A vector describes a direction of change. A covector measures that change. A tensor organises several such inputs without letting the coordinate system decide the result.
In Cartesian coordinates, vectors and covectors often appear as identical-looking lists of numbers. That convenience can hide an important distinction. Change to polar coordinates or an oblique grid, and the two lists obey different transformation rules. Treat them as interchangeable and a derivative, length or contraction can become wrong even when the matrix multiplication is performed perfectly.
This guide develops the distinction from first principles. We begin with velocities and directional derivatives, build the dual space of linear measurements, and then combine vectors and covectors into tensors. A complete nonorthogonal-coordinate example keeps every transformation visible. Polar coordinates provide a second test, especially for the difference between a differential and a gradient.
Read Smooth Manifolds and Maps first for charts and differentials. Use Curves and Surfaces for concrete tangent planes and metric measurements. The BTT Mathematics Hub provides the wider learning route.
Prerequisites: finite-dimensional linear algebra, dual spaces at an introductory level, partial derivatives, the chain rule and smooth coordinate changes. The article is university-level enrichment, not a school syllabus checklist. All manifolds and tensors here are real and smooth unless stated otherwise.
Reading route: tangent vectors → covectors → coordinate changes → tensor types → a complete calculation → metrics and gradients → fields and derivatives → practice → solutions.
The idea that holds the subject together
Let v be a tangent vector and α a covector at the same point. The number α(v) is the measurement of v by α. When coordinates change, the component list of v changes and the component list of α changes in the complementary way. The number α(v) does not change.
This is the simplest tensor calculation and the best check on the whole theory. Coordinate independence does not mean that every entry stays fixed. It means the entries change according to rules that preserve the underlying operation.
For example, a directional derivative of a scalar function is df(v). The vector v supplies an instantaneous motion; df supplies the local rate-measuring rule. Their pairing gives the rate of change along that motion. No metric is needed to define this pairing.
1. Tangent vectors as velocities of curves
Take a smooth curve γ(t) on a manifold M with γ(0)=p. In a chart x around p, the coordinate velocity is the derivative of x∘γ at zero. Two curves represent the same tangent vector when these coordinate velocities agree.
The definition does not depend on the chosen chart. Under a coordinate change, both velocities are multiplied by the same invertible Jacobian. Equality in one chart therefore implies equality in every compatible chart.
The set of tangent vectors at p is the vector space TₚM. If M has dimension n, then TₚM has dimension n. At a point of an embedded surface in R³, this agrees with the familiar two-dimensional vector plane of possible instantaneous velocities.
The curve-based interpretation and the derivative map are developed in Brian Conrad’s notes on derivative maps and velocity vectors. A tangent vector depends on the first-order motion at a point, not on the entire path that produced it.
A point and a tangent vector have different roles
On the unit sphere, p=(0,0,1) is a point, while v=(2,−1,0) is a possible tangent vector at p. Their coordinates both have three entries when represented in R³, but p lies on the sphere and v lies in a vector space attached to p.
Another vector at the south pole belongs to a different tangent space. In an ambient Euclidean picture we may compare their coordinate triples, but an abstract manifold does not provide a canonical identification of tangent spaces at different points. Comparing or transporting them requires additional structure or a specified map.
2. The same vector as a directional derivative
A tangent vector v at p acts on a smooth scalar function f by v(f)=d(f∘γ)/dt at t=0, where γ represents v. The result depends only on the tangent vector, not on which representative curve was chosen.
This operation is linear and satisfies the product rule v(fg)=f(p)v(g)+g(p)v(f). Conversely, a point-derivation with these properties defines a tangent vector. One may formulate this using germs of functions, which simply means that only the function’s behaviour near p matters.
In coordinates x¹,…,xⁿ, every tangent vector has the form v=Σvⁱ∂/∂xⁱ|ₚ. Its action is v(f)=Σvⁱ∂f/∂xⁱ at p. Here the partial-derivative symbols are basis vectors understood as operators on smooth functions.
