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6

There's also a cylinder. That's it. You can prove this by fully classifying 2-dimensional Lie groups. It's much easier to classify 2-dimensional Lie algebras, of which there are two up to isomorphism, and hence 2 simply connected 2-dimensional Lie groups up to isomorphism: $\Bbb R^2$ and $\text{Aff}(1)$, the affine transformations of the line. Now one ...


5

It looks like polar would be good for this, i.e. $x=u\cos v,\ y=u \sin v$ where $1\le u \le 5$ and $0 \le v \le 2\pi.$ [The last should technically have $v<2\pi$ but that doesn't matter in integration.] Then $z$ from your formula. $z=\log u^2$ Oops it's $1 \le u \le \sqrt{5}.$


3

By the Spectral Theorem, you can orthogonally diagonalize the first fundamental form matrix $A=\begin{bmatrix} E&F\\F&G\end{bmatrix}$, and from this it follows that $A\mathbf x\cdot\mathbf x \ge c\|\mathbf x\|^2$, where $c$ is the smaller eigenvalue of $A$. Since the first fundamental form is everywhere positive definite and varies continuously on ...


3

If a geometric object has a particular symmetry, generally parameterizations that reflect that symmetry result in easier computations. In our case, $S$ is the surface of a graph whose domain $A := \{1 \leq x^2 + y^2 \leq 5\}$ is an annulus centered at the origin. This suggests letting one of our parameterization variables be the distance $r$ of a point $p ...


2

First you need to find the point(s) where you have to give the tangent plane. For the gradient you have: $$\nabla u = \left( \frac{\partial u}{\partial x},\frac{\partial u}{\partial y} \right) = \left( \frac{1}{x+1/y},\frac{-1/y^2}{x+1/y} \right)$$ And you're looking for the points $(x,y)$ where this is equal to $\left( 1,-\tfrac{16}{9}\right)$, so: ...


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

Just try and calculate. The easiest example of a cusp is $\text{Spec }k[t^2,t^3]\to\text{Spec }k[x,y]$, where $x\mapsto t^2$ and $y\mapsto t^3$. The completion of the plane at the origin is $\text{Spec }k[[x,y]]$, formal power series in two variables. The fiber of the cusp over the completion is $k[[x,y]]\otimes_{k[x,y]}k[t^2,t^3]=k[[t^2,t^3]]$, formal power ...


2

Note that by setting $z = x + iy = re^{i\theta}$, your expression is equal to $$ f(x,y) = \operatorname{Re} \left( \frac{z^4}{|z|^2} \right) = \frac{\operatorname{Re} \left( (x + iy)^4 \right)}{x^2 + y^2} = \frac{x^4 - 6x^2y^2 + y^4}{x^2+y^2} = \frac{8x^4}{x^2+y^2} + y^2 - 7x^2. $$ Now, a priori $f$ is not defined at $(0,0)$. However, the polar expression ...


2

In a triangulation (specifically, a simplicial complex), the three vertices of a triangle are distinct. (Technically, the two 0-cells at the boundary of each 1-cell are distinct, the three 1-cells at the boundary of each 2-cell are distinct, et c. This leads to: the vertex set of a $k$-cell contains $k-1$ distinct vertices.) That is, if I tell you three ...


2

One technique that I found useful when I was teaching calc III this past semester is to parametrize a surface of the form $z=f(x,y)$ as $\mathbf{r}(x,y)=(x,y,f(x,y))$, get a double integral over some region, and then change coordinates in that integral. This is an alternative to directly choosing your parametrization in the "convenient" coordinates (in this ...


1

When you use the cross product method, the Jacobian takes care of itself. Along the surface the $\vec r=\langle x,y,z\rangle=\langle r\cos\theta,r\sin\theta,2\ln r\rangle$. Then the total differential along the surface is $$d\vec r=\langle\cos\theta,\sin\theta,\frac2r\rangle\,dr+\langle-r\sin\theta,r\cos\theta,0\rangle\,d\theta$$ Then we can get the vector ...


1

Before the demonstrations, remember a proposition: Proposition $\bigstar$: Let $S\subset\mathbb{R}^{3}$ an compact connected surface; then one of the values $2,0,-2,\ldots,-2n,\ldots$ is assumed by the Euler-Poincaré characteristic $\mathcal{X}\left(S\right)$. Furthermore, if $S'\subset\mathbb{R}^{3}$ is other compact surface and ...


1

Here's a sketch of a proof that a $4g+2$-gon with opposite sides identified is a surface of genus $g$. The first step is to show it is a surface: Every point has a neighborhood homeomorphic to an open disk. Around interior points of the polygon this is obvious, and around points on the interiors of edges this is also obvious: two half disks glue together to ...


1

Hint: What are the images of the corners of the big square in the quotient? See also this Q&A (and the comments), where the poster makes much the same mistake.


1

OK for the first part (and a product of two compact sets is compact). For the second, fix $y$ and take any smooth curve on $M$ passing through $x$ at $t=0$. Express the fact that the distance from $y$ to $x(t)$ has a maximum at $x = x(0)$. Use that to show that the vector from $x$ to $y$ is orthogonal to any vector tangent to the surface at $x$. Edit: a few ...


1

One way of seeing this surface is slicing it with a plane parallel to the plane $z=0$. That means setting $z=k$ and imagining what the section should look like. In your case, \begin{equation} z=k\quad\Longrightarrow\quad x^2y=3k\quad\Longrightarrow\quad y=\frac{3k}{x^2} \end{equation} so if $k>0$ the sections seen from "above" look like the function ...


1

Well, you say you want to show that $\langle\alpha'(u),\beta(u)\rangle = 0$. But, as you go on to say, you really want to choose a different directrix curve $\tilde\alpha(u)$ so that you'll have $$\langle\tilde\alpha'(u),\beta(u)\rangle = 0\tag{$\star$}.$$ So you want to choose a scalar function $\lambda(u)$ so that, setting $\tilde\alpha(u) = ...


1

I had seen usage of terms oblate/ prolate ovaloids in American literature.


1

Use "oval" or "ovoid" for such curves and surfaces respectively. Make sure to provide your own definition for these words though.


1

Three vectors $x,y,z$ in a plane are necessarily linearly dependent. Without loss of generality, we may assume $z$ is a linear combination of $x$ and $y$, i.e. $z=\lambda x+\mu y$. Then $$(x\wedge y)z+(y\wedge z)x+(z\wedge x)y=\lambda(x\wedge y)x+\mu(x\wedge y)y+\lambda (y\wedge x)x+\mu(y\wedge y)x+\lambda(x\wedge x)y+\mu(y\wedge x)y.$$ Using skew-symmetry ...



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