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5

My favorite way to interpret the trace is as the average value of an associated quadratic form. Here's how that works. Let $V$ be an $n$-dimensional vector space, and let $T$ be a tensor on $V$. First let's consider the case in which $T$ is a tensor of type $(1,1)$, which we can also interpret as a linear map from $V$ to itself. Choose an inner product ...


4

If $\mathfrak g$ is a Lie algebra, and $V$ a representation of $\mathfrak g$, then you can define a Lie algebra structure on $\mathfrak g\oplus V$ by the formula $$[(X,u),(Y,v)]=([X,Y],X\cdot v-Y\cdot u)$$ This applies to your example via the standard action of vector fields on functions by derivation. Here are the details. The bracket is clearly ...


3

Since $Y$ is smooth in the variable $p$, and now $Y_{\varphi_t(p)}$ (or maybe this notation: $Y_{p(t)}$) is just the restriction of $p$ to the integral curve, and therefore also smooth. Think of it as a composition of the smooth map $t\mapsto p(t)$ with $p\mapsto Y_p$. On the other hand, $(\varphi_{-t})_\star$ is nothing but a jacobian (linear map) ...


3

Given a symplectic structure $\omega$ on a $(2 n)$-manifold $M$---that is a closed $2$-form such that $d\omega^n \neq 0$---Darboux's Theorem guarantees that for every point $p \in M$ there are local coordinates $(x^i, y^i)$ in which the coordinate representation of $\omega$ is $$ \hat{\omega} = \sum_{i = 1}^n dx^i \wedge dy^i.$$ We call such coordinates ...


2

A smooth manifold $M$ equipped with a closed, non-degenerate two-form $\omega$ is called a symplectic manifold. It follows almost immediately that a symplectic manifold is even dimensional. Darboux's Theorem: Let $(M, \omega)$ be a symplectic manifold. For any $p \in M$, there is a coordinate chart $(U, (x^1, \dots, x^n, y^1, \dots, y^n))$ with $p \in ...


2

You can help your intuition by saying: an immersion is locally an embedding. But this does not need to be globally. So what can happen is, that your immersed image contains self intersetcions. E.g. the Klein bottle can only be immersed in your mind, since it thinks in 3d. But it is obvious that an $n$-manifold can easier be immersed into some $n+k$-manifold ...


2

You cannot have a smooth arclength parametrization unless the curve itself is geometrically smooth. The standard example $\alpha(t) = (t^2,t^3)$ illustrates the issue: the curve has a cusp, where no tangent line exists. The map $\alpha$ manages to be smooth because it crawls across the cusp with vanishing speed. But if you attempt to travel along the ...


2

No! For example, take one of the 27 exotic differential structures $X$ on $S^7$ and delete a point $p\in X$ : you will obtain a differentable structure on the manifold $X\setminus \{p\}$, homeomorphic to $\mathbb R^7$. But there is only one differential structure (up to diffeomorphism) on $\mathbb R^7$ (or for that matter on any $\mathbb R^n$with $n\neq ...


1

Unless I'm missing something, this has nothing to do with Airy Function or differential geometry. Here is my solution the way I understand this problem; $$(w^2) ''-(10z+5)w=0 \iff 2w w'-(10z+5)w=0 \iff w(10z-2w'+5)=0 \iff w=0 \; \textrm{or} \; w=\frac{5z}{2} (z+1)+c$$


1

The Möbius bundle over $S^1$ has this property.


1

Let $f(t)=\alpha(t).\alpha(t)$. We want to minimize this function i.e. $f'(t_0)=0$ and so $$2(\alpha'(t_0).\alpha(t_0))=0$$


1

Here's my take: $\mathbb{R}^+$, if we assume it is the additive semigroup of positive real numbers, cannot be the structure group. The most compelling reason for this is that $\mathbb{R}^+$ is not a group. Earlier in the post, he talks about rescaling. The set of all allowable rescalings is also $\mathbb{R}^+$, but it is a group under multiplication. These ...


1

Here's a partial answer, which gives a more direct analogue of the matrix rank example but (like my earlier comment) ignores the product structure. It follows partially from the subadditivity property $$\text{rank}(A + B) \leq \text{rank} A + \text{rank} B$$ for $m \times n$ matrices $A, B$ that the rank of a matrix $A$ is the minimum number $r$ for which ...


