Let $L(-1)$ be the tautological line bundle on $\mathbb{P}^n$. I would like to understand the proof of the following:  Fact. The only global section on $L(−1)$ is the zero section.   Proof. Let $s : \mathbb{P}^n \to L(−1)$ be a holomorphic section. For any $l ∈ \mathbb{P}^n$ we have $s(l)=(l,z_l)$ for some $z_l \in \mathbb{C}^{n+1}$ lying on the line $l$. Thus, $l \to z_l$ is a holomorphic map $\mathbb{P}^n \to \mathbb{C}^{n+1}$. By the maximum principle this map must be constant, so $z_l \equiv w \in \mathbb{C}^{n+1}$. On the other hand $s$ is fiber preserving, so $w \in l$ for each line $l$ through the origin of $\mathbb{C}^{n+1}$. Hence $w = 0$.  How does the maximum principle imply that the map is constant?  Thanks a lot.  EDIT: my question is slightly different from the linked answer as that result is for holomorphic functions (to $\mathbb{C}$), whereas here I have a holomorphic map (to $\mathbb{C}^{n+1}$).
123796
Is there a simple proof that every holomorphic function $M\to\mathbb{C}$ on a compact complex manifold $M$ is constant?
9974
i'm currently working my way through  and stuck at Exercise 7.9 b)     Let $f: \mathbb{P}^n \to \mathbb{P}^m$ be a morphism Prove:   If $ n &gt; m $ then $f$ must be constant   I found this old thread  and would like to proceed by showing that for an affine subset $\mathbb{A}^n \subset \mathbb{P}^n$ we know that the morphism $f$ restricted on the subset has the form $f(x) = (f_1(x),...,f_m(x))$ for homogeneous $f_i$. So i would like to show what was mentioned in the first commentar that the $V(f_1,...,f_m)\setminus \{ 0 \}$ is not empty and hopefully inside my affine subset. Would this be a possible solution for this exercise and how can i do it?
27340
Suppose that $f$ is analytic on a domain $D$, which contains a simple closed curve $\gamma$ and the inside of $\gamma$. If $|f|$ is constant on $\gamma$, then I want to prove that either $f$ is constant or $f$ has a zero inside $\gamma$  Here is my take: if $f$ is constant, i dont see a reason why $|f|$ wouldnt be constant :)  if $f$ is not constant, then the max/min modulus principle applies ... meaning $|f|$ can not have any local max/min on D now i am lost at this point ...
66273
Let $f$ be non-constant analytic in a neighborhood of the closed unit disk such that $|f|=1$ on the unit circle. Show that $f$ has a zero inside the unit disk. It has been suggested to argue this by contradiction and using the maximum principle. However, is it possible to argue this question using properties of conformal maps since doesn't $f$ map the unit circle to itself? Regardless, I don't understand what information the Maximum Principle would give me to argue by contradiction. Why couldn't $0&lt;|f|&lt;1$ inside the disk?
87908
Prove that $f(z) = e^z$ maps $\mathbb{C} \setminus B $ onto $\mathbb{C} \setminus \{0\}$ where $B$ is a bounded subset of $\mathbb{C}$.  I have no idea how to even attempt this question so I would greatly appreciate some hints.
5436
I'm confused about the following proposition  Proposition. Let $U,V$ are open sets in $\mathbf{C}$. If $f:U\to V$ is holomorphic and bijective, then  the inverse  $f^{-1}:V\to U$ is also holomorphic.  The proof of the proposition think that the  continuity of $f^{-1}$ is obvious, but I find it is really difficult to prove  using $\epsilon-\delta$ definition. Can anyone give some hints?
47412
I am reading the section on Castelnuovo-Mumford regularity from Lazarsfeld's Positivity in algebraic geometry. Theorem 1.8.3 in the book reads as follows: Let $F$ be an $m$-regular sheaf on $P^n$. Then for every $k \geq 0$ (i). $F(m+k)$ is generated by its global sections. (ii). The natural maps $$H^0(P^n, F(m))\otimes H^0(P^n,O_{P^n}(k))\rightarrow H^0(P^n,F(m+k))$$ are surjective. (iii). $F$ is $(m + k)$-regular. The proof claims (i) is a consequence of (ii): it says that, since for $l\gg 0$, $F(m+l)$ is globally generated, the surjectivities in (ii) imply that $F(m)$ itself must be globally generated. I am not able to understand this statement. Choose $l\gg 0$ such that $F(m+l)$ itself is globally generated. Now consider the morphism $$H^0(F(m))\otimes O_{P^n}\rightarrow F(m)\,.$$ We need to prove that this is surjective. Tensoring by $O(l)$, we get $$H^0(F(m))\otimes O_{P^n}(l)\rightarrow F(m+l)\,.$$ A priori we do not know the injectiveness or surjectiveness of the above morphism. But if we take global sections, by (ii), the global sections morphism is surjective. Does this mean $H^0(F(m))\otimes O_{P^n}(l)\rightarrow F(m+l)$ is surjective? If so, I can tensor back by $O(-l)$ and get my required result. Thanks in advance.
