is a local rescaling of the metric tensor $$ g_{ab}\rightarrow e^{-2\omega(x)}g_{ab} $$   maps to a theory under arbitrary differentiable coordinate transformations (Diffeomorphism is an isomorphism of smooth manifolds. It is an invertible function that maps one differentiable manifold to another such that both the function and its inverse are smooth.)   Question 1: Is Weyl transformation part of diffeomorphism? It seems that the answer would be yes,  if this $e^{-2\omega(x)}$ is arbitrary differentiable and if the starting manifold with a $g_{ab}$ is differentiable.  Question 2: Because  Is this correct to say that the gravitational anomaly capture also the anomaly due to  Weyl transformation? p.s. I asked more additional details in a previous post , but I got no answer. So let us zoom into a specific case. I hope someone can give a definite correct answer this time.
117340
I see that the weyl transformation is $g_{ab} \to \Omega(x)g_{ab}$ under which Ricci scalar is not invariant. I am a bit puzzled when conformal transformation is defined as those coordinate transformations that effect the above metric transformation i.e $x \to x' \implies g_{\mu \nu}(x) \to g'_{\mu \nu}(x') = \Omega(x)g_{\mu \nu }(x)$. but any covariant action is clearly invariant under coordinate transformation? I see that what we mean by weyl transformation is just changing the metric by at a point by a scale factor $\Omega(x)$. So my question is why one needs to define these transformations via a coordinate transforms. Is it the case that these two transformations are different things. In flat space time I understand that conformal transformations contain lorentz transformations and lorentz invariant theory is not necessarily invariant under conformal transformations. But in a GR or in a covariant theory effecting weyl transformation via coordinate transformations is going to  leave it invariant. Unless we restrict it to just rescaling the metric?  I am really confused pls help.
121779
I want to understand what exactly means to ask the question if the Lie bracket $[X,Y]$ of two vector fields $X,Y\in \mathcal{XM}$, where $\mathcal{M}$ is a differentiable manifold, is well defined.    What does it mean conceptually that the Lie bracket is well defined?  How could I prove that indeed it is well defined? Can I do it use it a specific basis, say, $e_x = \partial/\partial x$ and $e_y = \partial / \partial y$?
100412
Consider the graphs of the functions $f_1(x) = |x|$, and $f_2(x) = x$ under the subspace topology of $\mathbb{R}^2$.   Both of these graphs are smooth manifolds, just pick coordinate charts to be $(x, f_i(x)) \leftrightarrow x$.   Moreover, they are diffeomorphic via the map $(x, f_1(x)) \rightarrow (x, f_2(x))$.   This seems to clash with my intuition. For example, the graph of $f_1$ has a corner, so it "shouldn't" be smooth, much less diffeomorphic to $f_2$, which is just a straight line.   Can someone explain what's going on here?   In light of these examples, how should I visualize smooth manifolds and diffeomorphisms?
51362
Let $\Omega\subset\mathbb R^n$ and consider an arbitrary functional $$ S\colon C^k(\Omega)\to\mathbb R $$ that is local, $$ \phi\overset S\mapsto \int_\Omega L(\phi,D\phi,\dots,D^k\phi) $$ for $L$ some $n$-form taking values in the $k$-jet space $J^k\phi$. We may call $S$ the action functional, and $L$ its associated Lagrangian (although we do not assume that $\delta S[\phi]=0$, i.e., we work off-shell).  There are two notions of a symmetry that are typically discussed:   Lagrangian symmetries, that is, transformations of the form $\phi\to\phi'$ that satisfy $$ L-L'=\mathrm d\Theta $$ for some $(n-1)$-form $\Theta$, and where $L'(J^k\phi)=L(J^k\phi')$. Action symmetries, that is, transformations of the form $\phi\to\phi'$ that satisfy $$ S-S'=0 $$ where $S'[\phi]=S[\phi']$.   If we assume that the boundary conditions are such that $\Theta|_{\partial\Omega}=0$, then it is clear that Lagrangian symmetries are also action symmetries.  If we assume that $\Omega$ is "nice-enough" (e.g. star-shaped) then I believe that any action symmetry is also a Lagrangian symmetry, although I would like to have a more precise statement about this fact. In other words, my question is     Are these two notions of symmetry equivalent? Under what conditions?   Moreover, how much can we weaken the assumptions? For example, can we take $\Omega$ to be simply-connected instead of star-shaped? Can we take $\phi$ to only have weak derivatives instead of strong derivatives?
