# Tag Info

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Your metric matrix $G$ is also sometimes referred as the first fundamental form (for surfaces in $\mathbb{R}^{3}$), or, more generally, as metric tensor. This matrix is used to define inner product on all the tangent planes of $S$. As you can image, defining inner product opens a huge number of applications for $G$. Knowing how to define inner product ...

1

You can use $\displaystyle S=\int_1^2 2\pi R(y)\sqrt{1+(f^{\prime}(y))^2}dy$ where $R(y)=y$ is the distance from a typical point $(x,y)$ on the curve to the axis of rotation $y=0$, so this gives $\displaystyle S=\int_1^2 2\pi y\sqrt{1+y^2(y^2+2)}dy=2\pi\int_1^2y\sqrt{(y^2+1)^2}dy=2\pi\int_1^2y(y^2+1)dy$

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The action of $\Gamma$ on $S^1$ is ergodic with respect to the Lebesgue measure class on $S^1$, and the set of points fixed by elements of $\Gamma$ is countable. So no, there is no good notion of a fundamental domain for the action on $V$.

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Yes, it is true. In brief, the reason is that there are only finitely many possibilities for the surface $S \setminus \alpha$ that is obtained by cutting $S$ along $\alpha$. Let me give some details. Consider such an arc $\alpha$. Cutting along $\alpha$, let $S \setminus \alpha$ denote the resulting bordered surface, which inherits marked points on its ...

-1

$I=\int \int_Vxy\ dx\ dy = \int_{x=-1}^1\int_{y=-\sqrt{1-x^2}}^\sqrt{1-x^2} xy\ dy\ dx$ Change of coordinates - $x=r\cos \theta$ $y=r\sin \theta$ Jacobian - $I=\int \int r^2\cos\theta \sin\theta\ \left\|d\frac{x,y}{r,\theta}\right\|\ d\theta\ dr$ $\left\|d\frac{x,y}{r,\theta}\right\|=\left|\begin{array}{c}\cos \theta&\sin\theta\\ -r\sin ... 0 The first integral equality is not correct---you seem to be integrating with respect to$dv$but over the interval for$u$. Since the integrand does not depend on$v$, we can write $$\int_0^{2 \pi} \int_0^1 u \sqrt{u^2 + 1} du\,dv = \int_0^{2 \pi} dv \int_0^1 u \sqrt{u^2 + 1} du = 2 \pi \int_0^1 u \sqrt{u^2 + 1} du$$ 0 This problem will not in general have a nice closed-form analytic solution, because it depends on many input parameters (semiaxes and angles of all the ellipses, how many ellipses there are and so on). Using analytic final results would maybe make sense for special cases with less parameters (such as two ellipses at right angles), but for more ellipses the ... 1 Notice that, by the usual trigonometric identity, $$y^2+z^2 = 9x$$ i.e. $$\dfrac{y^2}{3^2}+\dfrac{z^2}{3^2} - x = 0$$ Which would normally be a circular paraboloid extending in the$x$-direction. (Easily seen by noting that, for any fixed$x$, we have a circle of radius$3\sqrt x$in the plane that is parallel to the$y,z$-plane at a distance$x$from ... 0 The interesting question is the action on the largest open subset of the reals on which$\Gamma$acts properly discontinuously (denote it by$\Omega$). Then$\mathbb{H} \cup \Omega/\Gamma$is a complex Riemann surface with boundary.$\Omega/\Gamma$is called the ideal boundary of$S$. It may be empty. It plays an important role in Teichmüller theory, see ... 2 There are two conventions for spherical polar coordinates. For concreteness, I will use the physicist version here: $$[0,2\pi] \times [0,2\pi) \ni (\theta,\phi) \quad\mapsto\quad \hat{n}(\theta,\phi) = (\sin\theta\cos\phi,\sin\theta\sin\phi,\cos\theta) \in \mathbb{R}^3$$ When we deform the unit sphere slightly, we can continue to use$(\theta,\phi)$as a ... 0 In cylindrical coordnates, you have, after some simplification $$z=r^{2}(\cos ^{2}\theta -\sin ^{2}\theta )=r^{2}\cos 2\theta$$ and $$r= a\sin \theta \cos \theta =\frac{\vert a\vert }{2} \sin 2\theta$$ where in the latter equation we have taken the positive root since$- \sin 2\theta = \sin 2(-\theta )$and so the graph is the same. The graph of the ... 2 You're right that there are a very large number of edge symbols that you can get by pairwise identifying the edges of an octagon. However, many of them yield homeomorphic surfaces. Here are a few facts that might help you in your thinking: The classification theorem for surfaces states that any compact 2-manifold is homeomorphic to either a sphere, a ... 1 Clearly one direction of the line is given by${\bf n} = (3,2,-\sqrt{2})$. If the plane is perpendicular to the line, then the normal direction to the plane, the gradient, must be parallel to our$\bf n$. You want$(x_0,y_0,z_0)$in the surface such that$\nabla f (x_0, y_0, z_0) = \lambda {\bf n}$for some$\lambda \in \Bbb R$. You want to solve ... 0 It's not a real answer, but may serve for you, since you just want to calculate the volume. Just doing the transformation you suggested, on can write:$u^2 + \frac{3}{4}y^2 + z^2 \leq 1$Whose Jacobian of this transformation is 1. So, you can calculate the volume of this ellipsoid, which is the same as the one you're looking for:$\frac{u^2}{1^2} + ...

