The divergence operator as an element of $T_pM$ (if it does make any sense)

Question

First of all, I'd like to mention "in my defense" that I'm actually a graduate student in physics, but I'm strictly related to math in terms of research area, then naturally, I realized that (in most cases) physicists get to learn some math concepts in a very superficial way.

Now let's get to what interests here. I had some contact with the idea of a tangent space $T_pM$ on a point over one n-dim manifold $M$, I also saw that in terms of a set of local coordinates $x^i$, we represent an element ${\bf v}\in T_pM$ as $\sum_{i=1}^{n}v_i\frac{\partial}{\partial x^i}$, meaning that each symbol $\frac{\partial}{\partial x^i} \hspace{0,2cm} (i=1,2,\ldots, n)$ would represent something similar to those L.I. canonical vectors $e^1 = (1,0,\ldots,0), \ldots, e^n = (0, \ldots, 0, 1)$ in $\mathbb{R}^n$. In this last context, we'd say that an inner product between two vectors $v = (\alpha_1,\ldots,\alpha_n) = \sum_{i=1}^{n}\alpha_i e^{i}$ and $w = (\beta_1,\ldots,\beta_n) = \sum_{i=1}^{n}\beta_i e^{i}$ is numericaly correspondent to $\sum_{i=1}^{n}\alpha_i\beta_i$, not difficult to verify if we consider $\langle e^i, e^j\rangle = \delta_{ij}$. Now, by transposing this same ideia to a space where the vectors are represented as ${\bf v}$ above, could I say that $\nabla = \left(\frac{\partial}{\partial x^{1}},\ldots,\frac{\partial}{\partial x^{n}}\right)$ is analogous to something like $(1,\ldots, 1)$? I just suspect I can't exactly consider it, the reason is that this wondering came up after one specific paragraph of a paper which says (given a first order polynomial differential system):

"[...] where $\mathcal{X}$ is the vector field associated to the system (1), i.e. $$\mathcal{X} = P_1(x)\frac{\partial}{\partial x_1} + \ldots + P_n(x)\frac{\partial}{\partial x_n}.$$ As usually the divergence of the vector field is defined by $$\textrm{div}\mathcal{X} = \sum_{i=1}^{n}\frac{\partial P_i}{\partial x_i}."$$

(From Liouvillian Integrability versus Darboux Polynomials - J. Llibre, C. Valls, X. Zhang)

Well, considering those observatios I mentioned, wouldn't $\langle \nabla, \mathcal{X}\rangle = \sum_{i=1}^{n} P_i$? There might be something I'm not taking into consideration, or maybe I'm trying to see some relation between unrelated things.

PS.: I vaguely remember hearing about $\nabla$ not being exactly a vector, but I never got it confirmed by myself. I also apologize for any breaches of mathematical rigor.

Answer 1

Firstly, let's address $\nabla$, which is the source of a lot of confusion. In vector calculus, one often does hear the phrase that the gradient "isn't a vector", and differential geometry makes this quite apparent. When we are given a smooth manifold $M$, we consider functions $f:M\to\mathbb{R}$. There is only one way to naturally define a derivative on such functions, and it is denoted by $d$, and is called the exterior derivative, or differential. In fact, for any map between smooth manifolds $F:X\to Y$, we can consider the differential $dF:TM\to TN$, and now we just take $N=\mathbb{R}$.

So we get $df:TM\to T\mathbb{R}=\mathbb{R}^2$. At the pointwise level, this means $df_p:T_pM\to T_{f(p)}\mathbb{R}=\mathbb{R}$, and this is a linear map of vector spaces. In other words, $df$ is a $1$-form: it it a linear function which, when given a vector field, gives you a function. Namely, when $X$ is a vector field on $M$, we get $df(X)=Xf$. In local coordinates where $X=\sum X^i\partial_i$, this is just $Xf=\sum X^i\partial_if$.

The gradient of $f$ is not $df$. In fact, the gradient of a function on a manifold cannot be canonically associated with the function $f$. First, one has to choose a Riemannian metric. That is, a pointwise inner product on the tangent spaces which varies smoothly. Then, one can define the gradient $\nabla f$ by the property that $g(\nabla f,Y)=df(Y)$ for all vector fields $Y$. You recover the gradient of vector calculus by taking $M=\mathbb{R}^n$ and taking the standard Euclidean metric. On a general manifold, there is no canonical notion of a gradient. But once you have chosen a metric, as well as local coordinates, you can represent the metric as an $n\times n$ matrix of functions $(g_{ij})$ with inverse matrix $(g^{ij})$. Then the gradient of $f$ is expressed as $\nabla f=\sum g^{ij}\frac{\partial f}{\partial x^i}\frac{\partial}{\partial x^j}$. So that is what the gradient is.

Consequently, it is not true that you can write "$\nabla=(1,\dots,1)$", in the way that you would like. The basis vectors $\partial_i$ depend only on the choice of coordinates, whereas $\nabla$ depends on an additional choice, namely the metric $g$. The criterion under which $\nabla f=(\partial_1f,\dots,\partial_nf)=\sum \partial_if\frac{\partial}{\partial x^i}$ in local coordinates on a Riemannian manifold $(M,g)$, is precisely that the (inverse) metric $g$ in those coordinates satisfies $g^{ij}=\delta^{ij}$, i.e. that the coordinate patch $(U,g)$ is indistinguishable from flat Euclidean space.

I do not know what $\langle\nabla,\mathcal{X}\rangle$ is supposed to mean, without further context.

