Curvature of Manifolds: Understanding Relative Concepts

[WannabeNewton]

Christoffel symbols rely on coordinates for their definition AFAIK. I believe that the coordinate independent notion equivalent to a Christoffel symbol is called a connection and/or a fiber bundle.

Indeed if $\nabla$ is a connection on a smooth manifold $M$ and $(U,(x^{i}))$ is a coordinate chart, the Christoffel symbols of the coordinate basis $(e_i) = (\frac{\partial }{\partial x^{i}})$ are defined by $\nabla_{e_{i}}e_{j} = \Gamma ^{k}_{ij}e_{k}$. Connections can be defined on fiber bundles, which are vast generalizations of tangent bundles (for the purposes of classical GR, one could think of the connection as a map on the space of smooth sections of the tangent bundle satisfying certain properties).

 

[Benjam:n]

Precisely, the classical Euclidean notion of parallelism only exists at a single point on curved manifolds. There is simply no notion of vectors at different points of space-time being parallel in the Euclidean sense because they live in different vector spaces which have different inner products. What we can do however is parallel transport a vector along some curve from one point to another and then compare the transported vector at the terminal point to vectors that live at the terminal point. What does it mean to parallel transport a vector along a curve? Well let's say $M$ is a smooth manifold and $\nabla_{a}$ is a derivative operator on $M$. If $\gamma$ is a curve and $V^{a}$ is a vector at some point $p$ that the curve passes through, we say it is parallel transported along $\gamma$ if $\xi^{a}\nabla_{a}V^{b} = 0$ where $\xi^{a}$ is the tangent vector to $\gamma$.
.

Can I just run past you my understanding of the covariant derivative and you can correct any misunderstandings if they're there. Take some vector field in a 2D space Vi(x1,x2), we can then quite easily take its derivative with respect to x1 and x2. Now produce a new coordinate system (y1,y2), such that the two are connected by a smooth transition function yi(x1,x2). Then if we want to see the covariant derivative relative to that new frame what we do is transform both the vector at the point and the vector infinitesimally close to that point as though they were both at that same original point (I think this is bringing them both into the same tangent space, but precisely what tangent space means I'm not 100% sure ). This means that we transform the vector into y, dyj/dxiVi(x1,x2) and then differentiate, but the transformation bit is now treated as though it were constant. So it becomes dyj/dxid/dxn(vi(x1,x2)) and then you've just contracted it with xi n, to get the change in the direction of the tangent vector. I am not to sure on the significance of this then being equal to zero. In the example above what it seems to mean is that there is some coordinate system in which the vectors will be seen as actually being parallel? But how does this work on the surface of a sphere. let's do the typical example of taking a pen at the north pole, following a great circle down to the equator moving along the equator and then going back to the north pole and the pen will point in a different direction. How do you actually calculate this derivative here and show for that path it was equal to zero?

 

[Benjam:n]

Because its different to in my example. In my example I just had two coordinates in the same space. Now we just have a curved space embedded in flat 3d space and I don't really understand how the concept can be used?

 

[WannabeNewton]

I'm afraid I don't quite understand what you're saying. The covariant derivative in and of itself is a coordinate independent map defined for smooth manifolds. Only the Christoffel symbols are relative to a coordinate system.

See here for a detailed calculation of parallel transport on $S^{2}$: http://www.scribd.com/doc/57524972/Parallel-Transport

 

[Benjam:n]

Another question. Why do we need the Riemann curvature tensor to measure curvature It took me a good few hours just to calculate a couple of components of it for a curved 2d space. It seems as if its just overly complex. Also the christoffel symbols alone could do the same job surely, since if your in flat space then the christoffel symbol of one coordinate system relative to itself is zero, which isn't true in curved space?

Also I've just about got an intuitive sense for the Riemann tensor. Could anyone shed some light on the Ricci tensor and the curvature scalar and what they mean intuitively (I know there maths definitions)

 

[WannabeNewton]

The Christoffel symbols don't tell us anything about curvature by themselves; they aren't even coordinate independent! How can you make any geometric statement about a manifold using Christoffel symbols as such? They are rather meaningless quantities as far as geometry goes. The Riemann curvature tensor is the most natural curvature tensor you can construct and it can be related to the Gaussian curvature. For 2d-surfaces, you have arguably the most beautiful theorem in geometry to aid you along: http://en.wikipedia.org/wiki/Theorema_egregium

As for the Ricci curvature, see here: https://www.physicsforums.com/showthread.php?t=698693&highlight=ricci+curvature

 

[WannabeNewton]

By the way, I really think you could learn a lot more efficiently if you picked up a textbook on Riemannian geometry. There are a couple of texts that assume the minimal amount of topology knowledge (e.g. Do Carmo) but I don't know your math background.

