1
$\begingroup$

I understand that the covariant and contravariant components of a vector are the same in Cartesian co-ordinate system, but they are generally different in some curvilinear co-ordinate system. When writing a vector, we write the components multiplied by the corresponding base vector.

What I don't understand is, when we write the covariant components of a vector we put the corresponding covariant base vector with it and not the contravariant, why is that?

$\endgroup$
1
$\begingroup$

Vector spaces have an axiomatic description; here, a basis can be chosen and with respect to this basis any vector can be expressed via a set of components; in this description there is no notion of a covariant & contravariant vector.

However, once we introduce an inner product:

$g: U \times V \rightarrow R $

a distinction does arise in a symmetric way because of the way we treat bases and change of bases here; suppose we have a basis $e^i$ of the vector space $U$ and a basis $f_j$ of the vector space $V$. We say that they are dual when

$g(e^i, f_j) =\delta^i_j$

Now suppose we transform the basis of $U$, then we can ask how does the basis of $V$ change in order to keep this equation invariant? Suppose, as it must be, the former transformation is effected by a linear matrix $T$ then the latter changes by $T^{-1}$; it's for this reason that we call the vector space $U$ covariant, and $V$ contravariant.

This is touched on, but not fully explained, in Diracs book on GR; and expanded upon in Weinbergs book on the same. It's common nomenclature in physics but not so much in mathematics, here a different notion of covariance and contravariance is given that is coordinate free and doesn't rely on bases.

Given a map between manifolds

$f: M \rightarrow N$

Then we have the tangent functor $T$

$Tf: TM \rightarrow TN$

and the cotangent functor $T^*$

$T^*f : T^*N \rightarrow T^*M$

Notice that the arrow for $Tf$ goes in the same direction as $f$, for this reason we say that $T$ is a covariant functor, or just a functor; and notice, that the arrow for $T^*f$ goes in the opposite direction to $f$, and for this reason we call it a contravariant functor, or just a cofunctor.

So the tangent functor is covariant and the cotangent functor is contravariant; this notion of covariance & contravariance is pervasive in mathematics and different from the physical one already described (though related).

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for the clarification but I cannot understand form this answer why we have to write a vector with covariant components when expressing in contravariant basis and vice-versa. Why we can't write a vector in covariant basis with corresponding covariant components. $\endgroup$ – Samapan Bhadury Oct 6 '17 at 5:01
  • $\begingroup$ @bhadury: see what I said in my first paragraph and then think about it. $\endgroup$ – Mozibur Ullah Oct 6 '17 at 8:16
  • $\begingroup$ @bhadury: I can't help you to think, you have to do that bit yourself...! $\endgroup$ – Mozibur Ullah Oct 6 '17 at 8:19

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.