Manifolds and Tangent Spaces

The abstract manifold and its tangent space are the working vocabulary of general relativity, where spacetime is modeled as a smooth $4$-manifold with no preferred embedding, and physical fields are sections of bundles built from tangent spaces at each event. In classical mechanics, the configuration space of a system with constraints — for instance the space of orientations of a rigid body, $SO(3)$ — is a manifold, and the phase space of the system is built from its tangent (or cotangent) bundle; Lagrangian and Hamiltonian mechanics are, formally, calculus on these tangent and cotangent spaces. In robotics and computer vision, the group of rigid motions $SE(3)$ and the rotation group $SO(3)$ are manifolds whose tangent space at the identity (their Lie algebra) linearizes rotational and translational velocities, exactly as in the $GL_2(\mathbb{R})$ example worked above. In statistics, families of probability distributions form manifolds (the subject of information geometry), with the Fisher information defining a Riemannian metric on the tangent spaces constructed exactly as here. In algebraic geometry and moduli theory, moduli spaces are frequently manifolds or orbifolds only in an intrinsic, chart-based sense, with no natural ambient Euclidean space, making the coordinate-free tangent space the only viable notion of infinitesimal deformation. Finally, machine learning applications such as manifold learning and Riemannian optimization use exactly this local-chart, tangent-space-at-a-point machinery to perform gradient descent on curved parameter spaces such as the space of covariance matrices or rotation matrices.

Ask the tutor: How does the abstract tangent space $T_pM$ relate to the cotangent space and to the definition of a Riemannian metric on $M$?

The single idea underlying everything is this: a manifold is a set with instructions for how to translate between overlapping coordinate patches, and as long as the translation instructions (transition maps) are smooth, all the calculus performed in one patch is guaranteed to be compatible with calculus performed in another. No patch is "the real one" and no ambient space is needed; the charts are the geometry, in the sense that any question about smoothness, derivatives, or tangent vectors can be answered by picking any convenient chart and computing, with the compatibility of transition maps guaranteeing the answer does not depend on the choice. Concretely, think of the circle $S^1$. No single angle coordinate $\theta \in \mathbb{R}$ covers all of $S^1$ continuously and injectively — go around once and $\theta$ jumps by $2\pi$. So instead we use (at least) two overlapping arcs, each mapped homeomorphically to an open interval in $\mathbb{R}$, and on the overlap the two angle coordinates differ by a locally constant integer multiple of $2\pi$, which is certainly a smooth (indeed constant-derivative) transition map. This is the prototype for every manifold: local coordinates that individually miss the whole space, but whose overlaps are stitched together smoothly. The tangent space at a point is reconstructed the same coordinate-free way. Rather than asking "which vectors in the ambient $\mathbb{R}^N$ are tangent to $M$ at $p$" — a question that presumes an embedding — we ask instead: which velocity vectors of curves through $p$ are indistinguishable from the point of view of any smooth function's directional derivative? Two curves through $p$ are declared tangent-equivalent if, in some chart, their images have equal derivative at the parameter value corresponding to $p$; smooth compatibility of charts guarantees this equivalence does not depend on which chart was used to check it. The resulting equivalence classes form an $n$-dimensional vector space $T_pM$, and any chart around $p$ gives a linear isomorphism $T_pM \cong \mathbb{R}^n$ by simply differentiating the chart representative of a curve.

