User henri - MathOverflow

Answer by Henri for Hyperbolic Riemann Surface

2013-05-03T11:49:06Z

Let $X'=X\setminus \overline {D(x,r_x)}$.

If $X$ is already hyperbolic, then the answer is yes (there is no entire curves in $X'$).

If $X=\mathbb C, \mathbb C^*$, then $X'$ has no entire curves by Picard's little theorem, so it is hyperbolic.

If $X=\mathbb C/\Lambda$, then $\pi_1(X')$ is non trivial, and different from $\mathbb Z$ or $\mathbb Z^2$, so $X'$ is hyperbolic. This works if $r_x$ is not too big. Else you could imagine that $X'$ becomes $\mathbb C^*$.

If $X$ is $\mathbb P^1$, then your surface is biholomorphic to $\mathbb D$ (it is simply connected and strictly realized inside the complex plane).

In conclusion, $X'$ is always hyperbolic.

Answer by Henri for When does $Aut(X)=Bir(X)$ hold?

2013-01-31T10:08:55Z

To complete the answer of Divierietti and the comment of Roy Smith, here is a statement which might interest you:

Theorem If $X,Y$ are varieties over a field $k$, assume $X$ is smooth and $Y$ proper containing no rational curves. Then any rational map $X\dashrightarrow Y$ is everywhere defined.

You can find that statement in Debarre's book Higher Dimensional Geometry, Corollary 1.44 p.31.

In particular, if $X$ is smooth projective and contains no rational curves, then its automorphism group is equal to the group of its birational endomorphisms.

Answer by Henri for restriction of sheaf

2012-11-21T08:03:54Z

In general, we have the socalled projection formula: if $f:X\to Y$ is a morphism of ringed spaces, $\mathcal F$ an $\mathcal O_X$-module, and $E$ be a locally-free $\mathcal O_Y$ module of finite rank, then $f_* (\mathcal F \otimes f^{*} E) \simeq f_{*}\mathcal F \otimes E$.

Edit (following Will's remark): The projection formula yields in the case of an open immersion $i:U \subset X$ the following identity : $i_* i^* F \simeq i_*\mathcal O_U \otimes F$. Therefore, if $U$ has codimension at least 2 in $X$, then $i_* i^* F\simeq F$ by normality of $X$.

Answer by Henri for Interesting results for open Riemann surfaces

2012-10-29T12:15:17Z

For example, a theorem of Grauert and Röhrl asserts that every holomorphic vector bundle on a non-compact Riemann surface is trivial.

You can find this result (and its proof) in the book of O. Forster, Lectures on Riemann Surfaces.

Answer by Henri for Does there exist a non effective divisor with positive degree?

2012-10-14T11:06:04Z

Another possible approach is the following one, using Abel-Jacobi map : if $g(X)\geq 2$, then Abel-Jacobi map gives an isomorphism $\varphi:Pic^0(X) \simeq \mathbb C^g/\Lambda$ for some lattice $\Lambda$. Now you can translate the map by some point $p\in X$; more precisely, define $\psi(D):=\varphi(D-(p))$ for every divisor $D$ of degree $1$ on $X$.

As $g\geq 2$, the image by $\psi$ of all points (seen as degree $1$ divisors) will be one-dimensional inside the $g$-dimensional variety $\mathbb C^g/\Lambda$. Therefore there exist on $X$ (a lot of) divisors of degree $0$ which cannot be written as $(q)-(p)$ for any point $q$.

Now consider such a divisor $D$, and define $D'=D+(p)$. It has degree 1, but if it were effective, it would be equivalent to $(q)$ for some point $q\in X$. Therefore one would have $D \sim (q)-(p)$ which is impossible.

Answer by Henri for Line bundles and rational singularities

2012-06-19T11:41:42Z

For any line bundle $F$, giving a map of sheaves $\mathcal O_X \to F$ is equivalent to giving a global section of $F$.

In your case, take $F=Q \otimes L^{-1}$ with the map given by your exact sequence tensorized by $L ^{-1}$. As $F$ has no non-zero global section, the aforesaid map is trivial, so $\eta \otimes Id_{L^{-1}}$ is an isomorphism, and therefore so is $\eta$ too.

