Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $\mathsf{C}$ and $\mathsf{D}$ two categories and $\mathcal F,\mathcal G$ two functors $\mathsf{C}\rightarrow\mathsf{D}$. A natural isomorphism from $\mathcal F$ to $\mathcal G$ is the datum of a isomorphism $\nu_X:\mathcal F(X)\rightarrow \mathcal G (X)$ for every $X\in Obj(\mathsf{C})$ such that for every $\alpha\in \operatorname{Hom}(X,Y)$ in $\mathsf{C}$ we have that

$$\mathcal G(\alpha)\circ\nu_X=\nu_Y\circ\mathcal F(\alpha)$$

Now, many books say that a linear isomorphism $f$ between vector spaces is a natural isomorphism if "$f$ doesn't depend from the choice of the basis". I have two questions:

1) What does formally mean the phrase "$f$ doesn't depend from the choice of the basis"?

2) How can I match the two definitions of natural isomorphism?

share|cite|improve this question

Formally, the meaning of 'does not depend on particular choices (e.g., of basis)' means precisely that the functions (or morphisms more generally) form a natural transformation. In the original article where Eilenberg and Mac Lane introduce categories they explicitly say that categories are introduced to define functors and that functors are introduced to define natural transformations.

It is very common in mathematics to use the term 'natural' about a construction (long before natural transformations existed) and in pretty much all of those cases it means that one actually constructs a natural transformation in the formal sense of category theory.

Now to answer your questions. A construction in linear algebra is said to be independent of the choice of basis if the construction uses a basis in the definition but choosing any other basis will yield the same final result. For instance, one can define the determinant of a linear transformation $T:V\to V$, on some finite dimensional vector space $V$, to be the determinant of a representing matrix relative to a basis $B$. One can then show that no matter which basis is chosen the determinant of the representing matrix is always the same. Thus, this concept is independent of the choice of basis.

Incidentally, the determinant gives an example of a natural transformation, see Why determinant is a natural transformation?

Now for question 2, quite often when you are able to construct a linear transformation that does not depend on any particular choice of basis it will be the case that this will actually be a general construction that will give you a whole family of linear transformation that together will form a natural transformation.

You might want to have a look at What is a natural isomorphism? for more examples.

share|cite|improve this answer
The fact that confuses me is that the definition of natural transformation involves functors, but then we use the adjective "natural" for morphisms. Clearly in the case of the bidual functor we show that this functor is naturally isomorphic to the identity functor, but when we use a natural isomorphism between vector spaces, where are functors? – Dubious Nov 20 '12 at 8:30
the functors are then hidden somewhere. Typically, if you have a construction between vector spaces that is 'natural' in some non-formal sense (i.e., no dependence on particular choices) then chances are there are some functors hiding there and that the construction is actually a natural transformation between the functors. Imagine that one would consider the vector space $\mathbb {R}^3$, construct its double dual and show it is naturally isomorphic to it. It's a particular result, but the identity functor and the double dual functor are lurking there. – Ittay Weiss Nov 20 '12 at 8:48
If $1$ is the category with one object and just the identity morphism, then any category $\mathbf{C}$ is isomorphic to the functor category $\hom(1, \mathbf{C})$; this makes every arrow a natural transformation in disguise. But we can say something more interesting. When we say "If $V$ is a vector space, then there is a map $V \to V^{**}$", $V$ is a variable that ranges over all vector spaces. .... – Hurkyl Feb 17 '13 at 12:33
... Rather than identify $V$ with a specific functor $1 \to \mathbf{Vect}$, we can instead make a statement about all vector spaces at once by identifying $V$ with the identity functor $\text{Ob}\mathbf{Vect} \to \mathbf{Vect}$. Even better, this fact is "functoral in $V$", so we can use all of $\mathbf{Vect}$ as the source category, rather than just its class of objects. So we identify $V$ with the identity functor $1_\mathbf{Vect}$, double dual with a functor $\mathbf{Vect} \to \mathbf{Vect}$, and the claimed arrow is a natural transformation between them. – Hurkyl Feb 17 '13 at 12:34

Consider the two real $\mathbb{R}^2 \to \mathbb{R}$ defined by

  • $f(x,y) = x + y$
  • $g(x,y) = x$

Clearly, different values for $y$ will give different values when plugged into $f(x,y)$, but will not give you different values when plugged into $g(x,y)$. We say that the value of $g(x,y)$ "does not depend on $y$", and we say the value of $f(x,y)$ does.

