At its coarsest level, spacetime is just a set. In other words, spacetime is just a collection of points, known as the elements of a set.1 However, this definition is not enough to talk about even the simplest notions we discuss in classical physics, namely continuity of maps. It is important that we are able to talk about and require continuity of maps as the motion of a particle is given by a map, and it is clear we want this map to be continuous. That is, we don’t want the position of a particle to all of a sudden ‘jump’ from one point to another:
So we need to introduce some new structure onto our set that allows us to talk about continuity. There are lots of things we could introduce in order to do this, however we need to be careful; we do not want to start adding additional properties to our set that will later come back to bite us. We therefore want to use the weakest structure we can. So what is it? Luckily the answer to this question is already known: it is a so-called topology.
The powerset \(\mathcal{P}(S)\) of a set \(S\) is the set of all subsets of \(S\).
Let \(\mathcal{M}\) be a set. A topology \(\mathcal{O}\) on \(\mathcal{M}\) is a subset \(\mathcal{O} \subseteq \mathcal{P}(\mathcal{M})\) satisfying:
\(\emptyset \in \mathcal{O}\) and \(\mathcal{M} \in \mathcal{O}\).
Given \(U,V\in\mathcal{O}\) then \(U\cap V\in\mathcal{O}\).
Let \(A\) be an arbitrary index set. Given \(U_{\alpha}\in\mathcal{O}\) then \(\bigcup_{\alpha\in A} U_{\alpha} \in \mathcal{O}\).
Remark 1.1.1. Conditions (ii) and (iii) look deceptively similar, but there is an important difference. Condition (ii) says that a finite intersection of elements is still in \(\mathcal{O}\), whereas (iii) says that an arbitrary union of elements is in \(\mathcal{O}\).
Example 1.1.2. Let \(\mathcal{M} = \{1,2,3\}\). Then we could choose to define \[\begin{split} \mathcal{O}_1 & := \big\{ \emptyset, \{1,2,3\} \big\} \\ \mathcal{O}_2 & := \big\{ \emptyset, \{1\}, \{2\}, \{1,2,3\}\big\} \\ \mathcal{O}_3 & := \big\{\emptyset, \{1\},\{2\},\{3\}, \{1,2\}, \{1,3\}, \{2,3\}, \{1,2,3\} \big\}. \end{split}\] The question is, which are topologies on \(\mathcal{M}\)? A quick check shows that \(\mathcal{O}_1\) and \(\mathcal{O}_3\) are, but \(\mathcal{O}_2\) is not as \(\{1\}\cup\{2\} = \{1,2\}\notin\mathcal{O}_2\).
So we see, given the set \(cM\) calculating whether something is a topology is rather easy (although perhaps a bit boring). What is we don’t know the form of \(\mathcal{M}\), can we still define specific topologies? The answer is yes and the example shows us what.
Let \(\mathcal{M}\) be a set. The chaotic topology on \(\mathcal{M}\) is defined as \[\mathcal{O}_{\text{chaotic}} := \{\emptyset,\mathcal{M}\}.\]
Let \(\mathcal{M}\) be a set. The discrete topology on \(\mathcal{M}\) is defined as \[\mathcal{O}_{\text{discrete}} := \mathcal{P}(\mathcal{M}).\]
So in Example 1.1.2, \(\mathcal{O}_1\) is the chaotic topology and \(\mathcal{O}_3\) the discrete topology. However, both the chaotic and discrete topology are utterly useless, they are simply the extreme cases of topologies with the least amount and most amount of elements, respectively. However, on \(\mathbb{R}^d = \{(p_1,...,p_d)| p_i\in\mathbb{R}\}\) there is a very important topology which we shall use throughout these notes.
Let \(\mathcal{M} = \mathbb{R}^d\). The standard topology on \(\mathcal{M}\) is defined as \[\mathcal{O}_s := \{ U \in \mathcal{P}(\mathbb{R}^d) \, | \, \forall p\in U \, \exists r\in\mathbb{R}^+ : B_r(p) \subseteq U \},\] where \[B_r(p) := \bigg\{ (q_1,...,q_d) \in \mathbb{R}^d \, \bigg| \, \sum_{i=1}^d (q_i-p_i)^2 < r^2\bigg \}\] is called a soft-ball of radius \(r\) about \(p\), also known as the neighbourhood of \(p\) with radius \(r\).
Show that the standard topology is in fact a topology, i.e. show it meets the conditions (i),(ii) and (iii).
Remark 1.1.3. To those familiar with vector space structures and normed spaces, you might be tempted to say ‘Ah the soft ball is just the Euclidean norm’. However, the definition above does not need a full vector space structure (which a norm does) in order for it to hold. All we require is that we know what \((q_i-p_i)^2\) means.
Let \(\mathcal{M}\) be a set and \(\mathcal{O}\) be a topology on \(\mathcal{M}\). We call the double \((\mathcal{M},\mathcal{O})\) a topological space.
Remark 1.1.4. In these notes if we talk about \(\mathbb{R}^d\) being a topological space (for example when saying a map is continuous, see below), if no specific topology is given it will be assumed that we equip it with the standard topology.
Remark 1.1.5. In this lecture course we might not always write down the topology and simply call \(\mathcal{M}\) a topological manifold. Obviously if we say that, there is an invisible \(\mathcal{O}\) kicking about, we just want to save some typing. However, we shall try to always be explicit in order to avoid confusion.
Intuitively we can think of the standard topology on \(\mathbb{R}^d\) as shapes that don’t include their boundaries and the soft ball as a circle2 of radius \(r\) that doesn’t contain the boundary. With this intuition we easily see the extension of the standard topology to general topologies: they are the sets of open sets within that set. In fact it actually works the other way around, we use a topology in order to define what we mean by an open set.
Let \((\mathcal{M},\mathcal{O})\) be a topological space. We call \(U\subseteq\mathcal{M}\) an open set if and only if \(U\in \mathcal{O}\).
Let \((\mathcal{M},\mathcal{O})\) be a topological space. We call a set \(V\subseteq\mathcal{M}\) a closed set is and only if \(\mathcal{M}\setminus V \in \mathcal{O}\), where \(\mathcal{M}\setminus V\) is known as the compliment of \(V\).
Remark 1.1.6. It is tempting to think that a closed set is simply a set that is not open. However, this is not true. In fact a set can be
Open and not closed, e.g. \((0,1)\) in \((\mathbb{R},\mathcal{O}_s)\),
Not open and closed, e.g. \([0,1]\) in \((\mathbb{R},\mathcal{O}_s)\),
Open and closed, e.g. \(\emptyset\) in any topological space,
Not open and not closed, e.g. \([0,1)\) in \((\mathbb{R},\mathcal{O}_s)\).
Show that the examples given in the above remark are correct.
Let’s first just recall the terminology/notation for a map. We say that \(f\) is a map from a set \(\mathcal{M}\), known as the domain, to another set \(\mathcal{N}\), known as the target, and we write this as \(f:\mathcal{M}\to\mathcal{N}\). We say that the element \(m\in\mathcal{M}\) is mapped to \(n\in\mathcal{N}\), which we write as \(f:m\mapsto n\). A map will take every element in its domain to some element in its target. It is possible that a two different elements of the domain are mapped to the same element of the target and it is not required that every element in the target is hit. Based on this we have the following definitions.
A map \(f:\mathcal{M}\to\mathcal{N}\) is said to be injective if it is one-to-one. That is \[f(m_1) = f(m_2) \iff m_1 = m_2 \qquad \forall m_1,m_2\in\mathcal{M}.\]
A map \(f:\mathcal{M}\to\mathcal{N}\) is said to be surjective if every element of the target is hit. That is \[\forall n\in\mathcal{N} \quad \exists m\in\mathcal{M} : f(m) = n.\]
A map \(f:\mathcal{M}\to\mathcal{N}\) is called bijective if it is both injective and surjective.
The answer to whether a map \(f:\mathcal{M}\to\mathcal{N}\) is continuous depends, by definition, on the choice of which topologies are chosen on the sets \(\mathcal{M}\) and \(\mathcal{N}\).