Answer by Henri for Does the induced map $\pi^*:H_{DR}^{k}(X)\rightarrow H_{DR}^{k}(\tilde{X})$ injective?

2012-03-29T07:34:38Z

If you have a morphism $\pi$ with finite generic fiber of cardinal $d$ between compact manifolds, then you have $\pi_* \circ \pi^* = d Id_{H^k(Y,\mathbb Z)}$, where $\pi_* $ is the Gysin morphism. In particular $\pi^*$ is injective.

Beside, one can still say something in the case where the generic fiber has positive dimension: if $\pi$ a surjective morphism between compact Kähler manifolds, then the induced map in cohomology is injective (this is proved in Hodge theory and Complex Algebraic Geometry by Voisin for instance, lemma 7.28 in the french version).

Answer by Henri for Holonomy group of calabi yau manifold

2012-03-20T08:19:24Z

Well, it depends on what you call a Calabi-Yau manifold (there are several possible terminologies indeed).

First of all, a compact Kähler manifold with trivial canonical class does not necessarily have holonomy group $SU_n$ (with respect to some Ricci-flat metric). More precisely, the holonomy group of some compact Kähler manifold $(X,\omega)$ is included in $SU_n$ iff there exists a non-zero parallel holomorphic $n$-form. As a consequence the restricted holonomy group $H_0$ is included in $SU_n$ iff $(X,\omega)$ is Ricci-flat.

Now the holonomy groups of a Ricci flat compact Kähler manifold can be smaller than $SU_n$: think about any torus (the holonomy is trivial) or any holomorphic symplectic variety (the holonomy is $SP(n/2)$.

However, there is a result, which was maybe what you had in mind:

Theorem. Let $(X,\omega)$ be a compact Kähler manifold of dimension $n\geq 3$ with holonomy group $SU_n$. Then $X$ is projective and $H^0(X, \Omega_X^p)=0$ for every $0 < p < n$ and $\chi(\mathcal O_X)=1+(-1)^n$.

A manifold with such properties is sometimes called Calabi-Yau, indeed. For a reference, see Beauville's article "Variétés Kähleriennes à première classe de Chern nulle".

As for the non-compact case, I don't know if such a results holds.

Answer by Henri for How to use Hirzebruch-Riemann-Roch to produce sections of a positive line bundle?

2012-02-23T20:03:24Z

I would suggest you two references; the first one is Demailly's survey on Hodge theory ([B3] here: http://www-fourier.ujf-grenoble.fr/~demailly/books.html ); exercise 15.11 explains exactly how to construct sections with desired vanishing order at some point.

This is a consequence of Nadel's theorem applied to singular metrics on ample line bundles; the idea is to impose the metric to have a logarithmic pole at your point, so that Skoda's lemma (the easy part) ensures you that any section of L twisted by the multiplier ideal of the metric has to vanish enough along P. As a consequence, you can deduce Kodaira's embedding theorem.

The second reference is the one already mentionned by Donu Arapura. More precisely, there is a lemma in Lazarsfeld' PAG I (edit: Proposition 1.1.31 p.23). It shows how elementary linear algebra (+ RR) can give a lower bound on the vanishing order of some section of a (say nef and big) line bundle with big enough top intersection.

Answer by Henri for Bound of dimension of $H^1$ of certain line bundle

2011-10-11T07:02:08Z

In general, for any line bundle $L$ over $X$, you have $h^i(X, L^{\otimes n})\leqslant C m^n$ where $m=\dim X$. (write $L$ as the difference of two very ample line bundles, and use the usual restriction exact sequence)

In the case where $L$ is nef, then one can say more: $h^i(X,L^{\otimes n})\leqslant C m^{n-i}$ using Fujita's theorem (cf Lazarsfeld, Positivity in Algebraic Geometry, 1.2.29 "growth of cohomology").

