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Index 1, 2

"these need to be rewritten or removed"

The following was commented out by JRSpriggs:

Riemann curvature tensor ${\displaystyle R^{\rho }{}_{\sigma \mu \nu }=dx^{\rho }(R(\partial _{\mu },\partial _{\nu })\partial _{\sigma })}$ Torsion tensor ${\displaystyle T^{c}{}_{ab}=\Gamma ^{c}{}_{ab}-\Gamma ^{c}{}_{ba}-\gamma ^{c}{}_{ab}}$ which follows from ${\displaystyle T=\nabla _{X}Y-\nabla _{Y}X-}$ where X and Y are vector fields and is the Lie bracket of vector fields. Levi-Civita tensor The covariant Levi-Civita tensor in an n-D metric space may be defined as the unique (up to a sign) n-form (completely antisymmetric order-n covariant tensor) that obeys the relation ${\displaystyle \left|\epsilon _{\alpha _{1}\dots \alpha _{n}}g^{\alpha _{1}\beta _{1}}\dots g^{\alpha _{n}\beta _{n}}\epsilon _{\beta _{1}\dots \beta _{n}}\right|=n!}$ The choice of sign defines an orientation in the space. The contravariant Levi-Civita tensor is an n-vector that may be defined by raising each of the indices of the corresponding covariant tensor: ${\displaystyle \epsilon ^{\alpha _{1}\dots \alpha _{n}}=g^{\alpha _{1}\beta _{1}}\dots g^{\alpha _{n}\beta _{n}}\epsilon _{\beta _{1}\dots \beta _{n}}}$

not that I agree/disagree with the commenting-out, just brought it to attention here so people can see it clearer. Unless anyone is up for rewriting these, if they're incorrect they may as well stay out of the article... F = q(E+v×B) ⇄ ∑_ic_i 10:08, 18 June 2012 (UTC)

Some specific comments

• I trust we all agree that the Riemann curvature tensor should be listed in any section listing notable tensors. The question is how we present it, and I'm not convinced that the current presentation is suitable for the article (it uses unexplained notation, for one). Another expression could be

${\displaystyle R^{\rho }{}_{\sigma \mu \nu }=\partial _{\mu }\Gamma ^{\rho }{}_{\nu \sigma }-\partial _{\nu }\Gamma _{\mu \sigma }^{\rho }+\Gamma ^{\rho }{}_{\mu \lambda }\Gamma ^{\lambda }{}_{\nu \sigma }-\Gamma ^{\rho }{}_{\nu \lambda }\Gamma ^{\lambda }{}_{\mu \sigma }}$

• The torsion tensor is just as notable as the the Riemann curvature tensor; anyone objecting to its presentation should state why.
• The Levi-Civita tensor in my mind is even more notable, only the Levi-Civita symbol gets used so often in its place as a mathematical shortcut that we forget that there is a perfectly good tensor for the job (provided that there is a metric). Some may have issues with the characterisation given thereof, and in some cases the pseudotensor is defined instead (IMO unnecessary when the tensor is as good or better); I guess it may make sense to find a good reference and go with their definition.
• The metric tensor is not quite right (you can't use the same symbol g to operate on both covariant and contravariant basis vectors). I would prefer eliminating the unexplained "operating on" notation.

In all, I think these do belong in the article, and objections should be explained so that we all understand them and can try to accommodate them. — Quondum 12:05, 18 June 2012 (UTC)

If the expression for the Riemann curvature tensor were rewritten to use the notation of this article it would become

${\displaystyle R^{\rho }{}_{\sigma \mu \nu }=\mathbf {e} ^{\rho }(R(\mathbf {e} _{\mu },\mathbf {e} _{\nu })\mathbf {e} _{\sigma })}$

which is not particularly informative, as I hope you will agree. I think we should provide a formula which shows how this tensor is the commutator of the covariant derivative with itself.

${\displaystyle A_{\nu ;\rho \sigma }-A_{\nu ;\sigma \rho }=A_{\beta }R_{\nu \rho \sigma }^{\beta }\,}$

as stated on pages 20 and 21 of "General Theory of Relativity" by P. A. M. Dirac. This can be generalized to get the commutators for two covariant derivatives of arbitrary tensors.

I really do not like the torsion tensor; it has no place in Einstein's theory. There is no evidence that such a tensor actually exists. It is a figment of the imagination of those who support the Einstein–Cartan theory.

