Vectors, covectors, and tensors

2.1 Vectors and their components#

A drone must move three metres east and four metres north. You can draw that displacement as an arrow from its starting point to its destination. Rotating the map changes the arrow’s horizontal and vertical components on the screen, but the destination stays the same.

We need to distinguish the displacement from the numbers used to describe it. Choose two basis vectors: $e_1$, one metre east, and $e_2$, one metre north. Any displacement in this plane can be written as a unique combination of these two arrows. Our drone’s displacement is

The numbers 3 and 4 are its components in this basis. The basis vectors need not have unit length or meet at right angles. In a plane, they must be nonparallel and nonzero, so that they can describe both independent directions.

For a general vector, write the components as $V^1$ and $V^2$:

The superscripts 1 and 2 label components; they are not powers. The last expression uses Einstein summation: when an index appears once upstairs and once downstairs in a term, sum over its allowed values. Here $i$ takes the values 1 and 2. The index is an instruction to add the two contributions.

Now keep the drone’s destination fixed but choose a longer first basis vector:

A prime labels the new description. How many copies of the new first arrow do we need? Each copy covers twice the distance, so we need half as many:

Thus $V'^1=3/2$ and $V'^2=4$. The vector has stayed the same while its components have changed. This compensating behavior is called contravariant transformation. We will calculate more general changes in §2.4.

Changing the basis while keeping $V$ fixed is a passive transformation. Rotating the actual displacement while keeping the basis fixed is an active transformation. In the first case the drone reaches the same place; in the second it generally does not.

2.2 Linear maps and matrices#

A linear map takes a vector as input and gives a vector as output, respecting addition and scaling:

Here $a$ and $b$ are numbers. Knowing what $A$ does to the basis vectors determines what it does to every vector. Suppose

For the drone’s vector, linearity gives

The matrix stores the components of $A(e_1)$ and $A(e_2)$ in its first and second columns:

Index notation expresses this same multiplication as

The summed index $j$ selects the input components. The unsummed index $i$ selects the output component: setting $i=1$ gives $2V^1+V^2$, while setting $i=2$ gives $3V^2$. We call $j$ a dummy index and $i$ a free index.

The identity map leaves each basis vector unchanged. Its matrix has ones on the diagonal and zeros elsewhere. Its components have a special name, the Kronecker delta:

A matrix can also represent a rule with a different job. The matrix in §0.6 accepted two vectors and returned a dot product. The matrix here accepts one vector and returns another vector. When a matrix appears, identify its inputs and output before using it.

2.3 Covectors measure vectors#

Suppose a measuring rule takes a vector’s first component and adds twice its second component:

For $V=(3,4)$, it returns $3+2(4)=11$. If we double the vector, the answer doubles. If we add two vectors, their answers add. A linear rule of this kind, taking a vector and returning one number, is called a covector.

More generally, write

The coefficients $\omega_1$ and $\omega_2$ specify the rule in this basis. In our example they are 1 and 2. This operation is called the pairing of a covector with a vector. It does not require an angle or a length.

There is a useful covector for each component: the rule that extracts just that component. Call these rules $\theta^1$ and $\theta^2$. Then $\theta^1(V)=V^1$ and $\theta^2(V)=V^2$. Applied to the basis vectors themselves, they obey

These component extractors form the dual basis. Any covector can be assembled from them:

In particular, $\omega_i=\omega(e_i)$: the coefficient tells us what the rule returns for one basis vector.

Now reuse the longer basis vector $e'_1=2e_1$. The same rule returns $\omega(e'_1)=2\omega(e_1)$, so its new first coefficient is $\omega'_1=2\omega_1$. The drone’s first component halved, but the measuring rule’s coefficient doubled:

The answer is unchanged. Lower indices mark this transformation behavior of covector components. The dual basis changes too: $\theta'^1=\theta^1/2$, because its job is to extract the new, halved component.

A scalar field assigns one value to each point: for example, the temperature $T(x,y)$ at each position on a plate. For this example, $x$ and $y$ are dimensionless labels counting steps on the plate’s grid. The first-order temperature change under a displacement with components $V^1,V^2$ is

The measuring rule $dT$, called the differential of $T$, is a covector. It is exactly linear in its input. Its prediction of a finite temperature change is an approximation when the temperature field varies nonlinearly.