But in the case $i=1$, you don't need Fujita, and this can be done more basically. (see e.g Debarre's "Higher dimensional Algebraic Geometry", Proposition 1.31 p21)

Answer by Henri for Analytic implicit function theorem

2011-08-22T09:12:38Z

In one variable, this is a trivial consequence of the standard local inversion theorem. Indeed, holomorphic functions are $C^1$ functions characterized by the fact that their differential is a similitude. And this property is stable by taking the inverse.

to be more precise, if $g$ is holomorphic on some open set $U\subset \mathbb C$, and its differential (as a function $U \to \mathbb R^2$) satisfies that it is invertible everywhere, with differential being a similitude. So the functions is a local diffeomorphism, and the differential of the inverse is the inverse of the differential, so is still a similitude. Therefore, $g$ is a local biholomorphism.

The statement of IFT is a direct consequence of the local inversion theorem then.

Answer by Henri for is the differential of the distance function holomorphic?

2011-08-13T10:30:10Z

Take the most simple example $M=\mathbb C$ and $X$ the unit circle. Then $d_X(z)=|z-\frac{z}{|z|}|^2$ is not holomorphic (so as its Laplacian $i\partial \bar \partial( d_X)$), as you can easily see by expanding the expression.

Answer by Henri for Complex line bundle over curves

2011-07-20T07:45:41Z

Here is the way you get a non-vanishing holomorphic section which will trivialize your line bundle.

Denote $U_1={[x:y]; x\neq 0 }$ and $U_2={[x:y]; y\neq 0}$, this is a covering of $\mathbb P^1$. The transition function of the bundle is $g_{12}([x:y])=e^{y/x}$ (you could also take $e^{x/y}$, this wouldn't change the argument), defined on $U_1 \cap U_2$.

Then you may defined the following holomorphic functions: on $U_1$, you put $s_1([x:y]):=e^{y/x}$ and on $U_2$, $s_2([x:y]):=1$. Thus $s_1$ and $s_2$ are non-vanishing holomorphic functions, and on $U_1 \cap U_2$, you have $s_1=g_{12}s_2$ so that they form a section of our line bundle.

EDIT: In particular the degree of the line bundle is 0!

Moreover, on a Riemann surface (or more generally on any smooth projective complex variety), any line bundle admits a meromorphic section, so that the correspondence your are talking about still holds.

Answer by Henri for Siu decomposition

2011-07-06T22:08:31Z

In my opinion, the point is not that $R$ is the biggest in whatever sense you may give to this, but that the decomposition $T=R+S$ with $S=\sum \lambda_j [A_j]$ ($A_j$ being a $p$-dimensional analytic set), and $R$ having zero Lelong number along any $p$-dimensional analytic set.

The uniqueness is clear because for $x$ generic in $A_j$, $\nu(R+S,x)=\lambda_j \nu([A_j],x)=\lambda_j$, which determines thus uniquely $\lambda_j$, and therefore $S$ and $R$.

Now, if you really want to see that $R$ is the biggest current (in the sense of positivity of currents) such that $R$ has zero Lelong number along any $p$-dimensional analytic set, then you can proceed this way: assume that $T=S'+R'$ is another such decomposition. Then for $x$ generic in $A_j$, $\nu(T,A_j)=\nu(T,x)=\nu(S',x)=\nu(S',A_j)$. But it is a classical fact (see e.g Demailly, Complex analytic and differential Geometry, Proposition 8.16) that $S'-\nu(S', A_j) [A_j]$ is a closed positive current, so that $S' \geqslant S$, and therefore $R' \leqslant R$, which concludes.

Answer by Henri for Zeros of holomorphic one-forms on Riemann surface

2011-06-16T07:03:16Z

On $\mathbb P^1$, there is no non zero holomorphic $1$-form, on any elliptic curves, the holomorphic forms are "constant" (the canonical bundle is trivial), so never vanish if they are not identically zero.

As for the other surfaces, namely if $g(X) \geqslant 2$, then $|K_X|$ has no base point (cf Hartshorne, IV, lemma 5.1), which amouts to saying that for all point $x\in X$, there exists a holomorphic form non-vanishing at $x$.

