Hassler Whitney
1. Tangent Vectors and Tangent Spaces
So we have manifolds, we have charts, and we have no idea what a vector is. The classic definition doesn’t work. Take two points on a sphere and draw the arrow between them: it cuts straight through the interior, leaving the surface entirely. And the interior is not part of the sphere. So we need to build vectors!
1.1 What works?
Forget arrows and ask what a velocity actually does for you. Suppose you're moving along a path, and there's a temperature at every point. Your velocity determines how fast the temperature changes as you go. Different velocity, different rate. Same velocity, same rate, regardless of what else the path does later. So a velocity is a device for turning functions into rates of change. And functions on a manifold are perfectly well defined without any ambient space.
1.2 Curves and their derivatives
A curve is a smooth map
\[ \gamma : (-\varepsilon, \varepsilon) \to M, \qquad \gamma(0) = p \tag{1} \]
Pick a chart around \(p\), so the curve has coordinates \(x^\mu(t)\). Take any smooth function \(f : M \to \mathbb{R}\) and ask how it changes along the curve. The chain rule gives
\[ \left. \frac{d}{dt} f(\gamma(t)) \right|_{t=0} = \left. \frac{dx^\mu}{dt} \right|_{t=0} \frac{\partial f}{\partial x^\mu} \tag{2} \]
summing over \(\mu\), as always. The function \(f\) appears only on the right of the expression; everything else is fixed by the curve. So the curve has handed us an operator, waiting for a function:
\[ X = \left. \frac{dx^\mu}{dt} \right|_{t=0} \frac{\partial}{\partial x^\mu} \tag{3} \]
And notice what this operator ‘remembers’ about the curve: only the numbers \(dx^\mu/dt\) at \(t = 0\). Two curves through \(p\) with the same initial velocity give the identical operator, no matter how wildly they differ later. So we define the tangent vector to be the operator.
1.3 Derivations
Which operators count? Two properties characterize them. Linearity:
\[ X(af + bg) = aX(f) + bX(g) \tag{4} \]
for constants \(a, b\). And the Leibniz rule:
\[ X(fg) = f(p)X(g) + g(p)X(f) \tag{5} \]
That's just the product rule, with the functions evaluated at the point because that's where the derivative is taken. An operator satisfying both is called a derivation at \(p\), and a tangent vector is defined to be exactly that. Here's a small consequence. Apply Leibniz to the constant function \(1\), using \(1 \cdot 1 = 1\):
\[ X(1) = X(1 \cdot 1) = 1 \cdot X(1) + 1 \cdot X(1) = 2X(1) \tag{6} \]
so \(X(1) = 0\), and by linearity \(X(c) = 0\) for every constant. Derivations kill constants, which is what any reasonable notion of derivative should do.
1.4 A basis, and the dimension
The operators \(\partial/\partial x^\mu\) are themselves derivations, one for each coordinate. Are there others? No! Take any smooth \(f\), and set up coordinates so \(p\) is at the origin. Taylor's theorem with remainder says
\[ f(x) = f(0) + x^\mu g_\mu(x) \tag{7} \]
for some smooth functions \(g_\mu\) with \(g_\mu(0) = \partial f/\partial x^\mu |_0\). Now apply a derivation \(X\). The constant term dies. On the product term, Leibniz gives
\[ X(x^\mu g_\mu) = x^\mu(p) X(g_\mu) + g_\mu(p) X(x^\mu) = 0 + \frac{\partial f}{\partial x^\mu} X(x^\mu) \tag{8} \]
since \(x^\mu(p) = 0\). Writing \(X^\mu = X(x^\mu)\), we get
\[ X(f) = X^\mu \frac{\partial f}{\partial x^\mu} \tag{9} \]
for every \(f\). So \(X = X^\mu \partial_\mu\), and the coordinate derivatives span everything. They're independent too, since applying \(X\) to the coordinate function \(x^\nu\) picks out \(X^\nu\). So the tangent vectors at \(p\) form a vector space of dimension \(n\), the tangent space \(T_pM\), with basis \(\{\partial_\mu\}\) and components \(X^\mu\). One thing to be careful about. \(T_pM\) is attached to \(p\) and nowhere else. There's no way to compare a vector at one point with a vector at another… for now! But let’s take a second to think about what the hell we just said here. Derivatives are vectors! In fact, the partial derivatives \(\partial_x, \partial_y, \partial_z\) form a basis for any tangent space isomorphic to \(\mathbb{R}^n\). Crazy! That’s part of why I love math, and especially geometry. Who knew we could use derivatives as vectors?
1.5 How components transform
Here’s where index notation gets very useful. Take two charts, \(x^\mu\) and \(\tilde{x}^\nu\), overlapping near \(p\). The chain rule relates the bases:
\[ \frac{\partial}{\partial x^\mu} = \frac{\partial \tilde{x}^\nu}{\partial x^\mu} \frac{\partial}{\partial \tilde{x}^\nu} \tag{10} \]
But the vector \(X\) itself is a single geometric object (an operator on functions) and doesn't care which chart we picked. So
\[ X = X^\mu \frac{\partial}{\partial x^\mu} = X^\mu \frac{\partial \tilde{x}^\nu}{\partial x^\mu} \frac{\partial}{\partial \tilde{x}^\nu} = \tilde{X}^\nu \frac{\partial}{\partial \tilde{x}^\nu} \tag{11} \]
and reading off the components,
\[ \tilde{X}^\nu = \frac{\partial \tilde{x}^\nu}{\partial x^\mu} X^\mu \tag{12} \]
That’s the transformation law for an upper index! This is where it comes from.
1.6 On the sphere
Use the stereographic charts from last post. The transition map was inversion,
\[ \tilde{u} = \frac{u}{u^2+v^2}, \qquad \tilde{v} = \frac{v}{u^2+v^2} \tag{13} \]
so let's see what happens to a basis vector. Differentiating,
\[ \frac{\partial \tilde{u}}{\partial u} = \frac{v^2 - u^2}{(u^2+v^2)^2}, \qquad \frac{\partial \tilde{v}}{\partial u} = \frac{-2uv}{(u^2+v^2)^2} \tag{14} \]
giving
\[ \frac{\partial}{\partial u} = \frac{1}{(u^2+v^2)^2} \left[ (v^2-u^2) \frac{\partial}{\partial \tilde{u}} - 2uv \frac{\partial}{\partial \tilde{v}} \right] \tag{15} \]
Check it at the point \((u,v) = (1,0)\), which sits on the unit circle and is therefore fixed by the inversion. There \(u^2+v^2 = 1\) and the formula collapses to
\[ \frac{\partial}{\partial u} = -\frac{\partial}{\partial \tilde{u}} \tag{16} \]
The radial direction has flipped. That's right: inversion sends \(r \mapsto 1/r\), so moving outward in one chart is moving inward in the other. Running the same computation for \(\partial/\partial v\) at that point gives \(\partial/\partial v = \partial/\partial \tilde{v}\), unchanged, since the tangential direction isn't reversed.
1.7 Bundles
Every point has its own tangent space. Bundling them together gives the tangent bundle
\[ TM = \bigcup_{p \in M} T_pM \tag{17} \]
which is itself a manifold, of dimension \(2n\). \(n\) coordinates saying where you are, \(n\) more saying which vector you have. A vector field is a smooth choice of one tangent vector at each point. Whether you can choose them smoothly and never have one vanish turns out to depend on the shape of \(M\). On a torus you can. On a 2-sphere you cannot, which is the hairy ball theorem, and it's why there's always a point on Earth with no wind.
A tangent vector eats a function and returns a number. Fix the function instead, and let it eat vectors: \(X \mapsto X(f)\) is a linear map from \(T_pM\) to \(\mathbb{R}\). That's a different kind of object living on the same manifold, and its components will transform with the Jacobian the other way up. Those are covectors!