What is the relationship between Ricci and K in two dimensions?

[stephen_weber]

Hi,

In two dimensions I am under the impression that the ricci tensor or the scalar curvature equals the negative of the fundamental tensor and the sectional curvature (K).
I'd have written it out with the proper symbols but I am new to this forum and this isn't at least a complex question.

I know that the sectional curvature is directly proportional to the Riemannian Tensor, and since I am only talking about two dimensions, the only term that is independent and nonzero is R 1212. OK with the symmetries there are dependent terms that are the positive and negative of that but all of the multiplicities cancel out in the definition of K.

I was wondering if there where was anyone out there who can walk me or US through how this equation is true?

 

[stephen_weber]

Ok I know Doodle Bob could help me. So to take me back to the top of the list again, I'll be more clear.

I am working on a problem in Schaum's "Tensor Calculus", chapter eight . problem 8.30(a). Without special symbols here if the characters in brackets are subscripts then the question is simply.
-------------------------------------
Show that in a Riemannian 2-space...
R[ij] = - g[ij]*K
--------------------------------------

 

[Doodle Bob]

Ok I know Doodle Bob could help me. So to take me back to the top of the list again, I'll be more clear.

I am working on a problem in Schaum's "Tensor Calculus", chapter eight . problem 8.30(a). Without special symbols here if the characters in brackets are subscripts then the question is simply.
-------------------------------------
Show that in a Riemannian 2-space...
R[ij] = - g[ij]*K
--------------------------------------

Good lord, why ever would you be studying out of that text? Track down Do Carmo's Riemannian Geometry or even Bishop and Goldberg for a much more intuitive treatment of the subject.

It's been a while since I've looked at this, but it should be just another low-dimensional phenomenon. Just put everything into its local components wrt an orthonormal basis and compare. K is necessarily R[1221]. The Ricci Tensor is given by:
(X,Y)--> trace(Z-->R_{XZ}Y). So, R[ij]=R[i1j]^1+R[i2j]^2, which then can be put in terms of g[ij] (which is just the Kronecker delta function in this case) and K. There might be a coefficient of 1/2 running around that I forgot. But that's more or less it. Every two-dimensional Riemannian manifold is necessarily Einstein.

 

[stephen_weber]

I am studying out of this text as I am studying alone and need actual worked out problems so that I can see the uses and methods. I find that most math books (as well as physics) are geared solely for the teacher student situation. Wonderfully worked out theory in the book, but there are no worked out problems. Since the teachers edition isn't for sale. It looks like the authors really have no idea what they are talking about. They give questions at the end of each chapter with no work, no answer. To me without the school and teacher behind the book it looks like the author is just writting down questions that he didn't know the answer too.

But I think your comment about working in a simple frame of orthonormal coordinates is the right step for me. I had K as R[1212] like you said but divided by the det of g, which could equal one...

R[ij] = - g[ij]*K
--------------
R[ij]= R[ijb]^ = -g[ij]*K

R[ij] = g^[ab]*R[aijb] = -g[ij]*K

g^[ab]*R[aijb] = - g^[ab]*R[aibj] = -g[ij]*K symmetry of R

g^[ab]*R[aibj] = g[ij]*K

g[ab]*g^[ab]*R[aibj] = g[ab]*g[ij]*K

R[aibj] = g[ab]*g[ij]*K

----------------------------------------
in two dimensions the non zero terms are:
R[1212]=R[2121]=-R[1221]=-R[2112]
----------------------------------------
R[1212] = K *(g[11]g[22]+g[22]g[11]-g[12]g[21]-g[21]g[12])

R[1212] = 2*K*( g[11]g[22] - g[12]g[21])

The correct defn of n=2 for K:

K= R[1212]/(g[11]g[22]-g[12]g[21])

So I am off by a factor of two now.
You brought up the factor of two. IS the Correct Definition wrong in the book? (The amazing thing about math books which leads me to my original point about why authors are incapable of including written out problems is that invariably there are typos and plain errors.)

 

[Doodle Bob]

It would seem that Schaum's is using the negative of the usual definition of sectional curvature as its sectional curvature. Typically, the sectional curvature of two unit vectors, X and Y, is given by K(X,Y)=R(X,Y,Y,X). The advantage of this definition is that gives a positive sectional curvature for spheres using the induced metric from R^3.

I'm not sure if I follow your reasoning precisely. Be careful that you don't assume what you're trying to prove. Start with the definition of the Ricci curvature and derive the desired results from there:

let e1 and e2 be an orthonormal basis of TM.
Then Ric(ej,ek)=R[1j1k]+R[2j2k]
=delta[2j]delta[2k]R[2121]+delta[1j]delta[1k]R[1212]
=-R[1221](delta[1j]delta[1k]+delta[2j]delta[2k])
=-K(delta[jk])

Done.

As for definitions and coefficient conventions, it all depends on the author; and it's not always due to typos. The biggest discrepancy has to do the wedge operator.

Some people like: a^b=(1/2)(a tensor b - b tensor a)
But other people don't put that (1/2) there. This difference has tremendous consequences with the d operator on forms.

