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Let $\phi:N\to M$ be a smooth mapping between Riemannian manifolds. We say a mapping $X:N\to TM$ is a vector field on $\phi$ if $X(x)\in TM_{\phi(x)}$ for each $x\in N$. Then we pick a neighborhood $V$ for $\phi(x)$, and take a frame $\{E_1,...E_n\}$ on $V$. Hence for each $y\in\phi^{-1}(V)$, we have

$$X(y)=\sum_i X^i(y)E_i(\phi(y))$$

for some functions $X^i$ on $N$. Let $D$ be the Levi-Civita connection on $M$, let $v\in TN_y$. We define the induced connection:

$$\tilde{D}_v(X)(y)=\sum_i (vX^i(y))E_i(\phi(y))+X^i(y)D_vE_i(\phi(y)).$$

It can be shown that this definition does not depend on the choice of the frame. My question is,

how do we show that for any vector field $Z,W$ on $N$, $$\tilde{D}_Zd\phi(W)-\tilde{D}_Wd\phi(Z)-d\phi[Z,W]=0?$$

I am aware that $d\phi[Z,W]=[d\phi Z,d\phi W]$, but not sure how to utilize it. Thanks.

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    Note that I asked here http://mathoverflow.net/questions/254651/does-symmetry-of-a-pullback-connection-should-be-obvious for a coordinate-free proof of the symmetry of an induced connection. By the way the connection you are talking about is a special case of a *pullback-connection*, which is a vet useful concept in general. – 2017-01-05

1 Answers 1

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Everything should be crystal clear on the level of local coordinates.

Let $Z=Z^i\partial_i, W=W^i\partial_i$ holds for a given local chart on $N$ and $\phi=(\phi^\alpha)$ for the given local coordinates on $N$ and $M$(Thus the number of $i$s is equal to the dimension of $N$ and the number of $\alpha$s is equal to the dimension of $M$). Then,

$\tilde{D}_Z (d\phi W)-\tilde{D}_W(d\phi Z)$

$=\tilde{D}_Z(W^i\partial_i \phi^\alpha\partial_\alpha)-\tilde{D}_W(Z^i\partial_i \phi^\alpha\partial_\alpha)$

$=Z^i\partial_i(W^j\partial_j\phi^\alpha)\partial_\alpha+W^i\partial_i\phi^\alpha\tilde{D}_Z\partial_\alpha-W^i\partial_i(Z^j\partial_j\phi^\alpha)\partial_\alpha+Z^i\partial_i\phi^\alpha\tilde{D}_W\partial_\alpha$

$=(Z^i\partial_iW^j-W^i\partial_iZ^j)\partial_j\phi^\alpha \partial_\alpha+Z^iW^j(\partial_i\partial_j\phi^\alpha - \partial_j\partial_i\phi^\alpha)\partial_\alpha+Z^iW^j\partial_i\phi^\alpha\partial_j\phi^\beta(D_{\partial_\beta} \partial_\alpha-D_{\partial_\alpha} \partial_\beta)$

$=d\phi[Z,W]+0+0=d\phi[Z,W]$

holds. Here the 2nd and the 3rd terms on the 4th line vanish due to the commutativity of partial derivatives and the torsion-freeness of $ D$. (Thus this proof also applies to anhy torsion-free connection on $TM$ to give the same conclusion.)

(Sorry for slight different notations from yours, but I think using partial derivatives rather than general frames is better to understand and do a practical calculation in this case.)

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    Thanks for your answer. If no one else gives a coordinate-free solution, I'll choose you as the best answer. – 2017-01-01