65962
I'm trying to grasp Levi-civita and Kronecker del notation to use when evaluating geophysical tensors, but I came across a few problems in the book I'm reading that have me stumped.   1) $\delta_{i\,j}\delta_{i\,j}$  2) $\delta_{i\,j}  \epsilon_{i\,j\,k}$  I have no idea how to approach evaluating these properties. Without knowing i, j, or k, how would I approach? I don't feel confident using the notation further until I can understand these properties.
86362
Any rank 2 tensor would do the same trick, right?  Then, which is the motivation for choosing the metric one?    Also, if you help me to prove that $g^{kp}g_{ip}=\delta^k_i$, I would be thankful too ^^
56066
I am reading Lee's book on Differential geometry. In Chapter 4, Lee has this exercise.  Let $M$, $N$ be smooth manifolds, and let $f:M\to N$ be a smooth map. Define $F : M\to M \times N$ by $F(x) = (x, f(x) )$. Show that for every smooth vector field $V$ of $M$ , there is a smooth vector field on $W$ on $M \times N$ that is $F$-related to $V$.  I can understand that we must have $W_{(x,f(x))}=V_x\oplus Df(x)V_p$ for all $x\in M.$ The set $\{(x,f(x)):x\in M\}$ is a closed set in $M\times N.$ Therefore, if we can show that for all $(p,f(p))\in M\times N$ there is a neighbourhood $U_p$ and a smooth vector field on $U_p$ extending $V_x\oplus Df(x)V_x$ we are done by partition of unity. But I cannot show that. Can someone help me out?
88935
I have a question in deriving Eqs. (3.6.15b) and (3.6.15c) in Polchinski's string theory vol I p. 105.   Given      $$\Delta (\sigma,\sigma') = \frac{ \alpha'}{2} \ln d^2 (\sigma, \sigma') \tag{3.6.6}$$   where $d(\sigma,\sigma')$ is the geodesic distance between points $\sigma$ and $\sigma'$.   It is said for the Weyl variation $\delta_W$,     $$\partial_a \delta_W \Delta(\sigma,\sigma')|_{\sigma'=\sigma}= \frac{ 1}{2} \alpha' \partial_a \delta \omega(\sigma) \tag{3.6.15a} $$   $$ \partial_a \partial_b' \delta_W \Delta(\sigma,\sigma')|_{\sigma'=\sigma}= \frac{ 1+ \gamma}{2} \alpha' \nabla_a \partial_b \delta \omega(\sigma) \tag{3.6.15b} $$   $$ \nabla_a \partial_b \delta_W \Delta(\sigma,\sigma')|_{\sigma'=\sigma}= - \frac{  \gamma}{2} \alpha' \nabla_a \partial_b \delta \omega(\sigma) \tag{3.6.15c} $$   Here $\gamma=-\frac{2}{3}$.   I know the reason for the factor of $1/2$ in the RHS of Eq. (3.6.15a) and I can derive it. But, putting $\nabla$ on LHS and RHS of Eq. (3.6.15a) will not lead to (3.6.15c). I guess it is because there is some higher-order terms in the geodesic distance where the treatment of (3.6.9)-(3.6.11) omitted. But I don't know how to get it. I tried $ds^2= g_{ab} dx^a dx^b= (\eta_{ab} + h_{ab})dx^a dx^b $ to separate the deviation of Minkowski metric, but I cannot get the factor $\gamma$   My question is, how to derive Eqs. (3.6.15b) and (3.6.15c) (especially there is a prime, $'$, namely $\partial'_b$ in the LHS of (3.6.15b) but not in the RHS)
77202
Einstein assumed in the derivation of his field equations that the covariant derivative (defined through the metric) of the energy momentum tensor of matter is zero, i.e. $$\nabla_{\mu}T^{\mu\nu}=0,$$  no matter how distorted the spacetime is.  It seems such a conclusion cannot be reached at via Nöther's theorem (since the metric is itself changing over spacetime). So my question is how did he have this insight then, and how would one justify the correctness of this assumption in retrospect from a Lagrangian point of view now?   If it's based on the existence of infinitesimal coordinate transformations that fix the boundary of some volume in spacetime, then would this also imply that the energy momentum tensor of any field will always have zero covariant derivative with or without gravity?    Edit: Previous form of the question had an abuse of language, viz., that I referred to the "vanishing of the covariant derivative of the energy-momentum tensor" as "energy momentum tensor is covariantly conserved". This was rightly pointed out by Ben Crowell in his answer and, I've edited the question and made it mathematically precise for the sake of clarity.
