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Given is a vector space $V$ and a dual vectorspace $V^*$. Elements of the vector space are typically decomposed as linear combinations of contravariant components times covariant basis vectors,

$$\vec v = v^i \vec e_i. $$

The elements of the dual vectorspace are linearcombinations of covariant components times contravariant basis vectors,

$$\hat w = w_i \hat e^i.$$

But couldn't i have switched all the names ? Wasn't the assignment of the prependices co/contra solely based on my decision to designate the first space as the "normal" vectorspace and the second as its dual?

Assuming i was given only $V^*$ could i determine that this is a dual space and that the elements should be decomposed in a contravariant basis instead of a covariant basis ?

Hans Wurst
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  • The elements of $V^$ are linear functionals: that is if $I \in V^$ then $I : V \to K$ where $K$ is the underlying field. This is true regardless of what you’ve determined to be named. – Taylor Rendon Jun 26 '20 at 20:46
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    @TaylorRendon On the other hand, the elements of $V$ can be seen as linear functionals $V^*\to K$. So which one is truly the dual is not, a priori, obvious. – Arthur Jun 26 '20 at 20:48
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    Exactly. I can do the whole math for of change of basis stuff, starting in either space, and treating the other as dual. Everything will work out the same when looking at components except that the terms co and contra seem to be switched. – Hans Wurst Jun 27 '20 at 07:18

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It is not easy to understand what you want to know.

The concepts of "covariant" and "contravariant" are related to the behavior under a change of basis of a vector space $W$. See my answer to Why is tensor from a vector space covariant, not contravariant?

These concepts only make sense for finite-dimensional vector spaces. If we are given a basis $\{e_i\}$ of $W$, the dual vectors $e^*_i$ only form a basis of $W^*$ if $\dim W < \infty$.

If we look at both $V$ and $V^*$, we consider bases for $V$ and their dual bases for $V^*$. These are the fundaments for denoting basis vectors resp. components as covariant and contravariant.

Of course we may change perspective and consider bases of the abstract vector space $W = V^*$. Then we get decompositions of vectors as linear combinations of contravariant components times covariant basis vectors. But this means that the starting point is no longer $V$ with its bases but $W = V^*$ with its bases. A basis transformation on $W$ is dual to a basis transformation on $V$, and this flips "covariant" and "contravariant".

Paul Frost
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    This is what i wanted know. Only when i introduce two spaces and start looking at the effect and the transformation laws when changing the basis of one space it becomes a meaningful distinction. It is not a "pure property" of one vector space. Is that about right ? – Hans Wurst Jun 27 '20 at 07:16
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    When i pick a space and a basis as reference point and apply my change of basis there, i determine which basis is called covariant. I could have done the same starting in the formerly labeled dual space, but treat it now as initial vector space and start with a change of basis there. Then the vectors of the dual space would be covariant. – Hans Wurst Jun 27 '20 at 07:24
  • Yes, that is correct. Also recall that $V^{*}$ is naturally isomorphic to $V$. That is, if you start with $W = V^$, then this induces dual decompositions on $V$. – Paul Frost Jun 27 '20 at 09:35