The Levi-Civita symbol#Tensor density is best understood as a tensor density, not as an ordinary tensor. JRSpriggs (talk) 16:16, 18 June 2012 (UTC)

Thank you for your clarification. I find I disagree on several points, though I suspect that each of us is a bit too fond of our own perspective – in my case, a coordinate-independent approach where possible.

Though on the Riemann curvature tensor, I agree that your expression has stripped the expression of any defining powers, as any order-4 tensor of that variance will satisfy your expression: it simply says what the components are. Though I'm not sure that I agree that this is the "notation of this article": I would like to remove any reference to the basis vectors except in a section defining the components of a tensor if possible; this goes for the section on the metric as well. You have not commented on my expression above that gives the components in terms of the Christoffel symbols, using what I would call the dominant notation of the article. I do however like your definition in terms of the commutator of the covariant derivative; my expression could be added as a derived consequence.

The torsion is generally zero. This is not the same as it being non-existent. It is thus a core tensor in all metric field theories.

On the Levi-Civita density I feel most strongly opposed, and I think I've said why. You clearly like tensor densities, though. — Quondum 17:35, 18 June 2012 (UTC)

On the subject of torsion and the related question of non-holonomic bases (Frame fields in general relativity) for the tangent bundle. These are unnecessary add-ons to the mathematical structure used in general relativity. They serve no purpose other than making the subject harder to understand. In reality, it is the irreducible representations of a group which are important. For example, second order covariant tensors which are neither symmetric (like g_αβ) nor antisymmetric (like F_αβ) do not appear because they are reducible. The Levi-Civita connection is irreducible; other connections (which include torsion) are not. Thus there is good reason (Occam's razor) to believe that torsion does not exist, beyond the mere absence of evident for it. JRSpriggs (talk) 07:51, 20 June 2012 (UTC)

To Quondum: If you think that you can get along without tensor densities, then tell me with what would you replace ${\displaystyle {\sqrt {-g}}\,}$? JRSpriggs (talk) 11:35, 21 June 2012 (UTC)

This is an inherently coordinate-dependent expression: a calibration of the local scale of the chosen coordinates against the metric. It has no other instrinsic meaning, at least as far as I can tell. In short, it is only needed to deal with non-tensors such as the Levi-Civita density; if one has only tensors (e.g. the the Levi-Civita tensor) as a starting point, its need completely vanishes — Quondum 13:08, 21 June 2012 (UTC)

Tensor densities arise unavoidably in calculating the action of a field theory. For example, see Hilbert action. JRSpriggs (talk) 23:54, 21 June 2012 (UTC)

Are they? I don't find this example compelling: it would be easily explained as the quantity ${\displaystyle d{\mathbf {x} }^{4}}$ transforming as a tensor density, not as a tensor, but by the quantity ${\displaystyle {\sqrt {-g}}\,d{\mathbf {x} }^{4}}$ being the tensor quantity. Not being familiar with actions nor differential forms I cannot comment with authority, of course, but it does have a marked resemblance to the covariant Levi-Civita tensor: the Riemannian volume form seems to be the natural quantity to use here: no tensor densities required other than to compensate for others. — Quondum 06:03, 22 June 2012 (UTC)

The Hilbert action is just ${\displaystyle \int RdV}$, where dV is the volume form generated by the metric. Not tensor densities involved. Tensor densities only become relevant if you try to formulate the Lagrangian density, which is obviously a scalar density.TR 16:42, 22 June 2012 (UTC)

It is not so: tensor densities are involved. In fact, ${\displaystyle dV={\sqrt {-g}}\,d{\mathbf {x} }^{4}={\sqrt {-g}}\,dx^{1}\wedge \cdots \wedge dx^{4}}$, ${\displaystyle R{\sqrt {-g}}\,}$ is the Lagrangian density of the Hilbert action. In this way, the Hilbert action ${\displaystyle I=\int RdV=\int R{\sqrt {-g}}\,d{\mathbf {x} }^{4}}$ is a scalar; geometrically well defined, invariant under (proper) coordinate transformations. Hence, the spacetime manifold must be oriented. Please cf. Lovelock, David (1989) . Tensors, Differential Forms, and Variational Principles. Dover. pp. 299–300. ISBN 978-0-486-65840-7. {{cite book}}: Unknown parameter |coauthors= ignored (|author= suggested) (help)Mgvongoeden (talk) 19:17, 22 June 2012 (UTC)

How does that contradict what I just said. The split ${\displaystyle dV={\sqrt {-g}}\,d{\mathbf {x} }^{4}}$ is not necessary to formulate the Hilbert action. Tensor densities only become relevant when you formulate the Lagrangian density. (Which is indicative of the fundamental non-covariantness of the used variational calculus).TR 20:56, 22 June 2012 (UTC)

Scalar densities arise unavoidably a priori in calculating the action of a field theory, for example, the Hilbert action, because in order to perform an integral on (pseudo) Riemannian manifolds one needs a volume element, i.e., a nowhere vanishing n-form ${\displaystyle \Omega }$ on ${\displaystyle M}$. In fact, on manifolds it is difficult to define integral of functions, so one prefers to define integral of n-forms. On an orientable (pseudo) Riemannian spacetime manifold with metric ${\displaystyle g}$, corresponding to an orientation of ${\displaystyle M}$, there is a uniquely determined 4-form ${\displaystyle \Omega _{g}=dV={\sqrt {-g}}\,d{\mathbf {x} }^{4}}$ which gives the orientation and which has the value ${\displaystyle +1}$ on every oriented orthonormal frame. Cf. Boothby, W.M. (1986). An introduction to differentiable manifolds and Riemannian geometry. Academic Press, Inc. p. 218. ISBN 0-12-116052-1. Just because you have this 'canonical' 4-form, a posteriori, you can 'forget' about scalar densities. We are telling the same story.Mgvongoeden (talk) 23:46, 22 June 2012 (UTC)

You can find this canonical 4-form without ever using scalar densities. (See for example J.M. Lee, Riemannian Manifolds, pg 29 and 30).TR 10:29, 23 June 2012 (UTC)

Which roughly corresponds to what I intended with "...may be defined as the unique (up to a sign) n-form that...", retained in the box above. — Quondum 12:52, 23 June 2012 (UTC)

I've encountered JRSprigg's peculiar prejudice against the torsion tensor elsewhere on Misplaced Pages. May I say that I do not find this prejudice helpful in building an encyclopedia. The torsion tensor has many applications to mathematics outside of general relativity. It even has applications within GR itself. Connections with torsion appear regularly in the work of people who study horizons, having no connection with JRSprigg's bogeyman of Einstein-Cartan theory. Sławomir Biały (talk) 19:39, 24 June 2012 (UTC)

Also, I find statement "There is no evidence that the torsion tensor actually exists" somewhat puzzling. Surely, the theory posits that not only does the torsion tensor exist, but that it is zero! The real question is, what evidence is there to support this much stronger assertion? (I don't know the answer. Perhaps JRSprigg's knows one.) Sławomir Biały (talk) 19:45, 24 June 2012 (UTC)

Perhaps we should rather be focusing on whether the torsion tensor should be mentioned in Ricci calculus#Notable tensors, regardless of when or whether it necessarily vanishes (and should not depend upon whether the chosen basis is holonomic). What criteria of notability should we use? Does the section belong in the article? — Quondum 20:53, 24 June 2012 (UTC)

Sławomir Biały says "the theory posits that not only does the torsion tensor exist, but that it is zero". This does not even make sense! This is like saying that GTR posits not only that The Purple People Eater tensor exists, but that it is zero. No theory should assume the existence of anything without some evidence that it has some non-trivial effect.

Not only do I reject the idea of a connection different from the Levi-Civita connection, I am also unsure that even the Levi-Civita connection exists (outside our mathematical imagination). If God has an action integral for the universe, how do we know that any covariant derivatives or Lie derivatives occur in it? All that seems clear so far is that it has something like partial derivatives and that it can be written most simply in a generally covariant form with the integrand being a scalar density. JRSpriggs (talk) 05:05, 25 June 2012 (UTC)

As Ayn Rand says, the burden of proof is on the one who asserts that something exists, not on the one who denies it. JRSpriggs (talk) 05:20, 25 June 2012 (UTC)

You seem to be getting tied up in semantic subtleties. Please substitute the term "defined" for "exists", and reconsider. — Quondum 06:40, 25 June 2012 (UTC)

