How can there be a non-covariant exterior derivative?

So, if I'm not wrong, the definition of the exterior derivative of a differential $$k$$-form is the differential $$k\text{+}1$$-form

$$\text{d}\omega = \omega_{i_1...i_k,i_{k+1}} \text{d}x^{i_{k\text{+}1}}\wedge\text{d}x^{i_1}\wedge\cdots\wedge\text{d}x^{i_k}.$$

When I was introduced to the concept of the covariant derivative, I was told that the partial derivative (the one used to define the exterior derivative of differential forms) doesn't produce objects whose components transform as tensor components would under a change of basis, so... how does that not contradict the idea of the exterior derivative mapping differential $$k$$-forms to differential $$k\text{+}1$$-forms? Am I missing something or am I using wrong definitions?

• The exterior derivative is well defined, ie independent of coordinate chart, because you are anti-symmetrizing the sum. – s.harp Feb 28 at 23:35
• Dear @s.harp could you explain intuitively why antisymmetrization leads to chart independence? – Arrow May 30 at 9:26
• @Arrow if you know the reason why $\partial_i f_j$ does not transform correctly under basis change its because you get an additional unwanted term of the form $\partial_{i'}( \partial_{j'}x^j) f_j$ which comes from the product rule. As you likely know $\partial_{i'}\partial_{j'}x^j(x^{k'})$ is symmetric in $i', j'$, so if you anti-symmetrize the indexes $i'$ and $j'$ this term dies and everything transforms correctly. You can do this with every index that appears, but you need to anti-symmetrize all of these with the $\partial_{i'}$. – s.harp May 31 at 13:54
• Now its possible I may have messed up where the $'$ go and so on, but thats the basic idea. Doing the calculation correctly and explicitly will make it clear. – s.harp May 31 at 13:55
• Dear @s.harp thank you for the proof sketch! I am more curious whether there is a geometric interpretation of antisymmetrization which might also make chart independence morally transparent. – Arrow May 31 at 13:56