Tangent space

The tangent space of a differentiable manifold M is a vector space at a point p on the manifold whose elements are the tangent vectors (or velocities) to the curves passing through that point p. The tangent space at this point p is usually denoted $$T_pM$$.

The tangent space is necessary for a manifold because it offers a way in which tangent vectors at different points on the manifold can be compared (via an affine connection). If the manifold is a hypersurface of $$\scriptstyle \mathbb{R}^n$$, then the tangent space at a point can be thought of as a hyperplane at that point. However, this ambient Euclidean space is unnecessary to the definition of the tangent space.

The tangent space at a point has the same dimension as the manifold, and the union of all the tangent spaces of a manifold is called the tangent bundle, which itself is a manifold of twice the dimension of M.

Definition
Although it is tempting to define a tangent space as a "space where tangent vectors live", without a definition of a tangent space there is no definition of a tangent vector. There are various ways in which a tangent space can be defined, the most intuitive of which is in terms of directional derivatives; the space $$T_pM$$ is the space identified with directional derivatives at p.

Directional derivative
A curve on the manifold is defined as a differentiable map $$\scriptstyle \gamma: (-\epsilon,\epsilon) \rightarrow M$$. Let $$\scriptstyle \gamma(0) \, = \, p$$. If one defines $$\scriptstyle \mathcal{F}_p$$ to be all the functions $$\scriptstyle f:M \rightarrow \mathbb{R}^n$$ that are differentiable at the point p, then one can interpret
 * $$\gamma'(0): \, \mathcal{F}_p \rightarrow \mathbb{R}$$

to be a linear functional such that
 * $$ \gamma'(0)(f) = (f \circ \gamma)'(0) = \lim_{h \rightarrow 0} \frac{f(\gamma(h)) - f(\gamma(0))}{h} $$

and is a directional derivative of f in the direction of the curve $$\scriptstyle \gamma$$. This operator can be interpreted as a tangent vector. The tangent space is then the set of all directional derivatives of curves at the point p.

Directional derivatives as a vector space
If this definition is reasonable, then the directional derivatives, must form a vector space of the same dimension as the n-dimensional manifold M. The easiest way to show this is to show that some n directional derivatives form a basis of the vector space, and in order to do so, one needs to introduce a coordinate chart (see differentiable manifold).

Let $$\scriptstyle \psi: \, U \, \rightarrow V$$ where $$ \scriptstyle U \subset M$$, $$\scriptstyle V \subset \mathbb{R}^n$$ be a coordinate chart, and $$\scriptstyle \psi \,=\, (x_1, \cdots,\, x_n)$$. The most obvious candidates for basis vectors would be the directional derivatives along the coordinate curves, i.e. the i-th coordinate curve would be
 * $$\gamma_i = \psi^{-1} (\psi(p) + te_i) \ $$

where $$\scriptstyle e_i = (0, \cdots, 0, 1 , 0, \cdots, 0)$$, the 1 being in the i-th position.

The directional derivative along a coordinate curve can be represented as
 * $$ \frac{\partial}{\partial x^i}\bigg|_{p} = \gamma_i '(0)$$

because
 * $$\frac{\partial}{\partial x^i}\bigg|_{p} (f) = (f\circ \gamma_i)'(0) = \frac{d}{dt} \bigg|_{t=0} f(\psi^{-1} (\psi(p) + te_i))$$

which becomes, via the chain rule,
 * $$\frac{\partial}{\partial x^i} (f \circ \psi^{-1})(\psi(p)).$$

For an arbitrary curve $$\scriptstyle \gamma, \, \psi \circ \gamma \, = \, (\gamma_1, \, \cdots, \, \gamma_n), $$ then
 * $$ \gamma'(0)(f) = \sum_{i=1}^{n} \frac{\partial}{\partial x^i} (f \circ \psi^{-1})(\psi(p)) \cdot \gamma_i'(0) $$

which is simply
 * $$ \sum^{n}_{i=1} \gamma'_{i}(0) \cdot \frac{\partial}{\partial x^i}\bigg|_{p} (f)$$

so
 * $$ \gamma'(0) = \sum^{n}_{i=1} \gamma'_{i}(0) \cdot \frac{\partial}{\partial x^i}\bigg|_{p} $$

as f is arbitrary.