4
$\begingroup$

Let $M$ be a smooth manifold. Let $(T_pM)_{alg}$ denotes the "algebraist's" tangent space at $p\in M$, that is the tangent space via derivations, and $(T_pM)_{kin}$ denote the "kinematic" or "geometer's" tangent space at $p$, that is the tangent space via curves.

We can define a map $\Phi:(T_pM)_{kin}\to (T_pM)_{alg}$ as $$\Phi([\gamma])=(f\mapsto d(f\circ \gamma)/dt|_{t=0}) \text{ for all } f\in \mathcal C^{\infty}(M)$$ for all $[\gamma]\in (T_pM)_{kin}$

It can be shown that this map is well defined and is an isomorphism.

Now here is my question. I recently came across the concept of a natural isomorphism. So I was wondering whether the isomorphsim above is a natural one?

I have read this at many places that an isomorphism between vector spaces is natural if it doesn't depend on the choice of basis. Here our map $\Phi$ is not dependent on a basis. So we would say that it is a natural isomorphism. I find this rather vague.

But how do we say it precisely in category theoretic terms? So far I only know about natural isomorphism between two functors, both taking objects of a category $C$ to a category $D$. I haven't seen a precise definition of a natural isomorphism between two vector spaces.

Can somebody please enlighten me on this?

$\endgroup$

2 Answers 2

4
$\begingroup$

When people say that two vector spaces are 'naturally isomorphic' (the most classical example is probably $V$ and $V^{**}$), that's just a somewhat sloppy way of saying that the two vector spaces are obtained as the output of two naturally isomorphic functors.

Let's have a look at what this means in the case of the two tangent space constructions.

In the following, we will let $\mathrm{Diff}_*$ denote the category of pointed smooth manifolds and smooth maps, and we will let $\mathrm{Vect}_\mathbb{R}$ denote the category of real vector spaces and linear maps. The two different ways of constructing the tangent space will then correspond to two different functors $\mathrm{Diff}_*\to \mathrm{Vect}_\mathbb{R}$.

The “algebraist's” functor. The algebraic way of constructing the tangent space gives rise to a functor $(T_\square\square)^\mathrm{alg}\colon \mathrm{Diff}_*\to \mathrm{Vect}_\mathbb{R}$, defined as follows:

  • It sends a pointed manifold $(M,p)$ to the tangent space $(T_pM)^\mathrm{alg}$ of derivations at $p$.

  • It sends a smooth map $\Phi\colon (M,p)\to (N,\Phi(p))$ to the differential $\mathrm{d}\Phi_p^\mathrm{alg}\colon (T_pM)^\mathrm{alg} \to (T_{\Phi(p)}N)^\mathrm{alg}$ given by $\mathrm{d}\Phi_p^\mathrm{alg}(X_p)(f)=X_p(f\circ \Phi)$ for every $X_p\in (T_pM)^\mathrm{alg}$ and every $f\in C^\infty(N)$.

(It's a good exercise to verify that this is a functor.)

The “geometer's” functor. The geometric definition of the tangent space gives rise to another functor $(T_\square\square)^\mathrm{geo}\colon \mathrm{Diff}_*\to \mathrm{Vect}_\mathbb{R}$, defined as follows:

  • It sends a pointed manifold $(M,p)$ to the tangent space $(T_pM)^\mathrm{geo}$ of equivalence classes of curves.

  • It sends a smooth map $\Phi\colon(M,p)\to (N,\Phi(p))$ to the differential $\mathrm{d}\Phi_p^\mathrm{geo}\colon (T_pM)^\mathrm{geo}\to(T_{\Phi(p)}N)^\mathrm{geo}$ defined by $\mathrm{d}\Phi_p^\mathrm{geo}([\gamma])=[\Phi\circ\gamma]$.

(Again, it's a good exercise to verify that this really gives a functor.)

