Fields

So far we have discussed a single tangent space and vectors lying in it. What we now want to study are vector fields, which is essentially a vector for every point on the manifold. We need to give a proper technical way to introduce vector fields, as simply saying ‘imagine a vector at every point’ isn’t good enough (two people might imagine differently). The answer to doing this is known as the theory of bundles.

6.1 Bundles, Fibres and Sections

As we shall see shortly, sections are the fields over our manifolds. The rough idea is that we make the fibres the tangent spaces to each point and then by taking a section, we pick one vector from each tangent space, giving us a vector field.

6.2 Tangent Bundle of Smooth Manifold

Let \((\mathcal{M},\mathcal{O},\mathcal{A})\) be a smooth manifold. We define the tangent bundle as the bundle whose base space is our smooth manifold and whose total space has the set² \[T\mathcal{M} := \bigcup^{\bullet}_{p\in\mathcal{M}} T_p\mathcal{M},\] where the dot means ‘disjoint union’. The projection map is given by \[\pi : X \mapsto p,\] where \(p\) is the unique point such that \(X\in T_p\mathcal{M}\).

We need to show how to turn the above set \(T\mathcal{M}\) into a smooth manifold (as we need for a bundle), but first two quick comments: it is important that we take the disjoint union above as this allows us to identify each vector with its base point \(p\). It is because we take the disjoint union that we can say the unique point; and the projection is surjective as we took the union over all \(p\in\mathcal{M}\) and so we hit every element in \(\mathcal{M}\).

We now need to make \(T\mathcal{M}\) into a smooth manifold, in such a way that \(\pi:T\mathcal{M} \to \mathcal{M}\) is a smooth map. So we need to define a topology on \(T\mathcal{M}\), the question is ‘how do we do this?’ With a little thought the answer becomes clear: we already have a topology on our base space and if we are going to require \(\pi\) to be smooth, why don’t we just use the coarsest³ topology on \(T\mathcal{M}\) such that \(\pi\) is continuous (as continuity is needed for smoothness). This topology is known as the initial topology w.r.t. \(\pi\). It is defined simply as⁴ \[\mathcal{O}_{T\mathcal{M}} := \{ \text{preim}_{\pi}(U) \, | \, U\in\mathcal{O}\}\] So far we have a topological manifold \((T\mathcal{M},\mathcal{O}_{T\mathcal{M}})\) and a continuous map \(\pi\) to a smooth manifold \((\mathcal{M},\mathcal{O},\mathcal{A})\). We now need to define a \(C^{\infty}\)-atlas for \((T\mathcal{M},\mathcal{O}_{T\mathcal{M}})\) in such a way that our map becomes smooth. As with the topology, we are going to construct this atlas from \(\mathcal{A}\).