 

[Benjam:n]

The way I understand the christoffel symbol is as the derivative of the basis vectors. If your in a curved space only certain vectors are allowed at each point (those in the tangent space) ergo my conclusion was that the basis vectors must change and that the christoffel symbol would not be zero?

 

[WannabeNewton]

Even in curved space I can find a coordinate system in which the Christoffel symbols vanish identically at a point (these are called Riemann normal coordinates) but the Riemann curvature tensor will still not necessarily vanish at that point. Like I said, they are coordinate dependent quantities so they by themselves can't be used to say anything unequivocal about the geometry of a (pseudo) Riemannian manifold and curvature is one of the most fundamental geometric properties of a (pseudo) Riemannian manifold.

From the physics side of things, in GR we want to abide by general covariance but explicitly incorporating Christoffel symbols into a physical equation does not abide by general covariance (see Wald 4.1)

 

[stevendaryl]

The way I understand the christoffel symbol is as the derivative of the basis vectors. If your in a curved space only certain vectors are allowed at each point (those in the tangent space) ergo my conclusion was that the basis vectors must change and that the christoffel symbol would not be zero?

There are two different notions: curved space (or spacetime) and curvilinear coordinates. If the Christoffel symbol is nonzero, that means that you are using curvilinear coordinates, but it doesn't mean that your space is curved. For example, on the flat 2D plane, you can use the curvilinear coordinates $r$ and $\theta$, and have nonzero Christoffel symbol.

The other way around is true, also. In curved spacetime, the Christoffel symbols may all be zero at a particular point, but that doesn't mean that there is no curvature.

The actual test for curvature is a particular quantity, the Riemann tensor, which is constructed from taking derivatives of the Christoffel symbols. (I should clarify that the components can be computed in terms of derivatives of the Christoffel symbols). In flat spacetime, the Riemann tensor is identically zero, no matter what coordinate system you use to compute the components.

 

[pervect]

There are two different notions: curved space (or spacetime) and curvilinear coordinates. If the Christoffel symbol is nonzero, that means that you are using curvilinear coordinates, but it doesn't mean that your space is curved. For example, on the flat 2D plane, you can use the curvilinear coordinates $r$ and $\theta$, and have nonzero Christoffel symbol.

The other way around is true, also. In curved spacetime, the Christoffel symbols may all be zero at a particular point, but that doesn't mean that there is no curvature.

The actual test for curvature is a particular quantity, the Riemann tensor, which is constructed from taking derivatives of the Christoffel symbols. (I should clarify that the components can be computed in terms of derivatives of the Christoffel symbols). In flat spacetime, the Riemann tensor is identically zero, no matter what coordinate system you use to compute the components.

The Riemann tensor measures the failure of parallel transport to commute.

To put it a bit more simply suppose you go "north" for 100 miles, and "east" for 200, in that order. Do you wind up at the same spot if you reverse the order, go east 200 , THEN go north 100?

If the answer is no, you are definitely on a curved surface. For instance, if you do that on the Earth's surface, you'll find that you don't wind up at the same spot - because it's curved.

To be able to pull this off, you need to know what "north" and "east" are. I've worded my above example for "ease of understanding", which means it'll be easier for some, and less clear and rigorous for others. The more rigorous approach would be to say that you travel north, parallel transporting an east vector , then go east along the vector you just transported. Similarliy, on the other leg of the trip, you parallel transport the north vector while you go east, then you go north. So in my example, I was envisioning the Earth, and starting out at a point on the equator.

Winding up in different spots when you change the order mean that the "north" and "east" vectors, do not commute.

North and east here are actually defined locally, then parallel transported as needed.

If all vectors commute where you're at, you are on a flat spot of whatever surface you're on (and all the components of the Riemann tensor are zero at wherever you're at).

IF all vectors commute everywhere, you're on a flat surface.

 

[Benjam:n]

I'm struggling with why covariant differentiation doesn't commute in curved spaces. To define a covariant derivative - you pick a coordinate system in which this new derivative is just going to equal the normal derivative in that coordinate system. Then from that you can define christoffel symbols of all other coordinate systems and the covariant derivative follows. From this definition I can see that what it effectively is is you take your infinitesimally close vector transport it keeping its components the same to the original point and then transform it into the new coordinates as though it were at that original point and then take the difference between the two vectors. I thought then that it would be this odd way of transforming the vectors that causes the non-commutatively, but then flat space would have the same problem. I presume then that it must be the problem that the vector that your transporting doesn't actually exist in the tangent space of your original point. So how do you actually pick that new vector - is it just the one with the greatest component in the same direction, i.e. that maximises the dot product?