The natural isomorphism. To specify a natural isomorphism $\eta\colon (T_\square\square)^\mathrm{geo}\Rightarrow (T_\square\square)^\mathrm{alg}$, we need to pick a vector space isomorphism $\eta_{(M,p)}\colon (T_pM)^\mathrm{geo}\to (T_pM)^\mathrm{alg}$ for every pointed manifold $(M,p)$. For this we use the map you described in the beginning of your post, i.e. we set $$\eta_{(M,p)}([\gamma])=\left(f\mapsto \tfrac{\mathrm{d}}{\mathrm{d}t}f(\gamma(t))\vert_{t=0}\right)\,.$$ In order for these maps to assemble into a natural isomorphism, we need the following diagram to commute for every smooth map $\Phi\colon (M,p)\to (N,\Phi(p))$: $$\require{AMScd} \begin{CD} (T_pM)^\mathrm{geo} @>{\eta_{(M,p)}}>> (T_pM)^\mathrm{alg}\\ @V{\mathrm{d}\Phi_p^\mathrm{geo}}VV @VV\mathrm{d}\Phi_p^\mathrm{alg}V \\ (T_{\Phi(p)}N)^\mathrm{geo} @>{\eta_{(N,\Phi(p))}}>> (T_{\Phi(p)}N)^\mathrm{alg}\,. \end{CD}$$ (To check that is really does commute is a straight-forward exercise.)

$\endgroup$
1
$\begingroup$

The intuitive definition is that the two objects are the same in every way, except in the notation in which you write it. For example, suppose $V$ is a vector space over $\mathbb{R}$, say finite dimensional, with a non-degenerate inner product. It is an easy exercise to see that $V$ and $V^*$ ($=\text{hom}(V,\mathbb{R})$) are naturally isomorphic. The way you do it, is send $v\in V$ to $v^*$, where $v^*$ is defined as the linear map $x\mapsto \left< v,x\right>$. Therefore, the elements in $V$ "look like" $v$, while the elements in $V^*$ "look like" $v^*$. So we say they are "naturally isomorphic". It is common for people to abuse notation and write $V=V^*$.

An example when the isomorphism is not natural, is say that $V$ is just a vector space, then it is still true that $V\simeq V^*$, as their dimensions are equal. However, the isomorphism is not natural.

A more precise meaning is in terms of preserving a commutative diagram. If you have a bunch of natural maps, fitting together in a diagram, it will be commutative. To clarify consider the following example,

Let $f:V\to V^*$ be an isomorphism, and $g:V^* \to V^{**}$ be another isomorphism, and $h:V\to V^{**}$ be a third isomorphism. It is not necessarily true that $h = gf$. However, if $V$ has inner product and $f$, $g$ are natural isomorphisms, and $h$ is the natural isomorphism of $V$ with its bidual then the diagram commutes.

$\endgroup$
6
  • $\begingroup$ Question: Is there a meaningful way in which the canonical isomorphism between $V$ and $V^*$ in the presence of an inner product on $V$ can be phrased in categorical language (in terms of the functors $\mathrm{id}$ and $\square^*)$? Since $\square^*$ is contravariant, I suppose the usual definition of a natural isomorphism won't really help us much here, but perhaps there is some other way of making sense of this? $\endgroup$ Commented Jan 13, 2020 at 6:01
  • $\begingroup$ I have trouble understanding Nicolas' answer : the isomorphism $x\mapsto \langle v, x\rangle$ between $V$ and $V^*$ depends on the choice of scalar product, so it cannot be canonical, am I wrong ? On the other hand, the one between $V$ and $V^{**}$ defined as $v\mapsto v^{**}:(w^*\mapsto w^*(v))$ clearly is since it doesn't depend on any choice. $\endgroup$
    – Johann
    Commented May 7, 2020 at 8:47
  • $\begingroup$ @Johann If $V$ is a finite dimensional space then $V^*$ has the same dimension and so is isomorphic to it. The map $V\to V^*$ is not given by the scalar product but by the dual. Every $x\in V$ gives rise to a linear function $\varphi_x:V\to \mathbb{R}$ by pointwise evaluation. $\endgroup$ Commented May 16, 2020 at 18:14
  • $\begingroup$ @NicolasBourbaki Thanks for your answer, but I still don't get it... Sure, $V$ and $V^*$ are isomorphic in the finite dimensional case, I understand that. But since the dual is spanned by the dual basis (independantly of the way the components are built using a chosen scalar product, I'll give you that), the map depends on the choice of that original basis... Hence not a canonical/natural isomorphism $\endgroup$
    – Johann
    Commented May 17, 2020 at 9:48
  • $\begingroup$ @NicolasBourbaki Alright, I think I get it : without any choice, the isomorphism is not canonical (depends on the basis). But once a scalar product is added (or another structure, hermitian, symplectic, etc.) then it does not depend on the basis anymore and is canonical. Amirite this time ? $\endgroup$
    – Johann
    Commented May 19, 2020 at 9:58

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .