# The curvature tensor

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Imagine in our manifold a very small closed loop whose four sides are the coordinate lines , , , .

**Figure 6.3:** Parallel transport around a closed loop *ABCD*.

A vector defined at *A* is parallel transported to *B*. From the parallel transport law

it follows that at *B* the vector has components

where the notation `` '' under the integral denotes the path *AB*. Similar transport from *B* to *C* to *D* gives

and

The integral in the last equation has a different sign because of the direction of transport from *C* to *D* is in the negative direction.

Similarly, the completion of the loop gives

The net change in is a vector , found by adding (93)-(96).

To lowest order we get

This involves derivatives of 's and of . The derivatives of can be eliminated using for example

This gives

To obtain this, one needs to relabel dummy indices in the terms quadratic in .

Notice that this just turns out to be a number times summed on . Now the indices 1 and 2 appear because the path was chosen to go along those coordinates. It is antisymmetric in 1 and 2 because the change would have the opposite sign if one went around the loop in the opposite direction.

If we use general coordinate lines and , we find

Defining

we can write

are the components of a 1/3 tensor. This tensor is called the Riemann curvature tensor .