# A Square of the exterior derivative

1. Oct 19, 2016

### spaghetti3451

Is $\text{d}^{2}=\text{d}\wedge\text{d}$ a definition of the exterior algebra, or can it be derived from more fundamental mathematical statements?

2. Oct 20, 2016

### Matterwave

I'm sorry if I'm mistaken as it has been a while since I've done differential geometry, but isn't $\text{d}^2=0$ one of the defining properties of the exterior derivative?

3. Oct 20, 2016

### spaghetti3451

What about the the Hodge star squared?

I know that $**=-1$, but is this a definition, or can it be proved in two to three lines?

Last edited: Oct 20, 2016
4. Oct 20, 2016

### Fightfish

That is neither a definition nor is it is true in general. For $n$-dimensional space and for a $p$-form, $** = -(-1)^{p(n-p)}$ in Minkowski space and $** = (-1)^{p(n-p)}$ in Euclidean space.

5. Oct 21, 2016

### lavinia

Yes although one can derive this from the assumption that d$^{2}=0$ on functions.

The condition d$^{2}=0$ makes the differential forms on a manifold into what is called a "Chain Complex". Chain complexes occur all over in mathematics. The definition is that d is linear and its square is zero.

BTW: Differential forms and exterior derivatives do not require the idea of a metric so they are not specifically restricted to Differential Geometry but rather to Calculus on Manifolds.

Last edited: Oct 21, 2016
6. Oct 21, 2016

### Staff: Mentor

.... and geometry, topology, and (homological) algebra.

7. Oct 21, 2016

### lavinia

I said it badly. The least amount of structure one needs to talk about differential forms is calculus in manifolds. A metric is added structure.

8. Oct 21, 2016

### Matterwave

I was not aware that differential geometry required a metric? Wouldn't that fall under Riemannian geometry, or Semi-Riemannian geometry?

9. Oct 21, 2016

### lavinia

Geometry to me means measurement of angles at least and usually also distance. These ideas are not needed to do calculus. Differential forms are just calculus. For instance one can integrate a differential form on a smooth manifold that has no shape and is just a bunch of smoothly overlapping coordinate charts..

Classical differential geometry always uses the metric induced by an embedding of a manifold in Euclidean space. It is the study of the induced metric relations.

One thing that is often missed because of the way calculus is taught using inner products in Euclidean space is that no inner product is needed and that the same derivatives can be taken with or without a metric. The subject of Differential Topology - not Differential Geometry - relies on calculus without using any metric.

Last edited: Oct 23, 2016
10. Oct 21, 2016

### spaghetti3451

This is what I have:

$*(*F)_{i_{1},i_{2},\dots, i_{n-k}}$

$= *\left(\frac{1}{k!}F^{j_{1},\dots,j_{k}}\sqrt{|\text{det}\ g|}\ \epsilon_{j_{1},\dots,j_{k},i_{1},i_{2},\dots, i_{n-k}}\right)$

$= *\left(\frac{1}{k!}F_{l_{1},\dots,l_{k}}g^{l_{1},\dots,l_{k},j_{1},\dots,j_{k}}\sqrt{|\text{det}\ g|}\ \epsilon_{j_{1},\dots,j_{k},i_{1},i_{2},\dots, i_{n-k}}\right)$

$= *\left(\frac{1}{k!}F_{l_{1},\dots,l_{k}}\sqrt{|\text{det}\ g|}\ {\epsilon^{l_{1},\dots,l_{k}}}_{i_{1},i_{2},\dots, i_{n-k}}\right)$

$= \frac{1}{k!}\left(*F_{l_{1},\dots,l_{k}}\right)\sqrt{|\text{det}\ g|}\ {\epsilon^{l_{1},\dots,l_{k}}}_{i_{1},i_{2},\dots, i_{n-k}}$

$= \frac{1}{k!}\left(\frac{1}{(n-k)!}F^{m_{1},\dots,m_{n-k}}\sqrt{|\text{det}\ g|}\ \epsilon_{m_{1},\dots,m_{n-k},l_{1},\dots,l_{k}}\right)\sqrt{|\text{det}\ g|}\ {\epsilon^{l_{1},\dots,l_{k}}}_{i_{1},i_{2},\dots, i_{n-k}}$

$= \frac{1}{k!}\left(\frac{1}{(n-k)!}F_{j_{1},\dots,j_{n-k}}g^{j_{1},\dots,j_{n-k},m_{1},\dots,m_{n-k}}\sqrt{|\text{det}\ g|}\ \epsilon_{m_{1},\dots,m_{n-k},l_{1},\dots,l_{k}}\right)\sqrt{|\text{det}\ g|}\ {\epsilon^{l_{1},\dots,l_{k}}}_{i_{1},i_{2},\dots, i_{n-k}}$

$= \frac{1}{k!}\left(\frac{1}{(n-k)!}F_{j_{1},\dots,j_{n-k}}\sqrt{|\text{det}\ g|}\ {\epsilon^{j_{1},\dots,j_{n-k}}}_{l_{1},\dots,l_{k}}\right)\sqrt{|\text{det}\ g|}\ {\epsilon^{l_{1},\dots,l_{k}}}_{i_{1},i_{2},\dots, i_{n-k}}$

$= \frac{1}{k!}\ \frac{1}{(n-k)!}\ F_{j_{1},\dots,j_{n-k}}\ |\text{det}\ g|\ {\epsilon^{j_{1},\dots,j_{n-k}}}_{l_{1},\dots,l_{k}}\ {\epsilon^{l_{1},\dots,l_{k}}}_{i_{1},i_{2},\dots, i_{n-k}}$

$= \frac{1}{(n-k)!}F_{j_{1},\dots,j_{n-k}}|\text{det}\ g|\ \delta^{j_{1},\dots,j_{n-k}}_{i_{1},\dots,i_{n-k}}$

$= \frac{1}{(n-k)!}F_{i_{1},\dots,i_{n-k}}|\text{det}\ g|$

How do I proceed next?

11. Oct 23, 2016

### lavinia

If you include as Differential Geometry the study of connections which are not compatible with any metric then yes you are right. Perhaps I should have said a connection rather than a metric. But the point is the same. Differential forms exist on any smooth manifold whether or not there is a connection or a metric. There is nothing about them that requires either. Differential Topology does not assume connections.

Some people have told me that they consider tensor fields as part of Differential Geometry. And maybe when one learns General Relativity tensors are introduced for the first time. But tensors like differential forms do not require a semi-Riemannain metric or in fact any metric or connection.