I have just started learning differential geometry and I am trying to answer the following exercise (it's not for homework):

Let $M$ be an $n$-dimensional manifold with chart $(U, \varphi_U)$ and let $(x^1, \dots, x^n)$ be the corresponding local coordinates. Define a vector field on $U$ by $X = X^i\frac{\partial}{\partial x^i}$. We will try to define a "derivative of vector fields" by $$d_Y X = d_Y X^i \frac{\partial}{\partial x^i}$$ By changing coordinates to $(\bar{x}^1, \dots, \bar{x}^n)$, show that unlike $df$, this operator is not independent of the choice of local coordinates.

I think I understand the method, but I am not certain of the details. I believe what I have to do is first apply $d_Y$ to $X$ and then change coordinates, and then change coordinates and apply $d_Y$ to $X$, and show that the two results are not then same. Here is what I have attempted:

Let $Y = Y^i\frac{\partial}{\partial x^i}$ (the question doesn't define $Y$ but presumably it's just another vector field). Then $$d_Y X = d_Y X^i \frac{\partial}{\partial x^i}$$

$$= dX^i(Y) \frac{\partial}{\partial x^i}$$

$$= \frac{\partial X^i}{\partial x^j}Y^j \frac{\partial}{\partial x^i}$$

This is just another vector field, so transforming it according to the contravariant transformation we get the following messy expression:

$$\frac{\partial \bar{x}^i}{\partial x^j} \frac{\partial X^j}{\partial x^k} Y^k \frac{\partial}{\partial \bar{x}^i},$$

which is $d_Y X$ in $\bar{x}^i$ coordinates (if I'm doing this correctly). Now I am trying to first change coordinates and apply $d_Y$ to $X$: first, in the new coordinates the vector field $X$ will be $\bar{X} = \frac{\partial \bar{x}^i}{\partial x^j}X^j \frac{\partial}{\partial \bar{x}^i}$ (contravariant transformation). So applying $d_Y$ to this we have $$d_Y \bar{X} = d_Y \left(\frac{\partial \bar{x}^i}{\partial x^j}X^j \right) \frac{\partial}{\partial \bar{x}^i}$$

Now I'm not sure what to do because it looks like $d_Y \left(\frac{\partial \bar{x}^i}{\partial x^j}X^j \right)$ will be terribly messy, and I don't really know how to evaluate it. Also, I don't know if I'm supposed to transform $Y$ to new coordinates as well (this would make the calculation even uglier) or if it's just $X$ that is transformed.

Could anyone help?

  • $\begingroup$ On second thought, I suppose $d_Y\left(\frac{\partial \bar{x}^i}{\partial x^j} X^j \right)$ won't be too messy - it's just that it involves second derivatives (I think?) $\endgroup$ – rt93 Aug 21 '12 at 14:54

You are basically on track. It is not $d_Y \bar{X}$, though you are supposed to calculate, rather still $d_Y X$, since $X$, as a vector field, is known to be independent of the representation. If you want to calculate $d_y($ugly expression$)$ in the new coordinate system then you will, in fact, have to express $Y = \bar{Y}^k\frac{\partial }{\bar{x}^k }$ And yes, you are supposed to then calculate $$d_{Y}\left(\frac{\partial \bar{x}^k}{\partial x^j}X^j\right) = \frac{\partial}{\partial \bar{x}^l}\left(\frac{\partial \bar{x}^k}{\partial x^j}X^j\right)\bar{Y}^l$$ and compare it to the other term you got earlier. It's just the product rule, though, nothing to be afraid of. If you want to dig deeper into differential geometry this will become daily business, so the earlier you start the further you may get ;-)

(Writing this kind of thing, esp if it get's even more complicated, with $\TeX$, can, admittedly, be kind of awkward).


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