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## 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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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?

Matterwave

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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?

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

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

lavinia

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

lavinia

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

Matterwave

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lavinia

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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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This is what I have: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.

##*(*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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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.

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