102740
Let $X$ be a complex manifold and consider a holomorphic vector bundle $\pi:E\to X$ of rank $r$. By the usual correspondence between vector bundles and locally free sheaves, we consider $E$ as a sheaf by abuse of notation. We denote with $\mathscr A_X^{p,q}$ the sheaf of $C^\infty$ $(p,q)$-differential forms on $X$ and we define: $$A^{p,q}(E):=\Gamma(X,\mathscr A^{p,q}\otimes_{\mathscr O_X} E)$$ $$A^k(E):=\bigoplus_{p+q=k}A^{p,q}(E)$$  Now suppose that $\nabla:A^\bullet(E)\to A^\bullet(E)$ is a linear connection on $E$, then the curvature operator is defined as: $$\nabla^2:A^0(E)\to A^2(E)$$ where $\nabla^2$ is the composition of the maps $\nabla: A^0(E)\to A^2(E)$   and $\nabla: A^1(E)\to A^2(E)$. One can show that $\nabla^2$ is $C^\infty(X)$-linear.    I don't understand the following sentence from the Griffiths-Harris book at page 75 (I rephrase it by using my notation):     Since $\nabla^2$ is $C^\infty(X)$-linear, then it is induced by a bundle map $E\to\bigwedge^2T^*X\otimes E$, or in other words, $\nabla^2$ corresponds to a global section $\Theta$ of the bundle $\bigwedge^2T^*X\otimes\operatorname{Hom}(E,E)$.   I absolutely don't understand how I can see the operator $\nabla^2$ as the section $\Theta$.
62345
I came across this question whilst doing some research into complex analysis, and I just can't see what to do!  Let $f(z)$ be a holomorphic function on $\mathbb{C}$. Show that $\overline{f(\overline{z})}$ is holomorphic, whilst $f(\overline{z})$ is holomorphic if and only if $f(z)$ is constant.  I know that holomorphic means that the function is differentiable everywhere, and I need to apply the Cauchy-Riemann equations somehow, but I'm not sure how to approach this.
55932
Let $f$ be a nonconstant element of ring $\mathbb Z[x]$. Prove that $\langle f \rangle$ is not maximal in $\mathbb Z[x]$.   Let us assume $\langle f \rangle$ is maximal. Then $\mathbb Z[x] / \langle f \rangle$ would be a field. Let $a \in \mathbb{Z}$. Then $a + \langle f \rangle$ is a nonzero element of this field, hence a unit. Let $g + \langle f \rangle$ be its inverse. Then $a g - 1 \in \langle f \rangle$, hence $ag(x)-1 = f(x)h(x)$ for some $h \in Z[x]$, hence $ag(0) + f(0)h(0) = 1$, thus $(a,f(0))=1$ for all $a \in \Bbb Z$, contradiction, hence the proof.  Is my argument correct? Is there any other method?
36023
I have to solve following problem     If $\Re (f)$ is bounded above or below for a function $f$ holomorphic on $\mathbb{C}$ then $f$ is constant.   My attempt: If there is $M$ such that $\Re(f) \le M$, then $\|e^{f}\|=e^{\Re(f)}\le e^{M}$. From Liouville's theorem the entire function $e^f$ is constant, that is $0=(e^f)'=f' e^f$. This means $f' =0$, so $f$ is constant. If $\Re(f)$ is bounded below, we consider $e^{-f}$ and proceed the same way.  Am I correct? Is there a solution using maximum modulus principle?
76647
Assume that $f(z)$ is holomorphic and satisfies the inequality $|f(z)-1|&lt;1$ in a region $|z|&lt;R$. Show that $$\int_\gamma\frac{f'(z)}{f(z)}dz=0$$ for all closed curves $\gamma$ in $|z|&lt;R$. (The continuity of $f'(z)$ is taken for granted.)   The anti-derivative of $\dfrac{f'(z)}{f(z)}$ is just $\ln(f(z))$, which is holomorphic as long as $\ln(f(z))$ avoids the log branch cut. The log branch cut is the negative real axis, $a\leq 0$.   I want to use the fact that: If $f$ is the derivative of a holomorphic function in the region $\omega$, then the integral $\int_\gamma fdz$ depends only on the endpoints of $\gamma$.  To use this, we only need the condition that $f(z)$ is not a non-positive real number. The condition $|f(z)-1|&lt;1$ seems like an overkill. (Of course it does imply that $f(z)$ is not a non-positive real number.) But am I misunderstanding something?
