# Uniqueness of induced covariant derivatives of tensor fields

I'm working on proving Lemma 4.6 in John Lee's "Riemannian Manifolds":

Lemma. Let $\nabla$ be a linear connection on a Riemannian manifold $M$. There is a unique connection in each tensor bundle $T_l^kM$, also denoted $\nabla$, such that the following conditions are satisfied:

1. On $TM$, $\nabla$ agrees with the given connection.
2. On $T^0M$, $\nabla$ is given by ordinary differentiation of functions: $$\nabla_X f = Xf.$$
3. $\nabla$ obeys the following product rule with respect to tensor products: $$\nabla_X(F \otimes G) = (\nabla_X F)\otimes G + F \otimes (\nabla_X G).$$
4. $\nabla$ commutes with all contractions: if "$\mathrm{tr}$" denotes the trace of any pair of indices, $$\nabla_X(\mathrm{tr}Y) = \mathrm{tr}(\nabla_X Y).$$
5. $\nabla$ obeys the following product rule with respect to the natural pairing between a covector field $\omega$ and a vector field $Y$: $$\nabla_X\langle \omega, Y\rangle = \langle \nabla_X \omega, Y \rangle + \langle \omega, \nabla_X Y \rangle.$$
6. For any $F \in \mathcal T_l^k(M)$, where $\mathcal T_l^k(M)$ is the space of smooth $(k,l)$-tensor fields on $M$, and for any vector fields $Y_i$ and 1-forms $\omega^j$, \begin{align} (\nabla_X F)(\omega^1, \ldots, \omega^l, Y_1, \ldots, Y_k) = &X(F(\omega^1, \ldots, \omega^l, Y_1, \ldots, Y_k)) \\ &- \sum_{j=1}^l F(\omega^1, \ldots, \nabla_X \omega^j, \ldots, \omega^l, Y_1, \ldots, Y_k) \\ &- \sum_{i=1}^k F(\omega^1, \ldots, \omega^l, Y_1, \ldots, \nabla_X Y_i, \ldots, Y_k). \end{align}

Lee suggests using 1-4 to prove 5 and 6 (and thus to prove uniqueness), and prove existence using 5 and 6 (ie using 5 and 6 to prove 1-4). The one part of this I'm having trouble with is proving 5 from 1-4 (I can prove 2-4 from 6 and 1 (note 5 follows from 6 also), and 6 follows from 3). Certainly 5 bears resemblance to 3, but 3 involves explicit tensor products, and 5 does not; besides, $\langle \omega, Y \rangle \in C^\infty(M) = \mathcal T^0(M)$, which seems to suggest 2 should be used, but again, the exact computation is lost on me. Any insights?

• Hint: Perhaps use 4 (along with 3) to get 5? May 15, 2017 at 18:28
• I'm not sure how to use 4, since 4 involves the trace of a vector field. Unless $Y$ is supposed to be a mixed tensor field instead...? (I had originally proved 6 with this assumption but now I'm doubting my original argument) May 15, 2017 at 18:36
• It's not trace of a vector field. It's a contraction (summing, for example, $\sum_i a^i_j b_i$, one upper index and identical lower index). This is specifically what you have in 5. May 15, 2017 at 19:02
• @DomVeconi: You're misinterpreting the statement about traces. I didn't say that $Y$ was a vector field -- in that statement, $Y$ can be any tensor field whose trace is defined, which means it has to have at least one upper index and one lower index. This doesn't make sense if $Y$ is a vector field. (Admittedly, using the letter $Y$ there was ill-advised. In my draft second edition, I've changed it to $F$ instead of $Y$.) May 15, 2017 at 23:37
• @JackLee I see, that makes more sense. Thanks for taking the time to help clarify the lemma! May 16, 2017 at 0:20

Let $(E_i)$ be a local frame and $(\varphi^i)$ be the dual coframe. First we note that given any 1-form $\eta=\eta_j\varphi^j$ (Einstein summation convention) and any vector field $Z=Z^iE_i$, we have $$\eta\otimes Z=\eta_jZ^i\varphi^j\otimes E_i$$ i.e., the coefficients of $\eta\otimes Z$ are $\eta_jZ^i$. Thus, by taking contraction, we have $$\text{tr}(\eta\otimes Z)=\eta_mZ^m$$ but the R.H.S. is just $\eta(Z)$. Therefore $$\text{tr}(\eta\otimes Z)=\left\langle\eta,Z\right\rangle$$ Now we can use 3 and 4, together with linearity of tr, to show 5 as follow: \begin{align} \nabla_X\left\langle\omega,Y\right\rangle&=\nabla_X(\text{tr}(\omega\otimes Y)) \\ &=\text{tr}(\nabla_X(\omega\otimes Y))\qquad\qquad\qquad\qquad\mbox{(By 4)} \\ &=\text{tr}((\nabla_X\omega)\otimes Y+\omega\otimes(\nabla_XY))\qquad\quad\mbox{(By 3)} \\ &=\text{tr}((\nabla_X\omega)\otimes Y)+\text{tr}(\omega\otimes(\nabla_XY))\qquad\mbox{(Linearity of tr)} \\ &=\left\langle\nabla_X\omega,Y\right\rangle+\left\langle\omega,\nabla_XY\right\rangle\qquad\mbox{($\nabla_X\omega$ is a 1-form while $\nabla_XY$ is a vector field)} \end{align} This completes the proof (of 5).