Example 1: the circle $S^1$ as a smooth manifold, and $T_pS^1$ computed two ways. Let $S^1 = \{(x,y)\in\mathbb{R}^2 : x^2+y^2=1\}$ with the subspace topology (Hausdorff and second countable, being a subspace of $\mathbb{R}^2$). Cover it with four charts using stereographic-style angle parametrizations, but let us use an explicit asymmetric two-chart atlas to make the transition map computation concrete. Let $U = S^1 \setminus \{(-1,0)\}$ and $\varphi: U \to (-\pi,\pi)$ be the inverse of $\theta \mapsto (\cos\theta,\sin\theta)$, i.e. $\varphi(x,y)=\theta$ where $\theta\in(-\pi,\pi)$ is the unique angle with $(\cos\theta,\sin\theta)=(x,y)$. Let $V = S^1\setminus\{(1,0)\}$ and $\psi: V\to (0,2\pi)$ be the analogous inverse using angles in $(0,2\pi)$. Both $\varphi,\psi$ are homeomorphisms onto open intervals (continuous bijections from a compact-minus-point... more directly, they are restrictions of the smooth, locally invertible covering map $\theta\mapsto(\cos\theta,\sin\theta)$, and standard analysis shows they are homeomorphisms onto their images). $U\cup V = S^1$, so $\{(U,\varphi),(V,\psi)\}$ is an atlas. The overlap $U\cap V = S^1\setminus\{(\pm1,0)\}$ has two components: the upper arc ($y>0$) and lower arc ($y<0$). On the upper arc, $\varphi$ takes values in $(0,\pi)$ and $\psi$ also takes values in $(0,\pi)$, and they agree: $\psi\circ\varphi^{-1}(\theta) = \theta$ there. On the lower arc, $\varphi$ takes values in $(-\pi,0)$ while $\psi$ takes values in $(\pi,2\pi)$, and

Both pieces of $\psi\circ\varphi^{-1}$ are smooth (indeed affine with derivative $1$) on their domains, so the transition map is $C^\infty$ with smooth inverse. Hence $\{(U,\varphi),(V,\psi)\}$ is a smooth atlas, and $S^1$ (completed to the maximal atlas it generates) is a smooth $1$-manifold. Now take $p = (0,1) \in S^1$ (note: this point lies in $U\cap V$, upper arc, with $\varphi(p)=\pi/2$, $\psi(p)=\pi/2$ as well). Consider two curves through $p$:

both smooth curves with $\gamma_1(0)=\gamma_2(0)=(0,1)=p$ (check: $\gamma_1(0)=(0,1)$ ✓; $\gamma_2(0) = (\cos(\pi/2),\sin(\pi/2))=(0,1)$ ✓). In the chart $\varphi$: $\gamma_1(t)$ near $t=0$ has angle $\theta_1(t) = \pi/2 - t$ (since $(-\sin t,\cos t) = (\cos(\pi/2+t)\cdot(-1)\dots$; more directly, differentiate the constraint or just note $(-\sin t,\cos t)=(\cos(\theta_1(t)),\sin(\theta_1(t)))$ forces $\theta_1(t) = \pi/2 + t$ up to sign — let us verify: $\cos(\pi/2+t) = -\sin t$ ✓, $\sin(\pi/2+t)=\cos t$ ✓. So $\theta_1(t)=\pi/2+t$.) Then

For $\gamma_2$, directly $\varphi\circ\gamma_2(t) = \pi/2 + 3t^3 - t$ (valid for $t$ small, staying in $(-\pi,\pi)$), so

So $D\varphi_p([\gamma_1]) = 1$ but $D\varphi_p([\gamma_2]) = -1$: these curves represent different tangent vectors at $p$, i.e. $\gamma_1 \not\sim \gamma_2$, even though both pass through $p$. Since $T_pS^1$ is $1$-dimensional (Theorem, part 3, with $n=1$), and $D\varphi_p$ is a linear isomorphism to $\mathbb{R}$, this says $[\gamma_1]$ and $[\gamma_2]$ correspond to the scalars $1$ and $-1$ respectively — antipodal, non-proportional-by-a-positive-scalar tangent directions, matching the geometric picture that $\gamma_1$ moves counterclockwise through $p$ at unit angular speed while $\gamma_2$ moves clockwise at unit angular speed (to first order). Let us confirm chart-independence using $\psi$ instead. On the upper arc near $p$, $\psi=\varphi$ (both equal to the same angle in $(0,\pi)$ as shown above), so trivially $D\psi_p([\gamma_1])=1$, $D\psi_p([\gamma_2])=-1$ as well — consistent, though this example does not yet showcase a genuinely different chart value because $p$ lies in the region of agreement. To see the general transformation rule at work, pick instead $q=(0,-1)$ on the lower arc, where $\varphi(q) = -\pi/2$ and $\psi(q) = 3\pi/2$, related by $\psi = \varphi + 2\pi$ near $q$; the transition derivative there is $D\tau = 1$ (a translation has derivative $1$), so $D\psi_q = 1\cdot D\varphi_q$ — the two charts assign the same real number to each tangent vector at $q$ in this example, illustrating that the transition derivative, while generally a nontrivial invertible linear map, can specialize to the identity, as it does for this particular angle-chart atlas on $S^1$.