In Cartesian calculus we also arrange these partial derivatives into a gradient arrow. Converting a covector into that arrow uses the lengths and angles of the axes. Chapter 4 develops this conversion; a covector by itself only supplies the measuring rule.

2.4 Changing coordinates with the chain rule#

Consider a path whose position is described by $x=x(s)$ and $y=y(s)$, with $s$ a dimensionless parameter marking progress along the path. Its tangent vector has components $V^1=dx/ds$ and $V^2=dy/ds$: the rates at which the position labels change along the path.

Relabel the same positions using $x'=x/2$ and $y'=y$. The chain rule gives $V'^1=V^1/2$ and $V'^2=V^2$. This is the longer-first-basis example again, expressed as a change of coordinates.

For a general smooth, invertible change, each new coordinate can depend on several old ones. We will use Greek indices for general coordinates. In a two-dimensional example their range has two values; in spacetime it will be $0,1,2,3$. The chain rule reads

The Jacobian $J$ is the matrix of first partial derivatives. Differentiating the inverse coordinate change gives its inverse matrix $K$:

For $x'=x/2$, $y'=y$, these are $J=\operatorname{diag}(1/2,1)$ and $K=\operatorname{diag}(2,1)$. The notation $\operatorname{diag}$ lists the diagonal entries; the other entries are zero.

To find how a covector transforms, require it to give the same answer on the same vector. Substitute $V^\mu=K^\mu{}_{\alpha}V'^\alpha$ into its pairing:

Since this must hold for every input vector, the new coefficients are

The vector uses $J$; the covector uses $K$. This is how their changes compensate.

For a scalar field $f$, the value at a given point is unchanged by relabeling: $f'(x')=f(x)$. Applying the chain rule to its derivatives gives the same covector law:

The differential is written

Here $dx^\mu$ denotes a coordinate component extractor: $dx^\mu(V)=V^\mu$. This connects two familiar uses of $dx$. Along a path, $dx/ds$ is a tangent component; the covector $dx$ measures exactly that component when applied to the tangent. In the illustration below, $dx$ counts the change in the $x$ label as we follow a vector through planes of constant $x$.

For a nonlinear coordinate change, $J$ and $K$ vary from point to point. The transformation rules above still work at each point. Chapter 6 will examine the extra term that appears when we differentiate components while the Jacobian itself changes.

SPACETIME LAB / 02

An arrow meets a family of planes

A covector measures how many level intervals a displacement crosses. Rotate the scene to see that this is a pairing, not a length.

ExploreDiagram

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Follow the displacement

StartThree intervals

$$\begin{gathered}\omega=dx,\quad v=(3,2,1)\\\omega(v)=3\end{gathered}$$

Read the scene. The numbers label the level sets of $x$ in three-dimensional Euclidean space. Small rings mark the exact intersections the arrow has reached. The control follows $\lambda v$ as the displacement fraction $\lambda$ runs from zero to one. Starting at level zero, three level intervals give $dx(v)=3$; the ring count is not a Euclidean length. At full extent, the arrow ends at $(3,2,1)$.

WORKED EXAMPLE

Change the coordinates. Keep the measurement.

If both the vector components and the measuring rule change, how can their answer stay the same?

Builds onPartial derivatives along a path Covectors and their pairing

See the idea

A moving sensor follows one path through a scalar field. One map uses $(x,y)$; another uses $u=x$ and $v=y+x^2$. The second map bends its coordinate grid, but it cannot change the sensor reading. We will calculate the same rate of change twice. All coordinates, the path parameter $\lambda$, and the scalar reading in this example are dimensionless.

Work it out

1. 01
   Give the experiment a point and a direction

   At $(x,y)=(1,2)$, let the path have tangent components $V=(dx/d\lambda,dy/d\lambda)=(2,-1)$. The scalar field is $f=3x-y$. Its differential $df$ is the covector that measures the first-order change along any tangent.

   $$df(V)=3(2)-(-1)=7.$$

   Why this step works The multivariable chain rule multiplies each partial derivative by the corresponding coordinate rate.