For example, at p=(1,2), let v=3∂ₓ−∂ᵧ and f(x,y)=x²y. Since fₓ=2xy and fᵧ=x², we obtain v(f)=3·4−1=11. The vector has supplied a direction and speed of motion; the function has supplied the quantity whose change is being measured.
3. The tangent bundle keeps the point attached
The tangent bundle TM is the collection of all tangent spaces, keeping each vector labelled by its basepoint. Locally it is described by pairs (x,v), with n coordinates for the point and n components for the tangent vector. Thus TM is a smooth manifold of dimension 2n.
On overlapping charts, the point coordinates transform by the coordinate change and the vector components transform by its Jacobian. Those transition rules assemble the local products into the tangent bundle.
A smooth vector field V chooses one tangent vector Vₚ at each point p, varying smoothly. In a coordinate chart it has smooth component functions Vⁱ(x). The bundle viewpoint explains why the vector at p must remain in TₚM rather than being an arbitrary vector in some unrelated space.
Local product descriptions do not mean every tangent bundle is globally a product. A single global frame need not exist. The definition asks for local compatible descriptions, not one universal list of basis vectors covering the whole manifold.
4. Covectors are linear measurements of tangent vectors
The cotangent space Tₚ*M is the dual vector space Hom(TₚM,R). Its elements are covectors: linear maps α:TₚM→R.
Given a tangent basis e₁,…,eₙ, the dual basis ε¹,…,εⁿ is defined by εⁱ(eⱼ)=1 when i=j and zero otherwise. A covector is α=Σαᵢεⁱ. If v=Σvⁱeᵢ, then α(v)=Σαᵢvⁱ.
In a coordinate basis eᵢ=∂/∂xⁱ, the dual basis is dx¹,…,dxⁿ. The expression dxⁱ(v)=vⁱ means that dxⁱ extracts the ith coordinate velocity. A smooth covector field is also called a differential 1-form.
A covector is not inherently a row vector. A row is one matrix representation of its action after choosing a basis. Likewise, a vector is not inherently a column. The distinction lies in the objects’ roles and transformation laws, not in the page layout of their components.
5. The differential df is a covector before it is a gradient
For a smooth function f, its differential at p is the covector dfₚ defined by dfₚ(v)=v(f). In coordinates, df=Σ(∂f/∂xⁱ)dxⁱ.
For f(x,y)=x²y, we have df=2xy dx+x²dy. At (1,2), this is 4dx+dy. Applied to v=3∂ₓ−∂ᵧ, it returns 4·3−1=11, agreeing with the directional derivative above.
In standard Cartesian coordinates with the Euclidean metric, the gradient happens to have the same component numbers as df. That coincidence can be misleading. The differential is a covector defined without a metric. The gradient is a vector obtained only after choosing a metric. Later examples will show their components separating.
The dual-space and multilinear foundations are developed in Conrad’s Tensor Algebras, Tensor Pairings, and Duality. The central operation is evaluation of a covector on a vector, not an assumed identification between the two spaces.
6. Fix the direction of a coordinate change before using a matrix
Let x denote old coordinates and u denote new coordinates, with x=x(u). Define J=∂x/∂u. Its columns tell us how the new coordinate basis vectors are expressed in the old basis.
Write Vₓ and Vᵤ for column lists of the same vector’s components. Then Vₓ=JVᵤ, so Vᵤ=J⁻¹Vₓ. This follows directly from differentiating x(u(t)) along a representative curve.
For a covector, store its coefficients as a column a, so its evaluation is aᵀV. To preserve that evaluation, aᵤ=Jᵀaₓ. Indeed, aᵤᵀVᵤ=aₓᵀJVᵤ=aₓᵀVₓ.
The transpose appears because a covector acts on vectors. The inverse appears in the vector component transformation because the basis itself has changed. Memorising “multiply by the Jacobian” without stating which coordinates are functions of which is unsafe.