Moreover, if $X$ has genus $g\geqslant 2$ as previously and $X$ is not hyperelliptic, then $K_X$ is very ample (cf Hartshorne, IV, proposition 5.2), which means that the linear system given by the (global) holomorphic $1$-forms induces an embedding into $\mathbb P H^0(X, K_X)^* \simeq \mathbb P^{g-1}$.

Answer by Henri for Non finitely generated graded ring of a divisor in dimension >2

2011-02-24T08:01:25Z

In their paper "Monge-Ampère equations in big cohomology classes", Boucksom, Eyssidieux, Guedj and Zeriahi give an example (Ex 5.4 page 46 here : http://arxiv.org/abs/0812.3674) of a nef and big line bundle over a smooth projective 3-fold which is not semi-ample. More precisely, every positive current in its cohomology class has poles along some subvariety.

Furthermore, it is well-known (Lazarsfeld, PAG e.g) that a nef and big line bundle has a finitely generated sections ring iff it is semi-ample.

In one word, their construction consists in using the famous example of Serre (and studied by Demailly-Peternell-Schneider) of a flat rank 2 vector bundle $E$ on some elliptic curve $C$, and considering on $V:=\mathbb P(E\oplus A)$ (for $A$ ample on $C$) the tautological line bundle $\mathcal O_{\mathbb P(V)}(1)$.

Answer by Henri for hesse matrix under diffeomorphism

2011-05-22T10:49:43Z

I guess your diffeomorphism is a biholomorphism, or at least an holomorphic map, otherwise there is no hope to say much about it.

Then the best way to look at these things is using differential forms. Indeed, $\imath \partial \bar{\partial} u =\imath \sum_{i,j} \frac{\partial^2 u}{\partial z_i \partial \bar z_j} dz_i \wedge d\bar z_j$. As is well known, as $\varphi$ is holomorphic (not necessarily bijective), $\varphi^* $ commutes with $\partial$ and $\bar \partial$. Therefore $\imath \partial \bar{\partial} u' = \varphi^* \imath \partial \bar{\partial} u$, which you can translate in terms of matrices then, if you want to:

$$Hess(u')_a(\xi)= \underset{j,k,l,m}{\sum} \frac{\partial^2 u(\varphi(a))}{\partial z_l \partial \bar z_m} \frac{\partial \varphi_l(a)}{\partial z_j } \xi_j \overline{\frac{\partial\varphi_m(a)}{\partial z_k } \xi_k} $$ if $\varphi= (\varphi_1, \ldots, \varphi_n)$, and $\xi$ is any tangent vector at $a$.

Answer by Henri for How should one think about pushforward in cohomology?

2011-05-12T09:31:18Z

In the case where $X,Y$ are smooth oriented (compact) manifolds, you may do this using currents. Indeed, currents are continuous linear forms on the space of (compactly supportly) smooth differential forms.

There is a natural way to push forward currents (by duality with the pull-back of smooth forms). Furthermore, the De Rham cohomology for currents is (naturally) isomorphic to the one of smooth forms. Therefore, you can push forward a smooth class considering it a current class, and using then the previous natural isomorphism.

Moreover, you have the projection formula: $\alpha, \beta$ are forms with appropriate degree, then $f_{\star} (f^{\star} \alpha \cup \beta) = (\alpha \cup f_{\star} \beta)$.

To sum up, the use of currents allows you to consider always duality on cohomology and not duality between homology and cohomology.

Answer by Henri for What is ample generator of a Picard group?

2010-12-20T23:10:12Z

First of all, the Picard group of a variety is not always monogen, so that the notion of "the ample generator" you are referring to surely concerns a restricted class of varieties.

Furthermore, an ample line bundle (or invertible sheaf) is a line bundle $L$ which satisfies any of the following properties :

For any coherent sheaf $\mathcal F$, the sheaf $\mathcal F \otimes L^{\otimes m}$ is generated by its global sections for every $m\geq m_0(\mathcal F)$.
The map $\Phi_{|L^{\otimes m}|} : X \to \mathbb P(H^0(X,L^{\otimes m})), x \mapsto [s_0(x): \ldots:s_N(x)]$ - where $(s_i)$ is a base of the space of global sections of $L^{\otimes m}$ - induces a embedding of $X$ in some projective space for every $m\geq m_0$.
If you are working over $\mathbb C$, which seems to be the case, this is also equivalent for $L$ to admit a smooth hermitian metric $h$ whose curvature $\Theta_h(L)$ is a (striclty) positive $(1,1)$-form.

