I like how they made an effort in showing the students' questions in text. Very sweet.
@CarlosGonzalez-rg6ht9 жыл бұрын
Wonderful explanation of the covariant derivative and the connections. This series of lectures are amazing, thank you for posting them !!
@wschadow8 жыл бұрын
A crystal clear series of lectures, thank you!
@feiwang47439 жыл бұрын
Crystal clear!
@aarontyrrell78038 жыл бұрын
Does the tensor product rule just follow from part 1 of the definition of a connection since the product of 2 tensors is just a function once it is evaluated at some covector and vector fields?
@danielm.mclaury32028 жыл бұрын
If you're talking about condition (iii) at 17:36, then the left-hand side is indeed determined by (i), but not every term on the right-hand side is. So it really does give a separate condition.
@aarontyrrell78038 жыл бұрын
Can someone please explain how the Leibniz rule implies the equality he writes @31.50
@danielm.mclaury32028 жыл бұрын
Note that each Y_m is a function in our neighborhood. Write T = Y_m * id. Then we're trying to determine ∇_X(T(d/dx_m)). Applying the Leibniz rule, we have ∇_X(T(d/dx_m)) = (∇_X T)(d/dx_m) + T(∇_X(d/dx_m)). Rewriting this, ∇_X(T(d/dx_m)) = (∇_X (Y_m * id))(d/dx_m) + Y_m ∇_X(d/dx_m). He also applies the linearity in X to expand the sum. (Note that he's using Einstein notation.)