Let X be a compact Riemann surface of genus $g$ corresponding to the equation $w^3 = P(z)$, P is a polynomial without multiple roots and $degP = 8$ It's known that there are $g$ holomorphic differentials on this surface. How can I construct them explicitly?  If my curve was defined as hyperelliptic one, $w^2=P(z)$, then I would use $\frac{z^k \cdot dz}{w}$ for $k = 0, \cdots ,g-1$. $\frac{z^k dz}{w} = \frac{2z^k dw}{P'(z)}$, we would check that the form is holomorphic at $z \to \infty$ and $\omega \to 0$. Mention here that $deg P = 2g+2$. (I don't understand to the full how we do this). Then $\frac{z^k dz}{w}$ is well defined when $z$ is a coordinate. and $\frac{2z^k dw}{P'(z)}$ is well defined when $w$ is a cordinate.  By implicit function theorem $w$ is a coordinate everywhere where $P'(z) \neq 0$ so the second representation works and we get over the case of $w=0$.  $z$ is a coordinate everywhere where $2w \neq 0$. so the first representation works and when $z \to \infty$ then we have a double pole from $dz$ and $g+1$-ble pole from $\frac{1}{w}$  Our curve equation means that $dw = \frac{P'(z)}{3w^2} dz$ so if we try the same way it is done for hyperelliptic curve, then we should seek the answer as $\frac{z^k dz}{w^2}= \frac{3z^k dw}{P'(z)}$.   How can we check the holomorphoucity of forms more rigorously and how can I construct the forms in the case $w^3 = P(z)$
120465
Let us have a surface $X$ which satisfy the equation $\omega^3 = (z-1)(z-2)\cdots (z-8)$ It's stated in the task, that there are exactly $g$ holomorphic differentials on the curve. How can I find $g$ and construct them?  The genus $g$ of this curve (I tried to use Riemann-Hurwitz theorem)  The map $X(z,\omega) \to CP^1(\omega)$ has degree 3. and we have 8 branching points with 3 leaves at each. $\chi(Domain) = DegreeOfMap * \chi(Target) - \sum_{points}(BranchingLeaves - 1) $ $2-2g = 3 *(2) - 8*2 $ hence $ 12 = 2g$ and  $g = 6$ I know that in case of hyperelliptic curve there are $g-2$ (If i'm not mistaken) holomorphic differetials and they are of sort $\frac{z^k dz}{w}$. But the form of the curve equation was heavily used. UPD: there is a branhing point at $\infty$, because if we go around all points $ 1 \cdots 8$ at once, we get that cubic root of the right side is multiplied by $\exp{4 \pi /3}$ $2-2g = 3 *(2) - 9*2 $ hence $ 12 = 2g$ and  $g = 7$ (I hope that this is true)
65230
$f(z)=xy+iy$  $f(z)=u(x,y)+iv(x,y)$  $u=xy$, $v=y$  $\frac{\partial u}{\partial x}=y$, $\frac{\partial v}{\partial x}=0$  $\frac{\partial u}{\partial y}=x$, $\frac{\partial v}{\partial y}=1$  We see that $\frac{\partial u}{\partial x} \ne \frac{\partial v}{\partial y}$, except for $y=1$  Also, $\frac{\partial u}{\partial y} \ne -\frac{\partial v}{\partial x}$, except for $x=0$  The Cauchy-Riemann equations are not satisfied anywhere in the z-plane except at the point $x=0$, $y=1$, i.e. $z=i$  $f^{'}(z)=\lim_{\delta z-&gt; 0}\frac{f(z+\delta z)-f(z)}{\delta z}$
8689
There is an exercise in Function Theory of One Complex Variable by Greene &amp; Krantz that is very similar to a Proposition in the book, but I am having trouble getting to the conclusion.      Let $f : \mathbb{C} \to \mathbb{C}$ be a polynomial. Suppose further that $$\frac{\partial f}{\partial z} = 0$$   and $$\frac{\partial f}{\partial \bar z} = 0$$    for all $z \in \mathbb{C}$    Prove that $ f \equiv$ constant.    Now, part of the proposition (1.3.2) proves that if $p$ is a polynomial, $$p(z, \bar z) = \sum a_{lm}z^l \bar z^m$$ with $\frac{\partial f}{\partial \bar z} = 0$, then $\frac{\partial ^{l + m}}{\partial z^l \partial \bar z^m}p$ evaluated at zero is $l!m!a_{lm}$.  Is this the constant that $f$ equals? It doesn't seem like I've used the hypothesis, or reached the conclusion!  Maybe I should use the definition of the partial... $$\frac{\partial f}{\partial z} := 1/2 (\frac{\partial }{\partial z} - i \cdot \frac{\partial }{\partial z})f$$  Any hints would be greatly appreciated.
