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mathwonk
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#4
Aug22-04, 03:02 PM
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Christoffel symbol as tensor


Hi Pete,

Input from the peanut gallery.

Well that quote is rather confusing to me, not least because i do not know what christoffel symbols are.

However, some people call a "tensor" anything that acts on families of vectors and covectors, sending them to other such things, with two defining properties:
1) the action is (multi) linear.
2) the value of the action depends only on the pointwise values of the arguments and not on their value in a nbhd.

From this perspective, even without knowing what a christoffel symbol is, one can deduce that Wald thinks his object is a tensor because of the first two sentences of your first quote from him:

"Thus, we have shown that *** defines a map of dual vectors at p (as opposed to dual vector fields defined in a neighborhood of p) to tensors of type (0, 2) at p. By property (1), this map is linear. Consequently *** defines a tensor of type (1,2) at p,"


So maybe the confusion is that when he changes coordinates, in order to have a tensor, he would have to transform both derivative operators by the tensor law, whereas he is merely taking one of the coordinate derivative operators as his delta a, rather than taking whatever he gets when he transforms the primed one.

So his definition of the christoffel symbols is not coordinate invariant, i.e. he says right there: (as you pointed out in red)

"as defined here, a christoffel symbol depends on a choice of coordinate system"
(roughly quoted).

But if it depends on the coordinate system it is not a true tensor, in the sense that it does not transform like one.

I.e. maybe the problem is something like this: maybe there IS a tensor asociated to a DIFFERENCE of two derivative operators, but when he tries to FIX one of the operators as differentiation in a coordinate direction, he steps out of bounds as far as defining a true tensor.

?????????
just a guess.


roy