Mathematically, a vector is an element of a vector space. Sometimes, it's just an n-tuple $(a,b,c)$. In physics, one often demands that the tuple has certain transformation properties to be called a vector (as opposed to e.g. a pseudovector). Sometimes its not a tuple, but e.g. an element of a hilbert space. In all cases, its possible to define a ordinary vector (or column vector, or contravariant vector): $$\vec v = \left(\begin{align}v_1\\v_2\\v_3\end{align}\right) \qquad x^\mu \qquad \left|\Psi\right\rangle $$

and a dual (row, or covariant) vector:

$$\vec v^T = \left(v_1, v_2, v_3\right) \qquad x_\mu \qquad \left\langle\Psi\right| $$

You can define a scalar product that gives you a number (an element of the field $K$ the vector space is over). If you're feeling fancy, you can think about the row vector as an element of a dual vector space. This consists of functions (1-forms) mapping vectors to numbers.

$$\vec v \in V_K$$ $$\begin{align} \vec u^T : V_K &\rightarrow K \\ \vec v &\mapsto x\end{align}$$

Applying this function $\vec u^T$ to the vector $\vec v$ is the same as to take the scalar product $\vec u \cdot \vec v$. This is more manifest in the co/contravariant notation, or in the bra-ket notation.

My question is why do we distinguish between both kinds of vectors (in concept, not in notation)? Is there any physical meaningful distinction, or couldn't we just teach that they are all "vectors" and skip the dual part? Could we say that the raised/lowered indicies are just a helpful notation to remember what is multiplied with what, or a shortcut for $x_\mu = g_{\mu\nu}x^\nu$, but that there is actually just the vector $\mathbf{x} = (x^0, x^1,\ldots)$

I mean, do we loose anything by thinking $\left|\Psi\right\rangle \equiv \left\langle\Psi\right|$ and the different forms are just shortcuts for different products: $$\left\langle\Phi\middle|\Psi\right\rangle = \left|\Phi\right\rangle \cdot \left|\Psi\right\rangle $$ $$\left|\Phi\right\rangle\left\langle\Psi\right| = \left|\Phi\right\rangle \times \left|\Psi\right\rangle $$

Is there for example any physical situation where there is a meaningful variable $A$ that can be either $A = \vec v$ or $A = \vec v^T$, and by giving up the distinction (conceptually) we can't tell the two situations apart?

  • 1
    $\begingroup$ There are many practical, ordinary situations where you don't have a metric. See this answer for some examples: physics.stackexchange.com/a/62564/4552 . When you don't have a metric, there is no way to convert back and forth between vectors and covectors. Also, vectors and covectors transform differently under a change of coordinates, e.g., a rescaling does opposite things to them. $\endgroup$
    – user4552
    Commented May 18, 2013 at 15:00

2 Answers 2


My question is why do we distinguish between both kinds of vectors

Take an arbitrary vector space. Then, the set of scalar-valued linear functions on that space inherits a linear structure in an obvious way and becomes a vector space in its own right.

We distinguish between these spaces because they are distinct.

Now, in the finite dimensional case it's easy to show that these spaces are isomorphic (ie have the same dimension) by introducing the dual basis.

However, the isomorphism depends on the choice of basis (ie it is not canonical), and for infinite-dimensional vector spaces, it need not exists at all.

We need additional structure like an inner product to single out a specific isomorphism that makes the vector space self-dual, which then allows us to conflate vectors and co-vectors.

The difference between vectors and covectors is also the reason why we end up with a minus sign in Hamilton's equations: The time derivative of a phase space trajectory yields a vector, whereas the differential of the Hamilton function yields a covector, which need to be connected with additional structure (the symplectic product, which manifests as the minus sign only if canonical coordinates are used).