Despite the bad analogy with "Purple People Eater", I think I understand your position. As soon as you have a rule for parallel transport, you have a connection. As soon as you have a connection, you have torsion (which might be zero). However, it could be that parallel transport is not something that is given a priori in the spacetime. Instead, agreeing on a parallel transport protocol if all we have is the metric might require something equivalent to setting the torsion to zero. In that way, the connection only "exists" in the sense that it defines an agreed-upon protocol for parallel transport, when the only datum that actually "exists" is the metric. So the torsion in spacetime isn't something real that can be measured, since the connection also is just a convenience. I wonder if there is a clear write-up somewhere? Sławomir Biały (talk) 16:47, 25 June 2012 (UTC)

Yes. As I see it, the metric should be used only when necessary. It is merely the potential of the gravitational field. So it should be used sparingly as the potential of the electromagnetic field is. Christoffel symbols are merely a device for keeping track of the ${\displaystyle {\frac {\partial y^{k}}{\partial x^{m}}}\,{\frac {\partial ^{2}x^{m}}{\partial y^{i}\partial y^{j}}}\,}$ terms which arise when we transform coordinates, because we need to make them cancel out when creating a scalar density to be the integrand of the action integral. JRSpriggs (talk) 18:27, 25 June 2012 (UTC)

The metric is not at all analogous to a potential. It's completely gauge invariant. Sławomir Biały (talk) 23:45, 25 June 2012 (UTC)

Spinors and van der Waerden notation...

Copied from here

I strongly disagree with tucking the statement of van der Waerden notation into a ref where people will not see it easily - it has been reinstated. On the contrary spinors and their notation are even more unclear and confusing than tensors, so the link should be obvious for the reader's awareness, and stated as soon as possible (why leave it to the end, even under the "see also section"?). Also that article should be expanded to be the spinor analogue of Ricci calculus (for tensors). F = q(E+v×B) ⇄ ∑_ic_i 23:11, 16 June 2012 (UTC)

Lie derivative...

is not (now) referenced. Chapter 4 in T.Frankel is about the Lie derivative but I don't recognize it... If anyone has this ref or any other then preferably add one (although it is linked, so it's not a desperate issue). F = q(E+v×B) ⇄ ∑_ic_i 09:52, 18 June 2012 (UTC)

There is now, thanks user:Mgvongoeden! In addtion to your other edits in the article. =) F = q(E+v×B) ⇄ ∑_ic_i 21:13, 19 June 2012 (UTC)

what are the Ricci identites?

In the very last section, which formulae are the Ricci identities?

• the commutator of the covariant derivative with itself

${\displaystyle A_{\nu ;\rho \sigma }-A_{\nu ;\sigma \rho }=A_{\beta }R_{\nu \rho \sigma }^{\beta }\,,}$

• since the connection ${\displaystyle \Gamma ^{\alpha }{}_{\beta \mu }\,}$ is torsionless, which means that the torsion tensor ${\displaystyle \Gamma ^{\lambda }{}_{\mu \nu }-\Gamma ^{\lambda }{}_{\nu \mu }\,}$ vanishes.

• This can be generalized to get the commutator for two covariant derivatives of an arbitrary tensor as follows

${\displaystyle {\begin{aligned}T^{\alpha _{1}\cdots \alpha _{r}}{}_{\beta _{1}\cdots \beta _{s};\gamma \delta }-T^{\alpha _{1}\cdots \alpha _{r}}{}_{\beta _{1}\cdots \beta _{s};\delta \gamma }=\,&-R^{\alpha _{1}}{}_{\rho \gamma \delta }T^{\rho \alpha _{2}\cdots \alpha _{r}}{}_{\beta _{1}\cdots \beta _{s}}-\cdots -R^{\alpha _{r}}{}_{\rho \gamma \delta }T^{\alpha _{1}\cdots \alpha _{r-1}\rho }{}_{\beta _{1}\cdots \beta _{s}}\\&+\,R^{\sigma }{}_{\beta _{1}\gamma \delta }T^{\alpha _{1}\cdots \alpha _{r}}{}_{\sigma \beta _{2}\cdots \beta _{s}}+\cdots +R^{\sigma }{}_{\beta _{s}\gamma \delta }T^{\alpha _{1}\cdots \alpha _{r}}{}_{\beta _{1}\cdots \beta _{s-1}\sigma }\,\end{aligned}}}$

