Recent content by HenryGomes

Of course, I can't, because it's not true. Only if v-w\in V_1. I don't know from where I got that idea... Thanks

Supposing we have a vector space V and a subspace V_1\subset V. Suppose further that we have two different direct sum decompositions of the total space V=V_1\oplus V_2 and V_1\oplus V_2'. Given the linear projection operators P_1, P_2, P_1', P_2' onto these decompositions, we have that...

Because it is not really the answer I am looking for, but to apply a different method to obtain adjoints when we do not have the Hodge star. Thanks for the answer!

Any good differential geometry book should have this. One I like, which is for physicist's is John Baes' "Knots, Gauge Theory and Gravity". You might also want to check out Bleecker's " Variational Principles in Gauge Theories". I re-read the post and it seemed a bit badly written. So just to...

Usually the adjoint to the exterior derivative d^* on a Riemannian manifold is derived using the inner product \langle\langle\lambda_1,\lambda_2\rangle\rangle:=\int_M\langle\lambda_1,\lambda_2\rangle\mbox{vol}=\int_M\lambda_1\wedge*\lambda_2 where \lambda are p-forms and * is the Hodge...

In the space of Riemannian metrics Riem(M), over a compact 3-manifold without boundary M, we have a pointwise (which means here "for each metric g") inner product, defined, for metric velocities k^1_{ab},k^2_{cd} (which are just symmetric two-covariant tensors over M)...