Equivalence of definitions for covariant derivative

differential-geometry

I understand that there seems to be many similar sounding questions on stackexchange but none of them seem to be able to address my confusion.

First of all, I am using the definition of covariant derivative found here (e.g. 21 min). Specifically, define $\nabla$ to be a map taking a smooth vector field $X$ and $(p,q)$-tensor $T$ to another $(p,q)$-tensor $\nabla_XT$ such that the following 4 axioms hold:

  1. $\nabla_Xf = Xf$ for all smooth functions $f\in C^{\infty}(M)$
  2. $\nabla_X(T+S) = \nabla_XT+\nabla_XS$ for tensors of the same ranks
  3. $\nabla_X (T(\omega,\dots,Y,\dots))=(\nabla_XT)(\omega,\dots,Y,\dots) + T(\nabla_X\omega,\dots,Y,\dots)+\dots+T(\omega,\dots,\nabla_XY,\dots)+\dots$
  4. $\nabla_{fX+Z}T=f\nabla_XT + \nabla_ZT$

First question: Later in the lecture (around 38 min), it was stated that condition 3 is equivalent to

3-alternate. $\nabla_X(T\otimes S) = (\nabla_XT)\otimes S + T\otimes (\nabla_XS)$

However, I cannot show that 3 and 3-alternate are equivalent. For simplicity, I was trying to prove the equivalence taking $T$ and $S$ to be (0,1)-tensors. Then I reduced the equivalence to basically showing that

$$\nabla_X(T(Y)) = T(\nabla_XY) + (\nabla_XT)(Y)$$
(and an analogous statement for $S$ with $Z$ instead of $T$ with $Y$). How do I know this is true? Axiom 1 gives me a grip on $\nabla_X$ on (0,0)-tensors but the RHS involves $\nabla_X$ on higher rank tensors. [Edit: Oh my, this just follows from axiom 3!]

Second question: I have seen in numerous physics textbooks where $\nabla$ is defined as a function from $(k,l)$-tensor fields to $(k,l+1)$-tensor fields. Are these definitions the same? The definition I am using seem to be instead a function from $(k+1,l)$-tensor fields to $(k,l)$-tensor fields.

Best Answer

I think that 3 implies 3-alternative but the converse is not true. You can actually view 3 as a definition for $\nabla_XT$, which only needs the action of $\nabla_X$ on vector fields and one-forms. (And the latter is reduced to the action of $\nabla_X$ on vector fields.) The resulting operation can be shown to be compatible with tensor products by a direct computation. In the case of $(0,1)$-tensor fields, this looks as follows. Using 3, you can expand $\nabla_X(T\otimes S)(Y,Z)$ as $$X((T\otimes S)(Y,Z))-(T\otimes S)(\nabla_XY,Z)-(T\otimes S)(Y,\nabla_XZ)=X(T(Y)S(Z))-T(\nabla_XY)S(Z)-T(Y)S(\nabla_XZ).$$ Using the product rule in the first term and collecting, this reads as $$(X(T(Y))-T(\nabla_XY))S(Z)+T(Y)(X(S(Z))-S(\nabla_XZ))=((\nabla_XT)(Y))S(Z)+T(Y)((\nabla_XS)(Z)),$$ and this proves the claim.

But in addition to 3-alternative, you need a condition that $\nabla_X$ is compatible with contractions. This also follows from 3 by direct computations. Having that, you can deduce 3 from 3-alternative via using that $T(\omega,\dots,Y,\dots)$ can be obtained as a "complete contraction" (which is a sequence of contractions) from $T\otimes\omega\otimes\dots\otimes Y\otimes\dots$.

Concerning your second questions, the interpretation as mapping $(k,\ell)$ tensor fields to $(k,\ell+1)$ tensor fields needs leaving the slot for the vector field $X$ free. So you define $\nabla T$ as mapping $(\omega_1,\dots,\omega_k,Y_1,\dots,Y_{\ell+1})$ to $(\nabla_{Y_1}T)(\omega_1,\dots,\omega_k,Y_2,\dots,Y_\ell)$. Rule 4 implies that this is indeed a tensor field.

There is no interpretation in terms of $(k+1,\ell)$ tensor fields. While you can form $X\otimes T$ and this is a $(k+1,\ell)$ tensor field, the covariant derivative $\nabla_XT$ cannot be obtained from $X\otimes T$. This is beause for a smooth function $f$, you get $(fX)\otimes T=X\otimes fT$ but $\nabla_X(fT)=X(f)T+f\nabla_XT$ while $\nabla_{fX}T=f\nabla_XT$.

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