Tag Info

Hot answers tagged

10

There is no Stokes's Theorem for general tensors. The general result is that $\displaystyle\int_M d\omega = 0$ for any compact oriented $n$-dimensional manifold $M$ without boundary and $(n-1)$-form $\omega$ on $M$. The divergence theorem you cite is for the case of $\omega = \star\eta$ where $\eta$ is the $1$-form corresponding to $X$ under the isomorphism ...


7

$S^6$ cannot even be a symplectic manifold. The reason is that on any compact symplectic manifold $X$ of dimension $2n$, the symplectic form represents a class in $H^2(X, \mathbb{R})$ whose $n^{th}$ power is a generator of $H^{2n}(X, \mathbb{R})$, and in particular it must be nonzero. But $H^2(S^6, \mathbb{R}) = 0$. In fact $S^{2n}$ cannot be a symplectic ...


5

On a Riemannian manifold, the divergence theorem applies to $1$-forms as well as to vector fields. The simplest way to see this is by using the "musical isomorphisms" between $1$-forms and vector fields. This are the inverse isomorphisms $\flat\colon TM\to T^*M$ and $\sharp\colon T^*M\to TM$ defined by raising and lowering indices. If $X$ is a vector field ...


5

In a coordinate chart $(U, (x^1, \dots, x^n))$ on an $n$-dimensional manifold, every one-form $\omega \in \Omega^1(U)$ can be written as $\omega = \omega_1 dx^1 + \dots + \omega_n dx^n$ where $\omega_1, \dots, \omega_n : U \to \mathbb{R}$ are smooth functions. So we can write $\omega$ as $$\omega = \sum_{i=1}^n\omega_i dx^i.$$ Using Einstein notation, we ...


4

The text here uses the Einstein summation convention, defined below: In any local coframe on a manifold, here we'll use $(dx^i)$, defined, say, on $U \subseteq M$, we can write (the restriction to $U$ of) a $1$-form $\omega$ as a linear combination of coframe elements, namely, as $$\omega = \omega_1 \,dx^1 + \cdots + \omega_n \,dx^n = \sum_{i = 1}^n \omega_i ...


4

Start with the standard Lorentzian metric $dx^2 - dy^2$ on the Euclidean plane. Restrict that metric to the unit square $[-1,1] \times [-1,1]$. Notice that the two gluing maps which are used to construct the Klein bottle, namely the maps $(x,y) \to (x+2,y)$ and $(x,y) \to (x,y+2)$, are isometries of the standard Lorentz metric. Therefore, the metric ...


4

Index notation can for the most part be understood using just the Einstein summation convention - you're just dealing with scalar equations between the components of various vectors, matrices etc. (This is not quite true when it comes to e.g. covariant derivatives in abstract index notation, but is probably the easiest way to understand things to start with ...


4

Regarding question (1): When $M$ is compact and $f$ is a positive $C^2$ function, such an $h$ exists if and only if $f$ is constant on each component of $M$. Proof: Suppose $M$ is compact and $f\colon M\to \mathbb R$ is a positive $C^2$ function. If $f$ is constant on each component, clearly $h\equiv 0$ works. Conversely, suppose $h\colon M\to \mathbb R$ is ...


3

I shall assume that your $$\gamma'(t)=\bigl(x'(t),y'(t)\bigr)\qquad(a\leq t\leq b)$$ is continuous. Furthermore it is no restriction of generality to assume $$P=(0,0),\qquad Q=(q,0),\quad q>0\ .$$ Then $$\eqalign{\int_a^b\|\gamma'(t)\|\>dt&=\int_a^b\sqrt{x'^2(t)+y'^2(t)}\>dt\cr &\geq\>\int_a^b\bigl|x'(t)\bigr|\>dt\geq\int_a^b ...


3

The idea of sectional curvature is to assign curvatures to planes - the sectional curvature of a plane $\Pi$ in the tangent space is proportional to the Gaussian curvature of the surface swept out by geodesics with starting directions in $\Pi$. Intuitively you're taking a two-dimensional slice through a given plane, and measuring the classical ...


3

A differential $k$-form on a smooth manifold $M$ is a skew-symmetric $C^{\infty}(M)$-linear map $\omega :\Gamma(TM)^k \to C^{\infty}(M)$. The exterior derivative of $\omega$ is a skew-symmetric $C^{\infty}(M)$-linear map $d\omega : \Gamma(TM)^{k+1} \to C^{\infty}(M)$ where $d\omega(V_0, \dots, V_k)$ is equal to $$\sum_{i=0}^k(-1)^iV_i(\omega(V_0, \dots, ...


3

It probably just means that it is a property the function might have, or might not have, but at least the curve, being a function, lies in the category of things for which it could make sense to determine one way or the other whether they're injective. It is to make sure you have clear that it is a function, as opposed to a set of points. If $n$ is a ...


3

Yes, $J$ here is definitely meant to be $d\varphi$. In particular, the equality $$ dy^1 \wedge \cdots \wedge dy^n = \det(d\varphi) dx^1 \wedge \cdots \wedge dx^n $$ holds, as you can check just by substituting $$ dy^i = \sum_j \frac{\partial y^i}{\partial x^j} dx^j $$ and noting that the matrix $d\varphi$ has $ij$-th entry $\partial y^i / \partial ...


2

You do not need the ambient space because formally Riemannian manifolds are defined via abstract glueings of Euclidean spaces, which you can think as little disks or cubes if you like. A human being living in a submanifold of the ambient space will not be able to "feel" the ambient space's existence. Rather to tell the distance and flatness in his/her space ...