7. Polar coordinates expose the difference
On a polar-coordinate patch away from the origin and an angular seam, x=r cosθ and y=r sinθ. Differentiating gives the basis transformations ∂ᵣ=cosθ∂ₓ+sinθ∂ᵧ and ∂θ=−r sinθ∂ₓ+r cosθ∂ᵧ.
The corresponding differentials are dx=cosθ dr−r sinθ dθ and dy=sinθ dr+r cosθ dθ. Solving for the new covectors gives dr=cosθ dx+sinθ dy and dθ=(−sinθ dx+cosθ dy)/r.
Notice the scale difference. The coordinate vector ∂θ has Euclidean length r, while the covector dθ includes a factor 1/r when expressed in Cartesian differentials. At a larger radius, the same physical tangential velocity produces a smaller angular rate.
The duality check remains exact: dθ(∂θ)=1, dr(∂ᵣ)=1, and the crossed pairings vanish. The objects do not need equal component lists to fit together correctly.
8. Push vectors forward; pull covectors back
For a smooth map F:M→P, the differential dFₚ sends a tangent vector at p forward to a tangent vector at F(p). A covector β at F(p) can be pulled back by composition: (F*β)ₚ(v)=β(dFₚ(v)).
For F(x,y)=(x²−y,xy), write target coordinates (a,b). If β=2da+db, then F*β=2d(x²−y)+d(xy)=(4x+y)dx+(x−2)dy.
At (1,2), F*β=6dx−dy. Its value on v=(3,−1) is 19. The differential sends v to (7,5), and β(7,5)=14+5=19. This verifies the pullback from both sides of its definition.
Pulling back a smooth covector field is available for any smooth F. Pushing forward an entire vector field to a well-defined field on the target is more delicate: different source points mapping to the same target point may supply different pushed vectors.
For F(x)=x² and V=∂ₓ, the points √y and −√y over y>0 produce pushed components 2√y and −2√y. There is no single target value specified by this field. A diffeomorphism avoids the ambiguity because every target point has one source point. More general cases require a compatibility condition.
9. Tensor types name the objects being combined
At a point p, a tensor of type (r,s) belongs to the tensor product of r copies of TₚM and s copies of Tₚ*M. Equivalently, it is a multilinear real-valued operation accepting r covectors and s vectors, with the slot order specified.
A vector has type (1,0); a covector has type (0,1); a bilinear form such as a metric has type (0,2); a linear endomorphism of the tangent space corresponds to type (1,1). A scalar is type (0,0).
The total number r+s is often called tensor order. It is not the same as the matrix rank of a particular component array. A type (0,2) tensor can have a matrix representation of rank zero, one or more. Keeping these uses of “rank” separate prevents a common terminology collision.
In n dimensions, the unrestricted space of type (r,s) tensors has dimension nʳ⁺ˢ. Symmetry or antisymmetry imposes additional relations and reduces the number of independent components. The tensor product and its relation to multilinear operations are treated in Conrad’s tensor notes.
10. Tensor product is not matrix multiplication
For covectors α and β, the tensor product α⊗β is the bilinear form (α⊗β)(v,w)=α(v)β(w). It has two independent vector inputs. The operation has not summed over a shared index.
Take α=2dx+dy and β=dx−3dy. Their tensor product has coefficient matrix with rows (2,−6) and (1,−3). It need not be symmetric. On v=(1,2) and w=(3,−1), the value is α(v)β(w)=4·6=24.
Swapping the inputs gives α(w)β(v)=5·(−5)=−25. Thus a general bilinear tensor does not ignore input order. Symmetric tensors and alternating tensors are special subclasses, not automatic properties of every tensor.
The alternating combination α∧β=α⊗β−β⊗α is a 2-form. In this example α∧β=−7dx∧dy. It vanishes when the two input vectors agree and changes sign when they are exchanged. This is the entry point to differential forms and oriented integration.
11. Two matrices of the same size can represent different tensor types
A bilinear form g evaluates two vectors by VᵀgW. With x=x(u) and J=∂x/∂u, its new matrix is gᵤ=JᵀgₓJ. This is a congruence transformation.
An endomorphism A takes a vector to a vector. Its matrix changes by Aᵤ=J⁻¹AₓJ, a similarity transformation. The difference arises because the output is a vector rather than a scalar evaluation on a second vector.
Consequently, the ordinary matrix trace is invariant for an endomorphism under basis change. The sum of diagonal entries of a bilinear-form matrix is not generally invariant under an arbitrary coordinate change. The same rectangular arrangement of numbers does not identify the mathematical object.
This is a practical reason to label tensor type before manipulating indices. A calculation can use the correct matrix arithmetic on the wrong kind of object.
12. A complete nonorthogonal-coordinate example
On the half-plane x>0, introduce coordinates x=u and y=uv, with u>0. The inverse is u=x and v=y/x, so this is a smooth coordinate change. Its Jacobian has rows (1,0) and (v,u), with determinant u.
The coordinate basis vectors are eᵤ=(1,v) and eᵥ=(0,u) in Cartesian components. Except at special points, they are not orthogonal. The Euclidean metric becomes gᵤ=JᵀJ, with rows (1+v²,uv) and (uv,u²).
Its determinant is u²>0, and its first diagonal entry is positive, so it is positive definite. The plane has not become curved because this matrix varies. We have described the same Euclidean geometry using a nonuniform, generally nonorthogonal coordinate grid.
Transform one vector and check its length
At (u,v)=(2,3), the Cartesian point is (2,6). The Jacobian has rows (1,0) and (3,2), while the metric matrix has rows (10,6) and (6,4).
Let the tangent vector have new-coordinate components A=(1,−1). Its Cartesian components are JA=(1,1). The Cartesian squared length is therefore 2.
Using the new metric gives AᵀgᵤA=10−12+4=2. The component list (1,−1) did not independently determine length; the metric specified what those components mean relative to the coordinate basis.
Transform a covector and preserve its measurement
Take α=2dx+dy. Since dx=du and dy=vdu+udv, we obtain α=(2+v)du+u dv. At (2,3), its new coefficients are (5,2).
In Cartesian coordinates, α(JA)=2·1+1=3. In the new coordinates, α(A)=5·1+2·(−1)=3. This equality checks the vector and covector transformations together.
The example supplies three independent consistency checks: the coordinate map has an inverse; the metric preserves the vector’s squared length; and the covector preserves its scalar pairing. A reliable tensor calculation should finish with an invariant check of this kind.
13. A metric creates an identification that was not there before
A Riemannian metric g assigns a positive-definite inner product gₚ to each tangent space, varying smoothly. It converts a vector v into a covector v♭ defined by v♭(w)=g(v,w). This is called lowering an index.
Because g is nondegenerate, the conversion is invertible. A covector α corresponds to a vector α♯ satisfying g(α♯,w)=α(w) for every w. In component columns, v♭=gv and α♯=g⁻¹α.
The symbols ♭ and ♯ emphasise that the identification uses the metric. They are not permission to change a vector into a covector without specifying g. A different metric generally produces a different associated vector.
Dual metrics and metric-induced operations are developed in Conrad’s Operations with Metrics. In the preceding plane example, gᵤA=(4,2), which is not the same coefficient list as A=(1,−1).
14. The gradient is the metric-dual of df
The gradient of f is defined by g(grad f,V)=df(V) for every tangent vector V. Thus grad f=(df)♯. In coordinates its components are Σgⁱʲ∂f/∂xʲ, where (gⁱʲ) is the inverse metric matrix.
For the Euclidean plane in polar coordinates, g=diag(1,r²). Hence grad f=fᵣ∂ᵣ+(fθ/r²)∂θ, while df=fᵣdr+fθdθ. The angular components differ by r² because ∂θ is not a unit vector.
Worked example: a Cartesian coordinate function in polar form
Let f(x,y)=x=r cosθ. Then df=cosθ dr−r sinθ dθ. Raising the index gives grad f=cosθ∂ᵣ−(sinθ/r)∂θ.
Substitute the polar basis vectors: cosθ(cosθ∂ₓ+sinθ∂ᵧ)−(sinθ/r)(−r sinθ∂ₓ+r cosθ∂ᵧ)=∂ₓ. The result is exactly the Cartesian gradient of x.
Using the differential’s coefficients directly as vector components would not give ∂ₓ. The metric is the missing conversion step. The calculation is a useful repair for the habit of treating every list of first partial derivatives as a vector in every coordinate system.
15. Contraction pairs one vector role with one covector role
Contraction reduces tensor order by evaluating a covector factor against a vector factor. The elementary pairing αᵢvⁱ is a contraction. For an endomorphism with components Aⁱⱼ, the trace ΣAⁱᵢ is another.
A type (0,2) tensor has two covector factors and no vector factor available for a natural contraction. Summing its diagonal entries in a general coordinate basis is not an intrinsic operation. A metric supplies the inverse tensor gⁱʲ, allowing the metric trace ΣgⁱʲTᵢⱼ.
For the metric itself on an n-dimensional manifold, Σgⁱʲgᵢⱼ=n. In our two-dimensional oblique-coordinate example, the ordinary matrix trace of g at (2,3) is 14, while its metric trace is 2. The latter records the dimension; the former depends on the coordinate basis.
This difference explains the upper and lower placement of indices. They are not merely stylistic notation. They identify which transformation rule applies and which contractions are natural.
16. Tensor fields must agree on chart overlaps
A tensor field assigns a tensor of the same type to every point, with smooth component functions in each chart. On overlaps, the components must transform according to the corresponding vector and covector factors.
Writing a smooth array in one chart is enough to define a local tensor field there. It is not automatically a global field. Extending the same numerical array to every chart without applying the transformation law usually defines inconsistent objects.
For a simple one-dimensional example, let y=2x. If V=∂ₓ, then V=2∂ᵧ. Declaring its component to be one in both coordinate systems would produce two different vectors. For the covector dx, the corresponding expression is dx=(1/2)dy. The pairing remains dx(∂ₓ)=1.
This is why a tensor is more than a multidimensional array. An array stores components in one representation. The tensor includes the meaning that determines how those components must change.
17. Ordinary derivatives of vector components are not automatically tensors
When vector components transform using a position-dependent Jacobian, differentiating those components introduces derivatives of the Jacobian. These extra terms prevent ordinary partial derivatives of component lists from transforming as a tensor in general coordinates.
Even a constant Cartesian vector field has changing polar components. For V=∂ₓ, the expression is V=cosθ∂ᵣ−(sinθ/r)∂θ. Differentiating only those coefficients does not account for the changing basis vectors.
A connection supplies a rule for differentiating vector and tensor fields while accounting for this change. A metric can determine a distinguished connection under additional compatibility conditions, but a connection is not part of the bare definition of a smooth manifold.
The same warning applies to the matrix of second partial derivatives of a scalar function. At a critical point, that matrix transforms as a bilinear form because the first-derivative correction vanishes. Away from critical points, a coordinate-independent Hessian requires additional structure such as a connection.
18. The Lie bracket is an intrinsic derivative comparison
Vector fields act on smooth functions. Define their Lie bracket by [X,Y](f)=X(Y(f))−Y(X(f)). The second-derivative terms cancel, leaving another vector field. This construction does not require a metric or connection.
In coordinates, [X,Y]ⁱ=Σ(Xʲ∂ⱼYⁱ−Yʲ∂ⱼXⁱ). Coordinate basis fields commute: [∂/∂xⁱ,∂/∂xʲ]=0. A general frame, however, need not commute.
In the Euclidean plane, the polar unit frame is eᵣ=∂ᵣ and eθ=(1/r)∂θ. Its bracket is [eᵣ,eθ]=−(1/r²)∂θ=−eθ/r. The coordinate frame commutes, but normalising one of its vectors has introduced a position-dependent factor.
The derivation viewpoint and Lie bracket are treated in Conrad’s Derivations and Vector Fields and Meinrenken’s chapter on vector fields. The bracket depends on the nearby fields, not only on their values at one point; it should not be confused with a bilinear tensor on a single tangent space.
