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If we have two volume forms $\alpha, \beta$ on a oriented manifold, is it true then, that

$(1-t)\alpha+t \beta$ for $t \in [0,1]$

is a volume form? It says so in lecture; the explanation is that $\alpha$ and $\beta$ give the same orientation, but I don't why this holds.

Can someone explain this to me?

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1 Answer 1

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Let's assume our manifold $M$ is connected.

Suppose $v_1,\dots,v_n$ is a positively-oriented basis for the tangent space of $M$ at some point $p$. Then, by assumption, $\alpha(p)(v_1,\dots,v_n)>0$ and $\beta(p)(v_1,\dots,v_n)>0$, so $[(1- t)\alpha+t\beta](p)(v_1,\dots,v_n)>0$. Then since a volume form is never $0$, the same holds at every point of $M$. (Here we used connectedness.)

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  • $\begingroup$ Ah I see, thanks!! $\endgroup$
    – User1
    Feb 7, 2020 at 19:23

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