- $Y$ is a connected object in $ \mathcal{FÉt}_{X} $.
- $Y$ is connected as a scheme.
Search found 284 matches
- Wed Feb 01, 2017 10:55 pm
- Forum: Algebraic Geometry
- Topic: Connect Object - Connected Scheme
- Replies: 0
- Views: 3749
Connect Object - Connected Scheme
Let $X$ be a scheme and let $ \mathcal{FÉt}_{X} $ be the category of finite étale schemes over $X$. Show that the following are equivalent:
- Sun Dec 25, 2016 6:34 pm
- Forum: Linear Algebra
- Topic: Tensors - Part 3
- Replies: 0
- Views: 3418
Tensors - Part 3
Definition 1 : Let $V$ be a finite-dimensional $\mathbb{R}$-vector space and let $k$ be a non-negative integer. A covariant $k$-tensor is a multilinear function $T \ \colon V \times \dots \times V \to \mathbb{R} $. Definition 2 : Let $V$ be a finite-dimensional $\mathbb{R}$-vector space. A 2-tensor...
- Sun Dec 25, 2016 6:25 pm
- Forum: Linear Algebra
- Topic: Tensors - Part 2
- Replies: 0
- Views: 2124
Tensors - Part 2
Definition 1 : Let $V$ be a finite-dimensional $\mathbb{R}$-vector space and let $k$ be a non-negative integer. A covariant $k$-tensor is a multilinear function $T \ \colon V \times \dots \times V \to \mathbb{R} $. Definition 2 : A covariant $k$-tensor is called alternating if its value changes sig...
- Sun Dec 25, 2016 6:17 pm
- Forum: Linear Algebra
- Topic: Tensors - Part 1
- Replies: 0
- Views: 2145
Tensors - Part 1
Definition 1 : Let $V$ be a finite-dimensional $\mathbb{R}$-vector space and let $k$ be a non-negative integer. A covariant $k$-tensor is a multilinear function $T \ \colon V \times \dots \times V \to \mathbb{R} $. Definition 2 : A covariant $k$-tensor is called symmetric if its value is unchanged ...
- Thu Dec 08, 2016 6:43 pm
- Forum: Algebraic Geometry
- Topic: On Plurigenera
- Replies: 0
- Views: 3639
On Plurigenera
Definition : Let $X$ be a smooth projective variety. For $m \in \mathbb{N}_{>0}$, the $m$-th plurigenus of $X$ is defined as $ P_{m}(X) = \mathrm{h}^{0}(X, \omega_{X} ) = \dim \mathrm{H}^{0}(X, \omega_{X} ) $, where $\omega_{X}$ is the canonical sheaf of $X$. Show that if $P_{1}(X) = 1 $ and $ P_{m...
- Sat Nov 26, 2016 7:09 pm
- Forum: Algebraic Geometry
- Topic: Nef and Big Divisors
- Replies: 0
- Views: 3787
Nef and Big Divisors
Let $X$ be an (irreducible) projective variety and let $D$ be a divisor on $X$. Show that the following are equivalent: $D$ is nef and big. There exists an effective divisor $N$ and a $k_{0} \in \mathbb{N} $ such that $D - \frac{1}{k} N$ is an ample $ \mathbb{Q} $-divisor for all $k \geq k_{0}$. Mor...
- Tue Nov 01, 2016 7:34 pm
- Forum: Algebraic Geometry
- Topic: (Very) Ampleness And Numerical Equivalence
- Replies: 0
- Views: 3962
(Very) Ampleness And Numerical Equivalence
Let $X$ be a non-singular projective surface over an algebraically closed field $ \mathbb{K} $. If $D$ is an ample divisor on $X$ and $ D \equiv D^{\prime} $ (numerically equivalent divisors), show that $D^{\prime}$ is ample too. Furthermore, give an example to show that the same statement does not ...
- Tue Nov 01, 2016 7:31 pm
- Forum: Algebraic Geometry
- Topic: Arithmetic Genus And Intersection Number
- Replies: 0
- Views: 3938
Arithmetic Genus And Intersection Number
Let $X$ be a non-singular projective surface over an algebraically closed field $ \mathbb{K} $. If $D$ is an effective divisor on $X$, $ p_{a} = 1 - \chi(\mathcal{O}_{D}) $ is its arithmetic genus and $ K $ is a canonical divisor on $X$, show that \[ 2p_{a} - 2 = D.(D+K) \]
- Sun Oct 23, 2016 5:42 pm
- Forum: Algebra
- Topic: A Projective Module
- Replies: 0
- Views: 4485
A Projective Module
To avoid confusion, let us give the following definitions: Definitions : Let $ \Lambda $ be $ \mathbb{R}, \mathbb{C} $ or the quaternions $ \mathfrak{Q} $ and let $G$ be a topological group. A $\Lambda G $ -space is a finite-dimensional vector space $V$ together with a morphism of topological groups...
- Thu Oct 20, 2016 6:05 pm
- Forum: Algebraic Geometry
- Topic: Constant Morphism - 3
- Replies: 0
- Views: 2976
Constant Morphism - 3
Let $V$ be a unirational variety and let $A$ be an abelian variety (over an algebraically closed field $k$). Show that any rational map $ V --> A $ is constant.