 

[stephen_weber]

If you just open this picture and enlarge it , I can be more clear

You are right. Schaums is contracting R on the third variable not the middle one like you and the internet are using.
I found one paper that describes your first step literally.
Ric(ej,ek)=R[1j1k]+R[2j2k]
(in...Conformal Flatness and Self-Duality of Thurston-Geometries, by Stephan Maier
Proceedings of the American Mathematical Society © 1998 ,page 1169)

So I could accept this step excepting that the two R's on the right should be multiplied by the diagonal of the inverse metric. Oh that paper had no connection to this reality.

My question is the use of basis 'ej' 'ek' etc. and turning them into deltas for the metric.

 

[Doodle Bob]

My question is the use of basis 'ej' 'ek' etc. and turning them into deltas for the metric.

The rigorous way to get around that is:

Let p be a point in the manifold. Let {e1,e2} be a local basis of TM which is also orthonormal at point p. Then, at T_pM,

[insert previous proof here]

Thus, at every point p, we see that Ric_p=-K_p g_p so that Ric=-Kg on the entire manifold.

 

[stephen_weber]

I forgot to mention that in your previous posts I thought you had a^b to mean b was a superscript of a. Schaums doesn't deal with wedge products or operators.

So I was using it to express superscripts and mixed tensors. Did you write
a^b=(1/2)(a tensor b - b tensor a)
to mean that my factor of two was lost in my expansion?

you have g[ij]= delta[1i]delta[1j]+delta[2i]delta[2j]=delta[ij]

so is
g[ab]*g[ij] the bit whereas it should be a wedge operator?

I apologize cause I don't exactly believe my question here. My knowledge of wedge operators is limited to their being the connection between basis like dx and dy in area. And the internet either can show me a carot sign or some use that is already twisted into seventeen different knots of other unknown beginnings, that I need to work through.

I do follow your last post. At the point of measurement p, is general enough to be any point. But then I believe my method (if you see superscripts for ^) and you wrote "[insert previous proof here]", which could mean your proof or mine.

IF you meant your proof. Then I am looking for how ric = R...+ R... without the metric included.
If the next line is the continuation of ric = with the metric being functions of deltas

Then Ric(ej,ek)=R[1j1k]+R[2j2k]
=delta[2j]delta[2k]R[2121]+delta[1j]delta[1k]R[1212]
=-R[1221](delta[1j]delta[1k]+delta[2j]delta[2k])
=-K(delta[jk])

then should it have been (2nd line)
ric(ej,ek)=delta[2j]delta[2k]R[1j1k] + delta[1j]delta[1k]R[2j2k]

 

[Doodle Bob]

Forget the business about the wedge product, it does not pertain at all to the problem at hand. The ^'s in my proof do indeed signify superscripts.

I'm a little unclear as to the problem at this point. Reviewing your previous proof, I did stumble on some things:

1. You're doing that high school trick of starting with the identity you want to prove and manipulating one side until you get the other. Stop doing that: it will only lead to madness. And one never can resist the temptation of manipulating both sides, which is verboten and which you do several times in your work. And, which means that it cannot be a correct proof: all of those "="s are not equal signs.

2. You are actually experiencing one reason why most sane people stay away from Schaum's. It teaches you how to manipulate indexed things beautifully, but it doesn't really teach what the hell those things mean. I like to call this "deck chair geometry." It's as if you're taught specific rules as to how to move chairs around on the deck of a ship, but you're never taught as to what the rest of the ship is like.

Anyway, in this case, the problem occurs here:

g^[ab]*R[aibj] = g[ij]*K

g[ab]*g^[ab]*R[aibj] = g[ab]*g[ij]*K

You have forgotten here, that a and b are just summation indices. So, it does not make sense to multiply by g[ab].

There are only 2 variables for the indices. Just write out all of the terms of both left sides (the top one will have 4 terms and bottom 8); and you'll see what the problem is.

 

[Doodle Bob]

The rule of thumb that Schaum's does not tell you is that, when you're dealing with a pointwise identity, i.e. anything dealing with the values of a tensor at a point (in this case, the 2-tensor given by Ric-Kg), then you can (and often should) use a basis that is nice at that point.

Maybe you haven't gotten to this part: Given any Riemannian manifold (M,g) and a point p in M. There is a local basis of TM {e1, ..., en} such that g[ij](p)=delta[ij]. Now, g is nice *only* at p with respect to this basis, i.e. if q is a point near p but not equal to p, then chances are that g[ij](q) is *not* equal to delta[ij].

 

[stephen_weber]

I knew you could help. But to continue your analogy a bit about the ship. You seem to forget that once you (YOU) own a ship and know every part you just wave your hand at the new deck hands to put the chairs there. And we sigh and realize that we will be moving chairs for awhile as you have hidden your vast knowledge of the ship into a miniscule curling of the fingers on your waving hand to signify the chairs exact expected future locations.
Not to dwell on it, but I can glance at the first page of that book you expect me to start with Riemannian Geometry by Manfredo. Let's see that first page has these words on it...Manifold,homeomorphism,injective,mapping,diffeomorphism. And the first page ends with "the unnecessary presence of R^3 is simply an imposition of our physical nature. "

Instead of arguing the point then let me just cut to the chase. I can afford the Schaum's book...

