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I am trying to figure out why the following equality is true :

$$f_*[X,Y]=[f_*X,f_*Y]$$ where $f:M\rightarrow N$ is a diffeomorphism, $M$, $N$ are smooth manifolds, $X$, $Y$ are smooth vector fields on $M$.

I have tried to write $$f_*[X,Y]=\dfrac{\partial f^i}{\partial x^j}\left( \chi^k \dfrac{\partial \psi^j}{\partial x^k}-\psi^k \dfrac{\partial \chi^j}{\partial x^k}\right)\dfrac {\partial}{\partial y^i}$$ where $$X=\chi^k \dfrac{\partial}{\partial x^k},Y=\psi^k \dfrac{\partial}{\partial x^k}, [X,Y]=\left( \chi^k \dfrac{\partial \psi^j}{\partial x^k}-\psi^k \dfrac{\partial \chi^j}{\partial x^k}\right)\dfrac {\partial}{\partial x^j}.$$ However, when it comes to write the second part of the equality: $$[f_*X,f_*Y]=\left( (f_*X)^k \dfrac{\partial (f_*Y)^j}{\partial y^k}-(f_*Y)^k \dfrac{\partial (f_*X)^j}{\partial y^k}\right)\dfrac {\partial}{\partial y^j}$$ where $y^j$ is a coordinate basis of N.

The problem I face is that I cannot differentiate $f_*Y, f_*X$ with respect to the basis $y^j$, in the above expression. Any help would be appreciated. ( I would prefer an answer which is based on the definition of Lie Bracket with coordinates, as I worked above)

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4 Answers 4

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Well, you see it is much simpler in coordinate independent form. As for diffeomorphism $ f : M \rightarrow N $ you have $ f_* : \mathcal{X}(M) \rightarrow \mathcal{X}(N) $ and hence for $p\in M $ it maps tangent spaces $T_p(M)$ to $T_{f(p)}(N) $ given by for $ g \in C^\infty(N) $ you have $f_* (X)(g)(f(p)) = X(g\circ f)(p)$ hence $ f_*(X)(g)\circ f = X(g\circ f ) $ Thus for $X,Y \in \mathcal{X}(M) $ we have for all $ g \in C^\infty(N) $ \begin{align*} & f_*[X,Y]_{f(p)}(g) = [X,Y]_p(g\circ f) \\ & = X_p(Y(g\circ f))-Y_p(X(g\circ f)) \\ & = X_p(f_*(Y)(g)\circ f) - Y_p(f_*(X)(g)\circ f) \\ & = f_*(X)_{f(p)}(f_*(Y)(g))-f_*(Y)_{f(p)}(f_*(X)(g)) \\ & = [f_*(X),f_*(Y)]_{f(p)} (g) \end{align*} Hence $ f_*[X,Y] = [f_*(X),f_*(Y)] $

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  • $\begingroup$ Thank you for your answer! Actually, I had already figured this solution , and I wanted a solution which uses the coordinates. If no one can provide an answer with using the coordinates, I will give you the Answer Points. $\endgroup$
    – Dimitris
    Commented Apr 18, 2013 at 17:23
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    $\begingroup$ @DimitrisDallas Well you can always replicate each of the above steps using coordinates if that's what you look for. $\endgroup$
    – smiley06
    Commented Apr 18, 2013 at 17:57
  • $\begingroup$ Well, I hoped to solve it without using a function $g\in C^\infty(N)$ and a point $P\in M$ , just with manipulating the coordinates, but it seems impossible... $\endgroup$
    – Dimitris
    Commented Apr 18, 2013 at 18:01
  • $\begingroup$ In the area of your problem, if $ \psi = (y^1, ..., y^n) $ is your chart then $ \partial/\partial y^k = \partial_k $ that is the usual partial derivative in $\mathbb{R}^n $ of $ f_*(X)\circ \psi^{-1} $, hence you can carry out the differentiation taking local expressions of $ f_*(X), f_*(Y) $. $\endgroup$
    – smiley06
    Commented Apr 18, 2013 at 18:07
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    $\begingroup$ How did you go from $X_p(Y(g \circ f)) - Y_p(X(g \circ f))$ to $X_p(f_*(Y)(g) \circ f) - Y_p(f_*(X)(g) \circ f)$? $\endgroup$
    – James C
    Commented Dec 14, 2020 at 2:59
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To do this computation in coordinates without using functions and points you have to adopt the physicist way of writing things which is messy and unplesant :-) However chain rule takes care of all the evaluation matters.