1

Let $\mathbf{y}:E\to M\subseteq\mathbb{R^3}$, $M$ a surface, be a proper patch. Then $\mathbf{y}^{-1}:\mathbf{y}(E)\to E$ is continuous. If $\mathbf{y}(E)$ contains any boundary points (i.e. it is not open), the function $\mathbf{y}^{-1}$ cannot be continuous, as a boundary point has no equivalent in an open set (every point in an open set has a ...

1


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It seems to me to be a thickened (and rather heavily stylized) Möbius strip, i.e., a torus with square cross section that is given a one-half twist. I made this image just now using the code from my math.SE question, Drawing a thickened Möbius strip in Mathematica

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It looks like an umbilic torus: From Wikipedia. Zev Chonoles pointed out another cover with the equation for the umbilic torus:

1

This is confusing. For i), I suppose that you speak about embeddings into an orientable manifold, such as $\Bbb R^n$? However, the Mobious strip embedds into $\Bbb R^3$ and is clearly not orientable. For closed hypersurfaces in $\Bbb R^n$, you may find these links useful: here, here and here.. Not sure if it can be proved by a completely elementary and ...

2

The restriction is not meant to avoid pathological behavior; I just put it there for convenience. For example, it would be perfectly reasonable to interpret $\langle a,b\mid aa^{-1}, bb^{-1}\rangle$ as a presentation of a disjoint union of two spheres, or $\langle a,b\mid ab, ab^{-1}\rangle$ as a presentation of a projective plane. But we don't need these ...

0

Another way to proceed, which is a bit different from polygonal presentations, is to think about connect sums. If $S$ and $T$ are surfaces, then we can form $S \# T$, a new surface, by cutting a disk out of each, and gluing together the resulting circle boundaries. For example, if $S$ and $T$ are both copies of $T^2$ (torus) then $S \# T$ is a copy of the ...

0

This is only a topology problem. Since that $M\subset \mathbb{R}^{3}$, $M$ have a relative topology induced by the topology on $\mathbb{R}^{3}$, $i.e.$, the open sets (or open neighborhood) $V$ on $M$ are open subsets on $\mathbb{R}^{3}$ intersection $M$ ($V=U\cap \mathbb{R}^{3}$). If $\chi: D\rightarrow M$ is a proper patch then this is a continuous ...

1

In a polygonal presentation, $\langle S | W_1, W_2, \cdots, W_n\rangle$, if any letter appears in any word $W_i$ more than twice (so that your word looks like $\cdots aaaa \cdots$, as an example), then in the geometric realization, you have to paste multiple consecutive edges in the same orientation, so that every neighborhood of any point on one of the ...

3

Interesting. That exercise got me thinking for awhile, I had no choice but to open everything in coordinates, but I nailed it. I wonder if there's a better way to do it. What you did so far is correct. Since we want to compare these forms in ${\bf x}(D)$ (and you should really use ${\bf x}$ instead of $x$ here, for reasons that'll be clear shortly), it is ...

1

The patch you found is good. Let's write everything neatly. We have: $${\bf x} \colon \Bbb R \times \left]0,2\pi\right[ \to M \\ {\bf x}(u,v) = (\cosh u \cos v, 3\cosh u \sin v, 2\sinh u)$$ This won't cover the whole surface: one meridian will be left out: the curve $(\cosh u,0,2\sinh u)$, which would correspond to ${\bf x}(u,0)$, but our domains must be ...