Then I guess that the ample generator of some Picard group is in some case one generator which besides is ample.

For example, the most basic example consists in taking the ample line bundle $\mathcal O_{\mathbb P^n}(1)$ over $\mathbb P^n$, which generates $\mathrm{Pic}(\mathbb P^n)$.

Finally, if your question is : if the Picard group of a projective variety is monogen, then can we choose an ample generator? Then the answer is yes because for every $m\geq 1$, $L$ is ample iff $L^{\otimes m}$ is ample.

Edit : I forgot to mention that in this case, the unicity of "the" ample generator is clear : indeed, if $L$ and $L^{-1}$ admit non trivial sections (say $s$ and $t$) then $st$ is a non-zero section of $\mathcal O_X$ thus is constant, so that $s$ and $t$ are both non-vanishing sections, which implies that $L$ is trivial. You can apply this to $L^{\otimes m}$ to get the unicity property.

Answer by Henri for Compactness properties of plurisubharmonic functions

2010-12-03T19:31:58Z

First of all, I would say that there exists $\textit{one}$ good topology for psh functions, that is the $L^1_{loc}$ topology. One of the main result is the following one :

Let $(u_n)$ be a sequence of psh functions on a connected open subset $\Omega \subset \mathbb{C} ^n$ with $u_n \not \equiv -\infty $. We suppose that $(u_n)$ converges to $u$ psh, in the weak topology of distributions. Then $(u_n)$ is locally upper bounded and $u_n \to u$ in $L^p_{loc}(\Omega)$ for every $p\in[1,+\infty[$.

Other similar results are :

-every bounded subset of $Psh(\Omega) \cap L^1_{loc}(\Omega)$ is relatively compact;

-if $(u_n)$ is locally upper bounded on $\Omega$, then either $(u_n)$ converges locally uniformly to $-\infty$ on $\Omega$ (for the $L^1_{loc}$ topology), either there exists some subsequence converging to a psh function on $\Omega$ (in $ L^1_{loc}$ - or $L^p_{loc}, p\geq 1$, this is the same-).

As for the references, the one I prefer is "Notions of convexity" by Hörmander, 94, around section 3.2. The online book of Demailly is good too, but far less detailed about this topic.

Answer by Henri for What are Central Limit Theorems and why are they called so?

2010-10-29T14:05:37Z

One of my teacher in Probability once told us that this name (Central Limit Theorem) was just used (at the beginning) to stress the importance of the result -which plays a central role in the theory. Besides, the ambiguity led to several different translations, corresponding to both interpretations of the term "central". (e.g in French, we can find "théorème central limite" and "théorème de la limite centrale")

Answer by Henri for Technique to prove basepoint-freeness

2010-09-03T13:17:11Z

I don't think that your assertion is true; for example, Lazarsfeld gives an example (PAG, 2.3.3) of a big and nef divisor on a surface such that its graded algebra is not finitely generated, so that the divisor can't be semiample.

But there are some close results for nef and big divisors, or even for good divisors (when the Kodaira dimensions equals the numerical dimension) as Mourougane and Russo showed : for example, Wilson's theorem asserts that for any nef and big divisor on an irreducible projective variety, there exists $m_0\in \mathbb N$ together with an effective divisor $N$ such that for all $m\geq m_0$, the linear system $|mD-N|$ has no base-point. (PAG, 2.3.9)

Answer by Henri for What are some open problems in algebraic geometry?

2010-08-30T17:10:24Z

We can also mention two other major open problems :

The abundance conjecture, stating that if a $K_X+\Delta$ is klt and nef, then it is semi-ample (a multiple has no base-point)
The Griffith's conjecture : if $E$ is an ample vector bundle over a compact complex manifold, then it is Griffith-positive. (this is known for line bundles of course)

Answer by Henri for Zariski closed sets in C^n are of measure 0

2010-05-21T18:04:41Z

Clearly, it is sufficient too show it for a closed set given by $f=0$ where $f$ is analytic. (write your set as included in a countable union of such described sets). Then, using the normal form of analytic germs as finite ramified coverings, you're done.

