Tangent and Cotangent Spaces

A tangent space is the vector space of infinitesimal directions available at one point of a manifold. A cotangent space is its dual space: the vector space of linear functionals that eat tangent vectors and return numbers.

The essential picture is local:

$$T_pM \quad \text{directions at }p, \qquad T_p^*M \quad \text{linear measurements of those directions}.$$

This page gives the coordinate formulas needed later for gradients, variational derivatives, Berry connections, Bloch-sphere motion, and phase-space language. It does not assume a full course in differential geometry.

Helpful background. Manifolds, First Look introduces charts; Vector Spaces and Dual Spaces explains linear functionals.

Local directions and linear measurements

Imagine moving a control parameter slightly and asking how an energy changes. The displacement is a tangent vector; the first-order energy change is obtained by applying the covector $dE$ to it. This distinction lets us reuse linear algebra at each point of a curved parameter space.

Tangent Vectors from Curves

Let $M$ be a smooth manifold and let $p\in M$. A tangent vector at $p$ can be represented by the velocity of a curve through $p$.

Take a smooth curve

$$\gamma:(-\epsilon,\epsilon)\to M, \qquad \gamma(0)=p.$$

If $x^1,\ldots,x^n$ are local coordinates near $p$, the coordinate velocity is

$$v^i = \left. \frac{d}{dt} x^i(\gamma(t)) \right\rvert_{t=0}.$$

The corresponding tangent vector is written

$$v = v^i \left. \frac{\partial}{\partial x^i} \right\rvert_p.$$

Here and below repeated upper-lower index pairs are summed. The vectors

$$\left. \frac{\partial}{\partial x^1} \right\rvert_p, \ldots, \left. \frac{\partial}{\partial x^n} \right\rvert_p$$

form a coordinate basis for $T_pM$.

The curve is only a convenient way to define the direction. Many different curves can have the same first-order velocity at $p$; they represent the same tangent vector.

Tangent Vectors as Directional Derivatives

There is an equivalent definition that is often more useful. A tangent vector acts on a smooth scalar function $f:M\to\mathbb R$ by taking a directional derivative:

$$v(f) = \left. \frac{d}{dt} f(\gamma(t)) \right\rvert_{t=0}.$$

In local coordinates this becomes

$$v(f) = v^i \frac{\partial f}{\partial x^i}(p).$$

This formula explains why the basis vectors are written as partial derivatives. They are not fractions; they are basis tangent vectors that differentiate functions.

For complex-valued scalar functions, the same formula is applied componentwise. Most parameter-space examples in quantum mechanics use real manifolds with complex-valued Hilbert-space data defined over them.

Coordinate Changes

A tangent vector is geometric, but its components depend on the chosen coordinates. Suppose $y^a=y^a(x)$ is a change of local coordinates. If

$$v = v^i \frac{\partial}{\partial x^i} = w^a \frac{\partial}{\partial y^a},$$

then the chain rule gives

$$w^a = \frac{\partial y^a}{\partial x^i} v^i.$$

Thus tangent components transform with the Jacobian of the coordinate change. This is why tangent vectors are sometimes called contravariant vectors.

The word “contravariant” is bookkeeping, not physics by itself. What matters is that the tangent vector is invariant while its coordinate components change in a controlled way.

Cotangent Spaces

The cotangent space $T_p^*M$ is the dual vector space of $T_pM$:

$$T_p^*M = \operatorname{Hom}(T_pM,\mathbb R).$$

An element $\alpha\in T_p^*M$ is a covector, also called a one-form at $p$. It takes a tangent vector $v\in T_pM$ and returns a number:

$$\alpha(v)\in\mathbb R.$$

In coordinates, the dual basis is written

$$dx^1,\ldots,dx^n,$$

with defining pairing

$$dx^i \left( \frac{\partial}{\partial x^j} \right) = \delta^i{}_j.$$

A covector is written

$$\alpha = \alpha_i\,dx^i.$$

Its action on a tangent vector is

$$\alpha(v) = \alpha_i v^i.$$

This pairing is the local model for many expressions that look like “dot products” but do not require a metric. A one-form integrates naturally along a path because it eats the path’s tangent vector.

