I want to show $\Delta x^{\alpha}\equiv x_{2}^{\alpha}-x_{1}^{\alpha}$, the difference between two points, is not a vector.

By definition, if $\Delta x^{\alpha}\equiv x_{2}^{\alpha}-x_{1}^{\alpha}$ was a vector, following a change of coordinates $x\rightarrow x'$ we would have $\Delta x'^{\mu}=\frac{\partial x'^{\mu}}{\partial x^{\alpha}}\Delta x^{\alpha}$.

Let's take, as an example of this not being true, a flat two-dimensional space with Cartesian coordinates $x^{\alpha}=(x,y)$ and spherical coordinates $x'^{\mu}=(r,\theta)$.

Let's just look at the first coordinate in $x'^{\mu}$, $r$, and assume it transforms like a vector. So by the transformation law, $\Delta r=\frac{\partial r}{\partial x}\Delta x+\frac{\partial r}{\partial y}\Delta y=\frac{x}{r}\Delta x+\frac{y}{r}\Delta y$.

We look at two points: $p_{1}=(x=0,y=1)=(r=1,\theta=\frac{\pi}{2})$ and $p_{2}=(x=1,y=0)=(r=1,\theta=0)$. By the transformation law, we get $\Delta r=0=\frac{x}{r}\cdot(-1)+\frac{y}{r}\cdot1=\frac{y-x}{r}$.

Now I am stuck. I want to say $\frac{y-x}{r}\neq0$, but I am not sure what $x$, $y$ and $r$ in this last expression are. Do I need to plug something in? There are two points, so what are $x$ and $y$? How do I treat this expression, yielding the contradiction I want?

  1. It is easy to see that local coordinates $x^{\prime\nu}=f^{\nu}(x)$ do not transform as vector field components under general coordinate transformations, so we can not associate local coordinates $x^{\mu}$ with a vector field, i.e. a contravariant (1,0) tensor.

  2. With pt. 1 in mind the situation only grows worse if we instead consider a bi-local expression of difference $\Delta x^{\mu}$ of local coordinates. $\Delta x^{\mu}$ is not a contravariant (1,0) tensor. This seems to have been OP's question.

  3. Finally it seems instructive to consider the following situation even if OP did not ask about it: Let there be given an $n$-dimensional manifold $M$, a vector field $X\in \Gamma (TM)$, and two points $P,Q \in M$. Furthermore let there be given two local coordinate charts $U,U^{\prime}\subseteq M$ that contain both points $P,Q \in M$. Let the corresponding local coordinates for the two charts be called $(x^{\mu})_{\mu=1,\ldots, n}$ and $(x^{\prime \nu})_{\nu=1,\ldots, n}$.

    One may wonder whether the difference $$\Delta X^{\mu}~:=~ X^{\mu}(P)-X^{\mu}(Q)$$ in vector components satisfies the transformation rule for a vector under coordinate transformations, i.e. whether $$ \Delta X^{\prime \nu} ~\stackrel{?}{=}~\left. \frac{\partial x^{\prime \nu}}{\partial x^{\mu}}\right|_R \Delta X^{\mu} $$ for some point $R\in U\cap U^{\prime} $?

    The answer is Not necessarily, as simple counterexamples show.

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  • $\begingroup$ Thank you, this is indeed what I am trying to show. I tried to give a counter example in my question. Does it work? $\endgroup$ – Whyka May 10 '17 at 0:04

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