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 $T_{p}M$.

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 submanifold of $\scriptstyle \mathbb {R} ^{n}$, then the tangent space at a point can be thought of as an n-dimensional 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 $T_{p}M$ is the space identified with directional derivatives at p.

Directional derivative

A curve on the manifold is defined as a differentiable map $\scriptstyle \gamma :(a,b)\rightarrow M$. Let $\scriptstyle \gamma (t_{0})\,=\,p$. 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

$\gamma '(t_{0}):\,{\mathcal {F}}_{p}\rightarrow \mathbb {R}$

to be an operator such that

$\gamma '(t_{0})(f)=(f\circ \gamma )'(t_{0})=\lim _{h\rightarrow 0}{\frac {f(\gamma (t_{0}+h))-f(\gamma (t_{0}))}{h}}$

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 the 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

${\frac {\partial }{\partial x^{i}}}{\bigg |}_{p}=\gamma _{i}'(t_{0})$

because

${\frac {\partial }{\partial x^{i}}}{\bigg |}_{p}(f)=(f\circ \gamma _{i})'(t_{0})={\frac {d}{dt}}{\bigg |}_{t=t_{0}}f(\psi ^{-1}(\psi (p)+te_{i}))$

which becomes, via the chain rule,

${\frac {\partial }{\partial x^{i}}}(f\circ \psi ^{-1})(\psi (p)).$