How Cotangent Components Transform

Under the coordinate change $y^a=y^a(x)$, write

$$\alpha = \alpha_i^{(x)}\,dx^i = \alpha_a^{(y)}\,dy^a.$$

Since

$$dy^a = \frac{\partial y^a}{\partial x^i}dx^i,$$

the components satisfy

$$\alpha_a^{(y)} = \frac{\partial x^i}{\partial y^a} \alpha_i^{(x)}.$$

Cotangent components transform with the inverse Jacobian. This is the basic reason lower indices transform oppositely from upper indices.

The invariant quantity is the pairing:

$$\alpha(v) = \alpha_i^{(x)}v^i = \alpha_a^{(y)}w^a.$$

The components changed; the measured number did not.

Differentials of Functions

For a smooth scalar function $f:M\to\mathbb R$, the differential $df$ is a cotangent vector at each point:

$$df = \frac{\partial f}{\partial x^i}dx^i.$$

Its action on a tangent vector is

$$df(v) = v^i \frac{\partial f}{\partial x^i}.$$

This is exactly the directional derivative of $f$ along $v$.

The differential is coordinate-independent. The component expression changes under a new chart, but the number $df(v)$ does not.

Differential versus Gradient

In ordinary Euclidean space, one often identifies the differential $df$ with the gradient vector $\nabla f$. On a general manifold, these are not the same kind of object:

$$df\in T_p^*M, \qquad \nabla f\in T_pM.$$

To turn the covector $df$ into a vector, one needs a metric $g_p$, a smoothly varying inner product on tangent spaces. The gradient is the unique vector satisfying

$$df(v) = g_p(\nabla f,v)$$

for every $v\in T_pM$.

Thus:

• the differential $df$ exists once the smooth structure is available;
• the gradient vector $\nabla f$ requires an additional metric;
• in Euclidean coordinates the distinction is hidden because the standard metric identifies vectors and covectors.

This is the geometric version of a familiar linear-algebra fact: a dual vector is not automatically the same thing as a vector until an inner product supplies an identification. See Inner Products for the finite-dimensional analogy.

Example: Euclidean Space

For $M=\mathbb R^n$, the tangent space at any point can be canonically identified with $\mathbb R^n$ itself:

$$T_p\mathbb R^n\simeq\mathbb R^n.$$

If

$$v = v^i\frac{\partial}{\partial x^i}, \qquad \alpha = \alpha_i dx^i,$$

then

$$\alpha(v)=\alpha_i v^i.$$

This example is useful, but it is also the source of a common overgeneralization. On a curved or topologically nontrivial manifold, tangent spaces at different points are separate vector spaces unless a rule is supplied for comparing them.

Example: The Circle

The unit circle can be locally parametrized by an angle $\theta$. A path $\theta(t)$ has tangent vector

$$v = \dot\theta \frac{\partial}{\partial\theta}.$$

The dual one-form $d\theta$ measures angular velocity:

$$d\theta(v)=\dot\theta.$$

Because $\theta$ is periodic, no single angle coordinate is a global one-to-one coordinate on all of $S^1$. Locally, however, the tangent and cotangent formulas work exactly as expected.

This is the simplest model for phase variables: infinitesimal phase changes are linear, even though phase itself is defined modulo $2\pi$.

Example: The Sphere

For the unit sphere

$$S^2 = \{\mathbf n\in\mathbb R^3:\mathbf n\cdot\mathbf n=1\},$$

the tangent space at $\mathbf n$ can be described extrinsically as

$$T_{\mathbf n}S^2 = \{\mathbf v\in\mathbb R^3: \mathbf n\cdot\mathbf v=0\}.$$

Thus allowed infinitesimal motions are perpendicular to the radius vector. This tangent plane is the local linearization of motion on the sphere.

For a qubit pure state, the Bloch Sphere Geometry uses this same sphere. A small change in the Bloch direction is a tangent vector to the sphere of pure states, not an arbitrary vector in $\mathbb R^3$.

Example: Berry Connection Along a Path

Let $R^i$ be local coordinates on a parameter manifold and let

$$C:t\mapsto R(t)$$

be a path. The path velocity is the tangent vector

$$\dot R = \dot R^i \frac{\partial}{\partial R^i}.$$

A Berry connection in a chosen local gauge has the one-form form

$$A = A_i(R)\,dR^i.$$

Evaluating it on the path velocity gives

$$A(\dot R) = A_i(R(t))\dot R^i(t).$$

The line integral is then

$$\int_C A = \int_0^T A_i(R(t))\dot R^i(t)\,dt.$$

This is the coordinate-free content behind the familiar notation $\oint \mathbf A\cdot dR$. The physical geometric phase is introduced in Berry Phase. This Toolkit page only supplies the tangent-cotangent language used to parse the formula.

Example: Variations as Tangent Directions

In a finite-dimensional configuration space with coordinates $q^i$, a path variation

$$q^i(t) \mapsto q^i(t)+\epsilon\eta^i(t)$$

uses $\eta^i(t)$ as an infinitesimal direction in the space of paths. In the simplest coordinate setting, the variation is a tangent vector to the space of possible histories.

The first variation of a functional is a cotangent-like object: it takes an allowed variation and returns the first-order change in the functional. The concrete Euler–Lagrange calculation is developed in Calculus of Variations.

Common Mistakes

• Treating a tangent vector as a nearby point of the manifold.
• Assuming tangent spaces at different points are automatically the same vector space.
• Calling $df$ a gradient vector before choosing a metric.
• Forgetting that covectors transform with the inverse Jacobian.
• Using a coordinate singularity, such as a pole of spherical coordinates, as if it were a singular tangent space.
• Thinking one-forms are ordinary vectors because they have the same number of components in Euclidean coordinates.
• Treating $\mathbf A\cdot dR$ notation as requiring a Euclidean dot product, rather than recognizing the underlying one-form applied to a tangent displacement.

Exercises

1. Let $\gamma(t)=(\cos t,\sin t)$ be a curve on the unit circle. Compute its tangent vector at $t=0$ as a vector in $\mathbb R^2$, and check that it is tangent to the circle.

Solution

Differentiate:

$$\dot\gamma(t)=(-\sin t,\cos t).$$

At $t=0$,

$$\gamma(0)=(1,0), \qquad \dot\gamma(0)=(0,1).$$

The tangent condition for the unit circle embedded in $\mathbb R^2$ is perpendicularity to the radius vector. Indeed,

$$(1,0)\cdot(0,1)=0.$$

Thus $\dot\gamma(0)$ is tangent to $S^1$ at $(1,0)$.

2. For $f(x,y)=x^2+y^2$ on $\mathbb R^2$, compute $df$ and evaluate it on $v=a\,\partial_x+b\,\partial_y$.

Solution

The differential is

$$df = 2x\,dx+2y\,dy.$$

Evaluating on $v$ gives

$$df(v) = 2x\,dx(v)+2y\,dy(v) = 2xa+2yb.$$

This is the directional derivative of $f$ in the direction $v$.

3. In one dimension, let $y=2x$. If $v=v^x\partial_x=w^y\partial_y$, find $w^y$. If $\alpha=\alpha_x dx=\alpha_y dy$, find $\alpha_y$.

Solution

For tangent components,

$$w^y = \frac{dy}{dx}v^x = 2v^x.$$

For cotangent components, use $dy=2dx$, so $dx=(1/2)dy$. Therefore

$$\alpha = \alpha_x dx = \frac{\alpha_x}{2}dy,$$

and

$$\alpha_y=\frac{\alpha_x}{2}.$$

The product is invariant:

$$\alpha_x v^x = \alpha_y w^y.$$

4. Why does $df$ exist without a metric, while the gradient vector requires one?

Solution

The differential $df$ is defined by its action on tangent vectors:

$$df(v)=v(f).$$

This uses only the smooth structure and the directional derivative. A gradient vector is a tangent vector $\nabla f$ satisfying

$$df(v)=g_p(\nabla f,v)$$

for all $v$. The right-hand side uses a metric $g_p$ to pair two tangent vectors. Without such a metric, there is no canonical way to convert the covector $df$ into a vector.

5. Let $A=A_R(R)dR$ be a one-form on a one-dimensional parameter space and let $R=R(t)$ be a path. Write the integral of $A$ along the path.

Solution

The path velocity is

$$\dot R = \dot R(t)\frac{\partial}{\partial R}.$$

Therefore

$$A(\dot R) = A_R(R(t))\dot R(t),$$

and

$$\int_C A = \int_0^T A_R(R(t))\dot R(t)\,dt.$$

This is the one-dimensional version of the line-integral formula used for Berry connections.

References

• T. Frankel, The Geometry of Physics, 3rd ed., Cambridge University Press, 2011.
• J. M. Lee, Introduction to Smooth Manifolds, 2nd ed., Springer, 2013.
• L. W. Tu, An Introduction to Manifolds, 2nd ed., Springer, 2011.