# Tag Info

11

The usual Euler characteristic is absolutely the correct choice, at least for closed manifolds. There are lots of reasons to believe this: $\chi(X \sqcup Y) = \chi(X) + \chi(Y)$, where $\sqcup$ is the disjoint union. (This is the coproduct in the category of, say, topological spaces.) Under some conditions we even have the "inclusion-exclusion" formula ...

4

Try the cone $M=\{z=\sqrt{x^2+y^2}\}$ (missing its vertex $(0,0,0)$. Holonomy is nonzero around any parallel $z=c$. This happens because that curve does not bound a region in $M$.

3

The statement Clearly $\widehat S$ contains exactly $r$ (-1)-curves is false. For example, blow up $\mathbf P^2$ in 2 points. Then the exceptional curves are (-1)-curves, but there is another (-1)-curve on this surface: namely the proper transform of the line that went through the two points. If we contract just that last (-1)-curve, we get the ...

3

The best reference I know for this is Expose XVII in SGA7 II by N.M. Katz "Pinceaux de Lefschetz: theoreme de monodromie" I am sure there are English texts by now (probably also written by Katz) which are even available online. Let me briefly comment on where to find the relevant results in SGA7. Let $k$ be an algebraically closed field. The ...

2

The intersection of the cylinder with the $x,y$ plane can be parametrized by $x(t)=1+\cos t,\ y(t)=\sin t.$ Now since the circle has radius $1$ the element $dt$ has unit differential length, and the height above the point $(x(t),y(t))$ is $$x(t)^2+y(t)^2=(1+\cos t)^2+ (\sin t)^2,$$ which integrated for $t$ from $0$ to $2 \pi$ gives the surface area of the ...

2

Any smooth surface in $\mathbb R^3$ that is a closed subset of $\mathbb R^3$ is complete as a metric space. Thus, if it is also connected, the Hopf-Rinow theorem guarantees that it is geodesically complete, which in turn implies that any two points can be connected by a minimizing geodesic. Therefore, the only ways that geodesic completeness can fail are ...

2

Punctures and boundary components both decrease the Euler characteristic, and the results are homotopy equivalent. For example, $\chi(S^2) = 2$. Adding a puncture gives $\chi(\mathbb{R}^2) = 1$. Adding a boundary component gives $\chi(D^2) = 1$. Puncturing the torus once gives a wedge of $2$ circles, with Euler characteristic $-1$, and puncturing it again ...

2

We have to reparametrize the meridian curve $t\mapsto\bigl(r(t),z(t)\bigr)$ in such a way that the metric tensor obtains the desired form. Therefore we consider the representation $$g(\tau,\phi):=f\bigl(p(\tau),\phi\bigr)=\bigl(r(p(\tau))\cos\phi,r(p(\tau))\sin\phi,z(p(\tau))\bigr)$$ of the same surface, where $\tau\mapsto t=p(\tau)$ is a monotonically ...

1

There are two curves. They can be parametrized as follows. $\gamma_1 :[0,2\pi]\to\mathbb{R}^3$ given by $\gamma_1(t)=(\cos t,\frac{1}{\sqrt{2}}\sin t,\frac{1}{\sqrt{2}}\sin t)$, and $\gamma_2 :[0,2\pi]\to\mathbb{R}^3$ given by $\gamma_2(t)=(\cos t,\frac{1}{\sqrt{2}}\sin t,-\frac{1}{\sqrt{2}}\sin t)$. So $C=\gamma_1([0,2\pi])\cup \gamma_2([0,2\pi])$. ...

1

From $x^2+y^2+z^2=1$ and $x^2+2z^2=1$ it immediately follows that $z^2-y^2=0$, or $z=\pm y$. Conversely, $x^2+y^2+z^2=1$ and $z=\pm y$ imply $x^2+2z^2=1$. Therefore the intersection of $S^2$ and the elliptical cylinder coincides with the intersection of $S^2$ with the two planes $z=\pm y$, i.e. with the union of two unit circles intersecting at $(1,0,0)$. ...

1

Let $f:X\to C$ be a relatively minimal projective family of curves with $X$ non-singular such that all (but a finite number of) fibres are isomorphic (analytically or algebraically...). Assume the fiber genus of $f$ is at least two. (In genus one you have to be a bit more careful.) In this case, $f$ is "isotrivial". So that $f$ becomes a trivial family ...

