# Tag Info

4

Your comment seems to be very much on the right track. The $\nabla f \ne 0$ condition tells us that the level curves are embedded submanifolds. We can exclude closed curves using the gradient condition along with the topological properties of $\mathbb R^2$ (e.g. Jordan curve theorem + extreme value theorem), but it's possible that the level sets can be a ...

4

If you fix a point $p \in U \subseteq \mathbb{R}^n$, then a differential $1$-form is a linear map from $T_pU$ (the tangent space to $U$ at $p$) to $\mathbb{R}$. The tangent space is spanned by the vectors $\frac{\partial}{\partial x_i}$, where $\frac{\partial}{\partial x_i}$ can be thought of as a vector pointing in the $x_i$ direction." Then you can ...

4

A coordinate mapping (such as polar coordinates in the plane) is only unique if the implicit function theorem is valid on the region. The implicit function theorem in n-dimensional Euclidean space states that if the Jacobian matrix, i.e. the matrix of partial derivatives, of the coordinatized mappings defining the graph of the function on an open subset of ...

3

There are topological constraints, even in the plane. For instance, an annulus cannot be continuously parametrized by the unit square or unit disk because an annulus is not simply connected. On the other hand, the Riemann mapping theorem says that every non-empty simply connected open subset of the plane can be nicely parametrized by the open disk. Here ...

3

Yes, if you have a vector bundle $E \to M$ with a fiber metric $h$, then there is a natural isomorphism induced by the usual identification of a vector space with its dual in the presence of an inner product (not necessarily of definite signature): The map $\Phi: E \mapsto E^*$ is given fiberwise by $$\Phi_p: v \mapsto h_p(v, \,\cdot\,),$$ which is ...

3

First, you need to stipulate that $M$ is connected for the distance function to turn $M$ into a metric space. When $M$ is connected, the answer is yes, the metric topology is the same as the given manifold topology. The proof boils down to showing that the Riemannian metric is uniformly comparable to the Euclidean metric in small coordinate balls. You can ...

3

As far as I can see the problem is that $\partial_{\theta_{i}} E_{\theta}[\ln(p)] \neq E_{\theta}[\partial_{\theta_{i}} \ln(p)]$ You wrote $E_{\theta}[\partial_{\theta_{i}} \ln(p)]$ correctly while the other term is $$\partial_{\theta_{i}} E_{\theta}[\ln(p)] = \int \left ( \partial_{\theta_{i}} \ln(p(x;\theta)) \right ) p(x;\theta) dx + \int ... 3 No, it involves projecting F(M)\to F(\partial M), i.e. observing that a smooth function on M restricts to a smooth function on \partial M. This seems to go the wrong directions, but now the derivations come into play: a derivation  F(\partial M)\to\mathbb R gives rise to a map F(M)\to F(\partial M)\to\mathbb R (that is also a derivation, as you may ... 2 In linear algebra it is important to understand the distinction between a linear transformation T:V\to W and a matrix A which represents T. Different choices of bases for V and W will give different matrices. The symbol T is "coordinate-free" but the matrix A is not. Another example is an inner product on a vector space. It exists ... 2 The error is here:$$ E_{\theta}\left [ \frac{\partial}{\partial \theta_{i}} \ln(p(x; \theta)) \right ] = \frac{\partial}{\partial \theta_{i}} E_{\theta} \left [ \ln(p(x; \theta)) \right ] $$This is incorrect as the density depends on \theta: when writing the derivative there is one additional term$$ E_\theta\left[\ln p(x;\theta) \frac{\partial ...

2

A $1$-form is a map $\omega:U\rightarrow T^*U$. Now, if $X:U\rightarrow TU$ is a vector field then you can consider $\omega_p(X_p)$, in that sense, you can apply the $1$-form to a vector field. $TU$ is a vector $bundle$ not a vector field.

2

For each $x \in U$, the tangent space $T_x U$ is a vector space. A "differential form at $x$" is a linear map $T_x U \to \mathbb R$; that is, it sends a vector to a scalar. Now when we say differential form we normally mean a field of such maps - i.e. a function that sends each $x$ to a differential form $\omega_x : T_x U \to \mathbb R$. Such a form can act ...

2

This is actually just the total derivative you probably learned in multivariable calculus. Remember this goes like so: if $F=F(x(t),t)$ then $dF/dt=\partial F/\partial t+\partial F/\partial x dx/dt.$ It's the same computation here.