54122
Still quite unsure about the chapter on orientation. For instance the Mobius map indeed preserves orientation, and I can show this rigorously, however I am still having difficulty understanding what it means for a holomorphic function $f$ to preserve orientation over some region in the complex plane. Which is why I asked a question regarding this $2$ days ago, but did not receive an answer.   The problem I am still thinking about is, I cannot show or provide a counter argument for why there does or does not exist some holomorphic function $f$ on a $\{z:|z|\le1\}$ such that it sends the unit circle with the counter- clockwise orientation into the unit circle with the clockwise orientation?  From the comment from this question asked earlier, one did suggest the use of the Argument principle which states that:     Argument Principle: If $f$ is meromorphic on a simply connected region $D$ and $\Gamma$ is a simply closed curve in $D$ not passing through the roots $z_j$ nor the poles $p_k$ of $f,$ then $$\frac{1}{2\pi i}\oint_{\Gamma}\frac{f'(z)}{f(z)}dz=n-m$$   Where $n$ is the number of roots of $f$(including multiplicities) in $\Gamma$ and $m$ is the number of poles of $f$ (including multiplicities) inside $\Gamma.$     I am really curious to find out why we can apply this to try and provide an explanation for my problem. I would appreciate some help because I am quite confused with orientation at the moment, and I think this problem is good in the sense it will help me understand some important notions on holomorphic functions which send or does not send certain orientation on a region to a different orientation of the same region.   I would really appreciate some rigorous argument for this problem. Much help will be appreciated.
53651
Let $\Omega$ be a non-empty, open subset of $\mathbb C$. Consider an holomorphic function $f:\Omega \setminus \{z_0\} \to \mathbb C$ and suppose we know $z_0$ is an essential singularity of $f$.   I am wondering what we can say about the function $\tilde{f}:=\frac{1}{f}$ and its singularity in $z_0$. Do you know any theorem that answers to this question?  Actually, I can't prove anything, since I do not know the answer: I've studied some examples. For instance, if you take $f(z)=e^{\frac{1}{z}}$ then $\tilde{f}$ will still have an essential singularity, hasn't it?   On the other side, if we take $f(z)=\sin(\frac{1}{z})$ then I think that $z_0=0$ becomes a limit point of poles for $\tilde{f}$ (so we can't classify it, because it isn't an isolated singularity).   Wha do you think? Do you know any useful theorem concerning this? Thank you in advance.
34430
I am studying some complex analysis and I have found this result. In fact it isn't a single theorem (I am reading Cartan's "Elementary theory of analytic functions" and I haven't found such a theorem, but maybe I just didn't find it), but there are many results that should give us this equivalence. However, I was wondering about two things.   The first is: is this equivalence "perfect"? I mean, could I exchange the terms "holomorphic" and "analytic" in theorems and various results without changing anything, or are there some hypothesis I should satisfy first? I am asking this because even though they should be equivalent, I always find theorems about (for instance) analytic prolongation that always talk about analytic functions, and there is no "holomorphic prolongation", if you understand what I mean. So I was not sure if this was simply a matter of tradition or if analytic functions are more general/poweerful than holomorphic ones.  The second one is: if I have a function from $\mathbb{R}^2$ to $\mathbb{R}^2$ that is differentiable in an open connected set $U$ and in that open set it satisfies Cauchy-Riemann conditions, then it should be holomorphic and analytic, if viewed as a complex function. So, for the prolongation theorem, there should be only one function like that (that is, differentiable and that satisfes Cauchy-Riemann)on $U$. So, are you telling me that if I have a differentiable function satisfying Cauchy-Riemann in an open connected set, not only it is the only function (unless you add/subtract constants obviously) that does that, but (being analytic in $\mathbb{C}$) it is also $C^{\infty}$. It would be so incredible I can't believe it!
92477
Let $f$ be holomorphic on $D(0,1)\cup \{1\}$, and $f(0)=0$, $f(1)=1$, $f(D(0,1))\subset D(0,1)$. Prove that $|f'(1)|\geq 1$.  I have no idea. Maybe $f'(\xi)=[f(1)-f(0)]/(1-0)$ for some $\xi \in (0,1)$, and the maximal modules principle applies? This is not exact.
45698
Let $D = B(0,1) \subset \mathbb{C} $ a disc, $f$ holomorphic on $D$. I want to demonstrate that if $$ 2|f^{'}(0)| = \sup_{z, w \in D} |f(z)-f(w)|$$  then $f$ is linear.   I know this is a well-known result. Where can I find the proof ?