## Thursday, 31 January 2019

### Commentary 2.9 Hodge star operator - in Euclidean space

 Swapping axes same as reversing an axis.
Carroll introduces Hodge duality and the Hodge star operator. The Hodge star operator is a map from $p$-forms on an $n$-dimensional manifold on to ($n-p)$ forms on the manifold. Thus: (his equation 2.82)
\begin{align}\large
{\left(*A\right)}_{{\mu }_1\dots {\mu }_{n-p}}=\frac{1}{p!}{\epsilon }^{{\nu }_1\dots {\nu }_p}_{\ \ \ \ \ \ \ \ \ \ \ \ \ \ {\mu }_1\dots {\mu }_{n-p}}A_{{\nu }_1\dots {\nu }_p} & \phantom {10000}(1) \\
\end{align}
which maps $A$ to "$A$ dual".

We have also used the Levi-Civita tensor which is, as the name suggests, a tensor not a mere number, defined as (Carroll's 2.69)
\begin{align}\large
{\epsilon }_{{\mu }_1{\mu }_2\dots {\mu }_n}=\sqrt{\left|g\right|}{\widehat{\epsilon }}_{{\mu }_1{\mu }_2\dots {\mu }_n} & \phantom {10000}(2) \\
\end{align}
where  $\widehat{\epsilon }$  is the Levi-Civita number and $g$ (also not a tensor) is the determinant of the metric.

Then at his 2.84 Carroll writes. "In three-dimensional Euclidean space the Hodge dual of the wedge product of two 1-forms gives another 1-form:
\begin{align}
*{\left(U\wedge V\right)}_i={\epsilon }^{\ \ jk}_iU_jV_k & \phantom {10000}(7) \\
\end{align}
(All of the prefactors cancel.) Since 1-forms in Euclidean space are just like vectors, we have a map from two vectors to a single vector. You should convince yourself that this is just the conventional cross product, and that the appearance of the Levi-Civita tensor explains why the cross product changes sign under parity (interchange of two coordinates or equivalently basis vectors.)"

Proving (7) itself was quite hard and the first step was to prove that the Levi-Civita tensor is completely antisymmetric, even when indices are up and down. Carroll had not mentioned this - he probably thought it was obvious. It was vital to the proof which followed with some tricky index swapping. The prefactors cancelling was easy.

The next part to show that (7) was the same as the cross product was very easy, including showing the dependence on the parity of the coordinate system which involves a lady flying over the north pole (diagram above) and the vexed question of fingering or screwing.

Read all four pages and 25 beautifully numbered equations in
Commentary 2.9 Hodge star operator - in Euclidean space.pdf