I'm reading a Quora answer on an intuitive explanation of covariant/contravariant components of vectors. If we have a coordinate system with straight coordinate axes, the geometric explanation given is that a vector's covariant components in such a system will be perpendicular projections on the axes, whereas its contravariant components will be parallel projections.

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However, I'm not sure how this translates to curvilinear coordinate systems. What's the geometric interpretation of co and contravariant components of a vector in such a system, and how is it related to the algebraic definition? (Picture a scenario like the following)

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Another thing that I don't understand: aren't covectors, and not vectors, the mathematical objects that have covariant components? Why do we say that a vector can have both covariant and contravariant components? From what I've read, vectors are rank $(0,1)$ tensors with contravarying components only, and covectors are rank $(1,0)$ tensors with covarying components only. I'm a bit confused.


2 Answers 2


However, I'm not sure how this translates to curvilinear coordinate systems.

In a curvilinear coordinate system, the "straight" axes are defined pointwise as the tangent lines to the coordinate lines at that point. In other words, and perhaps with greater intuitivity, the coordinate curves are curves on the manifold, which at each point $x\in M$ define a "linear" coordinate system of the linear algebraic kind in the tangent space $T_xM$ at that point.

So, they work the same way as they do with linear coordinate system, but you have to consider each tangent space separately.

Another thing that I don't understand: aren't covectors, and not vectors, the mathematical objects that have covariant components? Why do we say that a vector can have both covariant and contravariant components?

Using standard terminology, vectors have contravariant components and covectors have covariant components.

However there is a common "abuse of perspective" so to speak. If your manifold $M$ is equipped with a nondegenerate metric tensor $g:TM\times_M TM\rightarrow\mathbb R$, then, as OP is likely aware, the metric realizes a (strict) vector bundle isomorphism between the tangent bundle and the cotangent bundle (or algebraicly speaking, an inner product realizes an isomorphism between the vector space and its dual). Let us denote this isomorphism as $\sharp:T^\ast M\rightarrow TM$ ("raising") and its inverse as $\flat:TM\rightarrow T^\ast M$ ("lowering").

By the usual property of dual spaces, if $e_a$, ($a=1,..,n$) is a local frame for $TM$, then there is a unique local frame $\theta^a$ ($a=1,...,n$) for $T^\ast M$ that satisfies $\theta^a(e_b)=\delta^a_b$. This is the well-known dual frame.

But then we can apply "raising" to the $\theta^a$ and obtain $e^a:=\sharp\theta^a$. These are now local vector fields (rather than 1-forms/covectors) on $M$ that satisfy (by the definition of the metric isomorphism) $g(e^a,e_b)=\delta^a_b$.

If one's so willing, one may call $e^a$ to be a "reciprocal frame" rather than a dual frame, since the elements of the reciprocal frame are local vector fields, not covector fields.

We can then say that if a vector or local vector field $v$ is given, we may express it as $v=v^a e_a=v_a e^a$ in either frame. As it can be easily seen, the "covariant components" $v_a$ have the exact same properties and form as the components of $\flat v$ (the "lowering" of $v$) in the dual frame.

Thus one may dispense with covectors and bigraded tensors altogether, and consider all tensors to be graded by degree/rank/order (number of indices) alone with the understanding that all indices can be taken with respect to either a frame or its reciprocal frame, and then everything will work precisely the same way as if one had considered bigraded tensors from the get-go, but one does not have to also learn about dual spaces and whatnot.

However this way of looking at things is "unnatural", and requires a fixed metric tensor.

Another thing that I don't understand: aren't covectors, and not vectors, the mathematical objects that have covariant components?

OP didn't mean this but I feel compelled to also say here that the "traditional" terminology of contravariant and covariant is completely backwards with respect to the modern category-theoretic point of view. If $M,N$ are smooth manifolds and $\phi:M\rightarrow N$ is a smooth map, then tangent vectors (which are the ones with contravariant components, traditionally) are transported along the map $\phi$ via the tangent functor $T:\mathsf{Diff}\rightarrow\mathsf{VecBun}$, which is a covariant functor, and covectors (which are the ones with covariant components, traditionally) are transported backwards along the map $\phi$ via the cotangent functor $T^\ast:\mathsf{Diff}\rightarrow\mathsf{VecBun}$, which is a contravariant functor.

Since when $\phi$ is a diffeomorphism, this is essentially the coordinate-free version of the change of coordinates formula, tangent vectors should be called covariant and 1-forms contravariant.