Answer by Henri for Differentiable structures on R^3

2010-05-16T21:02:23Z

There are infinitely many differentiable structures on $\mathbb R$ : take any homeomorphism which is no diffeomorphism (such as $x\mapsto x^3$), and you get an non-usual differentiable structure on $\mathbb R$!

Even better : there exist uncountably many different real analytic structures on $\mathbb R$. But this example is general : for every $k\in \mathbb N \cup {\infty,\omega}$, the group Homeo($\mathbb R$) acts transitively on the set of $C^k$ differentiable structures on $\mathbb R$.

EDIT : If you want to show that up to $\textit{diffeomorphism of differentiable manifold}$, there exists only one differentiable structure on $\mathbb R$, just remember that (more generally) all $1$-dimensional non-compact manifolds are diffeomorphic to $\mathbb R$. [this is proved in Lafontaine e.g]

Answer by Henri for strick inequality for Fatou theorem

2010-05-15T18:01:41Z

This is just the definition of convergence for sequences of functions : $\forall x \in \mathbb R, \lim_{n \to+\infty} f_n(x) =0$, which is of course the case here, all sequences $(f_n(x))$ being stationary.

Answer by Henri for Contracting divisors to a point

2010-05-13T14:37:55Z

In the case of smooth surfaces, this is already a tough question : If E is a rational curve, then E can be contracted to a point in a smooth variety if the self-intersection $E\cdotp E=-1$ (thm of Castelnuovo). If you don't impose the smoothness of the variety, then a sufficient condition is $E\cdotp E<0$.

Moreover, there exists a theorem, due to Grauert, which states that if $\sum E_i$ is a divisor on a smooth surface such that the matrix of intersections $(E_i\cdotp E_j)$ is negative, then there exists $f:S \to S'$ birational such that the exceptional locus of $f$ is exactly $\cup E_i$. But this theorem is valid only in the analytic setting, which means that $S'$ is a priori not an algebraic variety.

Concerning the general question of contraction of curves in an algebraic variety, the main results come from Mori's theory (cf the book of Kollar and Mori, Birational geometry of algebraic varieties)

Answer by Henri for What's the difference between a real manifold and a smooth variety?

2010-05-08T21:26:16Z

I think there's some big difference concerning the metric approach too.

In fact, the Gram-Schmidt process (which is real analytic) enables us -in real differential geometry- to find some local orthonormal frames (for any hermitian bundle, and in particular for the tangent bundle), whereas in the holomorphic case, very subtle differences may occur there.

For example, in the Kähler case, we can find "orthonormal" frames for the tangent bundle at order 2, which is the key for the Kähler identities, leading to fundamental results like the equality of all Laplacians and thus the Hodge decomposition theorem in the compact case.

Answer by Henri for Positive vector bundles

2010-04-27T09:11:06Z

Vamsi, the answer to your question "Is positivity sufficient to ensure that the bundle is generated by finitely many sections?" is yes. If E is positive, E is in particular globally generated which means that the evaluation map H°(X,E)->E_x is onto for every x in X. But it is (well?)-known that in such a case, thanks to Baire theorem applied in those Frechet spaces, there exists V a finite dimensional subspace of H°(X,E) such that V generates all fibers E_x. Moreover, we can choose V such that dim V is less or equal to dim X+dim E.

More precisions on Demailly's online book, Complex analytic and algebraic geometry, chapter VII, prop. 11.2, avalaible at http://www-fourier.ujf-grenoble.fr/~demailly/books.html

Comment by Henri

2013-03-20T13:37:00Z

Yes you are right of course; I just had in mind the case where $M$ is a domain in $\mathbb C^n$. My point was just that the assumptions on $\Omega$ should not be too restrictive like pseudoconvexity e.g.; but on $M surely!