Now consider calculations we might do with a rational number $q$. Here are two examples:

  • Write $q=m/n$, define $r = (m+1)/n$
  • Write $q=m/n$, define $s = (m+n)/n$

While both calculations are phrased as the $r$ and $s$ being computed are functionally dependent on the value of $q$, that's not quite true; there is another, implicit dependency: a choice of the way we wrote $q=m/n$. In functional terms, we've defined two functions

  • $r(q,m,n) = (m+1)/n$ whenever $q = m/n$
  • $s(q,m,n) = (m+n)/n$ whenever $q = m/n$

However, the value of $s$ does not depend on the choice of $m$ and $n$ satisfying $q=m/n$! This is important, because it means the value we constructed really and truly can be expressed as a function of $q$ alone; i.e. we can express it as $s(q)$.

These same concerns apply to "higher" constructions; e.g. in the usual construction of an isomorphism bewteen finite dimensional vector spaces $T : V \to V^*$, the isomorphism we construct depends not only on $V$, but on a choice of bases $B$ for $V$ (and a basis for $V^*$, but I'll choose the dual basis to $B$) so rather than getting an isomorphism $T_V : V \to V^*$, we get an isomorphism $T_{V,B} : V \to V^*$, where I've added subscripts to $T$ to indicate the dependence.

It turns out, unfortunately that the notion of a natural transformation doesn't really have anything to do with the above concerns; however, it often comes up in the same context.

Consider the construction above. $V$ is a variable that denotes a finite-dimensional vector space and $B$ a basis on $V$; it is easy to interpret the above construction not as something you do for a particular choice of vector space and basis, but as something you do for all choices. If we define

  • S = { (V,B) | V is a finite dimensional vector space and B is a basis for V }
  • FinVect = the category of finite-dimensional vector spaces

then we can view $S$ as a discrete category and there are functors

  • $F : S \to \mathbf{FinVect}$ defined by $F(V,B) = V$
  • $G : S \to \mathbf{FinVect}$ defined by $G(V,B) = V^*$

and a natural transformation

  • $\eta : F \to G$ defined by choosing $\eta_{V,B}$ to be arrow constructed by the argument mentioned above.

While $\eta$ is indeed a natural transformation, we usually wouldn't say that $\eta_{V,B} : V \to V^*$ is natural, because the choice of $S$ was somewhat artificial and trivial.

e.g., since $V$ is a variable denoting a vector space, $S$ really should have had a component that looks like $\mathbf{FinVect}$; but to set everything up in terms of categories, functors, and natural transformations, we had to "forget" the vector space structure to get things to work.

The usual isomorphism $I_V : V \to V^{**}$, however, depends only on $V$. And it depends on $V$ in a way that is consistent with the vector space structure: we have

  • $F : \mathbf{FinVect} \to \mathbf{FinVect}$ given by $F(V) = V$
  • $G : \mathbf{FinVect} \to \mathbf{FinVect}$ given by $G(V) = V^{**}$
  • $\eta : F \to G$ given by $\eta_V = I_V$.

Because of this, we say that $V$ and $V^{**}$ depend functorally on $V$, and we say that $\eta$ is natural in $V$.

Constructions that depend on choices can still be natural in this sense, though. For example, the construction of homotopy groups of a nonempty topological space depend on choosing a point in that space, but it depends functorally on the space and the choice of point.

This fact is usually expressed by coming up with a new term for a space and choice of point (an "pointed space"), and expressing homotopy in terms of pointed spaces.

The "colloquial" usage of natural usually wants to imply both of the concerns mentioned above, and usually connotes some degree of aesthetics as well.

However, I'm not sure if the colloquial usage really means to imply the technical meaning of natural I described, or if they just turn out to be highly correlated because things that violate the technical meaning tend to also have some sort of dependence on choices, or are rather aesthetically displeasing.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.