19. What the distinction changes in modelling
A velocity is naturally a tangent vector. The differential of a scalar objective is naturally a covector. To turn that objective differential into a direction of steepest ascent, one must specify how the size of a direction is measured. That is a metric choice.
Suppose two coordinates use very different scales. Applying a Euclidean dot product to the raw component lists chooses a particular geometry in those coordinates. It is not automatically invariant under rescaling. Tensor notation makes the choice visible instead of hiding it inside an apparently innocent numerical operation.
The mathematics does not select the right physical or statistical model by itself. It provides a consistency test: once the geometric objects and metric are chosen, changing coordinates should not change the interpreted result.
20. A calculation routine that catches the common errors
First write the basepoint, coordinate domain and tensor type. Then state the direction of the coordinate map and define its Jacobian. Transform bases or components consistently, not both in the same direction by habit.
Before identifying a vector with a covector, name the metric. Before taking a trace, inspect the input and output roles. Before differentiating a field, decide whether the operation is an ordinary coordinate derivative, a Lie bracket or a covariant derivative.
Finally, check an invariant: a scalar pairing, a squared length, the trace of an endomorphism or the value of a tensor on specified inputs. A component transformation is unfinished until its intended meaning has survived a check.
21. Independent practice: twelve questions
All vectors and covectors in each question are based at the same stated point unless a map explicitly changes the basepoint. Keep coordinate and orthonormal bases separate.
- At (1,2), let v=3∂ₓ−∂ᵧ and f=x²y+y². Compute df and df(v).
- Let α=2dx−3dy, β=dx+dy, v=(1,2) and w=(3,−1). Find α(v), (α⊗β)(v,w) and (α∧β)(v,w).
- Under x=u,y=uv, find the coordinate bases ∂ᵤ,∂ᵥ and the covectors du,dv in Cartesian notation.
- At (u,v)=(2,3), transform the vector with components (1,−1) to Cartesian coordinates and check its Euclidean squared length using the transformed metric.
- Transform α=2dx+dy to (u,v) coordinates and evaluate it on the vector in Question 4.
- For the metric matrix with rows (10,6) and (6,4), lower the index of A=(1,−1). Raise the index of the resulting covector and check that A returns.
- In Euclidean polar coordinates, find df and grad f for f=r².
- In Euclidean polar coordinates, find df and grad f for f=r sinθ. Verify the Cartesian gradient.
- For F(x,y)=(x²−y,xy), pull back β=2da+db and evaluate at (1,2) on v=(3,−1).
- For X=∂ₓ and Y=x∂ᵧ on R², compute [X,Y].
- Compute [∂ᵣ,(1/r)∂θ] on a polar chart. Why does it differ from [∂ᵣ,∂θ]?
- Let x=2u,y=3v. Compare the transformed Euclidean metric matrix with the transformed identity endomorphism matrix. What are their ordinary matrix traces?
22. Worked solutions and interpretation
1–2. Respect the input slots
1. Since fₓ=2xy and fᵧ=x²+2y, df at (1,2) is 4dx+5dy. Thus df(v)=4·3+5·(−1)=7. The answer is a scalar rate, not a vector. Including the extra y² term changes the second coefficient, so copying the earlier example’s differential would be incorrect.
2. α(v)=2−6=−4 and β(w)=3−1=2, so (α⊗β)(v,w)=−8. Also α(w)=6+3=9 and β(v)=3. Therefore (α∧β)(v,w)=−8−27=−35. Equivalently, α∧β=5dx∧dy and det(v,w)=−7, again giving −35.
3–5. Verify vector–covector compatibility
3. Differentiating the coordinate map gives ∂ᵤ=∂ₓ+v∂ᵧ and ∂ᵥ=u∂ᵧ. The inverse coordinates are u=x,v=y/x, so du=dx and dv=−(v/u)dx+(1/u)dy. Applying these covectors to the two new basis vectors gives the identity pairing matrix.
