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Let $M$ be a smooth manifold (may be almost complex, almost Kahler, Kahler..). and Let $\phi : T^*M \rightarrow T^*M$ be a cotangent bundle automorphism. (the restriction of $\phi$ on the base $M$ is just the identity) Then it maps 1-forms to 1-forms. Moreover, $\phi$ can be canonically extended to the isomorphism of $\Lambda T^*M$, bundle of differential forms.

Question. Is there any criterion whether $\phi$ commutes with $d$ or not? ( I mean, I want to know when $\phi$ gives an isomorphism of de Rham cohomology ring $H^*(M)$)

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  • $\begingroup$ Maybe I am confused by your question, but a bundle isomorphism induces a diffeomorphism $f: M \to M$ over the base, and the exterior derivative $d$ always commutes with pullback by a diffeomorphism. Is there something else that you mean here? $\endgroup$
    – Spiro Karigiannis
    Commented Nov 11, 2010 at 11:46
  • $\begingroup$ Spiro: your choice of wording bothers me ("induces"). I would say that part of the data of a bundle isomorphism is a diffeomorphism of the base and that since the exterior derivative always commutes with a pullback by a diffeomorphism, you can use the base map to reduce the problem to that of an isomorphism of $T^* M$ covering the identity. But there's still a question to answer there since there can be several of those. $\endgroup$
    – Andrew Stacey
    Commented Nov 11, 2010 at 11:55
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    $\begingroup$ $\phi$ commutes with $d$ if and only if $\phi$ is the identity. Proof: If $\phi$ commutes with $d$, then $\phi(df)=d (\phi f)=d (\phi (f\cdot 1))$. Since $\phi$ is a bundle homomorphism, $d (\phi (f\cdot 1)) = d (f \phi (1))$. Since $\phi(1)=1$, this is $df$. Or did I miss something in your question? $\endgroup$
    – Johannes Ebert
    Commented Nov 11, 2010 at 13:21
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    $\begingroup$ Johannes: $phi$ is a vector bundle isomorphism, so on each fibre of $T^* M$ it is a linear map, but the induced map on $\Lambda^0 T^* M = M \times \mathbb R$ is necessarily the identity map. What you have shown is that for $\phi$ to commute with $d$, it must be the identity on exact $1$-forms. But that does not imply that it is the identity on all $1$-forms. $\endgroup$
    – Spiro Karigiannis
    Commented Nov 11, 2010 at 14:08
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    $\begingroup$ @Spiro: but any 1-form is a linear combination of exact ones. $\endgroup$
    – Michael Bächtold
    Commented Nov 11, 2010 at 15:09

2 Answers 2

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The question is local. Therefore it can be answered in coordinates. On functions, $\phi$ acts trivially. Let $x_1,\dots,x_n$ be local coordinates. Then $dx_i=d(\phi\circ x_j)=\phi(dx_j)$, which implies that $\phi$ is the identity.

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The answer is "never" (except for the trivial case of a multiplication by a constant). Suppose that you have coordinates $x_1, ..., x_n$ on M. Then $x_i dx_i$ is closed, hence $x_i \phi(dx_i)$ is also closed. Therefore, $dx_i \wedge \phi(dx_i)=0$, hence $\phi(dx_i) = f_i dx_i$. Since this is true for each linear combination $\sum \alpha_i dx_i$, the function $f_i$ is the same for all $i$. Therefore, $\phi(dx_i)=fdx_i$. However, since $\phi(dx_i)$ is closed, $f$ is a constant.

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