Your second email about g[ij]=delta[ij]. Fine I assume these are the normal coordinates . And on that note I have a question. I read that k1 and k2 the minimum and maximum curvature are orthogonal to each other. I haven't completely wrapped my belief around that, having only read it once as a statement. But if it is true then is that used as the normal coordinates usually? Such as it being the geodesic path to define one direction...
(Normal coordinates are such that ds^2=dx^2+dy^2+ k(xdy-ydx)... blah blah blah..)

What I was searching for originally is a general proof. Not one where we are using a local geometry. Why? Cause of what I started with.
In this book he outlines the sectional curvature with regards to any two vectors. You are saying that it is in regards to two unit vectors. Now a bit of the confusion you might see being clarified is that I originally need to have a term to make my any two vectors -> unit vectors. That I believe is the det(g). OR (I wrote a great email earlier and then lost it before it was sent ,,sorry)
Schaums has in n=2 K=R[1212]/(g11g22-g12g21)
-----------
Technically you told me never to take one side of the equation and turn it into the other side."manipulating one side until you get the other. Stop doing that". I am assuming you don't believe that a proper proof should deal with BOTH sides to get some truthful axiomatic identity. I have never had a class in proof's so I am just a barbarian to that particular edicate.
If the math is wrong, the math is wrong.
And on that note. You told me to write out each term on the left side of my two equations. I agree that there will be four terms in the first one , but you claim eight in the second. If g[ij] has four terms. Then g[ab] should have 4 and then I write out 16 terms not eight. And I could probably see where you wanted to point me. I suspect it will be that I could easily write R[1111]=g11*g11*K which since R[1111]=0 gives K=0.

I do understand now your proof with nice geometry, and I'll finish later.

 

[Doodle Bob]

Instead of arguing the point then let me just cut to the chase. I can afford the Schaum's book...

i understand completely. But always keep in mind that how much you learn about a topic is dependent always on how much you put in. If you want to learn this very deep topic well, you need to be dogmatic in finding the best material. If Do Carmo's Riem. Geom. is daunting, then check out his earlier book, The Geometry of Curves and Surfaces.

What I was searching for originally is a general proof. Not one where we are using a local geometry. Why? Cause of what I started with.
In this book he outlines the sectional curvature with regards to any two vectors. You are saying that it is in regards to two unit vectors. Now a bit of the confusion you might see being clarified is that I originally need to have a term to make my any two vectors -> unit vectors. That I believe is the det(g). OR (I wrote a great email earlier and then lost it before it was sent ,,sorry)
Schaums has in n=2 K=R[1212]/(g11g22-g12g21)

the principle i am using is that, given any vector X, you can find a unit vector in the same direction. Namely, X/sqrt(g(X,X)).

I am assuming you don't believe that a proper proof should deal with BOTH sides to get some truthful axiomatic identity.

that is absolutely correct. it's not a matter of belief at all: it's a matter of logic.

a proof starts with a known true statement (or statements) and lists true statements throughout until the very end, which should be the statement of the desired result.

experts will often fudge the rules, but in general every single statement of a proof needs to be a true statement or a true implication from a true statement. there must be no statements of indeterminate truth.

 

[stephen_weber]

Ok I am officially finished with this question. I have followed your (I mean THE) idea about proofs. Starting with ,

proof starts with a known true statement (or statements) and lists true statements throughout until the very end, which should be the statement of the desired result

Fact One : In n=2 ::: K=R[1212]/g where g=g11g22-g12g21

Desired Result ::: R[ij] = g[ij]*K (noting that the original negative is based on direction of curvature and Schaum's is in the minority )

Starting with::: R[ij] = R[ikj]^[k] = g^[hk]* R[hikj]
and using
g^[hk]= g[hk]^-1
as a substitution
R[ij] = g[hk]^-1 * R[hikj]

Some shortcuts:::: the only non zero R's are
R[1212]=R[2121]= -R[1221] = -R[2112]
and
g[11]^-1 = g22/g
g[22]^-1 = g11/g
g[12]^-1 = -g12/g
g[21]^-1 = -g21/g
where again g is the determinant g=g11*g22-g12g21

Writting out all the terms for the right hand side of
R[ij]= g[hk]^-1 * R[hikj]

R[ij]=g[11]^-1*R[1212]+g[22]^-1*R[2121]+g[12]^-1*R[1221]+g[21]^-1*R[2112]
All other Summation factors of R equal zero.

substitution of inverses and converting all the R terms to R[1212] gives

R[ij] = R[1212]*(g22/g+g11/g+g12/g+g21/g)

R[ij] = (R[1212]/g)*(g11+g22+g12+g21)

R[ij] = g[ij] * (R[1212]/g)

with fact one being K=R[1212]/g

I have my desired result

R[ij]=g[ij]*K

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I will look for those books next. Thank you for the time and effort.