Are all or some the identities? It just vaguely says after the last one "which are often referred to as the Ricci identities. ". F = q(E+v×B)⇄ ∑_ic_i 08:39, 21 June 2012 (UTC)

Mgvongoeden wrote it in a way which made it clear that it was the last (rather complicated) formula. I was trying to make the text briefer by changing his sentence into the subordinate clause which you find ambiguous. Obviously the formula involving A is a special case of the more general formula, so it would also be a Ricci identity. JRSpriggs (talk) 11:29, 21 June 2012 (UTC)

What was I thinking, the tensor-component expression represents a number of identites, not just one... thanks for clarification. F = q(E+v×B)⇄ ∑_ic_i 11:47, 21 June 2012 (UTC)

Please, cf. the following interesting paper

• Cabras, A. (1996), "The Ricci identity for general connections", Proc. 6th Int. Conf. Conf. Differential Geometry and Applications Brno, Czech Republic August 28 - September 1, 1995: 121–126

It is a pretty good article. Mgvongoeden (talk) 15:32, 21 June 2012 (UTC)

"Fact" tagging...

Do we really need a reference for this:

"The context should prevent confusion between letters used as indices (which take integer values) and as labels for coordinates"

as tagged by TR, which I hoped would state the blatent obvious? I will not remove it, but what is the reason?

F = q(E+v×B)⇄ ∑_ic_i 21:54, 24 June 2012 (UTC)

A general statement here: the article has chosen a particular convention and is falsely presenting that as the only convention (even in relativity theory). It's one thing for the article to be internally consistent. Some conventions are necessary for this. But let's not pretend that sources all use the same conventions as we do. That's just wrong. Sławomir Biały (talk) 22:33, 24 June 2012 (UTC)

Are you referring to the space-time split convention before this italic statement? What other conventions are missing/misleading in the article?

The sentence in italic here simply refers to notation like "A_x, A_y, A_z" or "A_r, A_θ, A_φ" or whatever making use of subscripts for "x, y, z" or "r, θ, φ" etc. which are not supposed to be indices but they would look like indices to a reader, i.e. the sentence and example is to prevent the reader thinking "A_x" etc. is the same as "A_i" which actually does comprise an index for "i = 1, 2, 3".

In any case, whatever the other conventions authors use, they should be summarized wherever appropriate, since notation is what the article is about, if we should'nt pretend this article's conventions are used by everyone else. If this is what TR is referring to, then I understand the citation tag... F = q(E+v×B)⇄ ∑_ic_i 22:57, 24 June 2012 (UTC)

Short answer, you need references for everything.TR 06:15, 25 June 2012 (UTC)

I think different people might be attributing different purposes to the statement. If it is intended as a guide to interpretation of something else in the article rather than as a referenceable fact in itself, the onus of reference seems excessive. Rephrasing may make this more apparent. For example, we may point out that "authors will usually make it clear whether a subscript is intended as an index or as a label". We are simply alerting the reader to be aware of what might otherwise be confusing. To find a reference for this statement, we'd need to find a text that is an analysis of typical notations, not merely examples of what is being described. — Quondum 06:53, 25 June 2012 (UTC)

I do agree with Qoundum, not with Mr. Index Man. Boys, this is an article about Ricci calculus, not Index Notation. Some others authors use latin indices in order to denote non holonomic bases, orthonormal frames, Lorentz indices (...). Mgvongoeden (talk) 11:38, 25 June 2012 (UTC)

To TR: If that's how "simple" it is, do we need references for "circles are round" or "petrol ignites when lit" or "rain makes you wet" or "an abacus can be considered a child's toy" also?

These silly examples are no different to the triviality of the above italic statement, and no references are needed. I don't have any reference which explicitly states the same thing but surley we all have books which use "A_x, A_y, A_z" and use index notation "A_i". Quondum and Mgvongoeden seem to disagree with the tagging also... F = q(E+v×B)⇄ ∑_ic_i 15:32, 25 June 2012 (UTC)

It has been re-phrased. F = q(E+v×B)⇄ ∑_ic_i 16:17, 25 June 2012 (UTC)

The wedge product returns...

