# Are different categories of manifolds non-equivalent (as abstract categories)?

Consider, for instance, the categories of $$C^k$$-manifolds, where $$k=0,1,2,...,\infty,\omega$$. ($$C^\omega$$ means real analytic.) Are these categories pairwise non-equivalent?

Of course, the obviuos forgetful functor is not an equivalence, because it is not full: for $$k, there are $$C^k$$ functions which are not $$C^l$$.

I was just thinking about possible ways of proving non-equivalence of categories and came up with this question. For example, one can prove non-equivalence by showing that limits or colimits of certain shape exist in one category but not in the other. I don't know whether one can deal with manifolds in this way.

What I can prove is that the category of complex-analytic manifolds is not equivalent to any of the above.

This follows from the fact that any real manifold other than the point (which is the terminal object) has non-trivial automorphisms. For $$C^k$$-manifolds with $$k=0,1,...,\infty$$ this is very easy to show, as there is a $$C^\infty$$-diffeomorphism of the ball $$|x|<1$$ which is the identity near its boundary sphere $$|x|=1$$. (One can construct such a diffeomorphism uning bump functions.) This does not work for the real analytic case. However, by a theorem of Grauert, for any real analytic manifold $$M$$ there is a closed analytic embedding $$i$$ of $$M$$ into some $$\mathbb R^n$$. Then one can choose a non-zero vector tangent to $$i(M)$$ at some point, extend it to a constant vector field along $$i(M)\subset\mathbb R^n$$ and project this vector field orthogonally to get a bounded vector field on $$M$$ which is not identically zero. As $$M$$ is complete in the induced metric (because $$\mathbb R^n$$ is complete and $$i$$ is closed), the flow of this vector field will be globally defined and thus it will contain a non-trivial analytic self-diffeomorphism of $$M$$.

On the other hand, there are lots of complex analytic manifolds with no non-trivial automorphisms. For example, a generic compact Riemann surface of genus $$\geq 3$$ is known to have this property.

Your method of looking for automorphisms of objects works for $$C^k$$ manifolds as well, $$0 < k \le \infty$$. This follows from a result of Filipkiewicz (which I found in Kathryn Mann's excellent survey):
If $$M, N$$ are smooth manifolds without boundary and $$\varphi: \text{Diff}^p(M) \to \text{Diff}^q(N)$$ is an isomorphism of groups (with $$1 \le p, q \le \infty$$), then $$p = q$$ and $$\varphi$$ is induced by a $$C^p$$ diffeomorphism $$w: M \to N$$.
(Recall that every $$C^p$$ manifold for $$p > 0$$ carries a compatible smooth structure, so the condition that $$M,N$$ are smooth manifolds is harmless --- cf. Hirsch, Differential Topology, chapter 2).
If $$F: \mathsf{Man}_p \to \mathsf{Man}_q$$ is an equivalence of categories, then $$F$$ induces an isomorphism $$\text{Diff}^p(M) \cong \text{Diff}^q(F(M))$$, and hence $$p = q$$ and $$M \cong F(M)$$. Not only do we have $$p = q$$, we also see that any autoequivalence of $$\mathsf{Man}_p$$ is the identity on isomorphism classes. Probably one can say something stronger (maybe they're all naturally isomorphic to the identity).
This handles all but the case where one of $$p,q$$ is zero or $$\omega$$. It seems plausible to me that Filipkiewicz' proof goes through when $$p = 0$$ (though I haven't read carefully enough to say this confidently). I don't know about the case of $$\omega$$, and Filipkiewicz asks about this as the final question in his paper.