My question is motivated by 2D CFT where one works in "complex coordinates". The question is the following: Suppose I am in 2D flat Euclidean space, i.e.  $$\mathrm{d}s^2 = \mathrm{d}x^2 + \mathrm{d}y^2$$  One of the crucial things here is that $x$ and $y$ are independent. Now it well known that the plane can also be described by one complex number. The point here is that it is ONE complex number, so once I write down:  $$z = x+iy$$  I have everything to describe the complex plane. And yet in every book, they then define  $$\bar z = x-iy$$  so the complex conjugate and write down the metric:  $$\mathrm{d}s^2 = \mathrm{d}z \mathrm{d}\bar z$$  Of course, the norm in the complex plane is given by $z\bar z$ but what I don't understand is that the complex conjugate is not an independent variable, so how does the above metric make sense. Of course, often in the books, they at some point mention that they complexify and essentially consider $\mathbb{C}^2$ in which case $z$ and $\bar z$ are to be considered as independent, then the metric makes more sense. But if you don't complexify, how am I supposed to understand the said metric?  Edit: It seems that the standard answer is that you consider the complexified $\mathbb{C}^2$, but that wasn't my question: If I don't want to go to $\mathbb{C}^2$, and just consider $\mathbb{R}^2$ as $\mathbb{C}$, what would the metric be?
120885
The Setup: Let's say we want to study a Euclidean $\mathrm{CFT}_2$ on $\mathbb R^2$ with coordinates $\sigma^1$ and $\sigma^2$ and metric   $ds^2 = (d\sigma^1)^2+(d\sigma^2)^2$.    It seems to me that in the usual discussion (e.g. di Francesco, Ginsparg, Polchinski), one proceeds to consider an analytic continuation of the CFT to $\mathbb C^2$ with coordinates $z^1, z^2$ and complex metric  $ds^2 = (dz^1)^2+(dz^2)^2$  and then, one performs the coordinate transformation $z = z^1+iz^2$ and $\bar z = z^1-iz^2$.  In this way the coordinates $z$ and $\bar z$ can be considered "independent" because they are coordinates on a complex two-dimensional manifold.  Also, in these coordinates the metric becomes  $ds^2 = dz\,d\bar z$  and it becomes clear that conformal mappings consist of mappings: $(z, \bar z)\to (f(z), g(\bar z))$.  My confusion is this: Since our original theory was on $\mathbb R^2$, books say that when we do calculations, we should consider the physical theory as living on the copy of $\mathbb R^2$ embedded in $\mathbb C^2$ given by the condition $\bar z = z^*$.  But consider the mapping $(z, \bar z)\to (z^2, \bar z)$.  This is a conformal mapping on $\mathbb C^2$, but it does not map the surface $\bar z = z^*$ to itself; for example the point $(z, \bar z)=(2,2)$ gets mapped to the point $(z^2, \bar z) =(4,2)$ and $2$ is clearly not equal to $4^*$.  In particular, it seems to me that analytic continuation to a CFT on $\mathbb C^2$ enlarges the set of mappings one can have, so what relevance does it really have to the original CFT on $\mathbb R^2$?
58154
We have a complex plane and there are inequality on it: $2|z| &gt; |1 + z^2|$. What is the locus of such set of $z$?  I tried to solve it by using polar coordinates and got: $r^4 - 2r^2\cos(2\phi ) - 1 = 0$.  As well I tried solve it using $z = x + iy$ and got: $4(x^2 + y^2) = (x^2 - y^2 + 1)^2 + 4x^2y^2$  By I still can`t figure out what is the explicit name or at least descritpion for such locus of points. Will be very grateful for any help.
16983
Note: I'm not too experienced with branches of mathematics like topology, so a lot of the terminology I use is used pretty loosely.   I think we all understand the fact that a point in $n$-dimensional space is described with $n$ parameters. In 2d space, you have $(p_1, p_2)$ (where $p_k \in \mathbb{R}$), in 3d space you have $(p_1, p_2, p_3)$, so in an $n$-dimensional space you have $(p_1, p_2, ..., p_n)$. I know the way I'm stating this with the notation I use isn't as robust as it could be, but you get the idea.  Going into the world of fractal geometry, we find non-integer dimensional spaces. Applying what I said previously, how would you even express a point in such a space? What does it even mean to be that kind of space? Are you unable to express a "point" in such a space?