A new User:Hublolly added the wedge product back, when we have decided through all this before (now archived). =( What to do? F = q(E+v×B)⇄ ∑_ic_i 00:42, 6 July 2012 (UTC)

As my summary stated - it SHOULD be included if you have the tensor product and tensor components and Koszul connection and so on.. Please leave it. Hublolly (talk) 00:46, 6 July 2012 (UTC)

If you have a problem with including all these things - why do you keep the section? Why did you add more to it today and in the past? And why the hell did user:JRSpriggs say in the archive that these are off-topic yet allow them??? (and user:Quondum and you did get things wrong, according to archive #1.. Hublolly (talk) 00:49, 6 July 2012 (UTC)

It was decided not to include it. If you insert this content, it dismisses and defeats the discussions (please see WP:consensus). F = q(E+v×B)⇄ ∑_ic_i 00:53, 6 July 2012 (UTC)

How ironic: you *summarily removed* my efforts and when it happens to YOU - you can't even take it?? (thouugh at least user:TimothyRias has eneogh decency to compromise ). :-( Reverted back. Again pretty please leave it. Hublolly (talk) 01:08, 6 July 2012 (UTC)

I should add that it's BECUASE of user:TimothyRias's comprimise that this article was created and now exists, from collaboration of others also, according to that link. It wasn't just your idea after all!? was it? See the type of thinking that gets things done and how progress is made and how we learn? Maybe you can learn from my example I lead right now - to be firm and correct in thinking and doing. How about that!? I have only just started out an hour or so ago and I'm already learning millions of times quicker than YOU. :-) Hublolly (talk) 01:33, 6 July 2012 (UTC)

And does your above non-specific and un-constructive and sarcastic comment (about the reference to index labels on round circles/lit petrol/wet rain/toy abacus) directed to user:TimothyRias himself NOT defeat WP:talk page guidelines??? Hublolly (talk) 01:35, 6 July 2012 (UTC)

Nope. And please concentrate on improving the article instead of looking for every opportunity to complain wherever someone gets it wrong. Thank you. F = q(E+v×B)⇄ ∑_ic_i 01:39, 6 July 2012 (UTC)

Hublolly, please refrain from being personally direct; direct your comments at the content of the articles exclusively, as per WP policy.
On the exterior product, I discovered through the discussion on the subject that there is more than one way to define the exterior product in terms of the tensor product – in particular, an exterior product may be defined with different scalar multiples of the fully antisymmetric tenosr product, and where two choices appear to dominate. If this product is to be included in the article, then the alternatives should be made clear. Thus, it should be stated along the lines of "An exterior product can be defined as a scalar multiple of the antisymmetrized tensor product... Common choices of the scalar are 1 and p!, where p is the number of vectors." (There are also other more complicated ways of defining an exterior product in terms of tensors, such as in geometric algebra.) And there is also the (as yet unaddressed here) problem of how to define the exterior product of arbitrary (higher-order, as well as non-antisymmetric) elements of the tensor algebra.
It was also unclear to me that this product was relevant in the contexts where the Ricci calculus was used. Notability in this context is to be established (through references). In short, the issues I have stated here should be resolved to justify inclusion. References are necessary in this instance. — Quondum 05:39, 6 July 2012 (UTC)

It’s in MTW - exactly where I said before (p.83). I asked this user if he has the reference or any other. To be fair, I can't tell anything incorrect with the equations (assuming scalar coefficient 1 not p!), and it does follow on from the tensor product admittedly nicely, and does illustrate an application of the index notation, but for reasons you point out it could potentially lead to extensive content much less on Ricci calculus and more on exterior algebra which is not helpful or purposeful at all. I reverted again for now. F = q(E+v×B)⇄ ∑_ic_i 06:46, 6 July 2012 (UTC)

It IS correct, but I see what you mean about the coefficient of 1 or p! for p vectors or covectors. But it wouldn't become an "extensive" discussion on the exterior algebra, no need for generalization just give the formulae for vectors/covectors (see what I mean)? Hublolly (talk) 07:49, 6 July 2012 (UTC)

Ok - using the reference quoted here and Quondum's words:

An exterior product can be defined as a scalar multiple of the antisymmetrized tensor product... Common choices of the scalar are 1 and p!, where p is the number of vectors.

Although why should Geometric algebra be relevent ? I notice Quondum's addiction to GA query at WP maths helpdesk or for that matter (also user:Rschwieb) in discussions on differential geometry! Hublolly (talk) 07:58, 6 July 2012 (UTC)

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