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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{equation} \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} \end{equation} 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{equation} \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} \end{equation}

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.

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  • $\begingroup$ 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 $\phi_* X$. $\endgroup$
    – Yuval
    Jan 10, 2021 at 21:52
  • $\begingroup$ There are many right hand sides... However, i do not see any $\phi_* X$ in any place. $\endgroup$ Jan 10, 2021 at 22:30
  • $\begingroup$ In the definition of $\tilde \nabla$, i.e. $\tilde{\nabla}_XY=\phi_*\left(\nabla_{\phi^*(X)}\phi^*(Y)\right)$. $\endgroup$
    – Yuval
    Jan 10, 2021 at 22:36
  • $\begingroup$ And for notation, why write $\phi^*(X)$ instead of $\phi_*(X)$? $\endgroup$
    – Yuval
    Jan 10, 2021 at 22:38
  • $\begingroup$ Because its not the same thing: $\phi^*X=(\phi^{-1})_*X\neq \phi_* X$ $\endgroup$ Jan 10, 2021 at 22:47

1 Answer 1

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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,

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

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  • $\begingroup$ thank you very much! $\endgroup$ Jan 11, 2021 at 12:48

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