43754
Given the Lagrangian density of a theory, are the representations on which the various fields transform uniquely determined? For example, given the  for a  $$ \mathscr{L} = \frac{1}{2} \partial_\mu \varphi \partial^\mu \varphi - \frac{1}{2} m^2 \varphi^2 \tag{1}$$ with $(+,-,-,-)$ Minkowski sign convention, is $\varphi$ somehow constrained to be a scalar, by the sole fact that it appears in this particular form in the Lagrangian? As another example: consider the Lagrangian $$ \mathscr{L}_{1} = -\frac{1}{2} \partial_\nu A_\mu \partial^\nu A^\mu + \frac{1}{2} m^2 A_\mu A^\mu,\tag{2}$$ which can also be cast in the form $$ \mathscr{L}_{1} = \left( \frac{1}{2} \partial_\mu A^i \partial^\mu A^i - \frac{1}{2} m^2 A^i A^i \right) - \left( \frac{1}{2} \partial_\mu A^0 \partial^\mu A^0 - \frac{1}{2} m^2 A^0 A^0 \right). \tag{3}$$ I've heard$^{[1]}$ that this is the Lagrangian for four massive scalar fields and not that for a massive spin-1 field. Why is that? I understand that it produces a  equation for each component of the field: $$ ( \square + m^2 ) A^\mu = 0, \tag{4}$$ but why does this prevent me from considering $A^\mu$ a spin-1 massive field?  [1]: From Matthew D. Schwartz's , p.114:  A natural guess for the Lagrangian for a massive spin-1 field is $$ \mathcal{L} = - \frac{1}{2} \partial_\nu A_\mu \partial_\nu A_\mu + \frac{1}{2} m^2 A_\mu^2,$$ where $A_\mu^2 = A_\mu A^\mu$. Then the equations of motion are $$ ( \square + m^2) A_\mu = 0,$$ which has four propagating modes. In fact, this Lagrangian is not the Lagrangian for a amassive spin-1 field, but the Lagrangian for four massive scalar fields, $A_0, A_1, A_2$ and $A_3$. That is, we have reduced $4 = 1 \oplus 1 \oplus 1 \oplus 1$, which is not what we wanted.
