If everything you are working with is in Euclidean 3-space (or $n$-space) equipped with the dot product, is there any reason to bother with distinguishing between 1-forms and vectors? or between covariant and contravariant tensor components? I'm fairly certain that if you do not, then none of your calculations or relations will be numerically wrong, but are they "mathematically wrong"?
Example: I'll write some basic tensor relations without distinguishing between vectors and 1-forms/dual vectors or between covariant/contravariant components. All indices will be subscripts. Tell me if any of the following is wrong:
I would often describe a rigid transformation of an orthonormal basis (in euclidean 3-space), ${\hat{\mathbf{e}}_i}$, to some new orthonormal basis ${\hat{\mathbf{e}}'_i}$ as $$ \hat{\mathbf{e}}'_i\;=\; \mathbf{R}\cdot\hat{\mathbf{e}}_i \;=\; R_{ji}\hat{\mathbf{e}}_j \qquad\qquad (i=1,2,3) $$ For some proper orthogonal 2-tensor ${\mathbf{R}\in SO(3)}$ (or whatever the tensor equivalent of $SO(3)$ is, if that's a thing). It's then pretty straightforward to show that the components, $R_{ij}$, of ${\mathbf{R}}$ are the same in both bases and $\mathbf{R}$ itself is given in terms of ${\hat{\mathbf{e}}_i}$ and ${\hat{\mathbf{e}}'_i}$ by
$$ \mathbf{R}=R_{ij}\hat{\mathbf{e}}_i\otimes\hat{\mathbf{e}}_j = R_{ij}\hat{\mathbf{e}}'_i\otimes\hat{\mathbf{e}}'_j = \hat{\mathbf{e}}'_i\otimes\hat{\mathbf{e}}_i \qquad,\qquad\quad R_{ij}=R'_{ij}=\hat{\mathbf{e}}_i\cdot\hat{\mathbf{e}}'_j $$
Then, given the basis transformation in the first equation, the components of some vector $\vec{\mathbf{u}}=u_i\hat{\mathbf{e}}_i=u'_i\hat{\mathbf{e}}'_i$ and some 2-tensor $\mathbf{T}=T_{ij}\hat{\mathbf{e}}_i\otimes \hat{\mathbf{e}}_j = T'_{ij}\hat{\mathbf{e}}'_i\otimes \hat{\mathbf{e}}'_j$ would transform as
$$ u'_i = R_{ji}u_j \qquad \text{matrix form: } \qquad [u]'= [R]^{\top}[u] \\ T'_{ij} = R_{ki}R_{sj}T_{sj} \qquad \text{matrix form: } \qquad [T]' = [R]^{\top}[T][R] $$ and for some p-tensor we would have $$S'_{j_1j_2\dots j_p} \;=\; \big( R_{ i_1j_1}R_{ i_2j_2} \dots R_{ i_pj_p} \big) S_{ i_1 i_2\dots i_p} $$
and if ${\hat{\mathbf{e}}_i}$ is an inertial basis and ${\hat{\mathbf{e}}'_i}$ is some rotating basis, then the skew-symmetric angular velocity 2-tensor of the ${\hat{\mathbf{e}}'_i}$ basis realtive to ${\hat{\mathbf{e}}_i}$ is given by
$$ \boldsymbol{\Omega} \;=\; \dot{\mathbf{R}}\cdot\mathbf{R}^{\top} \qquad,\qquad \text{componenets in } \hat{\mathbf{e}}_i \;: \qquad \Omega_{ij} = \dot{R}_{ik}R_{jk} $$
Or, in matrix form (in the $\hat{\mathbf{e}}_i$ basis) the above would be $[\Omega]=[\dot{R}][R]^{\top}$. The third equation can be used to convert to the $\hat{\mathbf{e}}'_i$ basis. The familiar angular velocity (pseudo)vector is then given by
$$ \vec{\boldsymbol{\omega}}= -\tfrac{1}{2}\epsilon_{ijk}(\hat{\mathbf{e}}_j\cdot \boldsymbol{\Omega}\cdot \hat{\mathbf{e}}_k)\hat{\mathbf{e}}_i \qquad,\qquad \text{componenets in } \hat{\mathbf{e}}_i \;: \qquad \omega_i = -\tfrac{1}{2}\epsilon_{ijk}\Omega_{jk} $$
where $\epsilon_{ijk}$ are the components of the Levi-Civita 3-(pseudo)tensor, $\pmb{\epsilon}$, which itself may be written in any right-handed orthonormal bases as
$$ \pmb{\epsilon} = \epsilon_{ijk}\hat{\mathbf{e}}_i\otimes\hat{\mathbf{e}}_j \otimes \hat{\mathbf{e}}_k = \tfrac{1}{3!}\epsilon_{ijk}\hat{\mathbf{e}}_i\wedge\hat{\mathbf{e}}_j \wedge \hat{\mathbf{e}}_k = \hat{\mathbf{e}}_1\wedge\hat{\mathbf{e}}_2 \wedge \hat{\mathbf{e}}_3 \quad,\quad \epsilon_{123}=1 $$
The time-derivative of some vector $\vec{\mathbf{u}}=u_i\hat{\mathbf{e}}_i=u'_i\hat{\mathbf{e}}'_i$ would then be given in terms of the components in the inertial and rotating bases by the familiar kinematic transport equation
$$
\dot{\vec{\mathbf{u}}} = \dot{u}_i\hat{\mathbf{e}}_i = \dot{u}'_i\hat{\mathbf{e}}'_i + \boldsymbol{\Omega}\cdot\vec{\mathbf{u}} \;=\; (\dot{u}'_i + \Omega'_{ij}u'_j )\hat{\mathbf{e}}'_i
$$
where $\boldsymbol{\Omega}\cdot\vec{\mathbf{u}} = \vec{\boldsymbol{\omega}}\times\vec{\mathbf{u}}$.
end example
question: So, I'm pretty sure that none of the above would give me numerically incorrect relations. But I called everything either a vector, 2-tensor, or 3-tensor. Nothing about forms, (1,1)-tensors, (0,2)-tensors, dual vectors, etc. Is the above formulation mathematically ''improper''? For instance, do I need to write ${\mathbf{R}}$ as a (1,1)-tensor, ${\mathbf{R}}=R^{i}_{\,j}\hat{\mathbf{e}}_i\otimes\hat{\boldsymbol{\sigma}}^j$, using the basis 1-forms $\hat{\boldsymbol{\sigma}}^j$? Does the angular velocity tensor need to be written as a 2-form or (0,2)-tensor?
context: My BS is in physics and I am currently a PhD student in engineering. Aside from a graduate relativity course I took in the physics department, I have never once seen raised indices or mention of dual vectors/1-forms in any class I have ever taken or in any academic paper, I have ever read. That was until I recently started teaching myself some differential geometry in hopes of eventually understanding Hamiltonian mechanics from the geometric view. So far, I have mostly only succeeded in destroying my confidence in my knowledge of basic tensor algebra involved in classical dynamics.
