[SOUND] [MUSIC] So let me illustrate what means, before going farther, let me illustrate what means parallel transport on some simple textbook type of exercise. So first of all, what does it mean that we make a parallel transport in flat, two dimensional flat space? Suppose we have a vector at some point, and then we want to parallel transport it along some closed curve, let it be triangle. So one way to this point, then here, and then back to this point. So during parallel transport, the angle between this vector and this pass should be conserved, so it shouldn't change. Why? Well, you will understand by starting many exercises. At this point, just believe it. So like, we parallel transport, so it goes like this, it appears here. Then I parallel transport with respect to this line again, I respect the angle. So it goes like this and then I end up here, and then again, I respect the angle and end up here. So in flat space, during the parallel transport along any closed path, it can be shown any closed path vector will come back to the same position. But now, let us do the same on a sphere. Suppose we have a sphere, like this. So it's like geometric sphere. And for simplicity, I choose to parallel transport the vector along the following path. So it's a triangle, consisting of three, like first of all, geodesic on the sphere. Their extremal path is obtained by taking a section of the sphere with the use of the plane, which is passing through the center of the sphere. Then geodesic between any two points is just a segment of equator, which is obtained by a section of the sphere by a plane, which is going through these three points. And this is just a segment of the equator. So let me take the following triangle which consists of three segments of different equators, which are orthogonal to each other. So it's like orthogonal here, diagonal between tangent vectors. Here and here is orthogonal, so here is also orthogonal, and here also orthogonal. And now let me take barrel of transport of a vector, which is sitting here. So again, I respect the angle within this vector and this curve means that within this vector and the tangent vector to this curve. So if I parallel it and do parallel transform with respect to this line, I obtain something like that. I don't have to take this vector, I can also take this vector. It doesn't matter, what I'm saying doesn't change. So now, I do parallel transport along this curve, again, respecting the angle. So I end up with something like this here. And then I parallel transport along this line, and I end up with a vector like this. And now we see that after parallel transporting along this curve, the vector is rotated by an angle pi/2, unlike here. So here, in flat space, it turns back, here, it is rotated. And that's exactly the point I am trying to tell you. So, suppose we have some vector field on the sphere, so it consists of some bunch of vectors at every point. And I want to take a covariant derivative. So I have a vector field value here. And I want to take a covariant differential. So I have two nearby points, like this and like this. I have a value of the vector field here. And I have a value of the vector field here. So, to take a covariant derivative, I have to make a parallel transport along the geodesic curve, say along the geodesic curve from here to here. So I take this geodesic and then parallel transport this guy respecting the angle. So I obtain this vector, which is different from this, and somehow transform this guy. So that's exactly what it has done, when I defined covariant derivative. A mu of (x + dx)- [A mu of (x) + delta A mu of (x)], where this guy, as I defined to you, delta A mu of (x), the result of the parallel transport is some rotation with the use of the connection. Of course, the connection depends on space. Here is concrete space, concrete curvature is defined. So it's like concretely curved space, sphere, two dimensional sphere. In ordinary situations, rotations can be more complicated, this is just to illustrate your example. So nu alpha of (x) A nu of (x) dx alpha, where this guy is a connection. So now, true from this formula, which I derived to you, another notation for the covariant derivative is that it's a A nu, this kind of notation, alpha dx alpha. This is just a different notation for the same quantity that I'm going to use. So now, to make this guy transform as a two tensor with two indices, alpha and mu, I have to define appropriate transformation rule for this connection. How do I do that? I use that formula. Remember that our ordinary differential transforms under coordinate transformations as follows. It's dx bar alpha / dx mu, dx nu / dx bar beta, A mu, nu(x) + d2x bar alpha / dx mu dx nu, and here, dx nu / dx bar beta times A mu (x). So, I have to define that this connection transforms such that this transforms appropriately, such that we compensate this term here. As a result, this defines the transformation rule for gamma. And let me write it right now for you. So the transformation rule for gamma is going to be as follows. The transformation rule for gamma is going to be as follows. So it's gamma mu nu alpha(x) = gamma bar beta gamma sigma of (x bar) dx mu / dx bar beta, dx bar gamma / dx nu, dx bar sigma / dx alpha, + d2 x bar gamma / dx nu dx alpha, dx mu dx bar gamma. So if this gamma transforms according to this rule, that means that this guy transforms according to this rule. And this transformation due to this term compensates in this, Guy, compensates this term. As a result, this guy transforms as a two tensor, tensor with two indices. It means appropriately under coordinate transformation. That was a goal to define covariant derivative. [SOUND] [MUSIC]