1

I'm putting what I have wrote in the comments together in this answer : Claim The polygon $abab$ with all the vertices identified is homotopy equivalent $\Bbb RP^2 \vee S^1$. First consider the polygon $abab$ with diagonally opposite pair of vertices identified, which is homeomorphic to $\Bbb RP^2$. Since all of the vertices must be identified, attach a ...

1

Answer to Q1) Nope. In order to use divergence theorem, the given area should be a boundary of some 3-dimensional object. Instead, you can use Stokes' theorem with $\partial S$ composed with 6 circles of radius 3. (The subset with $|x|,|y|,$ or $|z|=4$) $S$ is a smooth surface with boundary because in any point on the interior of $S$ you can find a local ...

1

Axis-symmetric surface of torus can be shear deformed by adding a torsion term to $z$ as $c \cdot \theta$. $$(x,y,z)= (a+ b \cos \phi ) \cos\theta,(a+ b \cos\phi )\sin\theta , b \sin \phi + c \cdot \theta$$ $a$ is spring radius, $b$ is coil radius, $c$ is torsional radius of curvature. The above is written with respect to $\theta$. To convert to arc ...

1

I reckon that the original reference for the example you mentioned is the paper by A. Martin, "Indefinite Einstein hypersurfaces with nilpotent shape operators". Hokkaido Math. J. 13 (1984) 241-250 - more precisely, Example 3.1 therein. In this example, the shape operator is nilpotent. More generally, Theorem 4.1 of this paper gives a characterization of ...

0

All that is required by the change of variables is that it is 1-1 and onto. Letting $v=y$ and $w=z$ will work just fine. In matrix form, this is $$\begin{bmatrix}u\\v\\w\end{bmatrix}=\begin{bmatrix}1 & 1/2 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}.$$ Since the determinant is nonzero, the change of ...

0

The variables $x, y, z$ are coordinates, so it doesn't make sense to speak of a "sphere of radius $\sqrt{1 - x y}$". One option is to eliminate the cross-term by making the linear change of variables, $$x = u + v, \qquad y = -u + v;$$ substituting gives that, in these coordinates, $S$ is given by $$u^2 + 3 v^2 + z^2 = 1.$$

2

A surface doesn't have a volume. Presumably what is meant is the volume of the region enclosed by the surface, namely $R = \{(x,y,z): x^2 + x y + y^2 + z^2 \le 1\}$. Hint: complete the square with respect to either $x$ or $y$. You'll find cross-sections that are disks...

0

Hint for the volume inside the surface $S$. Find a linear transformation of the space such that the image of the surface $S$ is the sphere centered on the origin with radius equal to $1$. Then use change of variables theorem for integrals. Giving some more details $$(x,y,z) \to x^2 + xy +y^2 + z^2$$ is an inner product. Trying to find an orthogonal basis, ...

1

Pretty much everything you wrote is technically correct - any open subset of $\mathbb R^3$ is a 3-manifold, and if you include the boundary then it is a 3-manifold with boundary; and it inherits the Riemannian metric from $\mathbb R^3$. The geometry you get is not very interesting, though - since curvature is a local notion, the fact that your set is open ...

1

A $3$-manifold, seen inside $\Bbb R^4$ is nothing more than a hypersurface. We have this generalization of the Gaussian curvature, called the sectional curvature, which for $2$-manifolds reduces to the Gaussian curvature that we already know. This indeed uses the Riemann curvature tensor. From this you can compute the scalar curvature using the Ricci tensor, ...

0

Maybe some programs will plot the intersection exactly as you've written it. Mathematica won't, but there is a trick to doing it here. For the spheres, you can work out the centre of the circle, it's the halfway point between the centres of the spheres. Then you need the plane containing the circle, and the radius of the circle. In general, suppose you ...

0

You are almost there, you just throw away too much: $$\begin{split} \int_M |K| &= \int_{K>0} K - \int_{K<0} K = \int_{K>0} K - \left(\int_{K<0} K + \int_{K>0} K - \int_{K>0} K\right)\\ &= \int_{K>0} K - \left(\int_M K - \int_{K>0}K \right)\\ &= 2\int_{K>0} K - \int_M K\ . \end{split}$$

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