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Suppose we have a a diffeomorphism between two manifolds, $f: M \rightarrow N$ and a volume form $\Omega$ on $N$. Then is it true that $f^{*}(\Omega) = \Upsilon$ will always be a volume form on $M$?

My thinking is that for any arbitrary vector fields $(X_1, X_2,...,X_N) \in T_pM$ then

$\Upsilon(X_1, X_2,..., X_N) = f^{*}\Omega(X_1, X_2,..., X_N) = \Omega(df_x(X_1),df_x(X_2),...df_x(X_N))$

is always non vanishing given that $\Omega$ is a volume form on $N$? This would be because $\Omega$ is acting on the tangent vectors $df_x(X_i) \in T_{f(p)}N$, which is non-vanishing given that $\Omega$ is a volume form on N.

I'm new to differential geometry, so any mistakes with notation, or comments which aren't accurate, please do let me know!

JMill
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1 Answers1

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Your agrument is basically coorect, the only observation you have not put in explicitly that that for a basis $\{X_i\}$ of $T_pM$ the vectors $df_x(X_i)$ form a basis of $T_{f(p)}N$, which is needed to conclude that the value of $\Omega$ on these tangent vectors is non-zero.

Andreas Cap
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  • Thank you for the answer. I'm wondering what conditions on $f$ are required so that a basis {$X_{i}$} of $T_{p}M$ implies {$df_{x}(X_{i})$} forms a basis for $T_{f(p)}M$? Is there a weaker condition than $f$ having to be a diffeomorphism? – JMill Dec 31 '20 at 12:21
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    What you need is that $f$ is a local diffeomorphism around $x$, but this is not the main point. What is missing in your describption is that you have to start with linearly independent tangent vectors (or vector fields whose values in $p$ are linearly independent) since this is needed for the volume form to have a non-zero value. – Andreas Cap Dec 31 '20 at 15:30