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## The Exterior Derivative is Nilpotent

Given a form A, its exterior derivative dA is defined as:
 Degree Expression for dA Traditional 0 dA = A[!i] dx[i] grad 1 dA = (A[j][!i] - A[i][!j]) dx[i] ^ dx[j] curl; i < j n-1 dA = A[i][!i] dx ^ ... ^ dx[n] div Generally dA = (d) ^ A where (d) = (d/dx[m]) dx[m]

In the n-1 case, the dx[j] product contains all dimensions and is the volume element.

From the equality of mixed partials it can be seen that ddA = 0 no matter what the class is of the form, i.e. scalar valued or any other class of value. Formally this arises because B^B = 0 for whatever form B, and every current (in particular, d) can be realized as the limit of a sequence of C-infinity forms.

Equations involving the ordinary derivative depend on the coordinate system, whereas an equation about tensors that involves the covariant derivative, if true for one coordinate system, is true for all. Thus we would like to see covariant derivatives used to construct the exterior derivative. When can the covariant derivative be substituted for the ordinary derivative in the definition of d? First consider scalar-valued forms. There are no covariant corrections for degree 0. In degree 1:

```    A[j][/i] - A[i][/j] =
+ A[j][!i] - A[m] g'[m][n] C[j][i][n]
- A[i][!j] + A[m] g'[m][n] C[i][j][n]
```

And because C[i][j][m] == C[j][i][m], the covariant and ordinary derivatives can be used interchangeably. The same kind of pairwise cancellation will happen in all degrees.

Now consider composite-valued forms. If the value represents something at a point that doesn't move with the argument point being differentiated, then covariant corrections do not occur, and the conclusion is the same as before: the covariant and ordinary derivative are interchangeable. However, if the value is at the same point being differentiated, covariant corrections will appear. For a specific example let's use 1-form-valued forms (the metric tensor is an example of degree 1). Definitely there are covariant corrections. At degree 0:

```    d(A[1.i]) = A[1.i][2!j] - A[1.m] g'[m][n] C[j][i][n]
```

At degree 1:

```    d(A[1.i][2.k]) =
+ A[1.i][2.k][2!j]
- A[1.m][2.k] g'[m][n] C[i][j][n]
- A[1.i][2.m] g'[m][n] C[k][j][n]
- A[1.i][2.j][2!k]
+ A[1.m][2.j] g'[m][n] C[i][k][n]
+ A[1.i][2.m] g'[m][n] C[j][k][n]
```

So the covariant corrections on the #2 subscript can be ignored; only the corrections on the #1 subscript survive. This conclusion is general for all degrees and all result classes.

So is ddA still zero if covariant derivatives are used? In general,

```    d(A[1.i][2/k]) =
= A[1.i][2/k][2/j] - A[1.i][2/j][2/k]
= -A[1.m] g'[m][n] R[n][i][k][j]
= A[1.m] g'[m][n] R[n][i][j][k]
```

Which in general is nonzero.

In the specific case of the Riemann-Christoffel tensor R, we want to know if dd*R = 0. (* is the metric dual, defined later.)

```    R[i][j][k][L] = C[j][L][i][!k] + C[j][k][m] g'[m][n] C[i][L][n]
- C[j][k][i][!L] - C[j][L][m] g'[m][n] C[i][k][n]
```

where [i] and [j] are skew symmetric and are the "extra" dimensions. Let G = positive square root of the determinant of the metric tensor.

```    *R = -1k+L G g'[k][p] g'[L][q] R[i][j][p][q] dx[1..n\k,L]
```

where the notation dx[1..n\k,L] means the product of basis elements dx^dx^...^dx[n] but omitting k and L. To greatly simplify the manipulations, let's do this in geodesic normal coordinates. Then G == 1 and g' = identity, so:

```    *R = -1k+L R[i][j][k][L] dx[1..n\k,L]

d*R = -1L (*R)[i][j][k][L][/k] dx[1..n\L]

dd*R = ( (*R)[m][j][k][L] g'[m][n] R[n][i][k][L]
+ (*R)[i][m][k][L] g'[m][n] R[n][j][k][L]) dx[1..n]
= ( R[m][j][k][L] R[m][i][k][L]
+ R[i][m][k][L] R[m][j][k][L]) dx[1..n]
```

In the latter step, the geodesic normal components of *R are substituted. Since R[m][i][k][L] = -R[i][m][k][L], the two terms cancel out and dd*R = 0.

This relation would not be true for a generic 2-form valued form. It is true only because the Riemann-Christoffel tensor is both the composite object being differentiated and the composite object that transforms it giving its covariant correction.

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