0
$\begingroup$

I'm trying to understand covariant derivative in the simplest case of $\mathbb{R}^n$ here.

On a Riemannian manifold $(M,g)$, for a connection $\nabla$ which is compatible and torsion-free, I have been shown that the Koszul formula

$$ 2g(\nabla_XY, Z) = \partial_X (g(Y,Z)) + \partial_Y (g(X,Z)) - \partial_Z (g(X,Y))+ g([X,Y],Z) - g([X,Z],Y) - g([Y,Z],X)$$

holds.

I have just started studying these thing, but I'm pretty sure $\mathbb{R}^n$ and the usuals operators have all the good properties (compatibility of the connection, etc...).

1) Is it true that for $\textbf{X, Y}: \mathbb{R}^n\rightarrow\mathbb{R}^n$ smooth vector fields, I can think of $\nabla_\textbf{X}\textbf{Y}$ as the field $\space p\mapsto\frac{\partial{\textbf{Y}}}{\partial\textbf{X}(p)}|_p$ ? (derivative of the field $\mathbf{Y}$ along the direction $\mathbf{X}(p)$ )

2) In demostrating the Koszul formula, I've written

$$ ...\space g(\nabla_XY,Z) = g([X,Y],Z)+g(\nabla_YX,Z) \space ... $$

Now with interpretation 1), since $[X,Y]=0$ in euclidean space, this piece should become $$ \langle \frac{\partial\textbf{y}}{\partial\textbf{x}},\textbf{z}\rangle = \frac{\partial\textbf{x}}{\partial\textbf{y}},\textbf{z}\rangle $$ that is to say $$ \frac{\partial\textbf{y}}{\partial\textbf{x}} = \frac{\partial\textbf{x}}{\partial\textbf{y}} $$ which is false. I guess 1) is false then, but I was pretty sure about this interpretation (since the Leibniz rule for connection seems to work perfectly). Do I have to throw away all this intuition?

$\endgroup$
1
$\begingroup$

You are right with one, but why should the Lie bracket of general vector fields be zero on $\mathbb R^n$. This is true for coordinate vector fields (and doesn't lead to a contradiction in this case) but not for general vector fields.

$\endgroup$
7
  • $\begingroup$ Ok, that's it. So, being $[X,Y]=XY-YX$, I interpret $XY$ as the operator $f\mapsto (X\circ Y)(f)$ where in euclidean space $(X\circ Y)(f)=X(Y(f))=\frac{\partial^2}{\partial X\partial Y}(f)$. This must be the point I get wrong, right? $\endgroup$ – roddik Apr 30 '20 at 11:39
  • $\begingroup$ I mean, if $X\circ Y=\frac{\partial^2}{\partial X \partial Y}$ then the Lie brackets would be $0$ for Schwartz's theorem. But I was sure $X(f)$ meant $\frac{\partial f}{\partial X}$ $\endgroup$ – roddik Apr 30 '20 at 11:49
  • 2
    $\begingroup$ I afraid that the notation you use carries a big danger of cofusion. If you view $X$ as a function to $\mathbb R^n$, then $X(f)(x)=Df(x)(X(x))$. So if you differentiate this again, you also differentiate the function $X$ and not only $f$. In fact, on $\mathbb R^n$, you can realize (in this language) the Lie bracket as $[X,Y](x)=DY(x)(X(x))-DX(x)(Y(x))$ without involving the action on a function $f$. But in any case the notation as kind of a partial derivative looks very dangerous to me. $\endgroup$ – Andreas Cap Apr 30 '20 at 13:54
  • $\begingroup$ This is definitely what's misleading me. Still, I don't grasp the right idea. Let me think about it a bit more, and thank you for your time! $\endgroup$ – roddik Apr 30 '20 at 14:46
  • 1
    $\begingroup$ There are two different descriptions here. Either you expand $X(Y(f))-Y(X)(f))$. This needs differentiating $Df(_))(Y(_))$ in direction $X$, which gives $D^2f(x)(X(x),Y(x))+Df(x)(DY(x)(X(x))$. The first summand is symmetric in $X$ and $Y$, and thus does not contribute to the difference. This computation basically also proves the fomula in my comment, in which one only differentiates the vector fields as functions with values in $\mathbb R^n$. $\endgroup$ – Andreas Cap May 1 '20 at 7:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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