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I believe problem is solved but I would like to receive some comments from the experts if I miss something. Set ##T## is given embedding in ##\mathbb{R}^4## and projection ##\pi## is very simple ##\pi(x_1,x_2,X_1,X_2)=(x_1,x_2)## For obvious reasons ##\pi(T)=S^1## Now I want to build the...

Yes. Using "physical intuition" everything is pretty clear but I would like to elaborate this simple example in a rigorous way. After even more thinking I come to the conclusion that this statement is not true. Providing such maps (##\pi## and ##\chi_\alpha##)implies that set ##T## is vector...

Book is Rudolph and Schmidt "Differential Geometry and Mathematical Physics"

First of all I would like to thank for all the answers. I think about this problem for a while again. On manifold there is homeomorphic map ##\kappa## from open subset ##U## to ##\mathbb{R}^n## and this is something touchable (don't know how to put it correctly. But it is quite clear how to...

Hello there! Reading the textbook on differential geometry I didn't get the commentary. In Chapter about vector bundles authors provide the following example Let ##M=S^1## be realized as the unit circle in ##\mathbb{R}^2##. For every ##x\in S^1##, the tangent space ##T_x S^1## can be identified...

To add some detail of my struggle. I represent my super matrices according to suggested in the proof way ##\mathcal{M}=\left(\begin{matrix}M_{00}+\beta_L A_{00} & M_{01}+\beta_L A_{01} \\ M_{10}+\beta_L A_{10} & M_{11}+\beta_L A_{11}\end{matrix}\right)##...

Hello! Reading Roger's book on supermanifolds one can find sketch of the proof for multiplicative property of super determinant. Which looks as follows All the words sounds reasonable however when it comes to the direct computation it turns out to be technical mess and I am about to give up. I...

One more time thanks for your help!

I do get the same results as you actually. But moving from line (8) to (9) you have used supercommutativity ##AB=(-1)^{|A|\, |B|}BA## but according to the textbook's definition algebra ##\mathbb{A}## is not necessarily super commutative it is just any super algebra. Perhaps author forgot to add...

Hello! I have some troubles with the definition of the so called super Lie module. In Alice Rogers' textbook "Supermanifolds theory and applications" definition goes as follows Suppose that ##\mathbb{A}## is a super algebra and that #\mathfrak{u}# is a super Lie algebra which is also a super...

I also believe that all this machinery is because of this very specific definition that derivation works only on function that are defined globally on the whole manifold. Book is "Differential Geometry and Mathematical Physics" by Rudolph and Schmidt (very hard to read (for me) but rigorous).

I do understand what you are talking about. But I believe the reason is simply formal. If ##f## is of class ##C^k(M)##. Then l.h.s. of (1.4.19) is ##C^k## and I can apply derivation that maps ##C^k## functions to real numbers. However the remainder in Teylor's formula is of class ##C^{k-2}##...

Do you mean that one can prove this isomorphism ##T_m M=D_m M## without assuming that manifold is of ##C^\infty##? Indeed. Sorry for that here is the full proposition with the proof Proof of this proposition also uses this ##h##-thing. I used to think that it was made just to make the proof...

Hello! According to the attached proposition on ##C^\infty## manifold space of derivations ##D_m M## is isomorphic to Tangent space ##T_m M##. Cited here another proposition (1.4.5) states the following 1. For constant function ##D_m(f)=0## 2. If ##f\vert_U=g\vert_U## for some neighborhood...

I think I found rigorous way to solve second excercise. Homeomorphism should map closed sets to closed sets. Every single point in ##\mathbb{R}## is closed while ##\pi((a,b))## is open in ##\mathbb{R}/\sim## so there is no homeomorphism.

I believe I do not understand your exercise. I need to provide open set ##U_a\subset\mathbb{R}/\sim## which contains equivalence class of ##a##. I can take ##U_a=\pi((a-\varepsilon,a+\varepsilon))##, ##\pi^{-1}(U_a)=(a-\varepsilon,b)##. Then ##U_a## is open due to properties of canonical...

Hello! I have two related exercises I need help with 1. Partition the space ##\mathbb{R}## into the interval ##[a,b]##, and singletons disjoint from this interval. The associated equivalence ##\sim## is defined by ##x\sim y## if and only if either##x=y## or ##x,y\in[a,b]##. Then...

Many thanks! Can you provide some easy readable reference on this? Currently it looks like a pure magic of definitions

Can I put it this way? If the inverse image of the set ##U## of the canonical projection to the quotient is open then ##U## is open.

I am slightly confused. What kind of ##U##s are considered here? If open, then I completely fine with this since canonical projection is continuous by definition. Nonetheless, I believe I've asked about another thing. I want to build a manifold from this quotient therefore I need open...

Hello! Reading a textbook I found that authors use the same trick to show that subsets of quotient topology are open. And I don't understand why this trick is valid. Below I provide there example for manifold (Mobius strip) where this trick was used Quote from "Differential Geometry and...

Authors did not provide any specific definition for ##\mathbb{K}_1## so I think ##\mathbb{K}_1=\mathbb{K}_1^1##. I believe it should be ##\min## in the definition of ##K_{\mathbf{x}}##.

I do take ##\mathbb{R}^2##. Suppose $$\mathbf{x}=\left(\begin{matrix} 1 \\ 0 \end{matrix}\right),\;\; \mathbf{y}=\left(\begin{matrix} 0 \\ 1 \end{matrix} \right)$$ then this ##l## function is equal to ##\sqrt{2}##. On the other hand ##\max## in the definition of ##K_{\mathbf{x}}## takes values...

Hello! I am reading "Differential Geometry and Mathematical Physics" by Rudolph and Schmidt. And they have and example of manifold (projective space). I believe that there is a typo in the book, but perhaps I miss something deep. Definitions are the following $$\mathbb{K}^n_\ast=\{\mathbf{x}\in...

Please, don’t get me wrong I do understand how analytic continuation works. And all functions in the expression are analytic. The question is why “above the cut”? Analytic continuation below the cut is also available but why it is considered to be unphysical?

Hello! I am currently reading Itzykson Zuber QFT book and on Chapter 7 where for the first time loops are considered. Particular method of dealing with divergences namely Pauli-Villars regularization is considered in section 7-1-1 considering vacuum polarization diagram. I do understand physics...

Hello all! I am trying to find self-consistent introduction to current state of quantum N=8 d=4 supergravity. Other books on supergravity are very welcome as well

Hello! There is a problem to write chemical reactions that goes with substances if they are not stored properly. For example theophylline should be saved from light and though I am trying to find its’ reaction with hv(light) but failed. Please help with some good reference Many thanks in advance

I don't know your actual background but this textbook helped me a lot then Pierre Ramond "Group Theory a Physicist's survey" It was also good in the following manner. You might get basic ideas and concepts on groups from any HEP textbook but for some reasons you want to enlarge your knowledge...

Hello Being a professional physicist(Quantum field theory and HS theory) I'd like to learn chemistry for some reasons. I've already tried to find a nice Chemistry textbook but failed to find physicist friendly one. My last class on chemistry was in high school like 11 years ago already, so my...

After some thinking and asking I believe that this identity may be true due different index structure of sigma matrices $$ \sigma_{\mu \alpha \dot{\alpha}}, \bar{\sigma}_\mu {}^{\dot{\alpha} \alpha}$$ If someone has a nice comprehensive refference on this spinor algebra issues I would be very...

Since it is involved in the contraction on r.h.s. of the identity it makes difference. This identity is true $$ A^\dagger \bar{\sigma}_\mu A=\bar{\sigma}_\nu\Lambda^\nu{}_\mu.$$ And it is kinda obvious. When I apply inverse SL(2,C) transformation it should result in inverse Lorentz...

Hi there! I am reading textbook "Supergravity" by Freedman and Van Proeyen and got stuck on a simple exercise (Ex 2.4). Usually I would proceed further marking it as a typo but I've checked the errata list on the website and didn't find this exercise there Exercise 2.4 Show that ##...

Homework Statement In Ex. 2.4 from textbook Supergravity by Freedman and Van Proeyen one needs to prove the following identity $$ A^\dagger \sigma_\mu A=\sigma_\nu \Lambda^\nu{}_\mu $$ Homework Equations It is easy to prove the other identity in this exercise $$ A\bar{\sigma}_\mu...

If Virasoro algebra has not central charge, Verma modules with $h=1$ and $h=0$ are in some sense equivalent $$ \vert 1 \rangle = L_+ \vert 0 \rangle, $$ where $$ L_0 \vert 0\rangle =0 \;\; L_0 \vert 1 \rangle=-\vert 1 \rangle $$ Applying lowering operators $L_-$ $$ L_- L_+ \vert 0\rangle = (L_-...

Hi there! I have som troubles with representation theory. It is obvious that bosonic strings fields $X^{\mu}$ has zero conformal dimension $h=0$. But when one builds Verma module (open string for example) highest weight state has the following definition $$ L_0 \vert h \rangle = 1 \vert h...

I know that ##\theta## anticommute, but how one deduce that ##d\theta## obey the same rule

Hi, everyone! I am trying to understand notation of this textbook http://arxiv.org/abs/hep-th/0108200 page 8, formulas 2.1.4 and 2.1.5 $$\int d \theta_\alpha \theta^\beta=\delta_\alpha^\beta$$ this could be found in any textbook the weird that from the above formula follows $$\int d^2...

Equal sign means that they are isomorphic. Equal algebras doesn't mean that group are isomorphic. O(3) and SO(3) have the same Lie algebra, but they are not isomorphic. Exponential map from algebra to group gives only simply connencted part. One can not build smooth curve from matrices with...

In string theory strings are fundamental, just electron or photon in SM

Lorentz group in three dimensions is SO(2,1) and it is NOT isomorphic to SL(2,R). SL(2,R) - is spin group in three dimensions with the signature mentioned above. SO(2,1)=SL(2,R)/Z_2 Majorana spinors are real because they are Majorana))) Definition of Majorana spinor

In string theory Lorentz invariance preserved only in 26 dimension.

By construction :-) You impose several conditions on your action and then get how auxilary fields should transform. Just like with YM gauge field, you want gauge invarianceand from this you get transformation law

Lagrangian should be invariant off-shell, you are right. I am not an expert in this. But probably you need to introduce new field with purely algebraic equations of motion, like F in WZ model, just to cancel this term.

Does the extra term vanish on the equations of motion?

I think I got the idea. Due to locality of Vertex Operators I can generate not just a state of a certain momenta but can actually put this state at a certain place. My problem was with understanding of constracting asymptotically free states which are needed in QFT as initial and final. In...

This point is not clear. Vertex operators generate single particle/string states without interaction, so this states are obviously free. And these states are our string |in> and |out> states. How to proceed to S-matrix from this point? If I understand you right. If we look from state...

Hi there! S-matrix is Path Integral with Vertex Operators inserted. I know how to compute Shapiro-Virasoro amplitude. So I don't have problems with calculations but with understanding. In this calculations formalism of 2-dimensional CFT is used. But there is no S-matrix in CFT, only...