# Square of the exterior derivative

## Main Question or Discussion Point

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?

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Matterwave
Gold Member
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?

• fresh_42
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?

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

lavinia
Gold Member
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?
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.

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fresh_42
Mentor
2018 Award
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.
.... and geometry, topology, and (homological) algebra.

lavinia
Gold Member
.... and geometry, topology, and (homological) algebra.
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.

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

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

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• dextercioby
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.
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?

lavinia
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