Efficiently Compute Trace of Differential Forms | Helpful Tips and Tricks

[DanM]

I need to compute the following thing

trace ( (dg + g^g)^2) = d (tr (g ^ dg + 2/3 g ^ g ^ g))

im messing up completely b/c I don't know how to use trace with differential forms

tr (dg + g^g) ^ (dg + g^g))

 

[Haelfix]

Just expand it out, there's only a few tricky problems

1) This is the Chern-Simons form, so you are dealing with Matrix valued forms.. So in general

tr (A ^ B) = -(1)^(rs) tr (B ^ A) (err 2nd wedge is a power)

Use the Bianchi identities

and of course

d (A ^ B) = dA^B + (-1)^rs (A^dB)

Note that
tr (A^A^A^A) = 0 b/c of the last identity

Expand out both sides, it should only take you a few lines.

 

[wais]

= tr (dg ^ dg + dg ^ g^g + g^g ^ dg + g^g ^ g^g)

To efficiently compute the trace of a differential form, there are a few helpful tips and tricks that can make the process easier.

First, it is important to understand the properties of the trace operator. The trace of a differential form is defined as the sum of the diagonal elements of the matrix representation of the form. This means that the trace of a 1-form is simply the function itself, while the trace of a 2-form is the sum of its coefficients.

In this particular problem, we are given the expression for the trace of (dg + g^g)^2. To compute this efficiently, we can use the properties of the trace operator and break down the expression into smaller, more manageable parts.

We can start by expanding the square of the expression using the distributive property, which gives us:

tr (dg + g^g)^2 = tr (dg + g^g) * tr (dg + g^g)

Next, we can use the fact that the trace of a product of two forms is equal to the product of their traces, as long as the forms are of the same degree. This means that we can rewrite the expression as:

tr (dg + g^g)^2 = tr (dg + g^g) * tr (dg + g^g) = (tr dg + tr g^g) * (tr dg + tr g^g)

Now, we can use the fact that the trace of a 1-form is simply the function itself, and the trace of a 2-form is the sum of its coefficients, to simplify the expression further:

tr (dg + g^g)^2 = (dg + g^g + g^g) * (dg + g^g + g^g) = dg * dg + dg * g^g + dg * g^g + g^g * g^g

Finally, we can use the fact that the trace of a product of two 1-forms is equal to the dot product of the two functions, and the trace of a product of two 2-forms is equal to the wedge product of the two forms. This gives us the final expression:

tr (dg + g^g)^2 = dg * dg + dg ^ g^g + dg ^ g^g + g^