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In principle, one can write quantities in a manifestly invariant - rather than covariant - fashion in e.g. special relativity. For example, rather than writing just $x^\mu$, we could write the basis explicitly, and ask that the basis transforms oppositely to the components, $$ x = x^\mu e_\mu^{(i)}, $$ such that $x$ is invariant. Why isn't such an invariant notation more common in e.g. special and general relativity? Do mathematicians also work with covariant language? or the invariant language?

Is it because the basis disappears in contractions if the basis is orthonormal? e.g. $$ x(y) = x^\mu e_\mu y_\nu e^\nu = x^\mu y_\nu \delta^\nu_\mu = x^\mu y_\mu $$

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It is used in differential geometry. For instance, you have a $1$-form $\omega=\omega_\nu dx^\nu$, a vector $v = v^\mu \partial_\mu$, where the $dx^\nu$ and $\partial_\mu$ are respectively basis of $1$-form and vectors. $\omega(v)$ is a complex quantity ($1$-form are dual quantities of vectors), and you have $\omega(v) = \omega_\nu v^\mu dx^\nu(\partial_\mu)=\omega_\mu v^\mu$ (because $dx^\nu(\partial_\mu) = \delta^\nu_\mu$) – Trimok Jun 24 '14 at 14:48

A physicist would write your first equation $x^a = x^\mu e_\mu^a$. The notation $x^a$ is invariant in your terminology. The $a$ is an abstract index. It is ostensibly not supposed to be thought of as ranging over a set of numerical values, but is just a marker that indicates that $x$ is a vector (i.e., rank 1,0 tensor.) Similarly for each $\mu$, $e^a_\mu$ is a vector. You can read more about abstract index notation on wikipedia.

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Wald is a nice reference with this notation. I believe that it depends on the nature of the problem: 1. theoretical, more abstract problems require manipulation of invariants and introducing coordinates often leads to confusion, 2. measurements, on the other hand, require introduction of a coordinate system, which is often done as a final step after statement of the theory. – auxsvr Jun 24 '14 at 17:08
@auxsvr physicists want to connect to experiments? so pick invariant notation? – innisfree Jun 26 '14 at 12:28
@innisfree The physical theory is developed in terms of invariants and the experiments require measurements, hence the introduction of coordinate systems. I'm not certain what your questions are. – auxsvr Jun 26 '14 at 13:36

Comments to the question (v5):

  1. In General Relativity (GR), the notation $x^{\mu}$, $\mu=0,1,2,3,$ usually denotes some (local) coordinates of a (spacetime) manifold $M$. Note that $x^{\mu}$ does in general not transform as a $(1,0)$ (contravariant) tensor in the sense that $$ x^{\prime \nu}~=~\frac{\partial x^{\prime \nu}}{\partial x^{\mu}} x^{\mu} \qquad (\leftarrow\text{Wrong in general!} )$$ under coordinate transformations $x^{\prime \nu}=f^{\nu}(x)$. Thus there is no useful physical concept of a vector basis in GR for the underlying local spacetime coordinates. [In Special Relativity (SR), the coordinate transformations are restricted to affine transformations $x^{\prime \nu}=\Lambda^{\nu}{}_{\mu} x^{\mu}+ a^{\nu}$ in an affine space.]

  2. On the other hand, if the notation $x^{\mu}$ is supposed to denote a $(1,0)$ (contravariant) tensor [rather than local spacetime coordinates], then one is of course welcome to introduce a corresponding dual basis $e_{\mu}$ that transform in the opposite way to keep $x=x^{\mu}e_{\mu}$ invariant. In general, it is probably a futile exercise to discuss which notation is best.$^1$


$^1$ However, have a look at e.g. this Phys.SE answer.

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