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§ Derivatives in Diffgeo

created 2021-11-18 · last edited 2022-01-02
  • A function of the form f:Ri→Rof: \mathbb R^i \to \mathbb R^of:Ri→Ro has derivative specified by an (o×i)(o \times i)(o×i) matrix, one which says how each output varies with each input.
  • Now consider a vector field VVV on the surface of the sphere, and another vector field DDD. Why is W≡∇DVW \equiv \nabla_D VW≡∇D​Vanother vector field? Aren't we differentiating a thing with 3 coordinates with another thing with 3 coordinates?
  • Well, suppose we consider the previous function f:Ri→Rof: \mathbb R^i \to \mathbb R^of:Ri→Ro, and we then consider a curve c:(−1,1)→Ric: (-1, 1) \to \mathbb R^ic:(−1,1)→Ri. Then the combined function (f∘c):(−1,1)→Ro(f \circ c): (-1, 1) \to \mathbb R^o(f∘c):(−1,1)→Ro needs only ooo numbers to specify the derivative, since there's only one parameter to the curve (time).
  • So what's going on in the above example? Well, though the full function we're defining is from Ri\mathbb R^iRi to Ro\mathbb R^oRo, composing with ccc "limits our attention" to a 1D input slice. In this 1D input slice, the output is also a vector.
  • This should be intuitive, since for example, we draw a circle parameterized by arc length, and then draw its tangents as vectors, and then we draw the normal as vectors to the tangents! Why does that work? In both cases (position -> vel, vel -> accel) we have a single parameter, time. So in both cases, we get vector fields!
  • That's somehow magical, that the derivative of a thing needs the same "degrees of freedom" as the thing in itself. Or is it magical? Well, we're used to it working for functions from R\mathbb RR to R\mathbb RR. It's a little disconcerting to see it work for functions from R\mathbb RRto Rn\mathbb R^nRn.
  • But how does this make sense in the case of diffgeo? We start with a manifold MMM. We take some curve c:(−1,1)→Mc: (-1, 1) \to Mc:(−1,1)→M. It's derivative must live as c′:(−1,1)→TMc': (-1, 1) \to TMc′:(−1,1)→TM. Now what about c′′c''c′′? According to our earlier explanation, this too should be a vector! Well... it is and it isn't, right? but how? I don't understand this well.
  • Looping back to the original question, W≡∇DVW \equiv \nabla_D VW≡∇D​V is a vector field because the value of W(p)W(p)W(p) is defined as taking D(p)∈TpMD(p) \in T_p MD(p)∈Tp​M, treating it as a curve dp:[−1,1]→Md_p: [-1, 1] \to Mdp​:[−1,1]→M such that dp(0)=pd_p(0) = pdp​(0)=p and dp′(0)=D(p)d_p'(0) = D(p)dp′​(0)=D(p), and then finally taking V()V()V().
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