111431
Letting $\mathbb{Q}$ be equipped with the Euclidean metric.  What I can work out is that it is bounded as its contained in the closed ball of radius ${\sqrt2}/{2}$ centred at ${\sqrt2}/{2} $.  Its not compact as it can be expressed as union of the two disjoint open sets $[0,{\sqrt2}/{2}) $and$ ({\sqrt2}/{2}, \sqrt2)$ (though I'm not sure if this makes it not compact in $\mathbb{Q}$ or just $\mathbb{R}$).  And its not closed as the sequence of truncations of $\sqrt2$ in $\mathbb{Q}$ converges to $\sqrt2$
40031
I know the chern classes-related theorem that states that $\mathbb{C}\mathbb{P}^2\# \dots \# \mathbb{C}\mathbb{P}^2$ ($k$ times) has no almost complex structure (hence no complex structure) if and only if $k$ is even.  I also know that $\mathbb{C}\mathbb{P}^2\# \mathbb{C}\mathbb{P}^2 \# \mathbb{C}\mathbb{P}^2$ has no complex structure, hence the almost complex ones you define on it are not integrable.  Where can I find a proof of this proposition? How do I see that the connected sum of three (or five, or seven, or every odd number) copies of $\mathbb{C}\mathbb{P}^2$ doesn't admit a complex structure?
15760
Since there are  complex numbers (2 dimensions) and quaternions (4 dimensions), it follows intuitively that there ought to be something in between for 3 dimensions ("triernions").   Yet no one uses these. Why is this?
111002
Define the following function on $\mathbb{R}$ by $d(x,y)=\dfrac{|x-y|}{\sqrt{1+x^2}\sqrt{1+y^2}}$. Prove that this is metric.  I proved the first two properties of metric. But how to prove that $d(x,y)+d(y,z)\geqslant d(x,z)$ for any $x,y,z\in \mathbb{R}$
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.
121140
Exercise: Let $f$ be entire and non-constant. For any positive real number $c&gt;0$ show that the closure of $\{ z:|f(z)|&lt;c\}$ is the set $\{z:|f(z)|\leq c\}$.   Solution: Let $\varepsilon&gt;0$ and define $A:=\{ z:|f(z)|&lt;c\}$. Then, because $f$ is continuous (via entire) and non-constant, there exists $z\in A$ such that   $$B(c;\varepsilon)\cap A \neq \varnothing .$$  That is to say that $c$ is a limit point of $A$. Since $A^-$ must contain all of its limit points, we have that $\{z:|f(z)|\leq c\}$.    Is my solution correct? I have another way of doing it using a sequence $z_n\to z$, but I would prefer to use the above if it correct. The reason I am feeling unsure about my solution is because I am not sure how we know (geometrically) that $|f(z)|$ goes all the way up to the boundary (i.e. $c$). Is it safe to say that $|f(z)$ gets infinitely close to $c$ but does not touch because $f$ is entire &amp; non-constant? FYI I am new at these concepts so please be a detailed as possible. Thank you in advance!  Note: I am using the notation $A^-$ to indicate the closure of $A$. This is the notation used in my text (Complex, Conway) and I suspect it is used as to not be confused with the complex conjugate.
118296
Suppose $f(z)$ is analytic in the open region $D$, and $C$ is a simple closed curve in $D$. For any $z_0\in D\setminus C$, prove:   $$\oint_C\frac{f'(z)}{z-z_0}\mathrm{d}z=\oint_C\frac{f(z)}{(z-z_0)^2}\mathrm{d}z.$$   It is a homework problem immediately after the introduction of complex integral and Cauchy's integral theorem.  When the region $S$ surrounded by $C$ is contained in $D$, the value of the above integral is either both $2\pi i f'(z_0)$ (when $z_0\in S$) or both $0$ (when $z_0\notin S$). But in the case when $S\not\subset D$, which means $C$ encircles some holes or singular point of $D$, I don't know how to handle it.  Any idea or hint will be appreciated, thank you!
50257
Identify $\mathbb{R}^2$, with coordinates $x, y$, with $\mathbb{C}$, with coordinate $z = x + iy$. Likewise, identify a copy of $\mathbb{R}^2$ with coordinates $u, v$ with $\mathbb{C}$ with coordinate $w = u+iv$. Let $f : \{\mathbb{R}^2 - (1, 0) - (-1, 0)\} \to \mathbb{R}^2$ be the function $f(z) = \frac{1}{z-1} - \frac{1}{\bar{z}+1}$. Show that $f$ extends to a smooth map $\tilde{f} : S^2 \to S^2$, where $S^2$ (or $\mathbb{CP}^1$ is the one-point compactiﬁcation of $\mathbb{R}^2$ (or $\mathbb{C})$.   Following FBD's comment below, I want to show:   1, $f$ is continuous, which is obvious;  2, $f$ is surjective;  3, $f^{-1}$ maps compact set to compact set   Correct?  Thank you very much.