2

Jonas Meyer's reading is right: O'Neill is talking about injectivity as a property that a curve could possibly possess, but may not. Admittedly, the wording is not the clearest, but I think the fuller context helps: "Since a curve $\alpha \colon I \to \mathbb{R}^3$ is a function, it makes sense to say that $\alpha$ is one-to-one; that is, $\alpha(t) = ...


2

Perhaps it's best just to look at what's going on here geometrically. I said in one of your other questions that the Riemmann tensor and sectional curvature can be written directly in terms of the exterior algebra. Namely, given a simple 2-vector $B$, $$K(B) = \frac{\langle R(B), B \rangle}{\langle B, B \rangle}$$ where the inner product ...


2

Take "coordinate" projections $\pi_X,\pi_Y$ from $M \times N$ to $M$ and $N$. Let $(p,q) \in M \times N$, so that you have a map $F : T_{(p,q)}(M \times N) \to T_p M \times T_p N$ sending $v$ to $\left(d(\pi_X)_{(p,q)}(v), d(\pi_Y)_{(p,q)}(v) \right)$. This map is a linear map that is an isomorphism with inverse given by the linear map $g : T_p M \times T_p ...


2

Actually, you can't. Maybe some background information is of interest. What you found is the special case of a parametrization parametrized 'proportional to arclength', that is the length of a segment of the curve is a constant times the difference of the parameter values. (That means that you can show the minimizing curve is a straight line, but you cannot ...


2

This is a problem involving similar triangles. Can you see the related ratios: $$\frac{L-\sqrt{L^2-r^2}}{r}=\frac{2 L}{\rho }?$$


2

Not using Einstein summation convention: \begin{align} \frac{d}{dt} \left( t \omega_j(tx) \right) &= \left( \frac{d}{dt} t \right) \omega_j(tx) + t \left( \frac{d}{dt} \omega_j(tx) \right) \\ &= \omega_j(tx) + \left( \sum_i t \frac{\partial \omega_j}{\partial x^i}(tx) \frac{d(t x^i)}{dt} \right) \\ &= \omega_j(tx) + \left( \sum_i t ...


2

Suppose you are considering the level set $$f(x,y)=c$$ and we want to study whether we can parametrize this curve near the point $(x_0,y_0)$, which is a point in it, $f(x_0,y_0)=c$. If one of the partial derivatives is non-zero at that point, say $\frac{\partial f}{\partial y}(x_0,y_0)\neq0$ then we can use the Implicit function theorem to argue that ...


2

As I said in my comment, isometric or conformal type of a surface in $R^3$ is independent of a parameterization. This is an answer in the conformal setting (isometric setting is hopeless). You need: Uniformization Theorem: Every simply-connected Riemann surface is conformal either to the unit disk, or to the complex plane or to the 2-sphere. In view of ...


2

I assume that you are on an orientable compact manifold. To show that $\int_M \omega \neq 0$, recall that $\int_M$ is defined using parition of unity: Let $\{U_i\}$ be an open cover of coordinate charts and $\phi_i$ is a partition of unity subordinate to $\{U_i\}$. Then $$\int_M \omega = \sum_i \int_M \phi _i \omega = \sum _i \int_{ U_i} \phi_i \omega . ...


2

I would call it a cylinder — a right cylinder if the movement is normal to the plane of the original two-dimensional region. However, cylinder is used in enough different senses that I’d probably explain my usage first.


1

In varying contexts, I've seen this type of object called a (generalized) cylinder; a tube; or a prism. If you're looking for a term to use in some piece of writing, the main things are to use something simple and familiar, and to be explicit about your meaning if the term is non-standard (or there's other danger of ambiguity).


1

Your uncle's explanation concerns mostly the tangent plane to a surface and the normal vector, rather than derivations. The definition quoted afterwards is indeed the (algebraic) definition of derivation. However, there is a family of derivations which can be obtained geometrically by tangent vectors. Namely, if we have a manifold $M$, a point $p\in M$ and ...


1

First, it's not quite true that the fact that the bundle is one-dimensional forces $v = fw$, since $w$ could be zero at some point. (One can assume $w$ is never zero by doing the calculation locally, but you should be careful about this.) Second, it appears that you've made a small mistake applying the Leibniz rule for connections when moving from the ...


1

You can prove that $\int_U \omega > 0$ for any oriented coordinate chart $(U, \varphi)$ of $M$ just by definition: We have $\varphi(U) \subseteq \mathbb{R}^n$ and, using the coordinate definition of integration of $n$-forms, $$ \int_U \omega = \int_{\varphi(U)} \sqrt{ \det g_{ij} } \; dx^1 \cdots dx^n, $$ which is positive (since we are now integrating ...


1

That definition is plain bad. Good students are in general very modest and when they stumble across a difficulty they think it is their fault. This is often not true: the greatest mathematicians, I'm thinking of Grothendieck for example, have made real mathematical mistakes. Bad choices in definitions, false statements, inelegant proofs, etc. are very ...


1

let $$u = e^{6t}.$$ then we have $$(u^2 + 1)T = (\sqrt 2 u, u^2, 1) \tag 1$$ differentiating $(1)$ with respect to $u,$ gives $2uT + (u^2+1)\frac{dT}{du} = (\sqrt 2, 2u, 0)$ which can be simplifies to give $$(u^2 + 1)^2\frac{dT}{du} = (u^2 + 1)(\sqrt 2, 2u, 0) - 2u(\sqrt 2 u, u^2, 1) =(\sqrt 2(1-u^2),2u,-2u) $$ therefore $$N = ...



Only top voted, non community-wiki answers of a minimum length are eligible