60453
The  of order $s$ are defined by $$\mathrm{Li}_s (z) = \sum_{k \geqslant 1} \frac{z^k}{k^s},  \quad |z| &lt; 1.$$ From the above definition, derivatives for the polylogarithms immediately follow. Here we have $$\frac{d}{dz} \mathrm{Li}_s (z) = \frac{\mathrm{Li_{s - 1}}(z)}{z}.$$ When $s = 0$ the following series which is geometric results $$\mathrm{Li}_0(z) = \sum_{k \geqslant 1} z^k = \frac{z}{1 - z}.$$ Starting with the expression for $\mathrm{Li}_0(z)$, repeated application of the derivative rule allows explicit expressions for the polylogarithms to all negative integer orders to be found. The first few are: \begin{align*} \mathrm{Li}_{-1} (z) &amp;= \frac{z}{(1 - z)^2}\\ \mathrm{Li}_{-2}(z) &amp;= \frac{z(1 + z)}{(1 - z)^3}\\ \mathrm{Li}_{-3}(z) &amp;= \frac{z(1 + 4z +z^2)}{(1 - z)^4}\\ &amp; \vdots \end{align*} All are rational functions in $z$. A closed form expression for $\mathrm{Li}_{-n}(z)$ where $n \in \mathbb{N}$ in terms of either  or  can be found. In terms of Eulerian numbers $\left\langle n \atop k\right\rangle$ the result is $$\mathrm{Li}_{-n} (z) = \frac{1}{(1 - z)^{n + 1}} \sum^n_{k = 0} \left\langle n \atop k\right\rangle z^{n - k}, \quad n \geqslant 1.$$ What I wish to know is how can such a result be proved. Of course once one knows what one is looking for, the result can be readily guessed, and I suppose induction used to prove the claim is true for all $n$, but is there a way to prove the result directly?
88421
It is, of course, one of the first results in basic complex analysis that a holomorphic function satisfies the Cauchy-Riemann equations when considered as a differentiable two-variable real function. I have always seen the converse as: if $f$ is continuously differentiable as a function from $U \subset \mathbb{R}^2$ to $\mathbb{R}^2$ and satisfies the Cauchy-Riemann equations, then it is holomorphic (see e.g. Stein and Shakarchi, or ). Why is the $C^1$ condition necessary? I don't see where this comes in to the proof below.   Assume that $u(x,y)$ and $v(x,y)$ are continuously differentiable and satisfy the Cauchy-Riemann equations. Let $h=h_1 + h_2i$. Then \begin{equation*} u(x+h_1, y+h_2) - u(x,y) = \frac{\partial u}{\partial x} h_1 + \frac{\partial u}{\partial y}h_2 + o(|h|) \end{equation*} and  \begin{equation*} v(x+h_1, y+h_2) - v(x,y) = \frac{\partial v}{\partial x} h_1 + \frac{\partial v}{\partial y} h_2 + o(|h|). \end{equation*} Multiplying the second equation by $i$ and adding the two together gives  \begin{align*} (u+iv)(z+h)-(u+iv)(z) &amp;= \frac{\partial u}{\partial x} h_1 + i \frac{\partial v}{\partial x} h_1 + \frac{\partial u}{\partial y} h_2 + i \frac{\partial v}{\partial y} h_2 + o(|h|)\\\  &amp;= \left( \frac{\partial u}{\partial x} + i \frac{\partial v}{\partial x} \right) (h_1+i h_2) + o(|h|). \end{align*}  Now dividing by $h$ gives us the desired result.   Does there exist a differentiable but not $C^1$ function $f: U \rightarrow \mathbb{R}^2$ which satisfies the Cauchy-Riemann equations and does NOT correspond to a complex-differentiable function?
