# Tag Info

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Theorem : A topological space $X$ has a simply connected covering space if and only if $X$ is path connected, locally path connected and semi-locally simply connected. We are interested in the "only if" part of the story. Let $(E, e_0) \stackrel{p}{\to} (X, x_0)$ be a based covering map. Pick a neighborhood $U$ of $x_0$ that is evenly covered by $p$. $U_i$ ...

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Given a holomorphic line bundle $L$ with trivialising open cover $\mathcal{U} = \{U_{\alpha} \mid \alpha \in A\}$ and trivialisations $\{\phi_{\alpha} \mid \alpha \in A\}$, Griffiths & Harris construct an element $t_L \in H^1(\mathcal{U}, \mathcal{O}^*)$. They then show that $t_L$ does not depend on the choice of trivialisations. In particular, if $L$ ...

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We need the hypothesis that $f$ is (either zero or) nonconstant. Assume WLOG that $f(x, y)$ is nonconstant in $x$. Write $f(x, y) = x^n g(y) + \text{lower terms}$ where $x^n$ is the highest order term in $x$ and $g(y)$ is a nonzero polynomial in $y$. For all but finitely many values $y_0$ of $y$, $g(y_0)$ is nonzero, so for all but finitely many values $y_0$ ...

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Does the tautological line bundle have a nowhere-zero section $\sigma$? (Consider $\sigma([x]) = f(x)x$ for some continuous function $f\colon S^n\to\Bbb R$.) The same argument will work in the case of $\Bbb CP^n$ if you think a bit.

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Let $\mathcal{O}_{T,t}$ be the local ring at $t\in T$. Then we want to show that $\mathcal{F_y}/\mathcal{O}_{C\times T,y}$ is flat over this ring, where $y\in C\times t$. It is well known that to check flatness of a coherent module $M$ over a Noetherian local ring $R$, it suffices to show that $Tor_1(M,k(R))=0$, where $k(R)$ is the residue field. So we ...

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The Plücker embedding is an isometry of $G(k,\Bbb C^n)$ to its image in $\Bbb P(\Lambda^k\Bbb C^n)$ with the standard Fubini-Study metric. In the moving frames notation, for example, the Kähler form on $\Bbb P^N$ is given by $$\frac i2\sum_{j=1}^N \omega_{0\bar j}\wedge\overline\omega_{0\bar j},$$ where $\{f_0;f_1,\dots,f_N\}$ is a unitary frame at the ...

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Notice that $f(T+1) = f(T+x) = f(T+x+1) = f(T)$. This gives you explicitly the Galois group and proves that the extension is Galois.

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Here are some things which make sense on complex manifolds, but not almost complex manifolds: Complex coordinates $(z_1, \ldots, z_n)$. In particular, vectors like $\frac{\partial}{\partial z_k}$ only make sense on a complex manifold. On complex manifolds, we have $\overline{\partial}^2 = 0$. This gives rise to the Dolbeault complex $$\Omega^{p,0}(M) ... 1 After getting the point of intersection of the given family of lines we need to simply calculate the distance between the given point (2,3) and the point of intersection of the family of lines i.e., (1,2) which is equal to$$d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$$so$$d= \sqrt{(2-1)^2+(3-2)^2}=\sqrt2$$Now by the definition of locus of a point (locus of a ... 1 First, assume that a polynomial p(t) \in \Bbb C[t] satisfies p(0) = p(1) = 0. Then there is a polynomial q(t)\in \Bbb C[t] such that p(t) = t(t-1)q(t). Expanding q(t) = c_1 + c_2t^1 + \cdots c_nt^{n-1}, we get that$$ p(t) = \sum_{i = 1}^n c_n t^n(t-1) $$In general, the term t^i(t-1)^j is the image of a monomial in x and y iff j \leq i \leq ... 1 An alternative description is that (x,y,z) belongs to the tangent surface if and only if the polynomial (in the variable s) 1+3sx+3s^2y+s^3z has a double root. The corresponding M2 commands to get the equation of the tangent surface are R=QQ[s,x,y,z] factor (discriminant(1+3*s*x+3*s^2*y+s^3*z,t)) The reason is that the twisted cubic corresponds to ... 1 It is. This is the Zariski Lemma. 1 The class group of U is zero, dear Fellow Mathematician. Indeed if you remove the line z=0 from \mathbb P^2, you are left with \mathbb A^2. And if you then remove the points where x=0 or y=0 you are left with the product U=(\mathbb A^1\setminus\{0\})\times (A^1\setminus\{0\}). This is an affine variety with coordinate ring$$A=\mathcal ...

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Contacting the author J.P. Demailly himself about this question has lead to the following conclusion: In the algebraic setting, $X$ is quasi-projective, so by the Noetherian property, we have Zariski-compactness, so we can choose $\alpha$ globally. In the analytic setting, (the maximal choice of) $\alpha$ can get arbitrarily small when approaching ...

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It's true that an algebraic map of Zariski-closed subsets of $K^n$ (this is not what "affine variety over $K$" means unless $K$ is algebraically closed) is an isomorphism iff the induced map on rings of functions is an isomorphism. The point is that an inverse map on rings of functions provides the components of an algebraic map which inverts the original ...

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Regarding your first question, there are certainly theorems ruling out the existence of such $f$ assuming $f$ is continuous or smooth or algebraic or whatever. (I'm thinking of proofs using (co)homological methods.) But I would expect that these proofs ultimately make use of linear mappings as a base case somewhere (especially when $f$ is assumed ...

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There is no such field. As $k[x,y]/(y) \cong k[x]$ is an integral domain for any field $k$, the ideal $(y) \subset k[x,y]$ is a prime ideal. Thus, the variety $V(y)$ is irreducible.

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Up to order of coordinates, and perhaps some signs (depending on the convention you follow), this looks like Plücker coordinates. I think about this as the line joining the point $\mathbf a$ with the point at infinity in direction $\mathbf b$. In homogeneous coordinates you could write both as the columns of a matrix: \begin{pmatrix} a_1 & b_1 \\ a_2 ...

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