The question is ‘how?’ Well we know that \(X\in T\mathcal{M}\) is described by two pieces of information: it is a vector and it has a base point. We can easily obtain the coordinates of the base point by simply projecting \(X\) down using \(\pi\) and then using our atlas on \(\mathcal{M}\) to find its coordinates. What about the vector part? Well, we have a chart on \(\mathcal{M}\) and so we can induce a chart on the tangent space and decompose \(X\) as \[X =: X^i_{(x)}\frac{\partial}{\partial x^i}.\] It is the components \(X^i_{(x)}\) that we want to use, but we want to get them by just using the chart \((U,x)\). The answer is very straight forward: just consider the gradient of the chart maps. That is \[X^i_{(x)} = (dx^i)_{\pi(X)}(X).\] So we construct the atlas \[\mathcal{A}_{T\mathcal{M}} := \{ (TU,\xi_x) \, | \, (U,x) \in\mathcal{A}\},\] where \[\xi_x : TU \to \mathbb{R}^{2\cdot\dim\mathcal{M}},\] given by \[\xi_x(X) = \big( \underbrace{(x^1\circ\pi)(X), ... , (x^d\circ \pi)(X)}_{(U,x)\text{-coordinate of }\pi(X)}, \underbrace{(dx^1)_{\pi(X)}(X), ... , (dx^d)_{\pi(X)}(X)}_{\text{Vector components w.r.t. } (U,x)}\big)\] We also need the inverse map: \[\xi_x^{-1} : \mathbb{R}^{2\cdot\dim\mathcal{M}} \to TU.\] With a bit of thought it is clear that it must satisfy \[\xi_x^{-1}(\alpha^1,...,\alpha^d,\beta_1,...,\beta^d) := \beta^i \bigg(\frac{\partial}{\partial x^i}\bigg)_{x^{-1}(\alpha^1,...,\alpha^d)} = \beta^i \bigg(\frac{\partial}{\partial x^i}\bigg)_{\pi(X)}.\] Now we need to check that these maps are smooth (as we need a smooth atlas). Consider another chart \((V,y)\) with \(V\cap U\neq\emptyset\), we have⁵ \[\begin{split} \big(\xi_y\circ \xi_x^{-1}\big) (\alpha^1,...,\alpha^d,\beta^1,...,\beta^d) & := \xi_y \bigg( \beta^m \bigg(\frac{\partial}{\partial x^m}\bigg)_{\pi(X)}\bigg) \\ & = \Bigg(..., (y^i\circ \pi) \bigg(\beta^i \bigg(\frac{\partial}{\partial x^i}\bigg)_{\pi(X)}\bigg),..., ..., (dy^i)_{\pi(X)}\bigg[\bigg(\beta^m \bigg(\frac{\partial}{\partial x^m}\bigg)_{\pi(X)}\bigg)\bigg], ...\Bigg) \\ & = \Bigg(..., (y^i\circ x^{-1})(\alpha^1,...,\alpha^d), ..., ..., \beta^m \bigg(\frac{\partial y^i}{\partial x^m}\bigg)_{\pi(X)}, ...\Bigg) \\ & = \Big(..., (y^i\circ x^{-1})(\alpha^1,...,\alpha^d), ..., ..., \beta^m \partial_m \big(y^i\circ x^{-1}\big)\big|_{(\alpha^1,...,\alpha^d)}, ... \Big), \end{split}\] where to go to the third line we have used the fact that \(\pi(X) = p = x^{-1}(\alpha^1,...,\alpha^d)\) and to get to the last line we have used the definition for the derivative fraction along with \((x\circ\pi)(X) = x(p) = (\alpha^1,...,\alpha^d)\). Now \((y^i\circ x^{-1})\) is smooth because \(\mathcal{A}\) is smooth and so the above result is smooth. We therefore have a smooth atlas \(\mathcal{A}_{T\mathcal{M}}\).

This all seems rather abstract and complicated, but the following example shows it’s actually rather natural and intuitive.

6.3 Vector Fields

We just put in a lot of work to check/prove that the set \(T\mathcal{M}\) can be made into a smooth manifold and so we have a bundle. It is reasonable to wonder why we did such a crazy calculation. The answer is that it allows the next definition.

The smooth part of the above definition is what all the work was for. Intuitively when we think of a vector field (a vector at each point) we see a smooth vector field, i.e. one where the vectors appear to naturally flow from one to another, rather then just pointing randomly at each point. Smooth vector fields will obviously play a vital rule in general relativity: the velocity of a particle is a smooth vector field over the manifold that is the worldline of the particle. If this vector field was not smooth, it would correspond to the particle’s velocity all of a sudden changing, which we know is not physical.

6.4 The \(C^{\infty}(\mathcal{M})\)-Module

So far we have a definition for a smooth vector field, but we have no way of adding them together or scaling them in any way. This is something we clearly want to be able to do, and so we want to try and make it into a vector space over some field. The addition is straight forward, just add the vectors in the tangent spaces together and take the result to be the new vector at that point. What about scaling? We don’t want to limit ourselves to only being able to scale the smooth vector field uniformly, i.e. by the same amount at all \(p\in\mathcal{M}\). So we need something that is defined all over \(\mathcal{M}\) but that can take different values at each point. This is just a scalar field.

