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Let us use the compact-open topology on the space $C^\infty(X,Y)$ of smooth maps between two given smooth manifolds $X,Y$. This induces subspace topologies on e.g. spaces of differential forms, vector fields, Riemannian metrics, etc.

I was wondering, for smooth manifolds $X,Y$, for an integer $k\geq0$, is the "differentiation" map $$ \begin{matrix} C^\infty(X,Y) & \rightarrow & C^\infty(\bigwedge^k T^*Y , \bigwedge^k T^*X) \\ f &\mapsto& f^* \end{matrix} $$ continuous, for the compact-open topology? Similarly, is $$ \begin{matrix} C^\infty(X,Y) & \rightarrow & C^\infty(TX , TY) \\ f &\mapsto& f_* \end{matrix} $$ continuous?

I.A.S. Tambe
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  • Do you mean the continuous compact open topology $CO^0$ or the smooth $CO^\infty$? With the smooth one this should work, doesn't it? – psl2Z Jun 12 '23 at 22:44

1 Answers1

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Consider $X=Y=\mathbb{R}$ and identify $T\mathbb{R}=\mathbb{R}\times\mathbb{R}$. There's a map $C^{\infty}(\mathbb{R}\times\mathbb{R},\mathbb{R}\times\mathbb{R})\rightarrow C^{\infty}(\mathbb{R},\mathbb{R})$, which is given by pushing forward along the projection $\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$ to the second factor and pulling back along the inclusion $\mathbb{R}\rightarrow\mathbb{R}\times\mathbb{R},\,x\mapsto(x,1)$. The composite $C^{\infty}(\mathbb{R},\mathbb{R})\rightarrow C^{\infty}(\mathbb{R}\times\mathbb{R},\mathbb{R}\times\mathbb{R})\rightarrow C^{\infty}(\mathbb{R},\mathbb{R})$ of this map and the differentiation map is then just the map $f\mapsto f^{\prime}$, taking the derivative in the sense of calculus. If the differentiation map were continuous, this map would be as well, since pushforward and pullback maps are tautologically continuous wrt the compact-open topology.

However, this map is infamously not continuous with respect to these topologies. Let's borrow an example from Matt Samuel. The sequence of functions $\left(\frac{\sin(nx)}{n}\right)_n$ in $C^{\infty}(\mathbb{R},\mathbb{R})$ converges uniformly on all of $\mathbb{R}$ to the $0$ function, which is even stronger than convergence in the compact-open topology (uniform convergence on all compact subspaces), yet its sequence of derivatives $(\cos(nx))_n$ in $C^{\infty}(\mathbb{R},\mathbb{R})$ does not even converge pointwise, which is weaker than convergence in the compact-open topology (becauses points are compact subspaces).

I wager if you want the differentiation map between spaces of $C^{\infty}$-maps to be continuous, you ought to use variation of a Whitney Topology.

Thorgott
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