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Suppose $(M,\gamma)$ is a Riemannian manifold, $g$ an isometry, $\mbox{d}x$ the Riemannian volume form on $(M, \gamma)$. I can't really understand these formulae, sometimes found in literature. First,

\begin{equation}\tag{1} g^*(\mbox{d}x) = \mbox{d}x\circ g^{-1}. \end{equation}

Next, let $f : M \to N$ be a smooth function (here, $g$ can be seen as an element of a group $G$ acting by isometries in some specified way on both $M$ and $N$); sometimes I find

\begin{equation}\tag{2} \vert{\mbox{d}(g \circ f)}\vert^2 = \vert \mbox{d}f\vert^2 \circ g^{-1}. \end{equation}

I can't catch the precise meaning of the rhs. I know what the operations involved are (pullback, exterior derivative, commutation property of pullback and exterior derivative, differential, chain rule...); my problem concern what one precisely means when writing a thing such as $g^*(\mbox{d}x) = \mbox{d}x \circ g^{-1}$ instead of

\begin{equation} g^*(\mbox{d}x) = \sqrt{|\gamma(g(x))|}J_g(x)\mbox{d}g^1(x) \wedge\dots \wedge \mbox{d}g^m(x) = \mbox{d}(g(x)). \end{equation}

Here, $m = \dim M$, $J_g(x)$ the jacobian of the coordinate transformation $g$ evalueted at $x$ while $\sqrt{|g^*(\gamma(x))|} = \sqrt{|\gamma(g(x))|}$ because $g$ is an isometry.

I'm not really into differential and Riemannian geometry, so I'm probably missing something entirely elementary.

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I suppose $(1)$ refers to the usual definition of pull-back for tensors (and hence differential forms): if $\omega$ is a $n-$form on $N$ and $F\colon M\to N$ is $C^{\infty}$, then $$ (F^*\omega)_p(v_1,\dots,v_n):=\omega_{F(p)}(dF_p(v_1),\dots,dF_p(v_n)). $$ which actually is what you have already written in the last part. So, actually it is $$ F^*\omega=\omega\circ F, $$ yet one needs to evaluate $\omega$ at vectors in $T_N$, and hence one uses the differential of $F$.

EDIT: $(2)$ means that $$ ||d(g\circ f)||_x^2=||df||^2_{g^{-1}(x)}, $$ where $||\cdot||$ is the norm induced by the bundle metric on $T^*_M$ (see Riemannian Geometry and Geometric Analysis, Jost Theorem 2.1.4) given in coordinates by $$ ||\omega||:=|g^{ij}\omega_i\omega_j|^{1/2}, $$ where $\omega=\omega_idx^i$ (Einstein summation is being used). Note that $(2)$ is equivalent to $$ ||\nabla(g\circ f)||_x^2=||\nabla f||^2_{g^{-1}(x)}. $$ Since $g$ is an isometry, this looks fairly reasonable. I suppose that some calculations (repeated use of the chain rule) would show that it is true.

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  • $\begingroup$ You can find (1) and (2), for instance, in Chapter V of Eells-Ratto, "Harmonic Maps and Minimal Immersions with Symmetries" but this way of writing is common in that branch of the literature. My guess is that (2) should not be context-dependent because they denote $\circ$ the usual composition of functions (the action of the group on $f$ is there denoted $g \cdot f$) - of course, $g$ in lhs of (2) must be seen as an isometry on $N$. $\endgroup$ – Federico Sep 1 '17 at 18:34
  • $\begingroup$ I edited my answer a little, even though I haven't yet found time to do the calculation. $\endgroup$ – Alessio Di Lorenzo Sep 2 '17 at 20:48
  • $\begingroup$ Thanks for your contribution. Actually, it looks somewhat customary see the rhs of (1) as the push-forward of the Lebesgue measure, although this is unsatisfactory because of the fact that the equality must be understood in the sense of the theorem of correspondence between volume forms and measures and the somewhat inconsistent behaviour of the two sides wrt composition. $\endgroup$ – Federico Sep 7 '17 at 15:32

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