continuous and has continuous inverse $y \mapsto (y^{2 / 3}, y)$, so it is a homeomorphism, and hence using it to pull back the (usual) smooth structure on $\Bbb R$ defines a smooth structure on $C$. Like ...


4

There's not going to be an associated vector bundle in general; you need your input data to have some connection to the topology on $\mathbb{R}$. Whatever construction you make, to have any chance of being "correct", it should definitely be functorial with respect to isomorphisms. But in the case $\mathcal{E}=\mathcal{O}_X$ (where the associated vector ...


3

If $$\omega = \sum_{i=1}^n\frac{(-1)^{i-1}x_i}{\|x\|^n}\,dx_1 \wedge\cdots \wedge \widehat{dx_i}\wedge \cdots \wedge dx_n,$$then: $$d\omega = \sum_{i=1}^n\sum_{j=1}^n\frac{\partial}{\partial x_j}\left(\frac{(-1)^{i-1}x_i}{\|x\|^n}\right) dx_j \wedge dx_1 \wedge\cdots \wedge \widehat{dx_i}\wedge \cdots \wedge dx_n.$$Now, the only surviving term is when $j = ...


3

The product rule tells us $$\nabla |\nabla u| = \frac{\nabla (g(\nabla u,\nabla u))}{2 |\nabla u|}=\frac{g(\nabla^2u,\nabla u)}{|\nabla u|}.$$ Taking the squared norm and estimating the numerator with Cauchy-Schwarz gives us $$|\nabla|\nabla u||^2 \le \frac{|\nabla^2 u|^2 |\nabla u|^2}{|\nabla u|^2} = |\nabla^2 u|^2$$ as desired.


3

Look at the tangent space to $S = g^{-1}(c)$ at $x_0$, say $S_{x_0}$. Then $S_{x_0} = \left[\nabla g(x_0) \right]^{\perp}$. Therefore, $S_{x_0}^\perp$ is the one-dimensional subspace spanned by $\nabla g(x_0)$. Now, $$\nabla f(x_0) = \lambda \nabla g(x_0)\text{ for some } \lambda \in \Bbb{R} \iff \nabla f(x_0) \in S_{x_0}^\perp \iff \nabla f(x_0).v = 0 ...


3

The transition functions between neighboring patches do not reveal that globally there is something wrong. Now locally there are two classes of coordinate patches. Two patches belong to the same class if the Jacobian between them is positive, and belong to different classes if the Jacobian between them is negative. Locally you can arbitrarily declare one ...


3

Don't forget that differentiability is a local property. We say that $f: S_1 \to S_2$ is differentiable at $p \in S_1$ if there is a parametrization $\sigma_1: U_1\subseteq \Bbb R^2 \to V_1 \subseteq S_1$ around $p$ and a parametrization $\sigma_2: U_2 \subseteq \Bbb R^2 \to V_2 \subseteq S_2$ around $f(p)$ such that $\sigma_2^{-1}\circ f \circ\sigma_1$ ...


3

The dependence of the curvature on $g$ is somewhat of a red herring here - it certainly makes solving the equation difficult, but assuming we have a smooth solution of Ricci flow defined on $M \times [0,T)$, the quantity $f(x,t) = r - R_{g(t)}(x)$ is a well-defined smooth function on $M \times [0,T)$. Thus the Ricci flow equation becomes ...


3

It's risky to give a blanket negative answer to so open-ended a question (particularly, one in which "canonical" is undefined), but with minor qualifications the answer appears to be "no". Here are a few easy observations, a bit too long for a comment: If for all $x$ and $y$ in $S$, there is a canonical parallel transport between $T_{x}S$ and $T_{y}S$, ...


2

Been a long time Justin! I don't know the literature, but here is an approach towards formulating a practical relaxation of the problem. If you relax the requirements of parallel transport a bit, you might be able to get something reasonable by formulating this as a regularized optimization problem over the whole tangent bundle at once. Imagine ...


2

Different choices of $\phi$ and $\psi$ would just lead to multiplying $J$ on the left or right by an invertible Jacobian. This is because of the chain rule and the fact that you're just doing a change of basis. Consider the map $\psi\circ F\circ \phi^{-1}$. If we have $\phi'$ and $\psi'$ maps for different charts (with appropriate domains), we have that ...


2

I think the confusion comes from using $x$ in two different ways. I will use $x = (x^1,.., x^n)$ as coordinates on $U$. Let $y = (y^1,...,y^n)$ be any particular point in $U$. Consider the curve $\gamma:[0,1]\rightarrow U$ with $\gamma(t) = p + t(y-p)$. In terms of the $x^i$ coordinates, we have $x^i(t) = p^i + t(y^i - p^i)$. Now, consider the ...


2

By simple computations, one checks $AB=BA$, $AC=CA$, $CB=BCA$ is a presentation. So in the abelianization $A=1$, $BC=CB$ the abelianized group is $\bf Z^2$. Note that $dy, dz$ are invariant by $G$ and form a base for the de Rham co-homology


2

Any point in the annulus $U$ is uniquely of the form $(t \cos \theta, t \sin \theta)$ for some real $t \in (0,\sqrt{\pi}), \theta \in [0,2\pi).$ Map this point to the point of the cylinder $(x,y,z)=(\cos \theta, \sin \theta, \cot t^2).$ This is clearly a subset of the cylinder as it satisfies $x^2+y^2=1.$ Also, because $\theta$ ranges in $[0,2\pi),$ for any ...


2

Yes $$\int_M d\omega =\int_{\partial M} \omega=0$$ since $\partial M=\emptyset$ beause $M$ is closed.