4. The Jacobian at (2,3) is the matrix with rows (1,0) and (3,2). Multiplying by (1,−1) gives (1,1). Its squared length is 2. The new metric gives 10−12+4=2, so the geometry agrees in both coordinate systems.
5. The transformed covector is (2+v)du+u dv, which at the point is 5du+2dv. Evaluated on (1,−1), it gives 3. In Cartesian components the same value is 2·1+1·1=3.
6–8. Use the metric, not visual similarity of components
6. Multiplying the metric by A gives the covector coefficients (4,2). The inverse metric has rows (1,−3/2) and (−3/2,5/2), since the determinant is 4. Multiplying that inverse by (4,2) returns (1,−1). Lowering and raising are inverse operations for this nondegenerate metric.
7. df=2r dr. The inverse metric is diag(1,1/r²), so grad f=2r∂ᵣ. There is no angular component. In Cartesian coordinates this is 2x∂ₓ+2y∂ᵧ, the gradient of x²+y².
8. df=sinθ dr+r cosθ dθ, while grad f=sinθ∂ᵣ+(cosθ/r)∂θ. Expanding the basis vectors gives ∂ᵧ. This is the expected gradient because r sinθ=y. The factor 1/r in the vector expression is required.
9–12. Distinguish maps, fields and tensor types
9. F*β=(4x+y)dx+(x−2)dy. At (1,2), this is 6dx−dy, which gives 19 on v. The pushed vector is (7,5), and β returns 2·7+5=19. The agreement is the defining pullback identity.
10. Acting on a test function f, X(Yf)=∂ₓ(xfᵧ)=fᵧ+xfₓᵧ and Y(Xf)=xfᵧₓ. The mixed derivatives cancel, leaving fᵧ. Thus [X,Y]=∂ᵧ. The changing coefficient x is responsible for the nonzero bracket.
11. Differentiate the coefficient 1/r along ∂ᵣ. The result is −1/r², so the bracket is −(1/r²)∂θ. The coordinate fields ∂ᵣ and ∂θ commute, but the normalised angular field has a variable coefficient.
12. J=diag(2,3). The Euclidean metric transforms to JᵀJ=diag(4,9), whose ordinary matrix trace is 13. The identity endomorphism transforms to J⁻¹IJ=I, whose trace is 2. Their matrices both started as I, but their tensor types require different transformation laws. The metric trace of the transformed metric, using its inverse, is still 2.
23. Questions that clarify the whole subject
Is every matrix a tensor?
A matrix can represent several different kinds of linear or multilinear objects once bases and transformation rules are specified. Its shape alone does not identify a tensor type. A data array without geometric meaning is not automatically a tensor field on a manifold.
Are vectors and covectors really different when the dimension is finite?
They have the same dimension and can be identified after a suitable choice, such as a metric. But equal dimensions do not make that identification canonical. Keeping them distinct explains why differentials, gradients and coordinate transformations behave as they do.
Why is the coordinate angular vector not a unit vector?
Changing θ by one radian travels a distance proportional to r. The coordinate vector records change per unit angle, so its length is r. Dividing it by r produces the unit angular vector, which is a different basis choice.
What should remain unchanged after a coordinate calculation?
The underlying point, vector, covector or tensor remains the same when only coordinates change. Scalar evaluations such as α(v) and g(v,v) agree. Individual components generally do not. This is the most useful verification principle to carry into later tensor calculus.
Sources and further study
Brian Conrad, Derivative Maps, Parametric Curves, and Velocity Vectors, explains tangent-space derivatives. His Tensor Algebras, Tensor Pairings, and Duality develops multilinear algebra; Operations with Metrics develops metric constructions; and Derivations and Vector Fields provides the operator viewpoint. Eckhard Meinrenken’s Differentiable Manifolds connects fields, flows, differential forms and bundles.
The next step is the alternating part of tensor calculus: differential forms, wedge products, pullbacks and integration. That route explains why the fundamental theorem of calculus, Green’s theorem and the divergence theorem belong to one broader structure. Return to the BTT Mathematics Hub.