Let f map $x$ to $y$. We will denote the Jacobian and inverse Jacobian by $\frac{\partial y^j}{\partial x^i}, \frac{\partial x^j}{\partial y^i}$

We will write $\tilde{Z}=f^*Z$ and $\tilde{W}=f^*W$ so the "components change as"

$Z^j = \tilde{Z}^i\frac{\partial x^j}{\partial y^i}$

$W^j = \tilde{W}^i\frac{\partial x^j}{\partial y^i}$

(here we really see Z as a pushforward of $\tilde{Z}$ by the inverse map etc)

Consider $f^*[Z,W]$

$=((Z^i\frac{\partial}{\partial x^i}(W^j) - W^i\frac{\partial}{\partial x^i}(Z^j))\frac{\partial y^l}{\partial x^j}\frac{\partial}{\partial y^l}$

$=((\tilde{Z}^k\frac{\partial x^i}{\partial y^k}\frac{\partial}{\partial x^i}(\tilde{W}^m\frac{\partial x^j}{\partial y^m}) - (\tilde{W}^k\frac{\partial x^i}{\partial y^k}\frac{\partial}{\partial x^i}(\tilde{Z}^m\frac{\partial x^j}{\partial y^m}))\frac{\partial y^l}{\partial x^j}\frac{\partial}{\partial y^l}$

The mixed derivative term is of the form

$((\tilde{Z}^k(\tilde{W}^m\frac{\partial}{\partial y^k}\frac{\partial x^j}{\partial y^m})-(\tilde{W}^k(\tilde{Z}^m\frac{\partial}{\partial y^k}\frac{\partial x^j}{\partial y^m}))$ =0

and the remaining terms give

$=((\tilde{Z}^k\frac{\partial}{\partial y^k}(\tilde{W}^m) - (\tilde{W}^k\frac{\partial}{\partial y^k}(\tilde{Z}^m))\delta_{lm}\frac{\partial}{\partial y^l}$

which is $[f^*Z,f^*W]$. The trick is to use chain rule when ever you want the derivation to be compatible with the function you are applying it to. So the middle expressions might not be completely sensible (just formal expressions to see the steps) but derivations are. Why this only works for diffeomorphisms is the component change rules given above is basically the pushforward expression in coordinates and that expression allows usage of chain rule at certain parts.

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  • $\begingroup$ That's exactly what I was looking for! It seemed too messy with all these indexes! One question: by $$\dfrac{\partial y^j}{\partial x^i}$$ we denote : $$\dfrac{\partial (\pi^j\circ\psi\circ f\circ \varphi ^-1)}{\partial x^i}$$ where $(U,\varphi)$ a chart in $M$ and $(V,\psi)$ a chart in $N $, right? $\endgroup$
    – Dimitris
    Commented Apr 30, 2013 at 16:26
  • $\begingroup$ yes, the Jacobian written in coordinates is precisely what you have written there. $\endgroup$
    – Sina
    Commented May 1, 2013 at 20:42
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I would give a basis-independent proof: $f:M\rightarrow N$ and $g:N\rightarrow\mathbb{R}$ \begin{eqnarray} f_*[X,Y](g)&=&\{[X,Y](g\circ f)\}\circ f^{-1}\\ &=&\{(X\circ Y)(g\circ f)\}\circ f^{-1}-(X\leftrightarrow Y)\\ &=&\{X[Y(g\circ f)]\circ f^{-1}\}-(X\leftrightarrow Y)\\ &=&\{X[Y(g\circ f)\circ f^{-1}\circ f]\circ f^{-1}\}-(X\leftrightarrow Y)\\ &=&\{X[f_*Y(g)\circ f]\circ f^{-1}\}-(X\leftrightarrow Y)\\ &=&(f_*X\circ f_*Y)(g)-(X\leftrightarrow Y)\\ &=&[f_*X,f_*Y](g). \end{eqnarray} Here one should keep in mind that: $X$ or $Y$ is a MAPPING from differential functions (on $M$) to differential functions (on $M$).

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Maybe the following plausibility argument will also help : [X,Y] = Lie-derivative of Y in direction X, which is e.g defined by a limiting process of 1-parameter-groups of differential equations (with X and Y the fields of direction); e.g. [F.Warner, Foundations of Differential Manifolds and Lie Groups, 1971. p.69]. By the uniqueness of solutions of diff.equs. the compatibility of Lie derivatives with diffeomorphisms should follow.

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