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Forum: Complex Analysis Topic: Best book(s) for Complex Analysis (undergrad) 
Tsakanikas Nickos 
Posted: Tue Jul 31, 2018 11:51 pm


Replies: 1 Views: 146

I would suggest that you read the following books on Complex Analysis: ⋅ J. Marsden, M. Hoffman  Basic Complex Analysis : Although it is developed in a rather slow manner, the exposition is in my opinion very nice, and the book contains lots of examples as well as exercises that help one... 


Forum: Algebraic Geometry Topic: Divisors and Picard Group 
Tsakanikas Nickos 
Posted: Wed Mar 07, 2018 10:53 am


Replies: 5 Views: 1010

I cannot see which is the Picard group of $X=Proj(\mathbb C[x,y,z]/(xyz^2)]\subset \mathbb{P}^{3}$ I don't know either the answer. I have just proved that $Cl(X)=\mathbb{Z}$. Maybe you could share your computations. You could also take a look at [Hartshorne / II / Ex. 6.3], which is related to you... 


Forum: Algebraic Geometry Topic: Divisors and Picard Group 
Tsakanikas Nickos 
Posted: Sun Jan 28, 2018 10:49 pm


Replies: 5 Views: 1010

Hi! Let me mention the following, which you may find helpful. ⋅ Recall the following general facts: On a variety \( X \), say, over \( \mathbb{C} \), it holds that \( \text{CaCl}(X) \cong \text{Pic}(X) \). Moreover, if \( X \) is normal, then Cartier divisors on \( X \) correspond to (are... 


Forum: Algebraic Geometry Topic: Divisors and Picard Group 
Tsakanikas Nickos 
Posted: Thu Jan 18, 2018 1:03 am


Replies: 5 Views: 1010

Hi!
How is your question related to [Hartshorne / II / 6.5.2]? Could you please explain exactly at which point of this particular example you are stuck? 


Forum: Algebraic Geometry Topic: Locally free sheaves 
Tsakanikas Nickos 
Posted: Tue Dec 19, 2017 12:40 am


Replies: 1 Views: 398

Hi!
You can find an answer to your question in the following reference: [Q. Liu  Algebraic Geometry and Arithmetic Curves  Chapter 6 / Lemma 4.1 & Corollary 4.2] 


Forum: Algebraic Geometry Topic: Geometric Genus 
Tsakanikas Nickos 
Posted: Sun Nov 12, 2017 1:01 am


Replies: 1 Views: 554

Let $ n = \dim X $. As $ X $ is rational, (by definition) $ X $ is birationally equivalent to $ \mathbb{P}^{n} $, and since the geometric genus $ p_{g}(X) (= P_{1}(X) = \dim H^{0} (X, \omega_{X} ) ) $ is a birational invariant, we have that $ p_{g} (X) = p_{g} (\mathbb{P}^{n}) $. But the canonical s... 


Forum: Complex Analysis Topic: Exercise On Cohomology of Complex Spaces 
Tsakanikas Nickos 
Posted: Sun Mar 05, 2017 12:49 am


Replies: 0 Views: 379

Assuming the following result THEOREM : Let $X$ be a complex space of dimension $n$ and let $\mathcal{S}$ be any sheaf on $X$. Then \[ \mathrm{H}^{q}(X, \mathcal{S}) = 0 \, , \, q > 2n \] prove the following results ⋅ LEMMA : Let $X$ be a complex space of dimension $n$ such that \[ \mathr... 


Forum: Functional Analysis Topic: An exercise on Fréchet Spaces 
Tsakanikas Nickos 
Posted: Sat Mar 04, 2017 10:16 pm


Replies: 1 Views: 494

Let $V,W$ be Fréchet spaces and let $T$ be a Hausdorff space. Consider the diagram \[ V \overset{f}{\longrightarrow} W \overset{i}{\longrightarrow} T \] where $i$ is a continuous, linear, injective map and $f$ is a linear map. Show that $f$ is continuous if and only if $ i \circ f $ is continuous. 


Forum: Algebraic Structures Topic: Isomorphism 
Tsakanikas Nickos 
Posted: Thu Feb 23, 2017 2:02 am


Replies: 2 Views: 610

Consider the composite map \[ R \overset{\varphi}{\to} R \overset{\pi}{\twoheadrightarrow} R/J \] Note that ⋅ $ \text{Ker}( \pi \circ \varphi ) = I $, as $ \text{Ker}( \pi ) = J $ and $ \varphi(I) = J $ ⋅ $ \text{Im}( \pi \circ \varphi ) = R/J $, as $ \pi $ is surjective and $ ... 


Forum: Algebraic Geometry Topic: Numerically Proportional 
Tsakanikas Nickos 
Posted: Mon Feb 20, 2017 10:42 pm


Replies: 0 Views: 489

Let $X$ be a smooth projective surface over $ \mathbb{C} $ and let $H$ be an ample divisor on $X$. Let $L$ and $M$ be two $\mathbb{Q}$divisors on $X$ which are not numerically trivial, and such that \[ L^2 = L \cdot M = M^2 = 0 \]Show that $L$ and $M$ are numerically proportional, i.e. \[ \exists \... 


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