118336
I come up with this conclusion after reading some books and review articles on conformal field theory (CFT).     CFT is a subset of FT such that the action is invariant under conformal transformation of the fields and coordinate but leave the metric unchanged.   Is this correct?   Let me explain further and take the $\phi^4$ theory in $4$-dim as an example (just discuss classical invariance, I know that loops break the invariance). In $4$-dim consider a scalar field with conformal weight $\Delta=1$ such that  \begin{align} x \to x' = \lambda x,\\ \phi'(x')=\phi'(\lambda x) = \lambda^{-1}\phi(x). \end{align} Then the action is unchanged \begin{align} S'&amp; = \int d^4 x' \sqrt{g}\left\{\frac{1}{2}g^{\mu\nu}\partial'_{\mu}\phi'(x')\partial'_{\mu}\phi'(x')-\phi'^4(x')\right\}\\ &amp;=  \int d^4 x \lambda^4 \sqrt{g}\left\{\frac{1}{2}g^{\mu\nu}\lambda^{-4}\partial_{\mu}\phi(x)\partial_{\mu}\phi(x)-\lambda^{-4}\phi^4(x)\right\} \\ &amp;=  \int d^4 x \sqrt{g} \left\{\frac{1}{2}g^{\mu\nu}\partial_{\mu}\phi(x)\partial_{\mu}\phi(x)-\phi^4(x)\right\}. \end{align} Note that I did not use $g'^{\mu\nu} = \lambda^2g^{\mu\nu}$, all metrics are unprimed. In this example we see that conformal invariance is realized without changing the metric. I was confused at the beginning since all textbooks and articles derive the conformal group and representations by considering the change of the metric.  If we use $g'_{\mu\nu}x'^{\mu}x'^{\nu} = g_{\mu\nu}x^{\mu}x^{\nu}$, the physical distance does not change at all and we are just choosing a new coordinate chart. My interpretation is, as what we meant by a physical scaling or transformation, we really change the distance between two points. Another reasoning is, the metric in CFTs are just background (not being integrated in the path integral) thus we do not change them. If we consider a theory including the metric as a dynamical field (we path integrate it and perhaps quantize it), the actions has to be invariant including the transformation of the metric.   Is the above correct? Please give me some comments and point out the wrong concepts if there is any. Thank you very much.  If you have time, could you please take a look at .
77251
It is well known that geodesics on some manifold $M$, covered by some coordinates ${x_\mu}$, say with a Riemannian metric can be obtained by an action principle . Let $C$ be curve $\mathbb{R} \to M$, $x^\mu(s)$ be an affine parametrization of $C$. (Using same symbol for coordinates and parametrization here, but it's standard). The action yielding geodesics is:  $$ S (C) =\int_CL ds $$ where  $$ L\equiv \sqrt{g_{\mu\nu} \dot{x}^\mu \dot{x}^\nu} $$  Now the popular  makes the claim that variation of $$F \equiv \frac{L^2}{2}$$  Will yield the exact same action principle solutions. However, the reference above ONLY shows that:  $C$ solves euler Lagrange equation for $L$ $\implies$ $C$ solves euler Lagrange equation for $F$  And this can be shown through a direct brute force computation.   My question: Is the converse of the above true, and how does one go about proving it. In particular, how do we show that $F$ yields no extraneous solutions to the Euler Lagrange equations.
116293
D. Tong's , chapter 5 (), feature the following in introducing the symmetries used in the Faddeev-Popov method:     We have two gauge symmetries: diffeomorphisms and Weyl transformations. We will schematically denote both of these by $\zeta$. The change of the metric under a general gauge transformation is $g \rightarrow g^{\zeta}$. This is shorthand for,   \begin{equation} g_{\alpha \beta} (\sigma) \rightarrow g_{\alpha \beta}^{\zeta} (\sigma)= e^{2\omega (\sigma)} \frac{\partial \sigma^{\gamma}}{\partial \sigma'^{\alpha}} \frac{\partial \sigma^{\delta}}{\partial \sigma'^{\beta}} g_{\gamma \delta} (\sigma) \end{equation}   In two dimensions these gauge symmetries allow us to put the metric into any form that we like — say, $\hat{g}$. This is called the fiducial metric and will represent our choice of gauge fixing.   Is there a proof available that shows that the combined symmetry allows us to "put the metric in any form (locally, of course)?" Is this property restricted to two dimensions?
