I would like to know what are the range of validity of the following statement:

Covariant vectors are representable as row vectors. Contravariant vectors are representable as column vectors.

For example we know that the gradient of a function is representable as row vector in ordinary space $ \mathbb{R}^3$

$\nabla f = \left [ \frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z} \right ]$

and an ordinary vector is a column vector

$ \mathbf{x} = \left[ x_1, x_2, x_3 \right]^T$

I think that this continues to be valid in special relativity (Minkowski metric is flat), but I'm not sure about it in general relativity.

Can you provide me some examples?

  • 1
    $\begingroup$ the gradient $\nabla f$ should be represented as a column vector as well - the dual row vector is given by the differential $\mathrm df$ $\endgroup$
    – Christoph
    Commented Sep 20, 2012 at 7:19
  • 2
    $\begingroup$ so why on wikipedia is the gradient represented as covariant vector? en.wikipedia.org/wiki/Covariance_and_contravariance_of_vectors So how should I modify my question in order for it to be more precise? $\endgroup$
    – linello
    Commented Sep 20, 2012 at 8:49
  • $\begingroup$ @Christoph: $\nabla f=\mathrm{d}f$. $\endgroup$
    – Siyuan Ren
    Commented Sep 20, 2012 at 9:29
  • 1
    $\begingroup$ @KarsusRen: $(\nabla f)^\flat=\nabla f\rfloor g=\mathrm df$; in practice, a bit of sloppiness doesn't hurt much (after all, we can always raise or lower the index as necessary by contraction with the metric tensor), but sometimes it does matter, eg when deriving the coordinate expression for the Laplace operator (or, more precisely, the Laplace-Beltrami operator) in curvilinear coordinates $\endgroup$
    – Christoph
    Commented Sep 20, 2012 at 10:32
  • $\begingroup$ so the usual gradient of a function in cartesian coordinates is or not a covariant vector representable with row vector? I've been lost... $\endgroup$
    – linello
    Commented Sep 20, 2012 at 11:34

4 Answers 4


Yes, the statement holds true in general relativity as well. However, as we need to deal with tensors of higher and in particular mixed order, the rules of matrix multiplication (which is where the idea of the representation via row- and column-vectors comes from) are no longer sufficiently powerful:

Instead, the placement of the index determines if we are dealing with a contravariant (upper index) or a covariant (lower index) quantity.

Additionally, by convention an index which occurs in a product in both upper and lower position gets contracted, and equations must hold for all values of free indices.

If the given metric is non-Euclidean (which is already true in special relativity), mapping between co- and contravariant quantities is more involved than simple transposition and the actual values of the components in a given basis can change, eg: $$ p^\mu = (p^0,+\vec p)\\ p_\mu = (p^0,-\vec p) $$ and in general: $$ p_\mu = g_{\mu\nu}p^\nu $$ where $g_{\mu\nu}$ denotes the metric tensor and a sum $\nu=1\dots n$ is implied.

  • $\begingroup$ Ok, so in general relativity with the index notation we can extend the usual matrices from linear algebra to (1,1) tensors, while for example (0,2) (completely covariant) tensors have no corresponding matrix in usual linear algebra right? I always have to use the metric tensor to raise/lower indices and get (1,1) tensors, is it right? $\endgroup$
    – linello
    Commented Sep 21, 2012 at 7:06
  • $\begingroup$ @linello: essentially correct; that's also what happens when you represent a bilinear map $A:(u,v)\mapsto\mathbb R$ as a matrix via $u^TAv$ $\endgroup$
    – Christoph
    Commented Sep 21, 2012 at 9:21
  • $\begingroup$ @Christoph the last equation $$p_\mu = g_{\mu\nu}p^\nu$$ means a row matrix equals a square matrix times a column matric. i.e. 1by4 matrix equals 4by4 times 4by1. But a 4by4 times a 4by1 should give a 4by1. Therefore, I believe it is not so simple to represent a covariant vector as a row matrix. $\endgroup$
    – MycrofD
    Commented Feb 9, 2018 at 11:33
  • $\begingroup$ @MycrofD If you want for this to be consistent you should represent $g_{\mu \nu}$ as a row vector of row vectors, then everything is fine. This is because $g_{\mu, \nu}$ is not really a matrix, it's a $(0,2)$ tensor. The matrices from linear algebra are $(1,1)$ tensors, i.e. they are from $V \otimes V^*$ not from $V \otimes V$. $\endgroup$
    – lightxbulb
    Commented Nov 25, 2023 at 10:20

It is meaningful in general, though it is a matter of convention, not of truth. But it never leads to incorrect results if you make this convention.

This is thoroughly discussed in the entry ''How are matrices and tensors related?'' of Chapter B8: Quantum gravity of my theoretical physics FAQ at http://arnold-neumaier.at/physfaq/physics-faq.html

Note that in multivariate analysis one generally defines the gradient is the transpose of the (exterior) derivative, so ''gradient'' and ''derivative'' are slightly different notions. The transpose makes sense only given a metric, as it essentially consists in replacing raised/lowered indices by lowered/raised ones.

Thus unlike a covariant exterior derivative, a gradient is no longer covariant but contravariant (and hence a column vector).