13856
We are asked to prove the following theorem found in page 88 of Differential Geometry: Curves, Surfaces, Manifolds by Wolfgang Kühnel.  Using standard parameters, calculate the Gaussian curvature and the mean curvature of a ruled surface as follows:  $K = -\dfrac {{\lambda}^2} {{{\lambda}^2 +v^2}^2}$   and   $H = -\dfrac {1} {2({\lambda}^2 +v^2)^{3/2}} (Jv^2 + \lambda' v + \lambda (\lambda J + F))$    In standard parameters, a ruled surface is $f(u,v) = c(u) + v X(u)$ and $||X|| = ||X'|| = 1$ and $\langle c', X' \rangle = 0$.  Thus, using standard parameters, a ruled surface is, up to Euclidean motions, uniquely determined by the following quantities:  $F = \langle c', X\rangle$  $\lambda = \langle c' \times X, X' \rangle = \det (c', X, X')$  $J = \langle X'', X \times X' \rangle = \det (X, X', X'')$  Also, the first fundamental form is given as follows:  $I = \begin {pmatrix} \langle c',c' \rangle + v^2 &amp; \langle c', X \rangle \\ \langle c', X \rangle &amp; 1 \end {pmatrix} = \begin {pmatrix} F^2 + {\lambda}^2  + v^2 &amp; F \\ F &amp; 1 \end {pmatrix}$ with $\det (I) = \lambda^2 + v^2$.  So far, I have that  $f_u (u,v)  = c' + vX'$  and  $f_v (u,v) = X$  Also,  $f_{vv} (u,v) =0$  I know the formula for the first fundamental form, the normal vector, and the second fundamental form. However, I don't know how to obtain the dot products and the cross products without having to isolate the components.
40763
I am stuck with the following problem. Every polynomial of degree $d$ can be expressed as $$ p(x) = p_d \binom{x}{d}+ p_{d-1}\binom{x}{d-1} + \cdots + p_0 \binom{x}{0} $$     What is the representation of    $$ \binom{x}{i} \binom{x}{j}  $$   in such a basis?   I have no idea how to solve this...
21083
The following appears as the second-to-last problem of Stewart's Complex Analysis:     Describe the Riemann surface of the function $y=\sqrt{z+z^2+z^4+\cdots +z^{2^n}+\cdots}$.   This problem intimidated me when I first saw it as an undergrad, as the series under the root isn't the expansion of any function I know (then or now). That renders the Riemann surface more subtle than usual, since any info about poles and zeroes must be found from this series alone. For instance, it has a zero at $z=0$ and therefore a branch point. But there should be other zeroes — indeed, infinitely many zeroes— since the number associated with any truncation of the series grows as $2^N$!. This suggests that the Riemann surface of this function must be exotic.  I'll give a sketch of an argument below, taking the pedestrian approach of hunting poles and zeroes of the function. However, I'd be glad to see an elegant and advanced perspective of the problem, particularly if it uses some ideas which I wouldn't have encountered as an undergraduate. The more food for thought, the better!  EDIT: As noted by O.L.'s answer below, the series above is the canonical example of a lacunary function with the unit circle as natural boundary. What that leaves open is the nature of the resulting Riemann surface, which would seem to hinge upon the quantity of the function's zeros within the unit circle. David Speyer has given an argument in his answer that this number is at least 10. Is there a more definitive result on the zeroes of this lacuuary function?
115211
Let $H^k(M,\mathbb R)$ be the De Rham cohomology of a manifold $M$.  There is a canonical map $H^k(M;\mathbb Z) \to H^k(M;\mathbb R)$ from the integral cohomology to the cohomology with coefficients in $\mathbb R$, which is isomorphic to the De Rham cohomology. As a  already revealed, the images of this map are precisely the classes of differential $k$-forms $[\omega]$ that yield integers when integrated over a $k$-cycle $\sigma$,  $$ \int_{\sigma} \omega \in \mathbb{Z}  \quad\text{ whenever } d\sigma = 0$$  Let us call them "integral forms".  Motivated by the cup product on cohomology, my question/request is the following:     Give a direct proof that the wedge product $[\omega\wedge\eta]\in H^{k+l}(M,\mathbb R)$ of two integral forms $\omega\in \Omega^k(M)$ and $\eta\in \Omega^l(M)$ is again an integral form.   This should be true because the cup product is mapped to the wedge product, but the point of the exercise is to prove this statement directly, without constructing the singular cohomology $H^k(M,\mathbb Z)$ or homology first.  Maybe I also have to make sure that the condition of being an integral form is something that can be "checked effectively" without singular homology; this might be subject to a new question.