1

Looking through the slides you linked, it looks like they can all be done without working in coordinates. Do you perhaps have specific identities in mind for which you mean ' possible to argue without coordinates.' You mentioned Cartan's magic formula can be proved without working in local coordinates, and so can $\mathcal L_{[X,Y]}\alpha=\mathcal ...


1

You seem to have switched notation in your question as there appears $\partial_i$ and $dx^i$, but you didn't start with a coordinate frame. If I understand your question correctly, then the answer is yes. What you are trying to find is $C_{ij}^k$ given by the formula $$\nabla_{E_i}\theta_j=C_{ij}^k\theta_k$$To find these functions evaluate both sides of this ...


1

It's been quite awhile since I've looked at a differential geometry book which covers this stuff, so this might not be exactly what is wanted here, but as I recall the connection's action on the dual basis $\theta_\mu$ may be found from its action in terms of the basis $E_\nu$, i.e., from the formula $\nabla_\gamma E_\nu = \sum_\kappa \Gamma_{\nu ...


1

Covectors are weighted averages. Keep the same data $\vec{x}$, the same style of inner product, $$\alpha_1 \cdot x_1 + \alpha_2 \cdot x_2 + \alpha_3 \cdot x_3 \ldots$$ but change the $\alpha_i$ in the above expression. This is "adjusting weights" in everyday parlance or "moving the covector" in mathematical words. People really do do this all the time, so ...


1

If you call $x=e^t$, with $x>0$, then $t=\log x$. Thus, the second coordinate is $y=t^2=(\log x)^2$. So the points in the curve are $(x,(\log x)^2)$, with $x>0$. If you want to write it as an equation, you may write $f(x,y)=0$, with $f(x,y)=y-(\log x)^2$, or $f(x,y)=x-e^{\sqrt y}$. In both cases you are limited to $x>0$, $y\geq0$.


1

Since $$ (\vec x-\vec x_0)^T\vec x'=0 $$ we have: $$ \frac{d}{dt}((\vec x-\vec x_0)^T(\vec x-\vec x_0))=2(\vec x-\vec x_0)^T\vec x'=0 $$ therefore: $$ (\vec x-\vec x_0)^T(\vec x-\vec x_0)=\text{constant} $$


1

This answer addresses the issue of the Laplacian on $S^1$ and not the issue of whether you are solving the differential equation correctly. Circles are completely classified by their radius $r$, or if you prefer, by their circumference $C$, where of course $C = 2 \pi r$. What I mean by this is that whether or not your circle is originally given to you as ...


1

HINT: I'm not sure how you discovered it was a helix. But once you know that it has constant curvature and torsion ($\kappa=1/4$, $\tau=-1/4$), the standard formulas for the curvature and torsion of a (circular) helix will tell you what $a$ and $b$ have to be.


1

Another way to describe the symplectic form $\omega$ on $S^2 \subset \mathbb{R}^3$ is as the volume form that the standard volume form $\text{vol} = dx \wedge dy \wedge dz$ on $\mathbb{R}^3$ induces on $S^2$ (using, as usual, the outward-pointing normal). At $p = (x, y, z)$, this normal is $(x \partial_x + y \partial_y + z \partial_z)_p \in T_p ...


1

No. Then we would have $a_1 e^{\lambda_1 k t} + a_2 e^{\lambda_2 k t} + a_3 e^{\lambda_3 k t}$ for all $t$. But the $e^{\lambda_i t}$ are linearly independent as functions (which follows from, say, the Vandermonde determinant, though for real $\lambda_i$ you can show this with a limit argument), so $a_1=a_2=a_3=0$.


1

A cylindrical helix is a curve on a generalized cylinder (take any space curve and form a cylinder by taking parallel lines through each point of the space curve) that makes a constant angle with the rulings (the parallel lines). A circular helix is such a curve on a right circular cylinder. Any such is congruent to a standard helix: $$\alpha(t)=(a\cos t, a ...


1

This is the so-called branching line (counter)example, which is related to a perhaps better-known example known as the line with two origins. To show that a space $X$ is not Hausdorff, it suffices to pick two points $x, y$ such that for every neighborhood $U$ of $x$ and $V$ of $y$, $U \cap V \neq 0$. We know that away from the points on the $y$-axis the ...



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