Let \((\mathcal{M},\mathcal{O}_{\mathcal{M}})\) and \((\mathcal{N},\mathcal{O}_{\mathcal{N}})\) be topological spaces. A map \(f:\mathcal{M}\to\mathcal{N}\) is called continuous with respect to \(\mathcal{O}_{\mathcal{M}}\) and \(\mathcal{O}_{\mathcal{N}}\) if and only if \[\forall V\in\mathcal{O}_{\mathcal{N}}, \qquad \text{preim}_{f}(V) \in\mathcal{O}_{\mathcal{M}},\] where \[\text{preim}_f(V) := \{ m\in\mathcal{M} \, | \, f(m) \in V\}.\] That is "the preimages of open sets in \(\mathcal{N}\) are open in \(\mathcal{M}\)".
Remark 1.2.1. Note the preimage of a map \(f\) is not the same thing as its inverse. For example, we can not define an inverse for a non-injective map; if two elements in the domain map to the same element in the target, there is no clear way to decide which element you get under the inverse map. However the preimage in the case is the collection of both points. So continuity does not require the map to be injective. Equally note that surjectivity is not required as any element that isn’t hit has preimage \(\emptyset\in\mathcal{O}_{\mathcal{M}}\).
Remark 1.2.2. Note is we choose the topology on \(\mathcal{M}\) to be the discrete topology then every map \(f:\mathcal{M}\to\mathcal{N}\) is continuous. This is easily seen because the preimage of any set in \(\mathcal{O}_{\mathcal{N}}\) is either a subset of \(\mathcal{M}\) or the empty set, both of which are in the discrete topology on \(\mathcal{M}\).
Let \(f:\mathcal{M}\to\mathcal{N}\) be a bijective map. Then the map is said to be a homeomorphism if both \(f\) and its inverse \(f^{-1}\) are continuous. They are the structure-preserving3 maps of topology.
Example 1.2.3. Let \(\mathcal{M}=\mathcal{N} = \{1,2\}\) and let \(f\mathcal{M}\to\mathcal{N}\) be given by \[f(1)=2 \qquad \text{and} \qquad f(2)=1.\] Now define \[\mathcal{O}_{\mathcal{M}} := \big\{ \emptyset, \{1\}, \{2\}, \{1,2\}\big\}, \qquad \text{and} \qquad \mathcal{O}_{\mathcal{N}} := \big\{ \emptyset, \{1,2\}\big\}.\]
To check whether \(f\) is continuous w.r.t. \(\mathcal{O}_{\mathcal{M}}\) and \(\mathcal{O}_{\mathcal{N}}\) we need to check the preimages of open sets in \(\mathcal{N}\) are open sets in \(\mathcal{M}\). \[\begin{split} \text{preim}_f(\emptyset) & = \emptyset \in \mathcal{O}_{\mathcal{M}}, \\ \text{preim}_f\big(\{1,2\}\big) & = \mathcal{M} \in \mathcal{O}_{\mathcal{M}}, \end{split}\] and so \(f\) is continuous.
How about the inverse map \(f^{-1}:\mathcal{N}\to\mathcal{M}\). Well it satisfies \[f^{-1}(1) = 2 \qquad \text{and} \qquad f^{-1}(2) = 1,\] however we have \[\text{preim}_{f^{-1}}(\{1\}) = \{2\}, \qquad \text{and} \qquad \text{preim}_{f^{-1}} (\{2\}) = \{1\},\] neither of which are in \(\mathcal{O}_N\), and so this map is not continuous.
Given two maps \(f:\mathcal{M}\to\mathcal{N}\) and \(g:\mathcal{N}\to\mathcal{P}\), we can define their composition as a new map \[g\circ f : \mathcal{M} \to \mathcal{P}\] where \((g\circ f)(m) := g\big(f(m)\big)\).
Let \(f:\mathcal{M}\to\mathcal{N}\) and \(g:\mathcal{N}\to\mathcal{P}\) be two continuous maps w.r.t. the relevant topologies. Then the composition map \(g\circ f : \mathcal{M}\to\mathcal{P}\) is also continuous.