  • $\begingroup$ So you are suggesting that is always true to treat covariant vectors as row vectors and contravariant as column vectors? $\endgroup$
    – linello
    Commented Sep 20, 2012 at 11:35
  • $\begingroup$ @linello: It is a matter of convention, not of truth. But it never leads to incorrect results if you make this convention. I added to my answer a clarifying statement. $\endgroup$ Commented Sep 20, 2012 at 12:45
  • $\begingroup$ @linello: also, be aware that while this convention can work pretty well for vectors and one-forms, it doesn't help you at all when distinguishing covarant and contravariant components of higher-rank tensors. ${T^{a}}_{b}$ is a 2x2 matrix just the same as $T_{ab}$. $\endgroup$ Commented Sep 20, 2012 at 14:29
  • $\begingroup$ @JerrySchirmer: No. Only $T^a_b$ is a matrix (linear self-mapping) on the space of column vectors, hence has a simple interpretation in linear algebra. on the other hand, 2-forms and bivectors need multilinear algebra or a distinguished metric for their proper interpretation as linear mappings. $\endgroup$ Commented Sep 20, 2012 at 16:06
  • $\begingroup$ @ArnoldNeumaier: yet people write 2-forms as matrices all of the time. Take the matrix way of writing $g_{ab}$, for examplle. Yes, the algebra doens't work, but that's kind of my point--the row vector/column vector thing breaks down immediately once you go to higher rank tensors. $\endgroup$ Commented Sep 21, 2012 at 15:02

This is not a full answer, but rather an attempt to clear up some misconception about the gradient: In particular, in my opinion saying that the gradient is a covector doesn't make much sense.

There are two ways to interpret the concept of vectors and covectors:

The first one is to say there is only a single entity - the vector - which has covariant and contravariant components. This is inspired by classical tensor calculus: when doing calculations, we often do not care about the placement of the indices of a particular tensor - after all, we can always lower or raise them (ie go from column vectors to row vectors and vice versa) by contraction with the metric tensor.

If you take this point of view, differential and gradient are two names for the same entity. It is somewhat misleading to say that the gradient is a covector, as what we really mean is that the gradient is a vector whose covariant components are given by the partial derivatives (whereas its contravariant components are given by contraction of the covariant components with the inverse of the metric tensor).

The second point of view - which is the one I prefer - is that vectors (or, more precisely as we're doing differential geometry, tangent vectors) are distinct from covectors (aka 1-forms). However, the scalar product gives an isomorphism between tangent vectors and 1-forms. The gradient is the (pre-)image of the differential under this isomorphism and an actual vector.

  • 2
    $\begingroup$ I completely disagree with your first point of view: There is no such thing as a single vector with both covariant and contravariant components, except in sloppy mode where concepts are not fully well-defined and confusion can easily arise. - I also disagree with your conclusion in that case: In coordinate-free notation, the only well-defined notion of gradient is the covector defined by $\nabla f^a=g^{ab}df_a$. $\endgroup$ Commented Sep 20, 2012 at 16:11
  • $\begingroup$ @ArnoldNeumaier: I disagree with my first point of view as well, but that's the intuition I got from visiting lectures in theoretical physics - quantities with upper or lower indices weren't discriminated, and in particular all tensors of order 1 were called 4-vectors; as to the last part of your comment: that expression defines a vector, not a covector $\endgroup$
    – Christoph
    Commented Sep 20, 2012 at 17:13
  • $\begingroup$ @ArnoldNeumaier The gradient is not a covector - it is a contravariant vector, this is even clear from your notation where you have put the index up and used the inverse of the metric tensor. The differential is a covector - for its definition it requires only a normed or topological vector space $V$ of functions, then $df\in V^*$ is from the continuous dual $V^*$. The definition of the gradient requires a Hilbert space structure and $\nabla f \in V$ is defined by $\langle \nabla f, v\rangle = df(v)$. $\endgroup$
    – lightxbulb
    Commented Nov 25, 2023 at 9:56
  • 1
    $\begingroup$ @lightxbulb: I consider df to be the covariant gradient = differential of f; it is always defined. The contravariant gradient given by my formula is defined only when an additional metric is given, hence has a less universal meaning. $\endgroup$ Commented Nov 25, 2023 at 11:11
  • $\begingroup$ @ArnoldNeumaier Now it makes sense, I guess it was a terminology mismatch. Typically with gradient in mathematics we refer to $\nabla f$ in the sense of $\langle \nabla f, v\rangle = df(v)$, where $df$ is the Frechet derivative. As you mentiomed a metric/inner product must be defined for $\nabla f$ to make sense. I have never seen the term covariant gradient used for the Frechet derivative, but maybe that's just me not having read enough, or likely a mismatch between terminology in physics and mathematics. $\endgroup$
    – lightxbulb
    Commented Nov 25, 2023 at 12:00

As from my experience, I had very hard time to understand the "physical" difference between contra- & cov- things, I really understood them only when I read differential geometry and get involved with one-forms, even more, some authors (like Shuch) argue that it is wrong to say that Covectors are really vectors, they are different objects, they are one forms!

  • $\begingroup$ TMS: you're mixing up co and contra - co-vectors are one-forms; one-forms are of course vectors insofar as they are elements of a vector space - they are just not tangent vectors $\endgroup$
    – Christoph
    Commented Sep 20, 2012 at 12:04
  • $\begingroup$ Yes sorry I miss typed it, and ofcourse they spans a vector space, anyway what Shuch meant that one-forms are duals to the usual vectors, and they can't be in the same vector space with them, thus he suggests to distinguish them, that's all. $\endgroup$
    – TMS
    Commented Sep 20, 2012 at 12:53

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