78557
If $g(w)$ and $f(z)$ are holomorphic functions, show that $g(f(z))$ is also holomorphic.   From the assumptions we have that for some $A,B$,  $$\lim_{h\rightarrow 0}\frac{g(f(z)+h)-g(f(z))}{h}=A,$$  $$\lim_{h\rightarrow 0}\frac{f(z+h)-f(z)}{h}=B.$$  We want to prove that there exists $C$ with  $$\lim_{h\rightarrow 0}\frac{g(f(z+h))-g(f(z))}{h}=C.$$  The second equation means, by definition of the limit, that for any $\varepsilon &gt; 0$, there exists $\delta$ such that $0&lt;|h|&lt;\delta$ implies $$\left|\frac{f(z+h)-f(z)}{h}-B\right|&lt;\varepsilon.$$  I don't know how to get to $\dfrac{g(f(z+h))-g(f(z))}{h}$ from here.
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.
19674
Let $d \geq 4$. I'm interested by know if there is a surface $S$ of degree $d$ in $\mathbb P^3_{\mathbb C}$ such that $S$ does not contains a line. I know  I have no idea how to do it.
52164
Let the degree of the polynomials $P(z)=a_0+a_1z+a_2z^2+\cdots+a_nz^n$   $a_n\neq0$ and $Q(z)=b_0+b_1z+b_2z^2+\cdots+b_mz^m$ $b_m\neq 0$ be such that $m\geq n+2$. Show that   if all the zeros of $Q(z)$ are interior to a simple closed contour C,   then $$\int_C\frac{P(z)}{Q(z)} \, dz=0$$   I am lost on how to show it $P(z)$ and $Q(z)$ are polynomials, then the ratio is analytic in all the points except where $Q(z)=0$. If $\frac{P(z)}{Q(z)}$ be analytical in all the points on the contour or inside, so I could tell straight that the integral is zero, but that is not the case.
64934
Here, $Q$ is a polynomial with distinct roots $\alpha_1, \ldots, \alpha_n$ and $P$ is a polynomial of degree $&lt;n$. Once again, the task is to show  $$\frac{P(z)}{Q(z)} = \sum_{k=1}^{n}\frac{P(\alpha_k)}{Q'(\alpha_k)(z-\alpha_k)}$$  For reference, this is page 32 exercise 2 in Complex Analysis by Ahlfors. I'm having a ton of trouble making connections between what is presented in the text and this problem. The section is on rational functions and is incredibly concise. There was never a mention of derivatives and their relationships to rational functions, so I'm guessing here that it just means that $Q'(a_k)$ is a polynomial of degree $n-1$.  I don't even know where to start here, but I do have a couple question which might motivate the proof. What does $P(\alpha_k)$ mean? I don't see what a root of $Q$ has to do with a completely different polynomial. What is the relationship between the roots of $Q$ and the roots of $Q'$? And finally, where does the derivative of $Q$ come in?