1

I interpret being 'in front of the $yz$-plane' as the first coordinate being non-negative, thus the surface $S$ in question $\{(x,y,z)\in \mathbb R^3\colon x\ge 0\land 5x^2 − 5y^2 − z^2 = 5\}$. It can easily be proved that this set equals $\left\{\left(\sqrt{1+y^2+\dfrac{z^2}5}, y,z\right)\colon y,z\in \mathbb R\right\}$, therefore S=\vec r\left[\mathbb ... 1 Usually, conic sections are parametrized best with some kind of cylindrical or spherical coordinate transformation. Given $$5x^2 - 5y^2 - z^2 = 5, \quad x > 0,$$ this suggests a cylindrical parametrization with $$(y,z) = (u \cos v, \sqrt{5} u \sin v),$$ which easily gives $$5x^2 = 5 + 5y^2 + z^2 = 5 + 5u^2 (\cos^2 v + \sin^2 v) = 5(u^2+1),$$ or x = ... 1 It seems to me that the argument about continuity is correct, because continuity and holomorphicity are local properties. So, if f|{U_r} is continuous for all r, then f is itself continuous. (It is enough to check on an open cover) The derivative of function f(z)=z^n is non-zero at all z\neq 0. So for all neighbourhoods of z\neq 0 we can apply ... 1 I think that part of the problem might be that you're trying too hard. First, you're using some definition to say that \begin{align} I &= \frac{1}{4} \int_S dS \\ &= \frac{1}{4} \int_u \int_v \| \frac{\partial \Phi}{ \partial u} \times \frac{\partial \Phi}{ \partial v} | ~du ~dv \end{align} where I've left the limits on u and v unwritten ... 1 If S can be defined over \overline{\mathbb Q}, then there exists a model S_0 for S over \overline{\mathbb Q} (with respect to some embedding of \overline{\mathbb Q} \to \mathbb C). Choose your Lefschetz pencil on S_0 and note that the critical points of this Lefschetz pencil lie in \mathbb P^1(\overline{\mathbb Q}); see the comment of Dori ... 1 If you did that, you'd just move the 2 to other places. For example, you'd have state the Poincaré-Hopf theorem as follows: Theorem. If S is a surface and \chi'(S)\neq2, then there is no nonzero tangent vector field to S. 1 Compact surfaces are completely classified. H_2(X,\mathbb{Z}) is either zero or \mathbb{Z} depending on whether your surface is orientable. There are non-orientable surfaces that don't satisfy your condition, for example \mathbb{R}P^2: H_1(\mathbb{R}P^2,\mathbb{Z}) = \mathbb{Z}_2 If you restrict yourself to compact orientable surfaces then ... 1 As krey says, this is false for compact real surfaces, and \mathbb{RP}^2 gives a counterexample. A general statement along these lines is that if X is any reasonable space (this includes all compact manifolds, not just surfaces), then there's an identification between the torsion part of H_i(X, \mathbb{Z}) and of H^{i+1}(X, \mathbb{Z}) (so for the ... 1 1020784 The minimum IS 9. The equation for a quadric surface, of which the ellipsoid is a special case, in terms of nine of its points [(x_1\mid y_1\mid z_1)\dots (x_9\mid y_9\mid z_9)] is \begin{vmatrix} x^2&y^2&z^2&y·z&z·x&x·y&x&y&z&1\\ ... 1 Let c \colon I \rightarrow \mathbb{R}^3 be a smooth enough curve with arclength parametrization and consider the associated Frenet-Serret frame. That is, e_1(t) = c'(t), e_2(t) = \frac{c''(t)}{||c''(t)||} and e_3(t) = e_1(t) \times e_2(t). How does the frame change if we change the orientation of the curve? Define \tilde{c}(t) = c(-t). Then ... 1 The "cooling area" is given by 2\pi r^2 + 2\pi rh = 2\pi r(r+h) So the optimal choice is the cup with the biggest value of r(r+h) and fixed volume V = \pi r^2 h \Rightarrow h = V/\pi \cdot r^{-2},r^{\text{opt}} = \mathop{\rm arg\,max}_{r>0} r^2 + V/\pi \cdot \frac1r = \mathop{\rm arg\,max}_{r>0} r^3 = +\infty\$ Answer: Take the biggest radius. ...

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