# Is the degree a sufficient condition such that a measure is the pullback of another one?

Consider $$M$$ a smooth compact connected manifold with $$w$$ a volum form. Take for example $$M=\mathbb{S}^n$$ with the uniform measure. For any smooth map $$f:M\rightarrow M$$ and its pullback measure $$\nu = f^* w$$ it is well known that we have $$\int_M \nu = \text{deg}(f)\int_M w$$with $$\deg(f)\in \mathbb{Z}$$. Is this condition sufficient? For any $$\nu$$ volum form on $$M$$ such that $$\int_M \nu / \int_M w \in \mathbb{Z}$$ does there exist $$f:M\rightarrow M$$ such that $$\nu = f^*w$$ ? Can we construct such an $$f$$ explicitly ?

• $S^n$ does not have a Haar measure unless $n=0,1,3$. Maybe $7$ if you stretch the meaning of Haar measure a bit. Ignoring this, most manifolds do not have self-maps of arbitrary degree; and if you want $\nu$ to also be nonvanishing then $f$ should be a covering map (even less common). It is a theorem of Moser that if $\int \omega = \int \nu$ for two nonvanishing top forms, then there's an oriented diffeomorphism $f$ with $f^*\nu = \omega$. The proof is what's usually called the Moser trick.
– mme
Sep 3 '20 at 13:09
• a note on terminology: usually measures are only pushed forward, and differential forms are pulled back. Defining pullback measures is tricky mathoverflow.net/q/122704/3948 . Sep 3 '20 at 16:08

No. Consider the $$2$$-sphere, and the standard volume $$\omega$$ form with volume $$4\pi.$$ Consider $$\nu=2\omega.$$ If $$f\colon S^2\to S^2$$ with $$f^*\omega=\nu$$ would exist, it would be a local diffeomorphism and by compactness a covering map. As $$S^2$$ is simply connected and $$f$$ would have degree 2, this is not possible.