# Is the space of smooth maps $C^{\infty}(M,N)$ with the Whitney $C^{\infty}$ topology locally compact, if $M$ is compact

The title says it all:

Let $M$ be a compact manifold and $N$ a (possible non compact) manifold. Equip the space of smooth functions $C^{\infty}(M,N)$ with the Whitney $C^{\infty}$ topology. (The strong an the weak one coincide, since $M$ is compact)

Hirsch's book on differential topology states without proof that this space is completely metrizable, second countable and locally contractible.

However, is it locally compact?

• Take $N = \mathbb{R}$. Then a locally compact Hausdorff topological vector space is necessarily finite-dimensional (terrytao.wordpress.com/2011/05/24/…). – Qiaochu Yuan May 22 '15 at 18:35
• @QiaochuYuan $\mathcal{C}^{\infty}(M,\mathbb{R})$ is not a topological vector space. The maximal topological vector space inside is $\mathcal{C}_c^\infty(M,\mathbb{R})$ – Kathrin L. Jun 21 '15 at 12:33