95562
I've consulted several books for the explanation of why   $$\nabla _{\mu}g_{\alpha \beta} = 0,$$  and hence derive the relation between metric tensor and affine connection $\Gamma ^{\sigma}_{\mu \beta} $  $$\Gamma ^{\gamma} _{\beta \mu} = \frac{1}{2} g^{\alpha \gamma}(\partial _{\mu}g_{\alpha \beta} + \partial _{\beta} g_{\alpha \mu} - \partial _{\alpha}g_{\beta \mu}).$$  But I'm getting nowhere. May be I've to go through the concepts of manifold much deeper.
91569
Suppose that $f:M \to N$ is a smooth map of two manifolds and $E$ is a vector bundle over $N$. Given a connecton $\nabla$ on $E$ one can pull it back to $f^*N$ (which is a bundle over $M$) and obtain a connection $f^* \nabla$.   If $E=TN$ is a tangent bundle one can consider the Levi-Civita connection $\nabla$, with respect to the chosen metric $g$ on $N$. It is characterised uniquely as torsion free metric connection. Given a map $f:M \to N$ one can pullback the metric $g$ to $f^*g$. If it happens that $f^*g$ is again a metric (which is the case for example for immersions) one can consider the Levi Civita connection on $M$ with respect to $f^*g$. I wonder how these two constructions are related, in particular      Is there a way to pullback affine connections to get again affine connection (i.e. connections on tangent bundle)?     Additionally I would like to know:     What is the relation between Christoffels symbols for the connection and for its pullback?
33743
I'm confused about the Kronecker delta. In the context of four-dimensional spacetime, multiplying the metric tensor by its inverse, I've seen (where the upstairs and downstairs indices are the same):  $$g^{\mu\nu}g_{\mu\nu}=\delta_{\nu}^{\nu}=\delta_{0}^{0}+\delta_{1}^{1}+\delta_{2}^{2}+\delta_{3}^{3}=1+1+1+1=4.$$  But I've also seen (where the upstairs and downstairs indices are different):  $$g^{\mu\nu}g_{\nu\lambda}=\delta_{\lambda}^{\mu}=\left(\begin{array}{cccc} 1 &amp; 0 &amp; 0 &amp; 0\\ 0 &amp; 1 &amp; 0 &amp; 0\\ 0 &amp; 0 &amp; 1 &amp; 0\\ 0 &amp; 0 &amp; 0 &amp; 1 \end{array}\right).$$  How can there be two different answers to (what appears to me to be) the same operation, ie multiplying the metric tensor by its inverse? Apologies if I've got this completely wrong.
57262
Let $X$ be a differentiable manifold. Let $\mathcal{O}_X$ be the sheaf of $\mathcal{C}^\infty$ functions on $X$. Since every stalk of $\mathcal{O}_X$ is a local ring, $(X, \mathcal{O}_X)$ is a locally ringed space. Let $Y$ be another differentiable manifold. Let $f\colon X \rightarrow Y$ be a differentiable map. Let $U$ be an open subset of $Y$. For $h \in \Gamma(\mathcal{O}_Y, U)$, $h\circ f \in \Gamma(\mathcal{O}_X, f^{-1}(U))$. Hence we get an $\mathbb{R}$-morphism $\Gamma(\mathcal{O}_Y, U) \rightarrow \Gamma(\mathcal{O}_X, f^{-1}(U))$ of $\mathbb{R}$-algebras. Hence we get a morphism $f^{\#} \colon \mathcal{O}_Y \rightarrow f_*(\mathcal{O}_X)$ of sheaves of $\mathbb{R}$-algebras. It is easy to see that $(f, f^{\#})$ is a morphism of locally ringed spaces.  Conversely suppose $(f, \psi)\colon X \rightarrow Y$ is a morphism of locally ringed spaces, where $X$ and $Y$ are differentiable manifolds and $\psi\colon \mathcal{O}_Y \rightarrow f_*(\mathcal{O}_X)$is a morphism of sheaves of $\mathbb{R}$-algebras. Is $f$ a differentiable map and $\psi = f^{\#}$?