Introduction to the Hodge Star Operator - 1 

Thank you for that!

Thank you for your question! The factor of 1/2! you're referring to typically appears when we express a 2-form using its components in differential geometry. However, let's clarify why it might not explicitly appear in the final result of the example in the video. Understanding the Factor of 1/2! : When we write a 2-form 𝜔 in local coordinates as 𝜔 = 𝜔_𝑖𝑗 𝑑𝑥^𝑖 ∧ 𝑑𝑥^𝑗 , the components 𝜔_𝑖𝑗 are typically defined to be antisymmetric in their indices. This antisymmetry means that if you swap the indices 𝑖 and 𝑗, you get −𝜔_𝑖𝑗. To account for this antisymmetry when summing over all combinations of 𝑖 and 𝑗, we often include a 1/2! factor. This factor corrects for counting each pair (𝑖,𝑗) twice. However, this normalization is already implicit in the definition of the wedge product for differential forms. Applying the Hodge Star: When applying the Hodge star operator ⋆ to a 2-form 𝜔, the result is an (𝑛 − 2)-form in an 𝑛-dimensional space. For example, in 𝑅^4 , applying the Hodge star to a 2-form yields another 2-form. The operation ⋆𝜔 does not inherently require an additional 1/2! factor because the wedge product and Hodge star already incorporate the necessary antisymmetry and normalization properties. So, when you see ⋆𝜔 in the final result, it's because ⋆ has already been applied to 𝜔, and the resulting form naturally reflects this operation. In the Context of the Video: In the video example, when we compute ⋆𝜔, the factor of 1/2! does not appear because the Hodge star operator is being directly applied to a fully formed 2-form. The notation ⋆𝜔 is concise and assumes all properties and symmetries of the forms and operations involved. If you have any more questions or need further clarification, feel free to ask! Thanks for watching and engaging with the content.