Does the coderivative of the wedge product of two differential forms satisfy Leibniz rule?

Question

Ohh you think you are so clever, aint you trolls? Trying to mess with my head with your bullshit.

Bah, i'll ask a real mathematician. Yahoo answers suck.

Answer 1

Not exactly. You have to take the index of the forms into consideration. Hence, if the index of v is K and the index of w is L, then
d(v^w)=(dv)^w+(-1)^K v^(dw)

Essentially, the d has to commute past a form of indexK, and that leads to a factor of (-1)^K.

Answer 2

Nope. Not without the stindapheral side divided by the fazar quotent of X times the coefficient of 5 ... or the coefficient of 6 IF the sneldendon factor is a prime divisible by powers of 2.

Answer 3

I take it you want to tutor that d(a^b) = (da)^b + (-a million)^deg(a) a^(db), if a is a p-form and b is a q-form. by linearity of d, and after relabling coordinates, it suffices to envision for forms of here form: a = f*dx_1^...^dx_p b = g*dx_{p+a million}^...^dx_{p+q} (be conscious that if a = f*dx_{J} and b = g*dx_{ok} and the multi-indices J and ok intersect nontrivially, a^b = (da)^b = 0, and so on.) properly, given the above form for: a, b, that's not perplexing to envision the Leibnitz rule by hand. below I record some references on differential kinds. Bott and Tu is considered the terrific source, John Lee is considered the main person-friendly. Illustrate greater needless to say? properly, ok. using fact the d-operator is a linear operator, all we would desire to constantly do is verify whether the formulation holds for a, b of the above written kinds. it incredibly is, a = f(x_1,...,x_n) dx_1^...^dx_p b = g(x_1,...,x_n)dx_{p+a million}^...^dx_{p+q} Then a^b = f*g dx_1^...^dx_{p+q} So d(a^b) = sum over i of: d(f*g)/dx_i dx_i^dx_1^...^dx_{p+q} = sum over i of: g df/dx_i dx_i ^ dx_1^...^dx_{p+q} + f dg/dx_i dx_i^dx_1^...^dx_{p+q}. be conscious that the 1st term in the sum is (da)^b. the 2d term in the sum isn't extremely a^(db). To make it look like a^(db), you will desire to pass the dx_i in the process the dx_1^...^dx_p words. this could reason you to p.c.. up a ingredient of (-a million)^p.

Answer 4

If the fourth cumulants of the respective commutators of the two differential forms are positive-definite when represented in matrix form, then yes. Otherwise, no.

But seriously, why are you asking online forums PhD-level math questions? Did you serious expect someone to know the answer?