Answer 2

The reason that this construction doesn't work for abstract manifolds $M$ is that, unlike Euclidean space, $\mathbb R^n$, they typically do not come with preferred frames $(e_i)$/Riemannian metrics/connections (i.e., directional derivatives).

Here's how we adapt the construction to the smooth manifold setting:

Definition 1. Given a smooth manifold $M$ equipped with a connection $\nabla$, we can encode $\nabla$ in any coordinate frame $(\partial_{x^i})$ by the Christoffel symbols, i.e., the functions $\Gamma_{ij}^k$ characterized by $$\nabla_{\partial_{x^i}} \partial_{x^j} = \Gamma_{ij}^k \partial_{x^k} .$$ The Leibniz Rule for connections gives that for a vector field $Y = Y^j \partial_{x^j}$ we have \begin{multline} \nabla_{\partial_{x^i}} Y = \nabla_{\partial_{x^i}} (Y^j \partial_{x^j}) = \partial_{x^i}(Y^j) \partial_{x^j} + Y^j \nabla_{\partial_{x^i}} \partial_{x^j} \\ = \partial_{x^i}(Y^j) \delta^k{}_k\partial_{x^k} + Y^j \Gamma_{ij}^k \partial_{x^k} = \left(\partial_{x^i}(Y^j) \delta^k{}_k + Y^j \Gamma_{ij}^k\right) \partial_{x^k} \end{multline} Since $\nabla$ is tensorial in its derivative index (filled by $\partial_x^i$ in the previous computation), we can leave that index free and so view $\nabla Y$ as an endomorphism field, that is, a section of $\Gamma(TM \otimes T^*M)$, namely, $$\nabla Y = \left(\partial_{x^i}(Y^j) \delta^k{}_k + Y^j \Gamma_{ij}^k\right) \partial_{x^k} \otimes dx^i .$$ Using the natural pairing $T_p M \otimes T_p^* M \to \mathbb R$, we can contract the indices (just as well, contract with the identity endomorphism $\delta^k{}_i$) to yield a function, $$\boxed{\operatorname{div} Y := \operatorname{tr}\nabla Y = \frac{\partial Y^i}{\partial x^i} + Y^j \Gamma_{ij}^i},$$ and we declare this function to be the divergence of $Y$ (with respect to $\nabla$). Evidently, this function depends on the Christoffel symbols, hence on the connection $\nabla$. In particular, unless the Christoffel symbols are all zero, the divergence depends not just on the coordinate derivatives of the components of $Y$ but on those components themselves. In this sense, we cannot expect that in general the divergence "act like an element of $T_p M$" in any reasonable sense.

Example For Euclidean space and the standard coordinate frame $(\partial_{x^i})$, we have the usual formula $$\nabla_{\partial_{x^i}} Y = \partial_{x^i} (Y^j) \partial_{x^j}$$ familiar from vector calculus, so the Christoffel symbols are all zero, and we recover the usual formula for the divergence, $\operatorname{div} Y := \frac{\partial Y^i}{\partial x^i}$, for that setting.

Definition 2. Alternatively, given an $n$-manifold $M$ equipped with a volume (or density) form $\operatorname{vol}$, we can define the divergence as follows. For any vector field $Y$, the Lie derivative $\mathcal{L}_Y \operatorname{vol}$ is another $n$-form, hence some multiple of $\operatorname{vol}$. We define the divergence of $Y$ to be this scale factor: $$\boxed{\mathcal{L}_Y \operatorname{vol} = (\operatorname{div} Y) \operatorname{vol}} .$$ This definition makes more apparent the property that divergence measures the growth or contraction of volume under the flow of a vector field. In particular, the flow of $Y$ preserves the volume form---that is, $Y$ is incompressible iff $\operatorname{div} Y = 0$.

In the common case that a connection $\nabla$ preserves a volume form or density $\operatorname{vol}$ (e.g., $\nabla$ is the Levi-Civita connection of a Riemannian metric and $\operatorname{vol}$ is its volume form or density), the two above notions coincide.

Definition 3. Following up on Ted Shifrin's helpful comment: Cartan's Magic Formula lets us rewrite the Lie derivative in the previous definition as $$\mathcal{L}_Y \operatorname{vol} = d(\iota_Y \operatorname{vol}) + \iota_Y d\operatorname{vol} = d(\iota_Y \operatorname{vol}),$$ where the second term in the second equality vanishes because $d\operatorname{vol}$ is an $(n + 1)$-form on an $n$-manifold. If $g$ is a (say, oriented) Riemannian metric $g$ with volume form $\operatorname{vol}$, we get for free Hodge star operators, which are certain bundle isomorphisms $\ast : \bigwedge^k T^*M \to \bigwedge^{n - k} T^*M$. For $k = 1$ this operator has the property that $\iota_Y \operatorname{vol} = \ast (Y^\flat)$, where $Y^\flat \in \Gamma(T^*M)$ is the $1$-form characterized by $Y^\flat(Z) = g(Y, Z)$. For $k = n$ we have $\ast (f \operatorname{vol}) = f$. Putting these facts together with the previous definition gives the formula $$\boxed{\operatorname{div} Y = \ast\, d \ast (Y^\flat)} ,$$ which makes apparent that, up to composition with canonical isomorphisms determined by $g$ (and its orientation), the divergence operator is just the last nontrivial exterior derivative, $d : \Gamma\left(\bigwedge^{n - 1} T^*M\right) \to \Gamma\left(\bigwedge^{n} T^*M\right)$.