  • $\begingroup$ Thank you! It'll take me a while to completely go through your answer, but could you elaborate on the first part more? Let's say I have a vector like this: i.sstatic.net/wHM0D.png (also edited the question to include the figure). You mentioned that we have to use the tangent space at each point as a pointwise coordinate system, but how? How would we specify the contra- and covariant compoenents of a vector using these pointwise coordinate systems? Would greatly appreciate further clarification, sorry if I sound dense $\endgroup$
    – Shirish
    Dec 26, 2019 at 20:18
  • $\begingroup$ @ShirishKulhari Can't use computer drawing so here is a shit hand-drawn figure: imgur.com/D81Eeut . The $\varphi=\text{const}$ lines are straight, so that's a bit less descriptive, but look at the $r=\text{const}$ line. The tangent vectors at the given point determine two "straight" lines. Those two straight lines are the linear coordinate lines. The contravariant comps of a vector at that point is given by parallel projection while the covariant comps are given by orthogonal projection, but since this system is orthogonal the two differ by scaling only. $\endgroup$ Dec 26, 2019 at 20:29
  • $\begingroup$ Ah I see! So then we're not calculating the components w.r.t. a global coordinate system, but instead the components depend on which point we choose for the tangent space? e.g. if you consider this figure: imgur.com/LivdZlQ , then if we want to find the coordinates of $P$, we first have to decide w.r.t. which point's tangent space. Suppose we choose $Q$, then find the coordinate curves passing through it, define unit vectors tangent to the curves at $Q$ to form a linear coordinate system, and then determine the covariant $u$ component to be $QB$... [cont'd in next comment] $\endgroup$
    – Shirish
    Dec 26, 2019 at 20:52
  • $\begingroup$ ...[cont'd] If we'd selected $R$, then we'd repeat the same thing - define a linear coordinate system in the tangent space at $R$ and do a perpendicular projection to find the covariant component as $RA$. So the values of co- and contravariant compoenents will depend on whether we'd chosen $Q$ or $R$ or any other point, and there's no such thing as "global" components. Does that sound correct? $\endgroup$
    – Shirish
    Dec 26, 2019 at 20:56

I think we need to define a reciprocal basis, besides the normal basis of vectors.

The simplest case of curvilinear is polar coordinates in the plane. A point $P = [X,Y]$ in Cartesian coordinates, will be $P(r,θ) = [r cos(θ) , r sin(θ)]$.

The basis vector is now a function of $P$. For the normal basis we take the gradient:

$$\mathbf e_r = ∂P/∂r = [cos(θ), sin(θ)]$$ $$\mathbf e_θ = ∂P/∂θ = [-r sin(θ), r cos(θ)]$$

While it is orthogonal, is not orthonormal. We define reciprocal vectors: $\mathbf e^r$ and $\mathbf e^θ$ so that: $$\mathbf e^r. \mathbf e_r = \mathbf e^θ. \mathbf e_θ = 1$$ $$\mathbf e^r. \mathbf e_θ = \mathbf e^θ. \mathbf e_r = 0$$

Following that requirements:

$$\mathbf e^r = [cos(θ), sin(θ)] = \mathbf e_r$$ $$\mathbf e^θ = (1/r)[-sin(θ), cos(θ)] = \mathbf e_θ / r^2$$

Now we can find the components of a small vector in the neighbourhood of $P$. First the contravariant ones, that are the dot product with the reciprocal basis (it is easy to prove that for straight coordinates it is equivalent to drawing parallel lines to the axis):

$$\mathbf dV^i = (\mathbf dV.\mathbf e^r)\mathbf e_r + (\mathbf dV.\mathbf e^θ ) \mathbf e_θ$$

If we call: $\mathbf dV.\mathbf e^r = dr\;$ and $\;\mathbf dV.\mathbf e^θ = dθ$ $$\mathbf dV^i = dr\mathbf e_r + dθ\mathbf e_θ$$

The covariant components are the dot product of the vector with the normal basis. But it has to be expressed as a linear combination of its reciprocal basis:
$$\mathbf dV_i = (\mathbf dV^i.\mathbf e_r )\mathbf e^r + (\mathbf dV^i.\mathbf e_θ ) \mathbf e^θ$$ $$\mathbf dV_i = ((\mathbf dre_r + \mathbf dθe_θ ).\mathbf e_r )e^r + ((\mathbf dre_r + \mathbf dθe_θ ).\mathbf e_θ ) e^θ = dr\mathbf e^r + r^2 dθ\mathbf e^θ$$

The vector length is: $dV^2 = \mathbf dV^i\mathbf dV_i = [dr\mathbf e_r + dθ\mathbf e_θ][dr\mathbf e^r + r^2 dθ\mathbf e^θ] = dr^2 + r^2dθ^2$

The components of the metric tensor are: $g_{rr} = \mathbf e_r.\mathbf e_r = 1; g_{rθ} = \mathbf e_r.\mathbf e_θ = 0; g_{θr} = \mathbf e_θ.\mathbf e_r = 0; g_{θθ} = \mathbf e_θ.\mathbf e_θ = r^2$


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