2. 02
   Convert the tangent using the local map

   The new address is $(u,v)=(1,3)$. Differentiate the coordinate change at the original point. The Jacobian is the matrix of these derivatives; it acts on the tangent, not on the point itself.

   $$J=\frac{\partial(u,v)}{\partial(x,y)}=\begin{pmatrix}1&0\\2x&1\end{pmatrix},\qquad V'=\left.J\right|_{x=1}\begin{pmatrix}2\\-1\end{pmatrix}=\begin{pmatrix}2\\3\end{pmatrix}.$$

   Why this step works The first new coordinate has the original horizontal rate. The second also includes the derivative of the squared horizontal coordinate.

3. 03
   Convert the measuring rule using the inverse map

   Solve for the old coordinates: $x=u$, $y=v-u^2$. The same scalar field now has the formula $f'=3u-v+u^2$. At $u=1$ its partial derivatives are $(5,-1)$.

   $$df=3\,dx-dy=(3+2u)\,du-dv.$$

   Why this step works A scalar keeps its value at each point. Substituting the inverse coordinate map gives its new formula before differentiating.

4. 04
   Let the two changes cancel

   Pair the new covector components with the new tangent components. The lists changed; the measured rate did not. No lengths or angles entered this calculation.

   $$df(V)=5(2)-1(3)=7.$$

   Why this step works Vector and covector transformation laws compensate because they describe the same directional derivative.

Go deeper

With vector components written as a column and covector components as a row, the transformation is $V'=JV$ and $\omega'=\omega J^{-1}$. Thus $\omega'V'=\omega J^{-1}JV=\omega V$. If you store covectors as columns instead, the rule is $\omega'=(J^{-1})^{\mathsf T}\omega$; the transpose comes from that storage convention. This is an exact algebraic proof for any invertible Jacobian. Here the inverse map exists on all of $\mathbb R^2$ and $\det J=1$ everywhere. The coordinate change is nonlinear even though its action on each tangent space is linear.

Further readingDavid Tong · Differential geometry, §§2.1–2.3

Test the idea

FIRST, PREDICT

A classmate transforms the column of covector components with $J$, just as for a vector, and obtains a different sensor rate. What needs repairing?

Multiply the final rate by $\det J$.Accept the new rate: a nonlinear map changes the physical experiment.Use the inverse transpose for covector columns.

Compare the reasoning

Multiply the final rate by $\det J$.

A directional derivative is a scalar, not a coordinate area. In this example $\det J=1$, so this would not even repair the error.

Accept the new rate: a nonlinear map changes the physical experiment.

This is a change of labels for the same path and field. An active physical deformation would be a different experiment.

Use the inverse transpose for covector columns.

Yes. A covector consumes a vector; its transformation must undo the vector’s transformation inside the pairing.

A hint

Write the desired identity $\omega'V'=\omega V$, then substitute $V'=JV$.

NOW CHANGE THE EXAMPLE

Keep $u=x$, $v=y+x^2$, but use $f=x+2y$. At $x=2$, a path has $(dx/d\lambda,dy/d\lambda)=(3,-2)$. Find $df/d\lambda$ using the new coordinates.

A hint

First obtain $V'=(3,10)$. Rewrite $f$ in terms of $u,v$ before finding its derivatives.

Work through the solution

The new formula is $f'=u+2v-2u^2$. At $u=2$, $df=(-7)\,du+2\,dv$, so $df/d\lambda=(-7)(3)+2(10)=-1$. Directly, $1(3)+2(-2)=-1$ as well.

A vector and a covector change their component lists together so that the measurement they make together stays fixed.

2.5 Tensors, products, and contractions#

Chapter 0 introduced a bilinear form: a rule that takes two vectors and returns a number, linear in each input separately. Write it as

There are two independent sums. The components are the values on pairs of basis vectors: $B_{\mu\nu}=B(e_\mu,e_\nu)$.

