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Definition

Suppose $X,Y$ are locally convex topological vector spaces and $U\subseteq X$ is open. We say a functional \begin{align} J:\ U &\longrightarrow Y \\ u &\longmapsto J[u] \end{align} is Gâteaux differentiable at $u$ if forall $h\in X$, there exist the limit \begin{align} \lim_{\varepsilon\to0} \dfrac{J[u+\varepsilon h]-J[u]}{\varepsilon} =: \delta J[u;h]. \end{align}

In the Riemannian geometry setting

Let $M$ be a Riemannian manifold, $Met(M)$ denotes the space of all smooth metric tensor fields on $M$ and $S^0_2(M)$ denotes the space of all smooth symmetric $(0,2)-$tensor field on $M$. Then the Levi-Civita Theorem shows that for each $g\in Met(M),\ h\in S^0_2(M)$ and for $\varepsilon\in\mathbb R$ small enough so that $g+\varepsilon h\in Met(M)$, there exists unique Levi-Civita connections $\nabla[g+\varepsilon h]$ and $\nabla[g]$.

Proposition (4.13 page 94, John M. Lee - Introduction to Riemannian Manifolds) asserts that for all $\varepsilon$, the difference $\nabla[g+\varepsilon h]-\nabla[g] $ forms a $(1,2)-$tensor field. Hence, for $\varepsilon\ne0$, $\dfrac{\nabla[g+\varepsilon h]-\nabla[g]}{\varepsilon}$ is also a $(1,2)-$tensor field.

Then, I interest that if the Gâteaux derrivative $\delta\nabla[g;h] = \displaystyle\lim_{\varepsilon\to0}\dfrac{\nabla[g+\varepsilon h]-\nabla[g]}{\varepsilon} $ is also a $(1,2)-$tensor field ? In this context, the spaces $U\subseteq X$ and $Y$ in the precede definition are $Met(M)\subseteq S^0_2(M)$ and $ C^\infty(M) $.

So I found in Lee's Introduction to smooth manifolds the Lemma 12.24 page 318 and Introduction to Riemmanian manifolds the Lemma B.6 page 398 states that the map $\delta\nabla[g;h]:\ \mathfrak X^*(M)\times\mathfrak X(M)\times\mathfrak X(M)\longrightarrow C^\infty(M)$ is a $(1,2)-$tensor field if and only if it is multilinear over $C^\infty(M)$.

I deduce that since $\dfrac{\nabla[g+\varepsilon h]-\nabla[g]}{\varepsilon}$ is multilinear over $C^\infty(M)$, and \begin{align} \left(\lim_{\varepsilon\to0}\dfrac{\nabla[g+\varepsilon h]-\nabla[g]}{\varepsilon}\right)(\omega,X,Y) &= \lim_{\varepsilon\to0}\left(\dfrac{\nabla[g+\varepsilon h]-\nabla[g]}{\varepsilon}(\omega,X,Y)\right) \end{align} for all $\omega\in\mathfrak X^*(M), X, Y\in\mathfrak X(M) $, it follows that $\displaystyle \lim_{\varepsilon\to0}\dfrac{\nabla[g+\varepsilon h]-\nabla[g]}{\varepsilon}$ is multilinear over $C^\infty(M)$, which is justified by the above Lemmas to be a $(1,2)-$tensor field on $M$.

Is my above argument correct ? And is this enough for one to be able to apply the covariant derivative for the Gâteaux derivative $\delta\nabla[g;h]$ ? I would appreciate any help. Thanks.

PermQi
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  • before speaking of Gateaux derivatives, you need to specify the topology on the space of tensor fields. But formally speaking atleast, this is right. The difference of two connections gives a $(1,2)$ tensor field, and this does “depend smoothly” on $g$, atleast if you look at the coefficients in local coordinates, in the sense that if you replace $g$ with $g+\epsilon h$, then the Christoffel symbols as functions of $(\epsilon,x)$ will be smooth. – peek-a-boo Aug 18 '25 at 20:40
  • @peek-a-boo So, if I follow this approach, then I don’t need to check the tensorial transformation law explicitly in order to conclude that $\delta\nabla[g;h]$ is a $(1,2)-$tensor field? – PermQi Aug 19 '25 at 03:58
  • No because you’re taking a limit of $(1,2)$ tensor fields. You just need to know the limit exists. – peek-a-boo Aug 19 '25 at 08:00
  • @peek-a-boo If in some previous steps, I showed that \begin{align} \delta\Gamma^k_{ij}[g;h] = \frac{1}{2}\sum_m g^{km} \Big[\big( \nabla_i h\big){jm} + \big(\nabla_j h\big){im} - \big(\nabla_m h\big){ij} \Big] \end{align} and \begin{align} \delta\nabla[g;h] = \sum{ijk} \delta\Gamma^k_{ij}[g;h] \partial_k\otimes dx^i\otimes dx^j, \end{align} then is it enough to point out the existence of the Gâteaux derivative $\delta\nabla[g;h] $ ? – PermQi Aug 19 '25 at 09:00
  • Yes, modulo you actually specifying the topology on the space of tensor fields. – peek-a-boo Aug 19 '25 at 09:05
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    @peek-a-boo to make everything valid, do I need to state the explicit topology for $S^0_2(M)$ and $Met(M)$, or do I just simply say that $S^0_2(M)$ is endowed with some appropriate topology such that $Met(M)$ is open ? – PermQi Aug 19 '25 at 09:16
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    There’s nothing “wrong” with either, but the second one is more natural, so is the one Id try to look up. – peek-a-boo Aug 19 '25 at 09:31

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