Let $(M,g),(N,h)$ be semi-riemannian manifolds, and $\varphi \in C^{\infty}(M,N)$ an isometry (hence a diffeomorphism). Let $(U,\psi = (x^1,\ldots,x^n))$ be a coordinate chart of a point $p \in U \subset M$, and $(\varphi(U),\psi \circ \varphi^{-1} = (y^1,\ldots,y^n))$ be a chart of $\varphi(p) \in \varphi(U) \subset N$. By a (I believe) standard-theorem, we get $$T_p\varphi\Big(\frac{\partial}{\partial x^i}\Big|_p\Big)$$$$ = \sum_{j = 1}^{n} \frac{\partial ((\psi \circ \varphi^{-1})^j \circ \varphi \circ \psi^{-1})}{\partial x^i}(\psi(p)) \cdot \frac{\partial}{\partial y^j}\Big|_{\varphi(p)}$$$$ = \sum_{j = 1}^{n} \frac{\partial x^j}{\partial x^i} \frac{\partial}{\partial y^j}\Big|_{\varphi(p)} = \frac{\partial}{\partial y^i}\Big|_{\varphi(p)}.$$

Now, by the definition of isometries, and the calculation above, we get $$g_{ij}(p) := g\Big(\frac{\partial}{\partial x^i}\Big|_p,\frac{\partial}{\partial x^j}\Big|_p\Big) = h\Big(T_p\varphi\Big(\frac{\partial}{\partial x^i}\Big|_p\Big),T_p\varphi\Big(\frac{\partial}{\partial x^j}\Big|_p\Big)\Big) = h\Big(\frac{\partial}{\partial y^i}\Big|_{\varphi(p)},\frac{\partial}{\partial y^j}\Big|_{\varphi(p}\Big) = h_{ij}(\varphi(p)).$$

Now, to my question: Why does this imply $$\frac{\partial}{\partial y^k}\Big|_{\varphi(p)} h_{ij} = \frac{\partial}{\partial x^k}\Big|_p g_{ij}?$$

I feel like I am missing something obvious. The reason we want to show this is to see that $${}^g\Gamma^k_{ij}(p) = {}^h\Gamma^k_{ij}(\varphi(p))$$ for the Christoffel-symbols on $U,\varphi(U)$ respectively.


1 Answer 1


This will essentially follow from the fact that $y^i = x^i \circ \varphi^{-1}$ (by your definition).

The claim now is to show that for any $f \in C^\infty(M)$, we have that $$ \frac{\partial}{\partial x^k}\Big|_{p} f = \frac{\partial}{\partial y^k} \Big|_{\varphi(p)} (f \circ \varphi^{-1}) \,,$$ but this is clear since you have already shown that $\frac{\partial}{\partial y^k} \Big|_{\varphi(p)}$ is the pushforward of $\frac{\partial}{\partial x^k}\Big|_{p}$.

More explicitly, we can unravel the definition of the pushforward/tangent/differential map to see that: $$\begin{align} \frac{\partial}{\partial y^k} \Big|_{\varphi(p)} (f \circ \varphi^{-1}) &= T_p \varphi\left(\frac{\partial}{\partial x^k}\Big|_{p}\right) (f \circ \varphi^{-1})\\ &= \frac{\partial}{\partial x^k}\Big|_{p}(f \circ \varphi^{-1} \circ \varphi)\\ &= \frac{\partial}{\partial x^k}\Big|_{p} f \end{align}$$

Then with $f = g_{ij}$, and using the fact that you showed $h_{ij} = g_{ij} \circ \varphi^{-1}$, your desired result follows.

  • $\begingroup$ I think I see where you are coming from, but I am not sure I follow this step $$T_p\varphi\Big(\frac{\partial}{\partial x^k}\Big|_p\Big)(f \circ \varphi^{-1}) = \frac{\partial}{\partial x^k}\Big|_p(f \circ \varphi^{-1} \circ \varphi)$$ $\endgroup$
    – Ben123
    Commented Aug 7, 2023 at 22:36
  • $\begingroup$ That is the definition of the tangent map. For some $F \colon M \to N$, how else are you defining the map $T_p F \colon T_p M \to T_{F(p)} N$? $\endgroup$
    – AlkaKadri
    Commented Aug 7, 2023 at 22:38
  • $\begingroup$ I want to accept your answer, but for me to accept it, I need to understand it. I don´t understand how you get from the LHS to the RHS here. I don´t see that it follows from the definition of the tangent-map, but it might be due to my own ignorance. $\endgroup$
    – Ben123
    Commented Aug 7, 2023 at 22:48
  • 1
    $\begingroup$ No worries, you definitely shouldn't accept the answer until you're satisfied. What book are you using? What is your definition of the tangent map? For reference, the definition I am using is that with $F \colon M \to N$ smooth, if $v \in T_p M$ then $T_p F (v)$ acts on $ f \in C^\infty(N)$ by $T_p F (v) (f) = v(f \circ F)$ $\endgroup$
    – AlkaKadri
    Commented Aug 7, 2023 at 22:49
  • $\begingroup$ Oh, my bad, yes, now I understand! $\endgroup$
    – Ben123
    Commented Aug 7, 2023 at 22:56

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