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    $\begingroup$ +1 for being distinct in the infinite-dimensional case. A good QM example would be how the delta function has no complex-conjugate analogue in $L^2(\mathbb{R})$. $\endgroup$
    – user10851
    Commented May 18, 2013 at 19:48

Short answer: we keep track of covariant and contravariant vectors because it gives us the freedom to use different quantities with respect to different bases and to find dot products of covariant and contravariant vectors without having to move everything into the tangent or cotangent basis and use the metric every time we want to find a physically meaningful dot product.

In terms of geometry, I think it is useful to distinguish between the two kinds of vectors because they will transform according to different (but related) laws under coordinate transformations.

That said, I've always found the idea that covectors are "linear functionals" meant only to act on vectors to produce numbers to be hogwash. Why not define vectors to be things that act on covectors to product numbers instead? There is an inherent primacy to vectors (over covectors) that is tacitly understood and yet not founded in the mathematics.

It can be helpful, however, to see that covectors arise in an $N$-dimensional vector space by taking $N-1$ basis vectors, forming a hyperplane with them, and then taking the normal to that hyperplane. As the hyperplane is reflected, rotated, or otherwise changed by coordinate transformations, the normal transforms as a covector. To me, this stimulates a thought: why deal with covectors at all? Why not deal directly with that hyperplane, which doesn't demand a definition of anything other than ordinary, tangent vectors?

(Of course, one could start with covectors and form hyperplanes from basis covectors, and the normals to those hyperplanes would be ordinary vectors.)

And naturally, in a metric space, we can freely talk about a quantity in terms of its contravariant or covariant components. This yields to an idea that quantities should be considered neither contravariant nor covariant at all, and it is just their components in terms of physical bases that change according to transformation laws. This is an idea I've seen in a gauge theory of gravity, where physical objects live in a third, flat spacetime halfway between tangent and cotangent space, transformed to them by the gauge field (or tetrad field). But this is only possible when there is a metric.

But typically in GR, it is not possible to talk about an object without talking about its components because you're always implicitly dealing with a coordinate basis--either the basis of tangent vectors or the basis of cotangent vectors--and you always have to remember which basis you referenced your description of the object to.

  • $\begingroup$ Why not define vectors to be things that act on covectors to product numbers instead? Because infinite-dimensional spaces need not be reflexive $\endgroup$
    – Christoph
    Commented May 18, 2013 at 16:02
  • $\begingroup$ Yes, that is true. I think what is more at issue is the choice that "vectors" mean tangent vectors and "dual vectors" mean cotangent vectors. The choice of what is original and what is dual is arbitrary. I agree that means we can't go in circles with our definitions, however. $\endgroup$
    – Muphrid
    Commented May 18, 2013 at 16:14
  • $\begingroup$ @Murphid: tangent and cotangent vectors only appear on equal footing if you use the coordinate-based definition via their transformation laws; there are two other common definitions of tangent vectors (directional derivates and equivalence classes of curves), and the latter one makes a lot of sense from a 'physical' point of view; I'm not aware of an intrinsic definition of cotangent vectors that's equally 'nice' and does not rely on the definition of tangent vectors $\endgroup$
    – Christoph
    Commented May 18, 2013 at 16:46
  • $\begingroup$ @Christoph Yes, tangent vectors can be defined through a family of curves through a point that all have the same direction. Is there something wrong with defining cotangent vectors via an equivalence class of hypersurfaces, containing a particular point, that all locally have the same linear approximation? $\endgroup$
    – Muphrid
    Commented May 18, 2013 at 17:20
  • $\begingroup$ (sorry for mangling your name in my last comment): the definition of hypersurfaces becomes easier if you already have a concept of tangent vectors, making this definition of cotangent vectors arguably less fundamental; there's actually a more severe problem: unless I'm mistaken, you need additional structure (eg a volume form) to make the hypersurface (or rather its tangent subspace of codimension 1) characterise a unique cotangent vector $\endgroup$
    – Christoph
    Commented May 18, 2013 at 18:15

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