# Proving a Covariant Derivative is Torsion Free

Let $$(M,g)$$ be a metric manifold and $$\phi:M\to N$$ a diffeomorphism, where $$N$$ is another manifold. Let $$\nabla$$ be the Levi Civita connection with respect to the metric $$g$$, and we define a connection in $$(N,\phi_*(g))$$ by:

$$\tilde{\nabla}_XY=\phi_*\left(\nabla_{\phi^*(X)}\phi^*(Y)\right)$$

I am trying to prove that $$\tilde{\nabla}$$ is the Levi Civita connection of $$(N,\phi_{*}(g))$$. So there is a step in which I evidently make a mistake and I am not being able do find it. Im trying to prove that it is torsion free (I don't want an alternative way to do this, I simply want to understand what am I doing wrong) so lets go. We want to see that

$$\tilde{\nabla}_XY-\tilde{\nabla}_YX=[X,Y]$$

So, since $$\phi$$ is a diffeo, we have that $$\phi^*=(\phi^{-1})_*$$ and thus,

$$$$\begin{split} \tilde{\nabla}_XY-\tilde{\nabla}_YX &=\phi_*\left(\nabla_{\phi^*(X)}\phi^*(Y)\right)-\phi_*\left(\nabla_{\phi^*(Y)}\phi^*(X)\right)\\ &=\phi_*\left(\nabla_{\phi^*(X)}\phi^*(Y)-\nabla_{\phi^*(Y)}\phi^*(X)\right)\\ &=\phi_*\left(\left[\phi^*(X),\phi^*(Y)\right]\right) \end{split}$$$$ where in the second line I used linearity of $$\phi_*$$ and in the third line that $$\nabla$$ is the LC connection and thus Torsion Free. So now comes the problem, lets act on a function $$f:N\to \mathbb{R}$$ to get:

$$$$\begin{split} \phi_*\left(\left[\phi^*(X),\phi^*(Y)\right]\right)(f)&=\left[\phi^*(X),\phi^*(Y)\right]\left(\phi^*(f)\right)\\ &=\phi^*(X)\left\{\phi^*(Y)(\left(\phi^*(f)\right))\right\}-\phi^*(Y)\left\{\phi^*(X)(\left(\phi^*(f)\right))\right\}\\ &=\phi^*(X)\left\{(\phi^{-1})_*(Y)(\left(\phi^*(f)\right))\right\}-\phi^*(Y)\left\{(\phi^{-1})_*(X)(\left(\phi^*(f)\right))\right\}\\ &=\phi^*(X)\left\{Y(f)\right\}-\phi^*(Y)\left\{X(f)\right\} \end{split}$$$$

where in the second line I have used that $$[X,Y](f)=X(Y(f))-Y(X(f))$$, in the third that $$\phi^*=(\phi^{-1})_*$$ and in the last line I used the definition of the pushforward of $$X$$ and $$Y$$ and the pullback of $$f$$ via $$\phi^{-1}$$ and $$\phi$$ respectively. However, in this last step there should be a mistake I am not being able to find since $$\phi^*(X)$$ should act on functions $$h:M\to \mathbb{R}$$ while $$Y(g):N \to \mathbb{R}$$.

Any help will be appreciated.

• Some notation issues. In the first eq you wrote, since $\tilde \nabla$ is supposed to be a connection on $N$, $X, Y$ should be tangent vectors on $N$; therefore on the right hand side one should write $(\phi^{-1})_* X$ to send $X$ backward to $M$ (in order to make use of the connection $\nabla$ on $M$), instead of writing $\phi_* X$. Jan 10, 2021 at 21:52
• There are many right hand sides... However, i do not see any $\phi_* X$ in any place. Jan 10, 2021 at 22:30
• In the definition of $\tilde \nabla$, i.e. $\tilde{\nabla}_XY=\phi_*\left(\nabla_{\phi^*(X)}\phi^*(Y)\right)$. Jan 10, 2021 at 22:36
• And for notation, why write $\phi^*(X)$ instead of $\phi_*(X)$? Jan 10, 2021 at 22:38
• Because its not the same thing: $\phi^*X=(\phi^{-1})_*X\neq \phi_* X$ Jan 10, 2021 at 22:47

## 1 Answer

The mistake is not only in the last line but also in the first line of your last mathematical paragraph. Let $$Z$$ be a vector field on $$M$$ and $$f \colon N \rightarrow \mathbb{R}$$ be a smooth function. In your derivation, you are using the "identity"

$$\left( \phi_{*} \left( Z \right) \right)(f) = Z(\phi^{*}(f))$$ (where $$Z = [\phi^{*}(Z),\phi^{*}(Y)]$$) but this identity clearly cannot hold as the left hand side is a function on $$N$$ while the right hand side is a function on $$M$$.

More explicitly, for $$p \in M$$ and $$q \in N$$ we have $$(\phi_{*}(Z)(f))(q) = df|_{q} \left( d\phi|_{\phi^{-1}(q)} \left( Z|_{\phi^{-1}(q)} \right) \right), \\ (Z(\phi^{*}(f)))(p) = d(\phi^{*}(f))|_{p}(Z_p) = d(f \circ \phi)|_{p}(Z_p) = df|_{\phi(p)} \left( d\phi|_{p} \left( Z_p \right) \right)$$ so a correct identity is $$\left( \phi_{*}(Z) \right)(f) = \phi_{*} \left( Z(\phi^{*}(f)) \right)$$ (this is an identity between functions on $$N$$ and you can also write another one in terms of functions on $$M$$).

Similarly, the identity

$$\phi^{*}(X) \left( \phi^{*}(f) \right) = X(f)$$

is wrong but

$$\phi^{*}(X) \left( \phi^{*}(f) \right) = \phi^{*} \left( X(f) \right)$$ is correct.

Thus,

$$\phi_{*} \left( \left[ \phi^{*}(X), \phi^{*}(Y) \right] \right)(f) = \phi_{*} \left( \left[ \phi^{*}(X), \phi^{*}(Y) \right] \left( \phi^{*}(f) \right) \right)= \\ \phi_{*} \left( \phi^{*}(X) \left( \phi^{*}(Y) \left( \phi^{*}(f) \right) \right) - \phi^{*}(Y) \left( \phi^{*}(X) \left( \phi^{*}(f) \right) \right) \right) = \\ \phi_{*} \left( \phi^{*}(X(Yf)) - \phi^{*}(Y(Xf)) \right) = X(Yf) - Y(Xf) = [X,Y](f).$$

• thank you very much! Jan 11, 2021 at 12:48