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I am posting this here, because in my experience, this sort of thing exists in physics-related works only.

Given a local frame $\{e_{(i)}\}$ on some $n$-dimensional manifold $M$, and given a local chart $(U;x^1,...,x^n)$, I can take the components $e^\mu_{(i)}$.

The index $\mu$ is obviously a tensor index, while the index $i$ is essentially a "counting index", and if I switch to a different local chart, $e^\mu_{(i)}$ will transform as vector components. If I use a $GL(n,\mathbb{R})$-valued field to transform the local frame, then obviously $e^\mu_{(i)}$ will transform "covariantly" in the $i$ index, indicating it is a one-form field for every value of $\mu$.

Obviously, it is possible to transform the local frame into the coordinate frame associated with the chart $(U;x^1,...,x^n)$, in this case $e^\mu_{(i)}$ will take the form of $\delta^\mu_{(i)}$, and if then the local frame and the coordinate frame is transformed together, this will transform like a tensor.

So basically, I can view $e^\mu_{(i)}$ as components associated with a set of vector fields, or as the components of the identity tensor field (expressed using two different sets of basis vectors).

Now, if there is a linear connection $\nabla$ on $M$, the two different interpretations of $e^{\mu}_{(i)}$ would yield two different results under covariant differentiation, for the first interpretation, it would covariantly differentiate the $e^\mu$-s "piecewise", the second interpretation would give back zero, since the identity tensor's covariant derivative is zero under any connection.

1) Obviously, the covariant derivative is well defined, so which of the two happenings is "real"?

2) From the abstract point of view, only the first interpretation (a set of vectors) make sense to me, but from the Ricci-calculus' "it is a tensor if it transforms like a tensor" point of view, the second interpretation is equally valid.

3) My main problem isn't this "tetrad ambiguity", just it was easier to present this, this issue can get to nightmarish levels when reading articles associated with Werner Israel or his students, who love to define the projection map to a hypersurface as a coordinate frame on the hypersurface expressed in a coordinate frame on the bulk spacetime (basically, as $e^\mu_{i}$ where $\mu$ is a 4-index, and $i$ is a 3-index), as opposed to a perhaps more usual convention of working with bulk indices on the hypersurface as well, with the usual $h_{ab}=g_{ab}\pm n_an_b$.

This usually implies that when they start to take covariant derivatives or variations, funky stuff happens that I am simply incapable of expressing in any "abstract" manner. It seems to yield good results, but if they are "proper" geometrical manipulations, then there should be a corresponding coordinate-, and index-free expression, which is something I am incapable of producing.

Basically, I don't have a very clear question, since the mere existence of this problem baffles me, but if I have to ask one question, it is regarding how to resolve this apparent or real paradox.

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The index $\mu$ is obviously a tensor index, while the index $i$ is essentially a "counting index".

That is wrong. Let $U$ be a chart on $\mathcal{M}$ and $T_m,\,m\in U$ the tangent space at a point on the chart. The manifold being finite dimensional implies the existence of a set of $n$ independent vectors $\{e_i\}, i=1,\ldots,n$. There is only one index labelling the basis and one may imagine such index as the position of the only non-zero entry in $ e_k = (0,0,\ldots,1,\ldots,0)$. Any vector $X\in T_m$ can surely be expanded upon that given basis (or any other of choice) as $X = X^{i}e_i$; notice that there is still only one index coming into play.

Let $f\colon U\to W$ be a diffeomorphism between two charts (i. e. a change of coordinates). Under such diffeomorphism the basis vectors $\{e_i\}\in U$ are mapped into some other basis vectors $\{e'_j\} \in W$ by means of the push-forward map $f^*$, whose representation in either of the coordinate frames is the usual change of basis matrix (namely the Jacobian of the transformation $\Lambda$): $$ e'_j = \Lambda^i_{\phantom{i}j}\,e_i $$ whereas the components of any vector $X = X^i e_i$ transform with the inverse matrix, i. e. $$ X'^j = {(\Lambda^{-1})}^j_{\phantom{j}i}\,X^i $$ in order for the vector to always be the same object, no matter what coordinates are being used for its description. Again, there is always only one index appearing.

Obviously, it is possible to transform the local frame into the coordinate frame associated with the chart $U$, in this case $e^{\mu}_i$ will take the form of $\delta$.

It is not the $e_i$ to take the form of $\delta$, rather the Jacobian matrix $\Lambda = \mathbf{1}$, and therefore $\Lambda^i_{\phantom{i}j} = \delta^i_{\phantom{i}j}$.

So basically, I can view $e^{\mu}_i$ as components associated with a set of vector fields, or as the components of the identity tensor field (expressed using two different sets of basis vectors)

No, because of the above argument.

All the other points 1), 2), 3) follow applying the aforementioned formalism correctly and give back the expected correct results. Please notice en passant that

it is a tensor if it transforms like a tensor

is false. A tensor of type $(r,s)$ is by definition any multilinear map $\tau\colon V^r\times {V^*}^s\to \mathbb{F}$, where $V, V^*$ is a vector space and its dual, respectively, and $\mathbb{F}$ is any field (also see this other answer of mine).

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  • $\begingroup$ I am well aware of modern differential geometry and what a tensor field actually is. In fact, what I am trying to do is to match that with some heavily Ricci-calulus based papers, mainly by Israel and Muhkoyama. And it really isn't wrong. If $\{e_{i}\}$ is a frame, then for any fixed $i$, $e^\mu_i$ are the components that appear in the expansion $e_i=e^\mu_i\frac{\partial}{\partial x^\mu}$, where $x^\mu$ is the $\mu$th coordinate of a chart. Under a diffeomorphism, $e^\mu$ will change the usual way. I am not changing the frame, only the chart. $\endgroup$ – Bence Racskó Dec 11 '15 at 0:49
  • $\begingroup$ What you call $\partial_{\mu}$ is actually the basis itself (by definition of vector as directional derivatives); you should in fact use the same latin index. The basis index is the only index that gets transformed and in your example the $i$ doesn't actually play any role (you can call the vector $X$ and expand it upon $\partial_{\mu}$). Moreover, changing the chart does change the frame, by definition of chart. $\endgroup$ – gented Dec 11 '15 at 8:21
  • $\begingroup$ If I have an anholonomic frame, it is simply a set of linearly independent vector fields defined on some open subset. It is completely independent of any chart. I am not talking about a chart frame, I am talking about a completely arbitrary frame. I am using the latin index to number the anholonomic frame and to denote components taken w.r.t said frame, and I am using greek indices to denote the holonomic frame and componenets taken respect to that. $\endgroup$ – Bence Racskó Dec 11 '15 at 10:44
  • $\begingroup$ No, you're not doing that if you use the notation $\partial_{\mu}$, in fact such notation means by definition that you are choosing a coordinate frame on a chart. If you want to start from an arbitrary set of fields that you reduce to the $e_i$ and all the above applies (unless I have misunderstood your terminology). $\endgroup$ – gented Dec 11 '15 at 10:56
  • $\begingroup$ I am using both a chart and a frame, and it is absolutely possible. Btw, although my primary problem is interpreting some of Muhkoyama's works, the tetrad example I have taken is from Carroll's Spacetime and Geometry (appendix J), and he does exactly this. He even mentions that some authors define the covariant derivative of the tetrad differently, which is what baffles me, since it should be unique. $\endgroup$ – Bence Racskó Dec 11 '15 at 11:01

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