So we want to try and turn the set of smooth sections over the tangent bundle into a \(C^{\infty}(\mathcal{M})\)-vector space. There is a problem, though. Recall the definition \[C^{\infty}(\mathcal{M}) := \{f:\mathcal{M}\to\mathbb{R} \, | \, f \text{ is a smooth function}\}.\] It is possible that a non-vanishing⁷ element of \(C^{\infty}(\mathcal{M})\) can vanish at some points, i.e. there are points \(p\in\mathcal{M}\) that are mapped to \(0\in\mathbb{R}\). We cannot turn \(C^{\infty}(\mathcal{M})\) into a field, then, as we don’t have an inverse under multiplication for every element (we can’t invert the points that vanish!). The best we can do, then, is to turn it into a ring. We clearly have a neutral element – the elements that just maps all points \(p\in\mathcal{M}\) to \(1\in\mathbb{R}\) – and we can define the commutativity pointwise, using the fact that \((\mathbb{R},+)\) is commutative. We therefore get a commutative, unital ring. If we build on top of this, we get a module.

This is the first point where we have introduced a new notation for the action of a field. Recall we have been denoting the action of a vector (at a point) on a \(C^{\infty}(\mathcal{M})\) function via standard brackets, \(X(f)\). In order to distinguish this from the action of a smooth vector field on \(f\), we use angled brackets for the latter \(\chi\langle f\rangle\).⁸ This might seem like a just a notational problem, however it actually encapsulates an important point: a smooth vector field is a map \(\chi:\mathcal{M}\to T\mathcal{M}\), so how does it act on a scalar field? The answer is obviously through the vectors that make up \(\chi\): \[\chi\langle f\rangle\big|_{p} := \big(\chi(p)\big)(f).\] That is, we first evaluate \(\chi(p)\), which gives us the \(X\in T_p\mathcal{M}\), and then we let this act on the scalar field, giving a real number. We do this for every point \(p\in\mathcal{M}\) and so get a map that associates to each point a real number, this is a scalar field. In other words we can think of smooth vector fields as maps \[\chi : C^{\infty}(\mathcal{M}) \xrightarrow{\sim} C^{\infty}(\mathcal{M}).\] It is because of this that we take the addition/multiplication on the right-hand side of the expressions in the definition to be those defined on \(C^{\infty}(\mathcal{M})\).

Now there is an important fact in set theory¹⁰ that every vector space has a basis. However, this incredibly useful fact does not apply to modules. That is, in general, we cannot simply take the subset \[\{\chi_1,...,\chi_d\} \subseteq \Gamma T\mathcal{M}\] such that any other \(\chi\in\Gamma T\mathcal{M}\) can be expressed as a linear combination of this subset \[\chi = f^i \odot \chi_i.\] Of course we can do this locally by simply decomposing our vector fields locally, but we cannot do it globally.

This failure to define a global, nowhere vanishing, smooth vector field is related to the fact that we can’t chart the space using only a single chart. That is, the minimal atlas for a sphere contains two charts. For example, if we used polar spherical coordinates on the surface to chart the sphere, the North and South poles will not be charted – whats the longitude value at these points?

We can repeat everything we did in order to define the tangent bundle but instead starting from the cotangent space \((T^*\mathcal{M},+,\cdot)\). In doing this we get the cotangent bundle and smooth covector fields. Finally we get the \(C^{\infty}(\mathcal{M})\)-module \((\Gamma T^*\mathcal{M},\oplus,\odot)\) where \[\Gamma T^*\mathcal{M} := \{ \alpha : \mathcal{M} \to T^*\mathcal{M} \, | \, \text{smooth section}\}.\]

6.5 Tensor Fields

We have the smooth vector fields and the smooth covector fields. We can, therefore, now construct smooth tensor fields.

The two definitions above don’t seem to quite match, though. We have seen that \(\alpha\in\Gamma T^*\mathcal{M}\) can map a \(X\in\Gamma T\mathcal{M}\) to a \(C^{\infty}(\mathcal{M})\) function, but the opposite isn’t true — vector fields map scalar fields to scalar fields — so how do does the tensor product definition work? The answer is simply that we interpret it as the following example highlights.