tag:blogger.com,1999:blog-4987609114415205593.post2365437661455753966..comments2024-03-28T13:40:26.497+00:00Comments on M-Phi: Sean Carroll on Leibniz EquivalenceJeffrey Ketlandhttp://www.blogger.com/profile/01753975411670884721noreply@blogger.comBlogger9125tag:blogger.com,1999:blog-4987609114415205593.post-65945399759958152432020-11-09T11:26:31.105+00:002020-11-09T11:26:31.105+00:00hi, your post is very helpful for me. Finally, I f...hi, your post is very helpful for me. Finally, I found exactly what i want. If you need information regarding yahoo mail then you can visit our site <a href="https://yahoo%20mail.helpdesktelefoon.com/" rel="nofollow">yahoo mail gratis mail</a><br />Rose Berryhttps://www.blogger.com/profile/01626196395843965757noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-77696630635954098222014-02-13T12:50:22.729+00:002014-02-13T12:50:22.729+00:00Mikael, thanks - I see. So, yes:
$\phi : M \to N$...Mikael, thanks - I see. So, yes:<br /><br />$\phi : M \to N$ is a diffeomorphism if and only if $N = \phi_{\ast}M$, if and only if $\phi: M \to N$ is an isomorphism.<br /><br />(A diffeomorphism is an isomorphism of differentiable manifolds. That is, a bijection between carrier sets which "preserves" the differential structure.)<br /><br />Cheers,<br /><br />JeffJeffrey Ketlandhttps://www.blogger.com/profile/01753975411670884721noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-27864039927414833772014-02-11T12:36:48.455+00:002014-02-11T12:36:48.455+00:00No. In your blogpost you write "Leibniz Equiv...No. In your blogpost you write "Leibniz Equivalence (LE)<br />If spacetimes (M,gμν,ψ) and (N,hμν,θ) are isomorphic, then they represent the same physical situation. This is equivalent to Wald's wording, which can be read here." <br /><br />Wald states that models (M, g, T) and (N, (push*g), (push*T)) are physically equivalent iff there exists a diffeomorphism F: M -> N. However he does not state that N has to be equal to (push*M). Thus it seems that for your version and Wald's to be equivalent you would have to show that M and N are diffeomorphic iff N = (push*N), which again would mean that they are isomorphic.<br /><br />MikaelAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-3173163131794543882014-02-10T21:03:21.719+00:002014-02-10T21:03:21.719+00:00Mikael, is the part that isn't clear this part...Mikael, is the part that isn't clear this part:<br /><br />"Let $\mathcal{M}_1 = (M,g,...)$ and $\mathcal{M}_2 = (N,h,...)$ be spacetime models, with carrier sets A and B respectively. Let $\pi : A \to B$ be a bijection such that<br /><br />(i) $N = \pi_{\ast}M$<br /><br />(ii) $h = \pi_{\ast}g$<br /><br />(iii) and $\dots$"<br /><br />Is this the bit that isn't clear? I.e., that this is equivalent to saying that $\mathcal{M}_1$ and $\mathcal{M}_2$ are isomorphic?<br /><br />JeffJeffrey Ketlandhttps://www.blogger.com/profile/01753975411670884721noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-61602859063289534672014-02-10T16:31:26.987+00:002014-02-10T16:31:26.987+00:00- Mikael- MikaelAnonymousnoreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-91190224272693444672014-02-10T16:29:35.936+00:002014-02-10T16:29:35.936+00:00Hi Jeff! Thanks so much for you reply.
