Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 1 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.