Change coordinates in each input, using $V^\mu=K^\mu{}_{\alpha}V'^\alpha$ and $W^\nu=K^\nu{}_{\beta}W'^\beta$. Then

so the new components are

Each lower index contributes one inverse Jacobian. For the linear map in §2.2, the output vector contributes a forward Jacobian and the input contributes an inverse Jacobian:

These are examples of tensors. A tensor of type $(r,s)$ has $r$ upper and $s$ lower indices in its component description, transforming with one $J$ for each upper index and one $K$ for each lower index. Equivalently, it can be viewed as a multilinear rule taking $r$ covectors and $s$ vectors and returning a number. A bilinear form is type $(0,2)$. A linear map is type $(1,1)$: apply it to a vector, then let a covector measure its output.

The total number of indices, $r+s$, is sometimes called tensor rank. This differs from matrix rank, the number of independent output directions of a linear map. A two-index tensor can have a matrix of rank one.

A tensor product builds a rule with separate inputs. For two covectors,

For example, let $\omega=(1,2)$ and $\eta=(2,-1)$ in our two-dimensional basis. If $V=(3,4)$ and $W=(1,3)$, the result is $11\times(-1)=-11$. Exchanging the inputs gives $7\times2=14$. The order of the inputs matters.

Contraction sums a matching upper and lower index. For a linear map, contracting its two indices gives its trace:

The matrix in §2.2 has trace $2+3=5$. Under a basis change, the $J$ and $K$ factors cancel in this sum, leaving the same answer. More generally, contraction removes one upper and one lower index and produces another tensor.

The indices also help check an equation before calculating:

A dummy index can be renamed without changing the sum: $V^\mu\omega_\mu=V^\alpha\omega_\alpha$. Free indices must match on both sides of an equation. A label must not appear three times in one product under this convention.

Finally, any two-input bilinear form can be split into a symmetric part and an antisymmetric part:

Parentheses denote symmetrization; square brackets denote antisymmetrization. Swapping the inputs leaves the first part unchanged and reverses the sign of the second. Adding the two parts recovers $B_{\mu\nu}$. For example, entries $B_{12}=5$ and $B_{21}=1$ give symmetric entry 3 and antisymmetric entry 2.

2.6 Approximations and oriented area#

The same notation lets us write Taylor’s formula for several inputs:

Here $x$ stands for the full coordinate list and $\delta x$ for a small change in that list. Repeated indices add the contributions from every direction. For $f(x,y)=x^2+3y$ near $(2,1)$, the expansion is

In this example the expression is exact. For a general smooth function, more terms remain. The notation $F=F_0+\varepsilon F_1+O(\varepsilon^2)$ means that the omitted remainder is bounded in magnitude by a constant times $\varepsilon^2$ as the chosen small parameter approaches zero, within the specified regime.

A small function need not have small derivatives. For dimensionless $x$, consider $f(x)=\varepsilon\sin(kx)$. Its magnitude is at most $|\varepsilon|$, but $f'(x)=\varepsilon k\cos(kx)$. If $\varepsilon=0.001$ and $k=10{,}000$, the function never exceeds 0.001 in magnitude while its slope can reach 10. This distinction will matter when we approximate fields and then differentiate them.

Further example: oriented area and differential forms

The antisymmetric rule

$$(dx\wedge dy)(V,W)=V^xW^y-V^yW^x$$

computes the signed coordinate area of the parallelogram spanned by two vectors. The symbol $\wedge$ is called the wedge product. For $V=(2,1)$ and $W=(1,3)$, the area is $2(3)-1(1)=5$. Swapping the vectors gives $-5$; parallel vectors give zero. This is the two-by-two determinant from linear algebra, now written as a rule for two inputs.

A smoothly varying antisymmetric covariant tensor is called a differential form. A one-form assigns a covector smoothly to each point; a two-form has two antisymmetric vector inputs, as in this area rule. Chapter 21 develops how these objects are integrated over curves and surfaces.

For now, practice expanding an indexed expression into its ordinary sums, and checking that a change of basis leaves a completed measurement unchanged. These operations will let us compare moving observers in the next chapter. David Tong’s differential-geometry notes provide a further formal treatment of vector and dual spaces.