Intuitive

The Hodge dual, also known as the Hodge star operator ($\star$), in $n$-dimensional space takes an antisymmetric rank-$p$ tensor and maps it to an antisymmetric rank-($n-p$) tensor.

Example #1: In Euclidean 3D space ($n=3$), the Hodge dual of the antisymmetric rank-2 tensor $a \wedge b$ is a vector, thus mapping rank $p=2$ to $p=3-2=1$. In this case, the resulting vector is just the ordinary cross product: $\star (a \wedge b) = a \times b$.

Example #2: In Euclidean 3D space ($n=3$), the Hodge dual of the antisymmetric rank-3 tensor $a \wedge b \wedge c$ is a scalar, thus mapping rank $p=3$ to $p=3-3=0$. In this case, the resulting scalar is just the scalar triple product: $\star (a \wedge b \wedge c) = a \times b \cdot c$.

Example #3: In 4D space ($n=4$), the Hodge dual of an antisymmetric rank-2 tensor is again an antisymmetric rank-2 tensor ($4-2=2$). This is the reason why self-dual and anti-self-dual tensors exist in 4D.

Concrete

Abstract

Why is it interesting?