2

When you write $dx^k$ in your standard form $\sum f_i dx^i$, you have $f_i = \delta_i^k$ (i.e., $1$ for $i=k$ and $0$ otherwise). Then $df_i=0$ and $d(dx^k)=\sum df_i\wedge dx^i =0$, as desired.

2

Yes, you've got it right. Remember the title of Halmos's book is Finite dimensional vector spaces! My guess is he probably thought the definition as the dual of bilinear forms was less abstract that either the quotient or the universal property definitions and used that one. As you've realized Halmos's definition only agrees with the others when the spaces ...

2

You have the crucial part of the argument. Fix $s_0\in S$, and consider the set $X = \{s\in S: s\in P\}$. By continuity, $X$ is closed, and by your proof, $X$ is open (there is a continuous, hence differentiable normal vector field on any simply connected open set). Thus, by connectedness, $X=S$.

2

One way to say in categorical terms what your remarks show is that there is a functor from the category $\bf{Man}^\simeq$ of smooth manifolds, where the morphisms are diffeomorphisms, to the category of sets; this functor assigns to a manifold the set of vector fields on it, and assigns to a diffeomorphism $F$ the pushforward function along $F$.

2

The curvature may of course still be defined if $t$ is not the arc-length along $\gamma(t)$, but the formulas will be different; we will not in general have $\kappa(t) = \Vert \gamma''(t) \Vert$. To find the correct expression for $\kappa(t)$, we recall the the curvature is defined in terms of the arc-length $s$ via the Frenet-Serret equation for $T'(s)$, ...

1

Yeah, it should really by the corresponding open intervals to be a genuine 'chart'. Moreover, all sorts of things are bad if we allow the boundary -- e.g. $\phi$ isn't bijective anymore since as soon as $r=0$, it no longer matters what the angle coordinates are. I think they're just doing a little abuse of language, since it's convenient to allow the ...

1

You know that if you integrate over a loop into which there is no singularity, as the form is exact the integral is 0. Then let us try with a loop containing 0 inside: let us take the circle $C(0,1)$. $$\int_{C(0,1)} -\frac y{x^2 + y^2}dx + \frac x{x^2 + y^2}dy = \int_0^{2\pi} -\frac {r\sin\theta}{r^2} (-r \sin\theta d\theta) + {r\cos\theta}{r^2} (r ... 1 Letting X_{1}, X_{2}, \ldots, X_{l} be a local frame for M_{1} and Y_{1}, Y_{2}, \ldots, Y_{m} be a local frame for M_{2}, the Levi-Civita connection \nabla of the product metric can be shown to satisfy the following: \nabla_{X_{i}}{X_{j}} =\tilde{\nabla}_{X_{i}}{X_{j}}, where \tilde{\nabla} is the Levi-Civita connection of M_{1}, ... 1 Your simplification is correct. It shouldn't be too surprising to find such a simple expression - since I is one-dimensional, every (0,2)-tensor on it is the product of g_I with some smooth function. I'm unsure why the authors didn't make this simplification - perhaps because the equation they give is slightly more general, while your simplification ... 1 If I remember correctly, what you are asking follows from Hopf-Rinow Theorm. Here is the link from wiki. But I recommend you to read another book by Do Carmo called Riemmanian Geometry. 1 In terms of any (C^1) parameterization X(u, v) of a surface, the area element is$$dA = \left\vert\left\vert X_u \times X_v\right\vert\right\vert \,du\,dv,$$so you can simply substitute. 1 Proving that the set of regular values of f is open is equivalent to proving that the set of critical values is closed. The set S of critical points of f is closed. (One way to show this is to show that the set of regular points is open. This can be done using a local parameterization.) Since X is compact, this means that S is compact. By ... 1 We only have to show that the condition in Wikipedia implies the one by Warner. Let$$ M_i = \{x \mid \varphi_i(x) \neq 0\}.  By assumption, this is a locally finite family. We want to show that this is also true of the family $\overline{M_i}$. This is true in general, because for each $x$, there is an (open) neighborhood $U$ with $M_i \cap U \neq ... 1 I am trying to answer your first question: "I was wondering if there is a simple explanation of the torsion and curvature in R3 of a curve". Torsion of a curve measures the planarity of the curve. That is, the curve is planar (i.e. it lies on a plane) if and only if its torsion is identically equal to zero. You can see here. 1 It converges uniformly in$M\times M$for each$t>0\$. If I remember correctly, a proof is given in Rosenberg's Laplacian on Riemannian manifolds. Chavel's book must also have something on this. In any case the proof is straightforward if you use Weyl's law (Please try).

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