2

This is not true as stated. Suppose $\alpha$ is a $d$-closed $(p, q)$-form, then $\partial\alpha = 0$, so $\mathcal{L}_X\partial\alpha = 0$. On the other hand, $$\partial\mathcal{L}_X\alpha = \partial(di_X + i_Xd)\alpha = \partial di_X\alpha = \partial(\partial + \bar{\partial})i_X\alpha = \partial\bar{\partial}i_X\alpha.$$ Now let $M = \mathbb{C}$, ...


2

Actually it's a bit to easy for an answer so I do not know if I missed something. From $\frac{d}{ds}\langle\frac{\partial f}{\partial s},\frac{\partial f}{\partial t}\rangle = 0$ you know that $\langle\frac{\partial f}{\partial s},\frac{\partial f}{\partial t}\rangle $ is constant in $s$. In addition you know $\langle\frac{\partial f}{\partial ...


2

You can simplify the calculations by working in an orthonormal frame so that contraction with the metric become standard traces. The only non-zero components of the Riemannian curvature are $$K := R_{1212} = R_{2121} = -R_{1221} = -R_{2112}.$$ Then $$R_{ij} = R_{1i1j} + R_{2i2j}= K\delta_{i2}\delta_{j2} + K \delta_{i1}\delta_{j1} = Kg_{ij}.$$ Taking the ...


2

The conjecture already fails at $3$-dimension. For $3$-dimension, consider the square anti-prism with vertices at $$\left( \pm \sqrt{3}, \pm \sqrt{3}, \sqrt{2} \right),\quad ( \pm \sqrt{6}, 0, -\sqrt{2} )\quad\text{ and }\quad( 0, \pm \sqrt{6}, -\sqrt{2})$$ It has volume $V = 16(\sqrt{2}+1)$ and diameter $d = \sqrt{20 + 6\sqrt{2}}$. Its "compactness" ...


2

The definition of the tensor product is $(f \otimes g)(u,v) = f(u)g(v)$. You should be able to check that whenever $f,g \in T^*$ (i.e. are linear), their product $f \otimes g$ is bilinear; and the product $f \otimes f$ is symmetric and weakly positive definite. Thus the action of $dx \otimes dx$ on a pair $u,v$ of tangent vectors is $$(dx \otimes dx)(u,v) = ...


2

If the surface is not orientable, then the unit normal vectors $N(p)$ are only defined up to sign. There is no way to consistently decide whether to take one normal vector or its negative.


2

The presence of a curl term in the Helmholtz decomposition is very much a three-dimensional phenomenon. While it is reasonable to define the curl of a vector field tangent to a surface, what you get is either a field normal to the sphere or a scalar field, so it's not going to help us decompose tangent vector fields. The Hodge decomposition tells us that ...


2

Hint Suppose $\sigma : U \to \Bbb R^3$ is a surface patch of $S$. Then, $\widetilde{\sigma} : \widetilde{U} \to \Bbb R^3,$ where $\widetilde{U} := \{(u, v) : (v, u) \in U\}$ and $\widetilde{\sigma}(u, v) := \sigma(v, u)$, is also a surface patch. How do the standard unit normals of $\sigma$ and $\widetilde{\sigma}$ compare at any point of $\sigma(U) = ...


2

You have a little more general result. You only need to be in a topological group not, a Lie group. Proposition : Let $G$ be a topological group, and $H$ a subgroup of $G$. If $H$ and $G/H$ are compact then $G$ is compact. Sketch of a proof : Consider a set $(F_i)_{i \in I}$ of closed sets of $G$. Suppose that for all finite $J \subseteq I$, $\cap_{j \in ...


1

Have a look at Gauss-Bonnet: $$ \int_G K \mathrm{dvol}_G + \int_{\partial G} \kappa_G (t) dt + \sum_i \alpha_i = 2 \pi \chi (G) \quad (1)$$ where $K$ is the Gaussian curvature, $\kappa_G$ the geodesical curvature and $\alpha_i$ the angle at the $i$th corner of the polygone in question. For your problem, let us suppose suche a simple region $G$ exists. The ...


1

The online notes on differential geometry by Balazs Csikos are really quite good for self-study: http://www.cs.elte.hu/geometry/csikos/ under Lecture Notes, then under BSM Lecture Notes. Assumed prerequisites are multivariate calculus, linear algebra, and just a little topology.


1

Write $i : \Bbb R^k \to \Bbb R^n$ for the inclusion $x \to (0,x)$. For a given vector $X \in \Bbb R^k = T_0 \Bbb R^k$ you get a vector $Y := T_{0}(x^{-1} \circ i) (X) \in T_q N$ and because of transversality, you can write $Y = Y' + Y''$, where $Y' \in T_q S$ and $Y'' \in T_p f (T_p M)$. Remark, that $T_q x$ takes the tangential plane of $S$ at $q$ to ...


1

The topics that you want to study use mostly the very essential ideas from Linear Algebra. Yes, over $\mathbb{R}$ and $\mathbb{C}$ is all you need. Since you seem to be a theoretically minded person with interest in geometry related subjects, I would recommend Gelfand's "Lectures on Linear Algebra". Strang's textbook is excellent, but probably not the style ...


1

It's true if the complex structure is invariant under the vector field X. In that case: $\mathcal{L}_X \partial a = \mathcal{L}_X (1 - \imath J) da = (1 - \imath J)\mathcal{L}_X da = \partial \mathcal{L}_X a$. (Note: This is written with $a$ being a scalar. For a $(p,q)$ form you just need to pick a suitable projector.)



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