 

[WannabeNewton]

You should really stop thinking in terms of coordinates. Geometry is much more important than coordinates; the latter are nothing but computational tools. The geometric concepts are much more fundamental; I'm not sure if this is a bias on my part from reading so many of Geroch's notes and papers on GR but still. Anyways, see the following screenshot from Wald's GR text: http://postimg.org/image/58tua25td/ The failure of the commutativity of the covariant derivative is related to the presence of curvature in the above manner.

 

[russelljbarry15]

A manifold is a local area that that looks Euclidean in a small area. When you drive around your neighborhood, mountains and hill not included, you only think about two dimensions. So locally, you consider it flat. Manifolds and curvature are two different things. Manifolds are really a fancy word for a surface. A manifold is nothing more than a continuous space of points that may be curved globally, but locally look like plain old flat space.

When you work with the standard vectors i, j, k, they do not change when you move around. But for example in spherical coordinates the basis vectors change when you move around. So in Euclidean geometry the Christoffel symbols do not change they are zero. But in spherical coordinates the Basis Vectors change and this is what the Christoffel symbols are. In spherical coordinates the basis vectors change so when you differentiate the vector you also have to differentiate the basis vector. The Christoffel symbols are the change in the basis vectors. So using the standard i,j,k vectors the Christoffel symbols are zero. Remember in spherical coordinates the basis vectors change, so they also need to be differentiated.

Curvature is something else. If you draw a triangle on a flat piece of paper the sides add up to 180 degrees. If you bend the paper a little, the triangle sides no longer add up to 180 degrees. Depending on the curvature it will be either less or more than 180 degrees. So the space is no longer flat. I hope this helps.

 

[WannabeNewton]

A manifold is a local area that that looks Euclidean in a small area. When you drive around your neighborhood, mountains and hill not included, you only think about two dimensions. So locally, you consider it flat. Manifolds and curvature are two different things. Manifolds are really a fancy word for a surface. A manifold is nothing more than a continuous space of points that may be curved globally, but locally look like plain old flat space.

A manifold is a much, much more general notion than this. A manifold is not necessarily a surface (a surface is a topological 2-manifold) nor does the notion of curvature even need to have any meaning for a manifold.

 

[stevendaryl]

I'm struggling with why covariant differentiation doesn't commute in curved spaces.

Think about traveling on the surface of the Earth. Suppose you have two sticks joined at right angles. Initially, you're at the equator, with one stick (Stick A) pointing North, and another stick (Stick B) pointing East. Let L be 1/4 the circumference of the Earth. If you travel a distance L in the direction of Stick A, and then a distance L in the direction of Stick B, you will end up distance L to the East of where you started. If instead, you travel in the direction of Stick B, and then in the direction of Stick A, you'll end up at the North Pole, a distance L to the north of where you started. So that it makes it clear that the order of traveling makes a difference.

In contrast, on a flat surface, you can take any two vectors, and if you travel in one direction and then the other, you get to the same spot as if you did it in the opposite order.

 

[Benjam:n]

Can someone please explain what the second fundamental form is. I know the first fundamental form is the metric, which is used to construct length/dot product of two vectors, so is the second fundamental form the way that you construct the area spanned by the two vectors i.e. cross product? So if you say wanted to construct an a surface area

 

[WannabeNewton]

The second fundamental form is given by $II_p(\xi,\eta) = -\left \langle dN_p (\xi),\eta \right \rangle, \xi,\eta\in T_p \Sigma$ where $\Sigma\subseteq \mathbb{R}^{3}$ is a regular surface and $N_p$ is the Gauss map. We call $dN_p$ the Weingarten map; the second fundamental form is what leads to the principal curvatures and principal directions of $\Sigma$ because the second fundamental form is symmetric implying that $dN_p$ is self-adjoint which implies by the spectral theorem that it is orthogonally diagonalize. Its eigenvalues are called the principal curvatures and its eigenvectors the principal directions. The product of the principal curvatures gives the Gaussian curvature and the average gives the mean curvature. One can also compute the Gaussian curvature directly from the coordinate representations of the first and second fundamental forms as $K_p = \frac{\det [II_p]}{\det [I_p]}$. So in a sense, the second fundamental form characterizes the curvature of $\Sigma$.

