Problem: What does it mean for a topological space \(X\) to be locally homeomorphic to a topological space \(Y\)? Hence, what does it mean for a topological space \(X\) to be locally Euclidean?
Solution: \(X\) is said to be locally homeomorphic to \(Y\) iff every point \(x\in X\) has a neighbourhood homeomorphic to an open subset of \(Y\). In particular, \(X\) is said to be locally Euclidean iff it is locally homeomorphic to Euclidean space \(Y=\mathbf R^n\) for some \(n\in\mathbf N\) called the dimension of \(X\).
Explicitly, this means every point \(x\in X\) is associated to at least one neighbourhood \(U_x\) along with at least one homeomorphism \(\phi_x:U_x\to\phi_x(U_x)\subset\mathbf R^n\); the ordered pair \((U_x,\phi_x)\) is an example of a chart on \(X\), and any collection of charts covering \(X\) (such as \(\{(U_x,\phi_x):x\in X\}\)) is called an atlas for \(X\). Hence, \(X\) is locally Euclidean iff there exists an atlas for \(X\).
(caution: there is a distinct concept of a local homeomorphism from \(X\to Y\); existence of such implies that \(X\) is locally homeomorphic to \(Y\), but the converse is false).
Problem: Give an example of topological spaces \(X\) and \(Y\) such that \(X\) is locally homeomorphic to \(Y\) but not vice versa.
Solution: \(X=(0,1)\) and \(Y=[0,1)\). Thus, be wary that local homeomorphicity is not a symmetric relation of topological spaces.
Problem: Let \(X\) be an \(n\)-dimensional locally Euclidean topological space. What additional topological constraints are typically placed on \(X\) in order for it to qualify as a topological \(n\)-manifold?
Solution:
- \(X\) is Hausdorff.
- \(X\) is second countable.
That being said, conceptually the most important criterion to remember is the local Euclideanity of \(X\). Topological manifolds are the most basic/general/fundamental of manifolds, onto which additional structures may be attached.
Problem: Let \(X\) be a topological \(n\)-manifold. What additional piece of structure should be attached to \(X\) so as to consider it a \(C^k\)-differentiable \(n\)-manifold?
Solution: The additional piece of structure is a \(C^k\)-atlas. A \(C^k\)-atlas for \(X\) is just an atlas for \(X\) with the property that all overlapping charts are compatible with each other. This means that if \((U_1,\phi_1),(U_2,\phi_2)\) are \(2\) charts of a \(C^k\)-atlas for \(X\) such that \(U_1\cap U_2\neq\emptyset\), then the transition map \(\phi_2\circ\phi_1^{-1}:\phi_1(U_1\cap U_2)\subset\mathbf R^n\to\phi_2(U_1\cap U_2)\subset\mathbf R^n\) is \(C^k\)-differentiable (equivalent to \(\phi_1\circ\phi_2^{-1}\) being \(C^k\)-differentiable).
An important point to mention is that a given topological \(n\)-manifold may have multiple different \(C^k\)-atlases. According to the above definition, that would seem to give rise to \(2\) distinct \(C^k\)-manifolds…but intuitively that would be undesirable; they should be viewed as really the same \(C^k\) manifold. This motivates imposing an equivalence relation on \(C^k\)-manifolds by asserting that two \(C^k\)-manifolds are equivalent iff their corresponding \(C^k\)-atlases are compatible…which just means that the union of their \(C^k\)-atlases is itself a \(C^k\)-atlas, or equivalently every chart in one \(C^k\)-atlas is compatible with every chart in the other \(C^k\)-atlas. This is sometimes formalized by introducing a so-called “maximal \(C^k\)-atlas”, but the basic idea should be clear.
Finally, note that topological manifolds are just \(C^0\)-manifolds because homeomorphisms are continuous. In physics, the most common case is to deal with \(C^{\infty}\)-manifolds like spacetime in GR or classical configuration/phase spaces in CM, or state space in thermodynamics, also called smooth manifolds.
Problem: Let \(S^1\) be the circle, a \(1\)-dimensional smooth manifold which is regarded as being embedded in \(\mathbf R^2\) via the set of points \(x^2+y^2=1\). Explain why \(S^1\) is locally (but not globally) homeomorphic to the real line \(\mathbf R\). Furthermore, explain whether or not the pair \((S^1,\theta)\) (where \(\theta:(\cos\theta,\sin\theta)\in S^1\mapsto\theta\in[0,2\pi)\) is the obvious angular coordinate mapping) is a chart on \(S^1\) or not.
Solution: \(S^1\) is locally homeomorphic to \(\mathbf R\) because, roughly speaking, every little strip of open arc in \(S^1\), upon zooming in, looks like a strip of straight line in \(\mathbf R\). However, \(S^1\) is not globally homeomorphic to \(\mathbf R\) because \(S^1\) is compact whereas \(\mathbf R\) is unbounded. The pair \((S^1,\theta)\) is not a chart on \(S^1\) because its image \([0,2\pi)\) isn’t open in \(\mathbf R\).
Problem: Hence, demonstrate an example of an atlas for \(S^1\).
Solution: To circumvent the problem above, one requires \(2\) charts to cover \(S^1\), thereby forming an atlas for \(S^1\). For example, a chart \(\theta_1:S^1-\{(1,0)\}\to (0,2\pi)\) similar to the above but with the point \((1,0)\) removed, and a second chart \(\theta_2:S^1-\{(-1,0)\}\to (-\pi,\pi)\) with the antipodal point \((-1,0)\) removed. Then the domain of these \(2\) charts overlap on the upper and lower semicircles respectively, so the transition function on this overlapping domain is:
\[\theta_2(\theta_1^{-1}(\theta))=\theta[\theta\in(0,\pi)]+(\theta-2\pi)[\theta\in(\pi,2\pi)]\]
which is indeed \(C^{\infty}\).
Problem: Define the real associative algebra \(C^k(X\to\mathbf R)\).
Solution: This is simply the space of all \(C^k\)-differentiable scalar fields on the \(C^k\)-manifold \(X\). To be precise, \(f\in C^k(X\to\mathbf R)\) iff for all charts \((U,\phi)\) in the \(C^k\)-atlas defining the differential structure of \(X\), the map \(f\circ\phi^{-1}:\phi(U)\to\mathbf R\) is \(C^k\)-differentiable.
Problem: What is a tangent vector to a \(C^1\)-manifold \(X\) at a point \(x\in X\)?
Solution: In the algebraic formulation, a tangent vector is simply any \(\mathbf R\)-derivation on \(C^1(X\to\mathbf R)\) at a specific point \(x\in X\) on the manifold. Concretely, this means that a tangent vector at \(x\in X\) is any real-valued function \(\partial_{\mu}|_x:C^1(X\to\mathbf R)\to\mathbf R\) which is linear and satisfies the product rule:
\[\partial_{\mu}|_x(\alpha f+\beta g)=\alpha\partial_{\mu}|_x(f)+\beta\partial_{\mu}|x(g)\]
\[\partial_{\mu}|_x(fg)=f(x)\partial_{\mu}|_x(g)+\partial_{\mu}|_x(f)g(x)\]
A more intuitive but equivalent formulation of what “tangent vector” means is to consider trajectories \(\gamma(t)\in X\) through the manifold \(X\) passing through the point \(x\in X\) at time \(t\). Then, choosing an arbitrary chart \((U,\phi)\) in a neighbourhood of \(x\in U\), one considers the coordinatized version of \(\gamma(t)\in U\subset X\), namely \(\phi(\gamma(t))\in\mathbf R^n\) which will be differentiable with some derivative \(\frac{d}{dt}\phi(\gamma(t))\in\mathbf R^n\). All curves \(\gamma(t)\) with the same derivative at \(x\) are considered to form an equivalence class, and the value of that derivative is then effectively the tangent vector. Unlike the algebraic formulation, here one has to check that this definition is independent of the choice of neighbourhood chart \((U,\phi)\) of \(x\in U\subset X\).
Problem: What is the tangent space \(T_x(X)\) to a \(C^1\)-manifold \(X\) at a point \(x\in X\)?
Solution: \(T_x(X)\) is simply the set of all tangent vectors to \(X\) at \(x\in X\). As the name of its elements (i.e. “tangent vectors“) suggests, \(T_x(X)\) is a real, \(n\)-dimensional vector space. By choosing a neighbourhood chart \((U,\phi)\) containing \(x\in U\), this provides \(n\) local coordinates \(\phi(x)=(x^0(x),…,x^{n-1}(x))\) on \(U\). An explicit basis for \(T_x(X)\) then consists of the \(n\) partial derivative tangent vector operators \(\partial_{\mu}|_x\in T_x(X)\) for \(\mu=0,…,n-1\) defined by:
\[\partial_{\mu}|_x(f):=\frac{\partial (f\circ\phi^{-1})}{\partial x^{\mu}}\biggr|_{\phi(x)}\]
(which is well-defined because \(f\in C^1(X\to\mathbf R)\)). Any tangent vector \(v_x\in T_x(X)\) at \(x\in X\) can then be written as a linear combination of these basis tangent vectors:
\[v_x=v^{\mu}\partial_{\mu}|_x\]
with real coefficients \(v^{\mu}\in\mathbf R\) called the contravariant components of \(v_x\in T_x(X)\) in the coordinate basis \(\partial_{\mu}|_x\).
Problem: A given tangent vector \(v_x\in T_x(X)\) may be written with contravariant components in \(2\) distinct coordinate bases:
\[v_x=v^{\mu}\partial_{\mu}|_x=v’^{\nu}\partial’_{\nu}|_x\]
simply from picking \(2\) different charts \((U,\phi),(U’,\phi’)\) in the \(C^1\)-atlas of \(X\) containing \(x\in U\cap U’\). Describe how the contravariant components \(v’^{\nu}\) may be obtained from the contravariant components \(v^{\mu}\).
