# Are covariant vectors representable as row vectors and contravariant as column vectors

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?

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the gradient $\nabla f$ should be represented as a column vector as well - the dual row vector is given by the differential $\mathrm df$ –  Christoph Sep 20 '12 at 7:19
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? –  linello Sep 20 '12 at 8:49
@Christoph: $\nabla f=\mathrm{d}f$. –  Siyuan Ren Sep 20 '12 at 9:29
@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 –  Christoph Sep 20 '12 at 10:32
so the usual gradient of a function in cartesian coordinates is or not a covariant vector representable with row vector? I've been lost... –  linello Sep 20 '12 at 11:34

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.

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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? –  linello Sep 21 '12 at 7:06
@linello: essentially correct; that's also what happens when you represent a bilinear map $A:(u,v)\mapsto\mathbb R$ as a matrix via $u^TAv$ –  Christoph Sep 21 '12 at 9:21

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://www.mat.univie.ac.at/~neum/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).

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So you are suggesting that is always true to treat covariant vectors as row vectors and contravariant as column vectors? –  linello Sep 20 '12 at 11:35
@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. –  Arnold Neumaier Sep 20 '12 at 12:45
@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 $T_{ab}$. –  Jerry Schirmer Sep 20 '12 at 14:29
@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. –  Arnold Neumaier Sep 20 '12 at 16:06
@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. –  Jerry Schirmer Sep 21 '12 at 15:02

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!

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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 –  Christoph Sep 20 '12 at 12:04
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. –  TMS Sep 20 '12 at 12:53

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.

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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 $\nabla f^a=g^{ab}df_a$. –  Arnold Neumaier Sep 20 '12 at 16:11
@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 –  Christoph Sep 20 '12 at 17:13