32283
This comes straight from Conway's Complex Analysis, VII.4, exercise 4.     Find an analytic function $f$ which maps $G:=$ {${z: |z| &lt; 1, Re(z) &gt; 0}$} onto $B(0; 1)$ in a one-one fashion.   $B(0;1)$ is the open unit disc.  My first intuition was to use $z^2$, which does the job splendidly, except for the segment $(-1,0] \subset B(0;1)$.  Under $z^2$, the pre-image for this segment is the segment $[-i,i]$, which is not in $G$.  My next thought is to modify $z^2$, something like $a(z-h)^2+k$.  I've yet to work out the details, but my gut tells me this isn't the right idea.  I've been teaching myself conformal maps in preparation for a qualifying exam.  So, if there's a shockingly basic, obvious solution... please patronize me.
123553
If complex numbers are in some sense 2 dimensional numbers, would it be useful or logical to extend the complex number system to a system of 3-dimensional or even n-dimensional numbers? If this does not make sense in general, why does it make sense only to extend from the 1st dimension to the 2nd dimension? What is special about 2-dimensional numbers that they are needed but n-dimensional numbers are not? I know that we have $R^k$, which is in some sense a field of $k$ dimensional numbers, but that is very different from the way in which complex numbers are constructed. So my question is could the complex numbers be extended to a 3rd dimension (or even an nth dimension) in the same way that the real numbers are extended to the complex numbers, in which the properties of complex numbers are preserved as a subset of this higher dimensional space?
8552
I wonder if this is correct: there is a holomorphic function on an open connected subset $G$ of $\mathbb{C}$ which maps $G$ onto a subset of a straight line, and I have to show that the function is constant.   I thought I can suppose that the straight line is the real axis (otherwise I can find a rotation and a traslation that will do so) and so using the Cauchy-Riemann equations I find that the function is constant since its imaginary part is zero. Is that correct? Thank you
34501
I would like a metric $d(\;)$ between pairs of (infinite) lines in $\mathbb{R}^3$, with these properties:   If two lines $L_1$ and $L_2$ are parallel and separated by distance $x$, then $d(L_1,L_2) = x$. $d(L_1,L_2)$ increases with the degree of skewness between the lines, i.e., the angle $\theta$ between their projections onto a plane orthogonal to the shortest segment connecting them (dashed below), where $\theta=0$ for parallel lines and $\theta=\pi/2$ for orthogonal lines. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;   Intuitively, I would like the metric to be related to the repulsive force between two electrically charged lines (but it need not be exactly the actual physical force).  Have such metrics been considered in the literature? If so, I'd appreciate descriptions and/or pointers&mdash;Thanks!  Update1. It turns out that the natural ad hoc definition (from the comments) fails to be a metric. For example, below, a sufficiently large value of $a$ ensures the triangle inequality will be violated: &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; So this question may be more difficult than it initially appeared...  Update2.  The  have revealed that in fact there is no metric that satisfies my two conditions (and is continuous w.r.t. $\theta$)!
87320
The unit sphere in $\mathbb{R}^3$ is $\{(x,y,z) : x^2 + y^2 + z^2 = 1 \}$. I always hear people say that this is closed and that it has no boundary. But isn't every point on the sphere a boundary point? Since for every point $x$ on the sphere, any open ball in $\mathbb{R}^3$ (defined by $\{ y : |y-x| &lt; r\}$) contains points on the sphere and outside/inside the sphere. Hence it is a boundary point.  I also sometimes see that people say instead of an open ball in $\mathbb{R}^3$, I should be looking at an open neighbourhood of the sphere but this doesn't make any sense I don't think as the definition of the open set is as I gave above.  Or is it that people mean it's a closed surface (analogous to a closed curve)?
95751
The SR invariance formula makes space-like intervals imaginary (e.g., the distance $x$ in a given frame has interval $ix$). Yet modern physicists consider it bad form to define the distance itself as $ix$, as Minkowski did. It seems to me that this distinction is trivial—what’s wrong with defining the distance as the value of its interval? An answer to a previous question of mine on this topic referred to a paper by Visser showing that there is a way to do this that can be generalized to GR.  So what’s the fuss about? Even if there weren’t such a way, I don’t see how that would invalidate Minkowski’s definition for SR. Since the interval measure is what’s important, what fundamental difference can it make whether one defines the distance as x or as the imaginary interval $ix$?  The whole idea of a LT being an imaginary rotation (important, because phi is additive) depends on the use of the invariant measure, which is the complex length between two points in C2, the space in which complex rotations exist, and so the space in which the x-t plane is being considered to be embedded when we talk of complex rotations. In C2 it can easily be shown that the angle between any two axes, real or imaginary, is a real right angle, and this explains the invariance formula—it is simply the Pythagorean theorem for that complex triangle. So all this ‘invariant interval geometry’ is an integral part of special relativity. If you insist that distances are real and use a Pseudo-Euclidean metric, which would imply entirely different values for these angles, you are throwing out an enormous portion of SR. It is hard to see how that can be billed as an advance in our understanding.