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It's easy to see in any DG course that if $X,Y$ are sufficiently smooth manifolds, then $$T_{(x,y)}(X\times Y )\cong T_xX \times T_yY.$$ Starting from this, can I recover some info about how $T_x( \varprojlim X_i)$ or $T_x(\varinjlim X_i)$ look like?

I'm assuming to work in a setting where "all is well behaved" (for example I'm not losing finiteness in doing colimits, nor I'm losing smoothness -if it can happen- in doing limits).

I'm also intentionally being vague in describing the category of spaces I'm working in: this is just because this problem arises quite naturally in at least three of them: smooth manifolds, Lie groups, algebraic geometry (where one can say $T_x(X)\simeq \operatorname{Ext}^1(k(x),k(x))$).

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If you know what happens with products and equalisers, then you know what happens for all finite limits, since every finite limit can be constructed as the equaliser of a pair of maps between finite groups. –  Zhen Lin Dec 5 '11 at 12:55
Yeah, I forgot to precise I would like to go beyond it. In any case, thanks for pointing it out! –  tetrapharmakon Dec 5 '11 at 13:16

2 Answers 2

up vote 2 down vote accepted

It seems to me that the nicest things happen when the tangent space functor is representable. Consider, for example, the opposite of the category of commutative $k$-algebras $A$ together with distinguished points $A \to k$; this is like the category of pointed affine $k$-varieties but more general. In this category the tangent space functor (take the tangent space at the distinguished point) is represented by $\text{Spec }k[x]/x^2$, so it follows that it preserves all limits. A similar argument should work for smooth manifolds in the context of smooth infinitesimal analysis.

Furthermore, limits in the pointed category have the obvious relation to limits in the unpointed category: the forgetful functor to the unpointed category is represented by $\text{Spec } k^2$ (equipped with the projection map $k^2 \to k$, say to the first coordinate) so also preserves all limits.

I don't understand what behavior you could be expecting with respect to colimits. It is not even clear to me whether or not coproducts exist in, say, the category of pointed smooth manifolds.

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I'm afraid, everything seems to be "in the wrong side".

By which I mean the following. You can look at the tangent space of a variety $X$ at a point $p$ as the vector space of derivations at $p$. That is, the linear maps

$$ l : {\cal C}^{\infty}(X,\mathbb{R}) \longrightarrow \mathbb{R} $$

which, for every pair of functions $f,g: X \longrightarrow \mathbb{R}$ verify

$$ l(fg) = f(p)l(g) + g(p)l(f) \ . $$

Forgetting for a moment about this restriction, what you are doing in order to construct the tangent space $T_pX$ is

$$ X \mapsto {\cal C}^\infty(X,\mathbb{R}) \mapsto {\cal C}^\infty(X,\mathbb{R})^* \ , $$

where by $E^*$ I mean the dual of the vector space $E$.

Writting this in a more "categorical" way, it looks like

$$ X \mapsto \mathbf{Var}(X, \mathbb{R}) \mapsto \mathbf{Vct}(\mathbf{Var}(X, \mathbb{R}), \mathbb{R}) \ . $$

Here $\mathbf{Var}$ is a "convenient" category of "varieties" and $\mathbf{Vct}$ the category of vector spaces.

Now, if you look at what happens with limits and colimits when you perform such operations, you get:

$$ \mathbf{Vct}(\mathbf{Var}(\mathrm{colim}_i X_i, \mathbb{R}), \mathbb{R}) = \mathbf{Vct}(\mathrm{lim}_i \mathbf{Var}(X_i, \mathbb{R}), \mathbb{R}) \ . $$

And I have no idea about what happens next: this limit is "in the wrong side" of the hom set.

For limits the problem is worse, I think:

$$ \mathbf{Vct}(\mathbf{Var}(\mathrm{lim}_i X_i, \mathbb{R}), \mathbb{R}) \ . $$

The limit is already "in the wrong side" right from the start.

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This argument doesn't hold water to me. What do you make of the example where $C$ is the category of finite-dimensional vector spaces over a field $k$? There $X \mapsto \text{Hom}(\text{Hom}(X, k), k)$ is an equivalence of categories so preserves all limits and colimits. –  Qiaochu Yuan Dec 5 '11 at 18:02
@Qiochu. I can't tell. I've just said what can be deduced from elementary category theory. But the problem, I guess, may be that both $\mathrm{Hom}$ functors are not the same in the original situation, are they? I mean: in general, $\mathbf{Var}(X, \mathbb{R})$ is not a finite dimensional vector space, is it? –  a.r. Dec 5 '11 at 19:53

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