Proof. Let \(V\in\mathcal{O}_{\mathcal{P}}\). Now we have \[\begin{split} \text{preim}_{g\circ f}(V) & := \{ m\in \mathcal{M} \, | \, (g\circ f)(m) \in V\} \\ & = \{ m \in \mathcal{M} \, | \, f(m) \in \text{preim}_g(V) \in \mathcal{O}_{\mathcal{N}}\} \\ & = \text{preim}_{f}\big(\text{preim}_g(v)\big) \in \mathcal{O}_{\mathcal{M}}, \end{split}\] where the second line follows from the continuity of \(g\) and the last line from the continuity of \(f\). ◻
Show that Theorem 1.3.1 extends to the composition of an arbitrary number of continuous maps.
So we know how to define a topology on a set \(\mathcal{M}\). We now want to ask the question: given some other set \(S\) is it possible to use the topology on \(\mathcal{M}\) to define one on \(S\)? The answer is obviously yes4 and it is known as the inherited topology. How you do this obviously depends on the situation. The way that is important for spacetime physics is the following.
Let \((\mathcal{M},\mathcal{O}_{\mathcal{M}})\) be a topological space and let \(S\subset \mathcal{M}\). We define the subset topology to be \[\mathcal{O}|_S := \{ U \cap S \, | \, U\in\mathcal{O}_{\mathcal{M}}\}.\]
Proof. We want to show that the subset topology is indeed a topology, i.e. need to check it meets the three conditions.
We have \(\emptyset = \emptyset\cap S\) and \(\emptyset\in\mathcal{O}_{\mathcal{M}}\) which tells us that \(\emptyset\in\mathcal{O}|_S\). Equally we have \(S = \mathcal{M}\cap S\) and \(\mathcal{M}\in\mathcal{O}_{\mathcal{M}}\) and so \(S \in \mathcal{O}|_S\).
Let \(A,B\in \mathcal{O}|_S\), then we know there exists \(\widetilde{A},\widetilde{B}\in\mathcal{O}_{\mathcal{M}}\) such that \(A=\widetilde{A}\cap S\) and \(B=\widetilde{B}\cap S\). From which we have \(A\cap B = (\widetilde{A}\cap S)\cap(\widetilde{B}\cap S) = (\widetilde{A}\cap\widetilde{B})\cap S\). Finally using \(\widetilde{A}\cap\widetilde{B}\in\mathcal{O}_{\mathcal{M}}\) we have \(A\cap B\in \mathcal{O}|_S\).
Let \(U_{\alpha}\in\mathcal{O}|_S\), which tells us that there exists \(\widetilde{U}_{\alpha}\in\mathcal{O}_{\mathcal{M}}\) such that \(U_{\alpha} = \widetilde{U}_{\alpha}\cap S\). As above this tells us that \(\bigcup_{\alpha\in A} U_{\alpha} = \big(\bigcup_{\alpha\in A}\widetilde{U}_{\alpha} \big)\cap S\) where \(A\) is a arbitrary index set. Finally using \(\bigcup_{\alpha\in A}\widetilde{U}_{\alpha}\in\mathcal{O}_{\mathcal{M}}\) we get \(\bigcup_{\alpha\in A} U_{\alpha}\in\mathcal{O}|_S\).
◻
We might wonder why on earth we would choose to define such a map, the answer is to do with the continuity of maps.
Claim 1.4.1. Suppose we have some continuous map \(f:\mathcal{M}\to \mathcal{N}\) between topological spaces \((\mathcal{M},\mathcal{O}_{\mathcal{M}})\) and \((\mathcal{N},\mathcal{O}_{\mathcal{N}})\). Then if we have some subset \(S\subset \mathcal{M}\) which we turn into a topological space with the subset topology \((S,\mathcal{O}|_S)\), then we are guaranteed that the restricted map \(f|_S : S\to \mathcal{N}\) is also continuous w.r.t. \(\mathcal{O}|_S\) and \(\mathcal{O}_{\mathcal{N}}\).
Prove Claim 1.4.1.