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On Carroll's spacetime and geometry book, page 67, the book gives the component form of vector field commutator $$[X,Y]^\mu=X^\lambda\partial_\lambda Y^\mu- Y^\lambda\partial_\lambda x^\mu \tag{2.23}$$ following sentence

Yet another fascinating exercise is to perform explicitly a coordinate transformation on the expression (2.23), to verify that all potentially nontensorial pieces cancel and the result transforms like a vector field.

I am trying to understand this sentence.I think

(1) $X^\lambda\partial_\lambda=X^{\lambda^\prime}\partial_{\lambda^\prime}$ is invariant under coordinate transformation;

(2) $Y^{\mu^\prime}=\frac{\partial x^{\mu^{\prime}}}{\partial x^{\mu}}Y^{\mu}$ is just the transformation of a vector field.

So where is the nontensorial pieces?

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    $\begingroup$ Well, $\partial_{\lambda'}$ will also hit the change of basis matrix. This is what gives rise to the non-tensorial part, but in the commutator it gets canceled. $\endgroup$
    – Gold
    Jan 16, 2023 at 14:27
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    $\begingroup$ @Gold, thank you very much! I think one critical point is that $\frac{\partial}{\partial x^{\lambda^{\prime}}}\frac{\partial x^{\mu^{\prime}}}{\partial x^{\mu}}=\frac{\partial^2 x^{\mu^{\prime}}}{\partial x^{\lambda^{\prime}} \partial x^{\mu}}=\frac{\partial^2 x^{\mu^{\prime}}}{ \partial x^{\mu} \partial x^{\lambda^{\prime}}}=\frac{\partial}{\partial x^{\mu}}\frac{\partial x^{\mu^{\prime}}}{\partial x^{\lambda^{\prime}}} $. So the two non-tensorial terms can cancel with each other. Would this be right? Thanks! $\endgroup$
    – Daren
    Jan 17, 2023 at 14:34
  • $\begingroup$ You are mixing derivatives in two different coordinate systems there. The argument is much simpler though, I have posted an answer with an outline. $\endgroup$
    – Gold
    Jan 17, 2023 at 15:33

1 Answer 1

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This is actually simple. Since as you point out $X = X^\lambda \partial_\lambda$ and $Y=Y^\lambda\partial_\lambda$ as vector fields (not their components) are themselves invariant, we can evaluate the transformation of the commutator as follows

$$[X,Y]^{\mu'}= X(Y^{\mu'})-Y(X^{\mu'})=X^\lambda \partial_\lambda Y^{\mu'}-Y^\lambda \partial_\lambda X^{\mu'}.$$

Then we just write $X^{\mu'}$ and $Y^{\mu'}$ in terms of $X^\mu$ and $Y^\mu$ in the original unprimed coordinates

$$[X,Y]^{\mu'}=X^\lambda \partial_\lambda \left(\dfrac{\partial x'^\mu}{\partial x^\nu}Y^\nu\right)-Y^\lambda \partial_\lambda \left(\dfrac{\partial x'^\mu}{\partial x^\nu}X^\nu\right).$$

Now use the product rule to get

\begin{eqnarray} [X,Y]^{\mu'}=\dfrac{\partial x'^\mu}{\partial x^\nu}[X,Y]^\nu+(X^\lambda Y^\nu-X^\nu Y^\lambda) \dfrac{\partial^2 x'^\mu}{\partial x^\nu \partial x^\lambda} .\end{eqnarray}

The second term is then noted to be the non-tensorial part. Observe, though, that it vanishes because $\frac{\partial^2 x'^\mu}{\partial x^\nu \partial x^\lambda}$ is symmetric while $X^{\lambda}Y^\nu - X^\nu Y^\lambda$ is anti-symmetric under the exchange $\lambda \leftrightarrow \nu$. As a result

\begin{eqnarray} [X,Y]^{\mu'}=\dfrac{\partial x'^\mu}{\partial x^\nu}[X,Y]^{\nu}.\end{eqnarray}

One can also prove that $[X,Y]$ defines a vector field without resorting to coordinates and this is quite instructive. The key thing about being a vector field is that $[X,Y]$ should obey the Liebnitz rule when acting on functions, i.e., we need

$$[X,Y](fg)=g[X,Y]f+f[X,Y]g.$$

This is a more abstract definition of a vector field: a vector field is a derivation acting on smooth functions. Using this definition it is straightforward to prove $[X,Y]$ is a vector field:

\begin{eqnarray}[X,Y](fg)&=&X(Y(fg))-Y(X(fg))\\&=&X(Y(f)g+fY(g))-Y(X(f)g+fX(g))\\ &=& [X(Y(f))g+Y(f)X(g)+X(f)Y(g)+fX(Y(g))]\\ &&-[Y(X(f))g+X(f)Y(g)+Y(f)X(g)+fY(X(g))]\\ &=& X(Y(f))g+ fX(Y(g))- Y(X(f))g-f Y(X(g))\\ &=& g[X,Y]f+f[X,Y]g.\end{eqnarray}

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