I do not t...Hi Jeff! Thanks so much for you reply. <br />I do not think your conclusion is neither obvious or it's subtlety appreciated by many physicists. Being a physicist myself I became interested in the subject of diffeomorphism invariance after reading Northon, Stachel and Rovelli. Also because I'm a physicists, I was a little uncomfortable with the proofs in your note. E.g. in Lemma 2 where you appealed to diffeomorphisms being isomorphisms between differentiable manifolds. It also seemed to me that you did not establish the equivalence of your version of LE (above in the blogpost) and Wald's. Would you not need the converse of Lemma 2 and 3 for that?<br /><br />Btw: Do you have any recommendations on literature for an interested physicist to read up on isomorphism theory, and the structuralism of Russel? I have some knowledge of topology, smooth manifold theory and Riemannian geometry.Anonymousnoreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-32947720911906946892014-02-09T15:34:09.842+00:002014-02-09T15:34:09.842+00:00This comment has been removed by the author.Anonymoushttps://www.blogger.com/profile/00418364146074319615noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-74515699660109375402014-01-29T00:54:23.664+00:002014-01-29T00:54:23.664+00:00Thanks for the comment - by structures $M_1$ and $...Thanks for the comment - by structures $M_1$ and $M_2$ being isomorphic I mean there's a bijection between the carrier sets which preserves all structure (relations, distinguished subsets, etc.). I use a modification of the definition of a what a manifold is from Robbin and Salamon,<br /><br />http://www.math.ethz.ch/~salamon/PREPRINTS/diffgeo.pdf<br /><br />Following them (roughly), I define a "manifold presentation" to be a pair $(A,C)$ where C is an atlas; and I define a manifold to be a pair $(A,C^{+})$, with $C^{+}$ the maximal atlas determined by $C$. So, distinct manifold "presentations" can give the same manifold, because you can get the same maximal atlas. (Yes, the topology is determined uniquely by the atlas $C$. The converse sometimes doesn't hold though for some weird manifolds.)<br /><br />Carroll's formulation -- which is fairly standard in a lot of GR literature -- is that if $\phi : M \to M$ is a diffeomorphism, then $(M,g,T)$ and $(M,\phi[g], \phi[T])$ represents the same physical world. I agree that this is correct - but is the condition "if $\phi : M \to M$ is a diffeomorphism" necessary? Do we have to move the points around in a smooth way? Or can we move them around in a crazy way?<br /><br />Wald's formulation implies something much stronger. Exposing the carrier set and the (maximal) atlas, spacetime models have the form $(A,C,g, T)$ where A is carrier set and C is a maximal atlas on A, g the metric and T maybe the energy tensor. Let $\phi : A \to A$ be any permutation of the carrier set whatsoever (may be as topologically crazy as you like), then $(A,C,g,T)$ and $(A, \phi[C],\phi[g],\phi[T))$ are isomorphic (in the sense meant in model theory). On Wald's formulation, they represent the same physical world. So, my conclusion is that the condition "$\phi : M \to M$ is a diffeomorphism" is unnecessary.<br /><br />Wald's formulation just amounts to "isomorphic spacetime models represent the same physical world".<br /><br />The mathematical point is kind of silly/obvious. If you have a structure $M = (A, R_1, ...)$ and take any bijection $f : A \to A$, then $(A, f[R_1], ...)$ is automatically isomorphic to the original. E.g., suppose $(\mathbb{R},<)$ is the ordered reals and $f : \mathbb{R} \to \mathbb{R}$ any bijection. Then $(\mathbb{R},f[<])$ is isomorphic to $(\mathbb{R},<)$.<br /><br />My note on this is here,<br /><br />https://www.academia.edu/4820228/A_Note_on_Leibniz_Equivalence<br /><br />But the responses I get to it in talks are "no, Jeff, that's wrong; it has to be a diffeomorphism of M to itself" or "yes, that's right, but obvious".<br /><br />JeffJeffrey Ketlandhttps://www.blogger.com/profile/01753975411670884721noreply@blogger.comtag:blogger.com,1999:blog-4987609114415205593.post-41592854927142574442014-01-27T20:24:38.385+00:002014-01-27T20:24:38.385+00:00From my reading of Wald he seem to be using the wo...From my reading of Wald he seem to be using the word 'diffeomorphic' for your (LE), not 'isomorphic'. In what sense are you using the word isomorphic here? <br /><br />If you use it synonymously with diffeomorphism I would still agree with you that Wald's statement is stronger than Carroll's, since if my understanding is correct two manifolds M and N can be diffeomorphic without having the same differential structure determined by their atlas, and thus not equal. (That is if one take the differential structure as being a criterion for determining manifold equality). Carrol seem to define M and N to be equal if they are diffeomorphic which might explain the apparent differences of what it means to be diffeomorphism invariant. <br /><br />Furthermore do you take the topology of M to be determined by the charts of it's maximal atlas? Anonymousnoreply@blogger.com