Comprehensive intro to diff geometry by Spivak Vol2.

[brown042]

I am reading a Vol2 of geometry book by Spivak. On page 220-221 he says that:
"Notice that the possibility of defining covariant derivatives depends only on the equation..."
The equation is some equality involving Christoffel Symbol.

If anyone has this book, could you explain why what he says is true? Is the equation derived from the equality which defines covariant derivative at the bottom of page 210?
Thanks.

 

[brown042]

Anyone has Spivak volume 2?

I aplogize that I can't write the equation using the regular typeset.

 

[mathwonk]

its sunday man, wait a while.

 

[brown042]

Comment

I found out that the equation comes from the condition that Riemannian metric tensor being well defined. If you have a Volume 1, look at chap 9 problem23. I suppose we can define covariant derivative if Riemannian metric tensor is well defined. But Spivak is saying that the possibility of defining covariant derivative depends only on the equation we are concerning.

Any comment about this would be appreciated.

 

[brown042]

?

It's Friday. So nobody has this book?

 

[mathwonk]

its at the office sorry, I am home. but i probably am no better at understanding it than you are. maybe pmb phy (pete) or some other specialiists will be.

hang in there, spivak volume two is the best differential geometry book on the planet.

 

[Chris Hillman]

Huh?

I am reading a Vol2 of geometry book by Spivak. On page 220-221 he says that:
"Notice that the possibility of defining covariant derivatives depends only on the equation..."
The equation is some equality involving Christoffel Symbol.

Can you give a more complete/exact quotation? Or at least the equation to which you claim Spivak is referring? What edition are you using? Are you sure you have the correct page numbers? Is this in the context of a Riemann normal chart?

 

[joe2317]

I am using third edition the page number is 220 221.
I don't know how to use latex. But the equation involves the relationship between $$\Gamma^{'\gamma}_{\alpha\beta}$$ and $$\Gamma^{k}_{ij}$$
In page 221, He also says:
A (classical) connection on a manifold M is an assignment if n^3 numbers to each oordinate system, wuch that equation (*) holds between the n^3 numbers $$\Gamma^{k}_{ij}$$ assigned to the coordinate system x and the n^3 numbers $$\Gamma^{'k}_{ij}$$ assigned to the coordinate system x'.

 

[Chris Hillman]

Behold the classical connection!

I don't know how to use latex.

LaTeX is a set of macros for tex, which is a markup language for text processing mathematics, which was developed by mathematician Donald Knuth. One variant of Latex happens to be due to Michael Spivak, the author of your textbook! Employing LaXeX is essential skill for any kind of mathematical discussion, as well as for writing papers and books. If you Google you can probably find a tutorial with examples which will show you now to make a matrix, symbols like ${\Gamma^a}_{bc}$, etc. Try clicking any formatted math at PF to see if you get a pop-up showing the markup.

Hmmm... Found it! It's all on 232-233 in my edition.

Spivak has discussed Riemann's lecture and now he is explaining Ricci's "absolute differential calculus", or "index gymnastics" style tensor calculus. Just before the passage you cited, he introduces the covariant derivative in the way Christoffel did, then asks what this new derivative operation means. He shows that in a Riemann normal chart, it reduces to ordinary partials, which reassures us that it really is a derivative. He explores some other formal properties and proves that the metric is covariantly constant (under what we now call a Christoffel connection--- at this point in Spivak this is still the only kind of connection we know about). He proves the Ricci identities and he proves that the Riemann tensor is indeed a tensor. Then he gives an application: if the Riemann tensor of M vanishes, then M is locally isomorphic to "$R^n$ with the usual metric", i.e. to $E^n$, with metric
$$ds^2 = \sum_{\alpha, \, \beta=1}^n \eta_{\alpha \beta} \, dx^\alpha \, dx^\beta = \sum_{\alpha=1}^n \left( dx^\alpha \right)^2$$
where the components of ${\eta^i}_j$ spell "identity matrix". (Here we are dealing with Riemannian manifolds; Lorentzian manifolds are not yet a certain glint in the eyes of Helmholtz and Gauss.) That is, the Riemann tensor is measuring the distortion of M from flat space, which goes back to surface theory where he has shown that at each point the curvature of a surface measures the quadratic distortion from a plane, at that point.

