Well, a Balarka, we agree that counting is for the birds :P
Actually, @robjohn (if you're thinking about it), I'm now confused. Since he's applying the derivative only to the vector field $X$ itself, that looks a lot like the integrated form of $\mathscr L_X X = 0$. Hmm.
that's effectively the same as writing down the Leibniz formula
for the record, the fact that the identity functor is covariant and the dual functor is contravariant is the trivial, but perhaps unsatisfying, reason that they are not naturally isomorphic
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there is a generalized version of natural transformations that allows you to compare covariant with contravariant functors and even that does not make identity and dual isomorphic
that is to say, their not being isomorphic is more than just a formality issue
why are we discussing functors at all? an isomorphism V -> W^* is choice of a nondegenerate pairing V x W -> R. There is no natural nondegenerate form on a vector space, so there is no natural isomorphism. When W = V^*, there is: it's evaluation
Welcome back. But you were here in the room yesterday, so I didn't know :)
Yeah, I think his result is actually correct. Just because you're applying to the vector field $X$ itself. But now I cannot figure out what's wrong with my counterexample.