# Regarding confusion of basis tensors and the usage of tensors.

Let us for example give a tensor example of following: $X = X^i \partial_i$. According to mny knowledge, in this case $\partial_i$, basis, is treated as tensor (otherwise, $X$ as tensor won't be invariant - am I wrong here?) - but this brings the question of what the "scalar part" $X^i$ would be. It seems to me at this point that this component ("scalar") part would depend on the coordinate system chosen to "interpret" $\partial_i$ - but this is definitely wrong, as then tensors will be useless. Any help here?

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Your tensor is a vector field. $X =X^i\partial_i$ implicits a sum over the dimension of the space (manifold) in which you're working. If $x^j$ for $j=1,2,\dots n$ are coordinate mappings on this space then $\partial_i = \frac{\partial}{\partial x^i}$ are the induced partial derivatives of these mappings. In particular, we can derive $\partial_i x^j= \frac{\partial x^j}{\partial x^i} = \delta_{ij}$ which nicely extends the known calculus of $\mathbb{R}^n$ to the manifold (in this case a particular coordinate patch suffices for the dicussion). The coefficients or "scalar" part are found from letting the vector field act on the coordinate functions: $$X(x^j) = X^i\partial_i x^j = X^i\delta_{ij} = X^j.$$ This formula shows these scalar-valued functions (there are $n$ of them) depend on the coordinate system in a very particular manner. Indeed, in view of the chain-rule they depend in a inverse manner in comparison with the coordinate derivatives. Let me elaborate. Suppose $\bar{x}^j$ forms another coordinate system on the set in question. Then: (by the chain-rule lifted to manifolds) $$\frac{\partial}{\partial \bar{x}^i} = \frac{\partial x^j}{\partial \bar{x}^i}\frac{\partial}{\partial x^j}$$ But $X = \bar{X}^j\bar{\partial }_j$ and so: $$X = \bar{X}^j\bar{\partial }_j = X^i\partial_i \ \ \Rightarrow \ \ \bar{X}^j\frac{\partial x^i}{\partial \bar{x}^j}\frac{\partial}{\partial x^i} = X^i\frac{\partial}{\partial x^i}$$ We derive: $$\bar{X}^j\frac{\partial x^i}{\partial \bar{x}^j} = X^i$$ In contrast: $$\frac{\partial}{\partial \bar{x}^i} = \frac{\partial x^j}{\partial \bar{x}^i}\frac{\partial}{\partial x^j}$$ The "scalars" transform inversely to the coordinate-derivative vector fields. The term "scalar" refers to the fact that $X^i$ for a particular point on the manifold gives you a number. However, it is not an invariant like mass, charge or color. The term "scalar" as it is sometimes used in physics more aptly attaches to an object like $X$ itself. A scalar in physics, in my limited experience, is an object for which there is some conservation of indices. You have an equal number of covariant and contravariant indices, or for a spinor the proper sort of up/down structure in the case of Weyl-spinors.