Comment by Henri

2013-03-20T00:47:16Z

No, not even. You just need $\Omega$ relatively compact, and $M\setminus \overline{\Omega}$ connected. (btw, you should have defined $u$ on $M\setminus \overline{\Omega}$ only if you want to make sense of pluriharmonicity)

Comment by Henri

2013-03-19T14:24:17Z

Oh, I just realized that I misunderstood your question; I thought that $u$ was only defined on $\Omega$. In fact, if $M\setminus \Omega$ has $b_1=0$, then every pluriharmonic function on $M\setminus \Omega$ is the real part of a holomorphic function; so $u=Re(f)$ for $f \in \mathcal O(M\setminus \Omega)$. Now, if $M\setminus \Omega$ is connected, Hartog's extension theorem tells you that $f$ extends (uniquely) to $M$, and then $Re(f)$ extends $u$ has a pluriharmonic function.

Comment by Henri

2013-03-19T14:00:15Z

Yes; if you want consider $\Omega = B(0, 1/2)$; by uniqueness of the pluriharmonic extension (ph functions are real analytic), $u$ won't extend to $\mathbb C^2$ either.

Comment by Henri

2013-03-19T00:40:29Z

How about $\Omega=B(0,1)\subset \mathbb C^2$ and $u(z,w)=Re(e^{1/(z-1)})$? (cf remark of Alexandre)

Comment by Henri

2013-01-31T10:50:13Z

Yes, Y. Consider the two projection $p_1$, $p_2$ from the graph to $X$ and $Y$. We know that the exceptional locus of $p_1$ contains rational curves $C \subset X\times Y$. By assumption, $p_2$ must contract $C$, so that $C$ is contracted by $p_1$ and $p_2$, which is absurd.

Comment by Henri

2012-11-21T10:11:30Z

In particular, as $X$ is smooth hence normal, the desired property holds as soon as $U$ has codimension at least $2$.

Comment by Henri

2012-11-21T10:06:18Z

Thanks, you are perfectly right! (I had in mind a morphism with connected fibers) So here here one can just say that $i_*i^*F= i_* \mathcal O_X \otimes F$.

Comment by Henri

2012-11-10T09:14:47Z

Indeed, there is no flat metric on the 2-sphere with only two singularities. Another way to see it is that if $D=\sum a_i (p_i)$ with $a_i$ the cone angles, then we must have $K_{\mathbb P^1}+D \simeq 0$. This is equivalent to $\sum a_i=2$.

Comment by Henri

2012-10-25T15:53:47Z

In dim>1, an ample divisor is always connected. This follows by Lefschetz theorem.

Comment by Henri

2012-09-30T13:37:16Z

Edit for the previous comment: of course $Z_t=\{xy-tz^2=0\} \subset \mathbb P^2$!

Comment by Henri

2012-09-30T12:36:20Z

Concerning the Picard group, look at the example where $Z_t = \{xy-t^2=0\} \subset \mathbb P^2$. Then for $t\neq 0, Z_t \simeq \mathbb P^1$, but $Z_0$ is a degenerate conic whose Picard group surjects onto the Picard group of its normalization (which is 2 copies of $\mathbb P^1$. Therefore $\mathrm{Pic}(Z_0)$ is bigger than $\mathrm{Pic}(Z_t)$.

Comment by Henri

2012-05-14T09:15:16Z

As for rational cohomology, the injectivity of $f$ is automatic as soon as $f$ is surjective (not necessarily finite), cf the second paragraph of mathoverflow.net/questions/92532/…

Comment by Henri

2012-04-25T07:16:26Z

For the second statement, I guess you mean $H^i(U, \mathcal O)=0$, right?

Comment by Henri

2012-04-23T09:03:25Z

And to see that the rational map you obtain is birational, take a non-zero section $s_E$ of $E$ whose divisor is $E$, and then the rational map $x\mapsto [s_0\otimes s_E:\cdots:s_N\otimes s_E]$ defined outside of $E$ ($(s_i)$ being a basis of $H^0(X,kA)$ for $k$ large enough) is clearly birational onto its image whose dimension is $\dim X$ as John explained above.