Example 2: the general linear group $GL_2(\mathbb{R})$ and its tangent space at the identity. Let $M_2(\mathbb{R}) \cong \mathbb{R}^4$ be the space of $2\times2$ real matrices with coordinates $(a,b,c,d)$ for $\begin{pmatrix}a&b\\c&d\end{pmatrix}$, and let

Since $\det: M_2(\mathbb{R})\to\mathbb{R}$, $\det(a,b,c,d)=ad-bc$, is continuous (a polynomial), $GL_2(\mathbb{R}) = \det^{-1}(\mathbb{R}\setminus\{0\})$ is open in $M_2(\mathbb{R})\cong\mathbb{R}^4$, hence Hausdorff, second countable (as a subspace of $\mathbb{R}^4$), and it is itself an open subset of $\mathbb{R}^4$, so the single chart $(GL_2(\mathbb{R}), \mathrm{id})$ — identity inclusion into $\mathbb{R}^4$ — is a smooth atlas consisting of one chart (transition-map compatibility is vacuous with only one chart). Thus $GL_2(\mathbb{R})$ is a smooth $4$-manifold, an open submanifold of $\mathbb{R}^4$. We compute $T_I GL_2(\mathbb{R})$ at the identity matrix $I = \begin{pmatrix}1&0\\0&1\end{pmatrix}$, i.e. $\varphi(I) = (1,0,0,1)$ in the single chart $\varphi=\mathrm{id}$. By the theorem, since there is one global chart, $D\varphi_I : T_IGL_2(\mathbb{R}) \to \mathbb{R}^4$ is a linear isomorphism, and every tangent vector is represented by the velocity of some curve. Consider the specific curve

Check $\gamma(0) = I$ ✓. Check $\gamma$ maps into $GL_2(\mathbb{R})$ near $t=0$: $\det\gamma(t) = (1+2t)(1-t^2) - (-t)(3t) = (1+2t)(1-t^2)+3t^2$; at $t=0$ this is $1\neq 0$, and by continuity of $\det$ it stays nonzero for $t$ in a neighborhood of $0$. In the (only) chart,

so

i.e. $D\varphi_I([\gamma]) = (2,-1,3,0)$, which under the identification $\mathbb{R}^4 \cong M_2(\mathbb{R})$ corresponds to the matrix

This matches the general fact (whose proof is exactly this kind of curve computation) that $T_I GL_2(\mathbb{R}) \cong M_2(\mathbb{R})$ canonically, by sending $[\gamma] \mapsto \gamma'(0)$ computed entrywise in the ambient $\mathbb{R}^4$ — here realized concretely as $X$ above, obtained from a genuinely asymmetric (not diagonal, not scalar) curve, illustrating that any matrix $X\in M_2(\mathbb{R})$ arises this way, e.g. via $\gamma(t) = I + tX + O(t^2)$ (our $\gamma$ is not exactly $I+tX$ because of the $-t^2$ term in the $(2,2)$ entry, but that term contributes $0$ to the first derivative at $t=0$, confirming that different curves with the same first-order behavior represent the same tangent vector, exactly as guaranteed by the theorem).