 

[Benjam:n]

Is the gauss map just a parameterization of the surface, i.e. each point on the surface is mapped to the point on the unit sphere with the same normal vector as that point on the sphere - or is there more to it than that? Is the second fundamental form then basically just the derivative of the normal vectors field on the surface?

Also I understand principle curvatures in terms of osculating circles - so that seems to have a nice link as presumably normal vectors determine the centre of the circle?

Finally is there an equivalent concept to the metric for area, i.e. eijkdyj/dxndyk/dxm. Presumably not because you'd need two of them contracted together to get the modulus and square rooted along with the vectors which they're later going to be contracted with.

 

[WannabeNewton]

Yes to all your questions up till the last. As for the last question, I'm not sure what you're asking (mainly because I can't really read what you wrote down there). Are you asking how we use the natural volume form $\epsilon_{abcd}$ associated with the metric $g_{ab}$ to find volumes?

 

[Benjam:n]

WannabeNewton - Could you show me the computation of the Riemann curvature tensor for the paraboloid z=x^2+y^2 at the point (1,1,1). I thought I knew what I was doing but I'm not sure I do. Does parallel transport mean project your vector into the tangent space of the infinitesimally close point and scale it up so it still has the same magnitude. Thanks

 

[George Jones]

Benjam:n, what are you using for references. Are you trying to work systematically through a particular reference? If not, you should be.

 

[stevendaryl]

WannabeNewton - Could you show me the computation of the Riemann curvature tensor for the paraboloid z=x^2+y^2 at the point (1,1,1). I thought I knew what I was doing but I'm not sure I do. Does parallel transport mean project your vector into the tangent space of the infinitesimally close point and scale it up so it still has the same magnitude. Thanks

Are you asking for how to do it, conceptually, or how to actually compute the components of the curvature tensor? The latter is an awful lot of tedious work, in my experience, even though there is nothing conceptually hard about it.

The first thing you'd need to do for your particular example is to compute the components of the metric tensor. It's a 2D surface, so the metric tensor has 4 components, which we can call
$g_{x x}, g_{x y}, g_{y x}, g_{y y}$
(Because the tensor is symmetric, $g_{x y} = g_{y x}$)

To figure out the components, you take two infinitesimally removed points:
$(x, y, z)$ and $(x + \delta x, y + \delta y, z + \delta z)$
and you compute the square of the distance between them:

$\delta s^2 = \delta x^2 + \delta y^2 + \delta z^2$

Since $z = x^2 + y^2$, we can eliminate $\delta z$ via
$\delta z = 2 x \delta x + 2 y \delta y$

Now write $\delta s^2$ in terms of $\delta x$ and $\delta y$, and then you can just read off the components of the metric tensor, because

$\delta s^2 = g_{x x} \delta x^2 + g_{x y} \delta x \delta y + g_{y x} \delta y \delta x + g_{y y} \delta y^2$

For your example, I think you end up with:

$g_{x x} = 1 + 4x^2$
$g_{x y} = g_{y x} = 2 x y$
$g_{y y} = 1 + 4 y^2$

Next, you need the inverse of the metric tensor, as well. The components are written using raised indices: $g^{x x}, g^{x y}, g^{y x}, g^{y y}$.

Next, you take derivatives of the metric tensor components to get the connection coefficients, $\Gamma^i_{j k}$ where $i, j, k$ are each either $x$ or $y$. These are computed in terms of the metric tensor as follows:

$\Gamma^i_{j k} = \frac{1}{2} \sum_{l} g^{i l} [\partial_j g_{lk} + \partial_k g_{lj} - \partial_l g_{jk}]$

Finally, you take derivatives of $\Gamma^i_{j k}$ and compute the Riemann tensor components using

$R^i_{jkl} = \partial_k \Gamma^i_{jl} - \partial_l \Gamma^i_{jk} + \sum_m \Gamma^i_{km} \Gamma^m_{lj} - \sum_m \Gamma^i_{lm} \Gamma^m_{kj}$

 

[Mentz114]

WannabeNewton - Could you show me the computation of the Riemann curvature tensor for the paraboloid z=x^2+y^2 at the point (1,1,1). I thought I knew what I was doing but I'm not sure I do. Does parallel transport mean project your vector into the tangent space of the infinitesimally close point and scale it up so it still has the same magnitude. Thanks

Does the point (1,1,1) lie on z=x^2+y^2 ?