dsaun777 said:
So the contraction with the differential 1 forms are not invariant objects necessarily?
You really need to be more specific. The scalar ##T^{\mu \nu} = T^{ab} (\mathrm{d}x^{\mu})_a (\mathrm{d}x^{\nu})_b##
is invariant, but by this we mean it's the ##\mu \nu## component of ##T^{ab}## in a
specified co-ordinate chart ##x^{\mu}##.
If you change to a different set of co-ordinates ##x'^{\mu}##, then ##T'^{\mu \nu} = T^{ab} (\mathrm{d}x'^{\mu})_a (\mathrm{d}x'^{\nu})_b## is the ##\mu \nu## component of ##T^{ab}## in this new chart. And ##T^{\mu \nu} \neq T'^{\mu \nu}## in general. That's the sense in which
@PeterDonis means that contractions with differentials are co-ordinate dependent.
But more generally, you're being too vague.
A contracted product of any tensors is always an invariant object, precisely because both tensor products and contractions are co-ordinate independent operations. See post #4. If you contract over
all of the remaining free indices you get a scalar, whilst if you leave some un-contracted you'll get a higher order tensor (which, as a geometric object is independent of co-ordinates, but has co-ordinate dependent
components).
Contracted products with arbitrary one-forms are no different; depending on what you contract it with you might get a scalar, vector, etc., which are all invariant objects geometrically (although, again, remember that tensors of rank ##>0## have co-ordinate dependent
components). If ##\omega_a## and ##\eta_a## are any two
arbitrary one-forms, then ##T^{ab} \omega_a \eta_b## is a scalar, and invariant.