# How to define a partial derivative invariantly?

Let $M$ be a smooth manifold and $f$, $g$ be smooth functions in some neibourhood of a point $x_0\in M$, $\nabla g\ne0$.

1) How to define $\displaystyle \frac{\partial f}{\partial g}$ invariantly? If $M$ is a domaqin in $\mathbb R^n$ then the derivative in the direction of $\nabla g$ seems to give an answer: $\displaystyle \frac{(\nabla f,\nabla g)}{|\nabla g|^2}$. But to calculate $\nabla g$ and $|\nabla g|$ on a manifold one needs a metric. From the other hand, if we consider smooth coordinates $(g_1=g, g_2,\ldots,g_n)$ in some neibourhood of $x_0$, then partial derivatives $\displaystyle \frac{\partial f}{\partial g}$ seem to be defined in the standard way. But the question arises, would the value $\displaystyle \frac{\partial f}{\partial g}$ be independent from the choice of $g_2,\ldots,g_n$? If no, what are the correct way do do it? Is there some reference?

2) Let $f_1, f_2,\ldots,f_n$ be a smooth coordinates in some neibourhood of $x_0$. What is the object $(\displaystyle \frac{\partial f_1}{\partial g},\ldots,\frac{\partial f_n}{\partial g})$? Would it be by some chance a section of some good fiber bundle? Is there some reference where such objects are considered?

Thanks in advance!

Addition of may 27

Suppose that now there is a riemannian metric on $M$. Then what would be most natural definition of partial derivative $\displaystyle \frac{\partial f}{\partial g}$? For expample, to take $\displaystyle \frac{\partial f}{\partial g}=df(\nabla g)=(df,*dg)$ seems not right since that would mean $\displaystyle \frac{\partial f}{\partial g}=\displaystyle \frac{\partial g}{\partial f}$. If we take $\displaystyle \frac{(\nabla f,\nabla g)}{|\nabla g|}$ it would be sort of directional derivative. So the right question seems to me here is what value of $a$ it is best to take for $\displaystyle \frac{(\nabla f,\nabla g)}{|\nabla g|^a}$ to be a partial derivative? Or is there no the "best" choice? I tried to apply to dimensional analysis and it seems $a=1$ is the choice to have the result be like $\nabla f$ but I'm not sure because may be one have to assign some dimension to metric.

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## 3 Answers

The partial derivative $\frac{\partial f}{\partial g}$ is not invariant. One way to think about partial derivatives (the way relevant to how you've phrased the question) is that you are picking out a particular coefficient in the Jacobian $df_p : T_p(M) \to T_p(\mathbb{R}) \cong \mathbb{R}$ presented as a linear combination of some basis for the cotangent space, and this construction depends on the entire basis (not just $dg_p$). If $M$ is Riemannian then the cotangent space can be equipped with an inner product, so then you don't need the entire dual basis, just a particular cotangent vector.

A nicer notion of partial derivative is to pick a tangent vector $v \in T_p(M)$ and consider $df_p(v)$. This is invariant in the sense that it comes from the evaluation map $T_p(M) \times T_p(M)^{\ast} \to \mathbb{R}$.

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This is only a partial answer, but that might be considered appropriate for this question. :)

Partial derivatives can only be defined if you have a coordinate system: vary one coordinate, keeping all the other ones fixed. So yes, it definitely depends on the choice of $g_2,\dots,g_n$.

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I had to think about this recently. My conclusion was that it can't be done: the invariant object is $dg$, and to pass from $dg$ to $\dfrac{\partial}{\partial g}$ you need to choose an isomorphism from the cotangent space to the tangent space. This can be done if you, for example, choose a basis for your cotangent space or have a (pseudo-)Riemannian metric.

In full: Let $M$ be a smooth manifold (without boundary) and let $g_1, \ldots, g_n : U \to \mathbb{R}$ be a family of smooth functions defined on some open set $U \subseteq M$. Then, if at $p \in U$ the differentials $dg_1, \ldots, dg_n$ form a basis of the cotangent space $T^*_p M$, by the inverse function theorem, $(g_1, \ldots, g_n) : U \to \mathbb{R}^n$ is locally invertible with smooth inverse. We may therefore assume $U$ is small enough that $(g_1, \ldots, g_n) : U \to \mathbb{R}^n$ is a diffeomorphism onto its image, and the dual basis $\dfrac{\partial}{\partial g_1}, \ldots, \dfrac{\partial}{\partial g_n}$ is just what you expect under this chart. If we choose a different chart $(\tilde{g}_1, \ldots, \tilde{g}_n)$, then we have the relation $$\dfrac{\partial}{\partial \tilde{g}_j} = \sum_{k=1}^{n} \dfrac{\partial g_k}{\partial \tilde{g}_j} \dfrac{\partial}{\partial g_k}$$ so in particular even if you set $g_1 = \tilde{g}_1$, in general $\dfrac{\partial}{\partial g_1} \ne \dfrac{\partial}{\partial \tilde{g}_1}$, even though $dg_1 = d\tilde{g}_1$!

As for your second question, I'm not sure what $\dfrac{\partial}{\partial g} (f_1, \ldots, f_n)$ could be. I mean, it's the coefficients of $\dfrac{\partial}{\partial g}$ with respect to the basis $\dfrac{\partial}{\partial f_j}$, but interpreting it that way doesn't seem to yield anything useful.

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