88809
I refer to , which I understand says that the branch points of a nonconstant holomorphic map $F: X \to Y$ between Riemann surfaces $X$ and $Y$, both of which are not necessarily compact, form a discrete subset of range $Y$ of the map $F$. (Not sure if 'connected' is part of the definition of Riemann surfaces in other textbooks, but it is here.) Question 1: Do I understand correctly, and is this indeed true? What else I understand:  $A$ is a discrete subspace/subset (I just assume everything here is the usual subspace topology) of $B$ if and only if every singleton subset of $A$ is open in $A$ if and only if every point of $A$ is isolated.  This overflow question () seems to suggest this is not true but is true if $F$ is proper. However, I'm not sure the definitions of Riemann surface are the same (maybe some answers use definitions where Riemann surfaces are not necessarily connected or something).  2.1. I know $F$ is 'discrete' (meaning that its fibres are discrete; here, i refer to a ): see definition before Lemma 3.1, on p. 7), by Proposition II.3.12, and open, by open mapping theorem, which is Proposition II.3.8, but I'm not sure these help show $F$ is proper or are otherwise helpful. Ostensibly 'discrete' maps don't map discrete subsets to discrete subsets.   This stackexchange question () seems to suggest this is true.  I think this is true if $F$ is injective or at least something like locally injective. (Please don't make me type up all the stuff I did for this part.) However based on the question in (3) and based on the , which I think uses the term 'branch points' to mean the same thing as what Miranda means by 'ramification points' (and thus is different from Miranda's 'branch points'), it seems that no neighbourhood $U$ of a ramification point $p \in U \subseteq X$ of $F$ is such that the restriction $F|_U$ is injective.  Kinda related:    This page () says    Now, the ramification and branch points must form a discrete set  I don't think the page assumes anything like $F$ is proper.  The  text in (2.1) actually seems to assume proper based on the '3.7. Proper holomorphic maps' part, but said part refers to 'Lemma 3.17', so I think proper can be replaced with closed. Maybe $F$ is closed or something.  Kinda related (see Question 2 below):     says    If $F$ is proper then the image $\Delta = F(R)$ is discrete in $Y$.   This seems to have all the same assumptions as with Miranda. Ostensibly, either Miranda is wrong or there's some higher level machinery that allows us to omit $F$ proper.  Question 2: To possibly generalise this, what are the minimum requirements on $F, X$ and $Y$ to make $F$ map discrete subsets to discrete subsets? I mean, I'm not sure we really need 'holomorphic' here. I have a feeling this applies perhaps to maps that are just open/closed, continuous and discrete and spaces that are just Hausdorff and locally compact or something.  (again) Kinda related:
15050
Prove that $$\frac {1} {2\pi i} \oint \frac {{\rm d}z} {P(z)} $$ over a closed  smooth simple curve that contains all of the roots of the polynomial $P(z)$ is either zero if $n \geq 2$ or equals $\frac{1} {a_0} $ if it's degree is $1$.   Proof: $n\geq 2$ $$\frac {1} {P(z)=(z-z_0)(z-z_1)\dots(z-z_n)}$$ So after partial decomposition of the fraction I'll have something like $$\frac {A} {z-z_0} +\frac{B} {z-z_1}+\cdots$$ where $A,B \in \mathbb{C}$ (Is this fact true that $A,B$ wont be polynomials and just coefficients I can see it why in an intuitive way but cannot write a full proof of it for the decomposition trying to get a linear system or something) then use Cauchy's theorem and integrate all of these on  different circles with their center at each root  and argue that  the integral is the same as the sum of those circle integrals. Now since $\frac {1}{z-z_n}$ will be analytic inside the circle when I'm integrating on circles that do not contain $z_n$  some  of those integrals  will be zero. and now since my curves are of the form $\gamma(t)=z_n+e^{it}$ , $t \in [0,2\pi]$ the rest of the integrals  will look like $$\int_0^{2\pi}  \frac{Aie^{ti}} {e^{it}}\,{\rm d}t $$  and all of them will be $A_i$.   Is my proof correct? Only part I cannot show is that I won't have any problem with decomposing the fraction. And I'm also using the assumption that after the decomposition I'll have as numerators simple complex numbers not not functions of any kind.
71108
I am trying to understand the fundamentals of Riemann surface theory and so far I have the following:  --Definition 1. A conformal structure on a Riemann surface $\Sigma$ is an equivalence class of metrics $$ [g]=\{e^{2u}g \colon u\in \mathcal{C}^\infty(\Sigma)\} $$  --Definition 2. A complex structure on a Riemann surface $\Sigma$ is an equivalence class of complex atlases, where two atlases are considered equivalent iff their union forms a new complex atlas.  Note 1. Each complex structure has a canonical representative given by the maximal atlas  ---Riemann's Uniformization Theorem. In any given conformal structure, there exists a unique metric with constant curvature of either $1,0$ or $-1$.   Note 2 This gives a means of choosing a canonical representative for each conformal structure.  My questions are these:  (I) Which theorem tells us that there is a bijection between these two definitions of conformal and complex structures?   (II) What is meant by the "Riemann moduli space"?  Many Thanks, A.  EDIT: In establishing the claim (I), it appears to be neccessary to take as assumption that a surface is orientable.