Let $M$ be a smooth manifold and $I:= [0, 1]$. Let $\mathfrak{X}(M)$ be the $C^\infty(M)$-module of vector fields on $M$ and $\mathfrak{X}(M\times I)$ be the $C^\infty(M\times I)$-module of vector fields on $M\times I$. Is it true that there is an isomorphism of $C^\infty(M\times I)$-module $$\mathfrak{X}(M\times I)\simeq (C^\infty(M\times I)\otimes_{C^\infty(M)} \mathfrak{X}(M))\oplus C^\infty(M\times I)?$$

The $C^\infty(M)$-module structure on $C^\infty(M\times I)$ is induced by the morphism of algebras $\textrm{pr}_1^*:C^\infty(M)\longrightarrow C^\infty(M\times I)$ where $\textrm{pr}_1:M\times I\longrightarrow M$ is the projection on the first component and the $C^\infty(M\times I)$-module structure on $C^\infty(M\times I)$ is given by the pointwise product of functions.


  • $\begingroup$ What would be your candidate for the isomorphism? $\endgroup$ – Daniel Robert-Nicoud Mar 29 '17 at 14:52
  • $\begingroup$ I got to that conclusion reasoning locally: A vector field on $M\times I$ would have components in $TM$ and in $TI$. As to the components in $TM$ we would have sums of things like $f_i\cdot X_i$ where $X_i\in \mathfrak{X}(M)$ and $f_i\in C^\infty(M\times I)$, as to the components in $TI$ we would have $g\cdot \frac{\partial}{\partial t}$ with $g\in C^\infty(M\times I)$. $\endgroup$ – PtF Mar 29 '17 at 14:57
  • $\begingroup$ The fact that $T(M\times I)\cong TM\times TI$ is true globally. For the rest, I don't think it is true. I will try to think about it. $\endgroup$ – Daniel Robert-Nicoud Mar 29 '17 at 14:59
  • $\begingroup$ Maybe the candidate would be $(1\otimes X, f)\longmapsto (X\circ \textrm{pr}_1, f\cdot \frac{\partial}{\partial t}\circ \textrm{pr}_2)$ where $\textrm{pr}_1:M\times I\longrightarrow M$ and $\textrm{pr}_2:M\times I\longrightarrow I$ are the projections. $\endgroup$ – PtF Mar 29 '17 at 14:59
  • $\begingroup$ Ok, thanks =)... $\endgroup$ – PtF Mar 29 '17 at 15:00

This is how I would do it. It is not really a complete answer to the question, but it is far too long for a comment.

Suppose the tangent bundle of $M$ is trivial, i.e. it is isomorphic to $M\times\mathbb{R}^n$ where $n$ is the dimension of $M$. Then it admits a global basis, which we denote by $\partial_1,\ldots,\partial_n$. The tangent bundle of $I$ is canonically trivial and has basis $\partial_t$, and $T(M\times I)\cong TM\oplus TI$ via the projection maps. Therefore, a vector field $X\in\mathfrak{X}(M\times I)$ can always be written as $$X(m,t) = X^1(m,t)\partial_1+\cdots+X^n(m,t)\partial_n + X^t(m,t)\partial_t$$ for functions $X^i\in C^\infty(M\times I)$. It follows that we have an isomorphism $$\mathfrak{X}(M\times I)\cong C^\infty(M\times I)\otimes_{\mathbb{R}}\mathbb{R}^{n+1}\ .$$ In particular, this is always true if $M$ is an open subset of Euclidean space, and thus it is always true locally for any manifold.

More generally, we can do the following. A vector field $X$ on $M\times I$ can be decomposed into $X=X^M + X^I$ with $X^M:M\times I\to TM$ and $X^M:M\times I\to TI$. Since the tangent bundle of $I$ is trivial, we can see $X^I$ as a smooth map, $$X^I:M\times I\longrightarrow\mathbb{R}$$ so that we have an isomorphism $$\mathfrak{X}(M\times I)\cong\Gamma(M\times I,TM)\oplus C^\infty(M\times I)$$ where $\Gamma(M\times I,TM)$ denotes the maps $M\times I\to TM$ such that the image of $(m,t)$ is in $T_mM$. i don't know if more can be said in general.

  • $\begingroup$ I didn't understand why $X^I$ could be seen as an element of $C^\infty(M)$. Notice $X^I$ has the form $X^I(p, t)=(t, \alpha(p, t))$ with $\alpha: M\times I\longrightarrow \mathbb R$ so $X^I$ should be seen as in $C^\infty(M\times I)$? $\endgroup$ – PtF Mar 29 '17 at 22:23
  • $\begingroup$ @PtF You're right. Dumb mistake on my part. Thanks. $\endgroup$ – Daniel Robert-Nicoud Mar 30 '17 at 6:07

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