Then he says

Notice that the possibility of defining covariant derivatives depends only on the equation (*)
$${{\Gamma^\prime}^\gamma}_{\alpha \beta} = \sum_{i,j,k} {\Gamma^k}_{ij} \, \frac{\partial x^i}{\partial {x^\prime}^\alpha} \, \frac{\partial x^j}{\partial {x^\prime}^\beta } \, \frac{\partial x^k}{\partial {x^\prime}^\gamma } + \sum_{\mu=1}^n \frac{\partial^2 x^\mu}{ \partial {x^\prime}^\alpha \, \partial {x^\prime}^\beta } \, \frac{\partial {x^\prime}^\gamma}{\partial x^\mu}$$

IOW, we only need that this transformation law should hold in order to be able to define a derivative in the manner of Christoffel. This suggests that (*) is what really matters, and that we should generalize Christoffel's definition by using (*) to define a more general "connection".

Notice that (*) looks like a tensor transformation law which has been marred by the addition of the term on the right, so whatever a connection is, it is not a tensor field. The abstract definition he proposes simply says that any three index object which satisfies this transformation law is a "classical connection".

Does this help?

If not, at least you now know the markup

{{\Gamma^\prime}^\gamma}_{\alpha \beta} =
\sum_{i,j,k} {\Gamma^k}_{ij} \, 
\frac{\partial x^i}{\partial {x^\prime}^\alpha} \,
\frac{\partial x^j}{\partial {x^\prime}^\beta } \,
\frac{\partial x^k}{\partial {x^\prime}^\gamma }
+ \sum_{\mu=1}^n 
\frac{\partial^2 x^\mu}{ \partial {x^\prime}^\alpha \, \partial {x^\prime}^\beta } \,
\frac{\partial {x^\prime}^\gamma}{\partial x^\mu}

See? It's easy!

 

[brown042]

I appreciate your comment. But still, why is possibility of defining covariant derivative depends only on the equation (*)?

 

[quetzalcoatl9]

i can't help but interject here..

I appreciate your comment. But still, why is possibility of defining covariant derivative depends only on the equation (*)?

seriously, is this the best that you can do?!? look at the wonderful response that Chris Hillman gave you, did you even read it?

all you can do is repeat your same ill-posed and lazy question? maybe you should consider another area of study

 

[Chris Hillman]

Keep reading, you'll be glad you did!

I appreciate your comment. But still, why is possibility of defining covariant derivative depends only on the equation (*)?

That's simply Spivak's way of saying that property (*) can be used to define a more general notion of covariant derivative. This is actually an important theme: "by its properties shall ye know it". If you keep reading you'll see that he discusses a good many possible definitions of "covariant derivative"; the numerous plot twists are a a major theme in the story of (differential) geometry!

Spivak's five volume Comprehensive Introduction is a long but tightly constructed book (in true "helical" style), so someone dipping into it at random points will probably find many remarks cryptic. But if you read it through and carefully familiarize yourself with his notation and style it should become clear, perhaps after some struggle. I trust you realize that his aim is not simply to teach students modern differential geometry, but to give some genuine historical context and to discuss not only the great ideas of the subject but also how they developed. This is what makes this book such an ambitious--- and impressive-- classic of late twentieth century literature.

Other factors have intervened, so I'll duck out of this thread here by repeating in other words what mathwonk said: while Spivak's five volume epic might not be the shortest route to mastery of differential geometry, his book is a true tour de force which rewards the hard study which is required to read it through!

 

[MathematicalPhysicist]

on spivak's 5 volumes, you would say it's cumpolsary to have chris/mathwonk?
every volume costs something like 20 dollars, i would say it's a bargain.

 

[mathwonk]

well they are all different. vol2 is the main one for nineteenth century (gauss + riemann) differential geometry to me. vol 1 is a treatise on manifolds includiong weil'sprof of the de rham theorem. and the others are also specialized, e.g. vol 5 seems to be on the gauss bonnet theorem. i only own vols 1 and 2, but aspire to own the others someday, and the income benefits my friend mike, as he well deserves.