Solution: Act on an arbitrary \(f\in C^1(X\to\mathbf R)\) to obtain:
\[v_x(f)=v^{\mu}\frac{\partial (f\circ\phi^{-1})}{\partial x^{\mu}}\biggr|_{\phi(x)}=v’^{\nu}\frac{\partial (f\circ\phi’^{-1})}{\partial x’^{\nu}}\biggr|_{\phi'(x)}\]
Now insert a “resolution of the identity” \(f\circ\phi’^{-1}\circ\phi’\circ\phi^{-1}\) and because \(X\) is a \(C^1\)-manifold, the transition map \(\phi’\circ\phi^{-1}\) will be differentiable and in particular, by the chain rule:
\[\frac{\partial (f\circ\phi^{-1})}{\partial x^{\mu}}\biggr|_{\phi(x)}=\frac{\partial x’^{\nu}}{\partial x^{\mu}}\biggr|_{\phi(x)}\frac{\partial(f\circ\phi’^{-1})}{\partial x’^{\nu}}\biggr|_{\phi'(x)}\]
where \(\phi'(x)=(x’^0(x),…,x’^{n-1}(x))\). This simple chain rule identity can by itself already be viewed as a passive change of \(T_x(X)\)-basis:
\[\frac{\partial}{\partial x’^{\nu}}\biggr|_{x}=\frac{\partial x^{\mu}}{\partial x’^{\nu}}\biggr|_{\phi'(x)}\frac{\partial}{\partial x^{\mu}}\biggr|_x\]
Or equivalently, as an active change of the contravariant components via the Jacobian matrix:
\[v’^{\nu}=\frac{\partial x’^{\nu}}{\partial x^{\mu}}\biggr|_{\phi(x)}v^{\mu}\]
Problem: What is a vector field \(v\) on a \(C^1\)-manifold \(X\)?
Solution: The more intuitive way to understand it is as a map \(v:X\to TX\), where \(TX:=\bigsqcup_{x\in X}T_x(X)\) is the so-called tangent bundle of \(X\). This basically corresponds to assigning a tangent vector \(v_x\in T_x(X)\) rooted at each point \(x\in X\) across the manifold \(X\). Alternatively, a more algebraic definition of a vector field is as a map from scalar fields to scalar fields \(v:C^1(X\to\mathbf R)\to C^1(X\to\mathbf R)\), defined so that for a scalar field \(f:X\to\mathbf R\) on the manifold \(X\), the new scalar field \(v(f):X\to\mathbf R\) can be thought of as a global directional derivative map \(v(f)(x):=v_x(f)\) giving the directional derivative \(v_x(f)\in\mathbf R\) of the scalar field \(f\) at each point \(x\in X\) according to some assignment \(v_x\in T_x(X)\) of tangent vectors (in the spirit of the first definition).
Problem: Given a \(C^1\)-manifold \(X\), and a neighbourhood of a point \(x\in X\) with local coordinates \((x^0,…,x^{n-1})\), explain how a vector field \(v\) may be expressed in this coordinate basis over said neighbourhood.
Solution: \(v=v^{\mu}\partial_{\mu}\) is essentially a directional derivative operator. Just to be clear, there is a standard abuse of notation going on here where the underlying chart \(\phi\) defining the local coordinates \((x^0,…,x^{n-1})\) is swept under the rug, so for instance \(v(f)(x)=v^{\mu}(x)\frac{\partial f}{\partial x^{\mu}}(x)\) where the shorthand \(\frac{\partial f}{\partial x^{\mu}}(x)=\frac{\partial (f\circ\phi^{-1})}{\partial x^{\mu}}(\phi(x))\) is being used.
Problem: Given \(2\) vector fields \(v,w\) on the same \(C^2\)-manifold \(X\), explain why the composition \(v\circ w\) (and hence also \(w\circ v\)) is not a vector field on \(X\).
Solution: Although it’s linear, it fails to satisfy the product rule.
Problem: Explain how the above situation can be remedied.
Solution: Consider instead the commutator \([v,w]:C^2(X\to\mathbf R)\to C^2(X\to\mathbf R)\). This is now both linear, but also importantly obeys the product rule:
\[[v,w](fg)=f[v,w](g)+g[v,w](f)\]
Indeed, as with any commutator this makes the space of all vector fields not only a real vector space but also a Lie algebra. In a common coordinate basis \((x^0,…,x^{n-1})\) where \(v=v^{\mu}\partial_{\mu}\) and \(w=w^{\nu}\partial_{\nu}\), the commutator has the representation:
\[[v,w]=\left(v^{\mu}\partial_{\mu}w^{\nu}-w^{\mu}\partial_{\mu}v^{\nu}\right)\partial_{\nu}\]