I'm reading the book (by A. Kostrikin) on linear algebra and I feel like I'm really missing something about this idea.

I understand the formal proofs of: a) isomorphism between vector space $V$ and its dual space $V^*$ b) natural isomorphism between $V$ and $V^{**}$.

What I don't understand is the idea behind it:

1) Why is it so important for isomorphism to be "natural"? Does it make it better, more usable is some way, or?..

2) It is said later in the book, that established natural isomorphism between $V$ and $V^{**}$ makes them "totally equal" (my translation of Russian "совершенно равноправные"). But wait, isn't it enough to establish any kind of isomorphism between two sets to make them "equal"? What is it in natural isomorphism that makes the sets more equal than with any other isomorphism?

3) Is there a way to strictly define natural isomorphism (at least for finite vector spaces) without using category theory?

Thank you!

  • $\begingroup$ I really am not sure about the "totally equal" thing (though I think your translation is accurate), just as I'm not sure it is very important the isomorphism is natural. It is, of course, convenient and easier to manage since it doesn't depend on choosing basis in $\;V^{**}\,,\,V\;$ , as is usually done with vector spaces, and this makes things less hard, but for that... $\endgroup$
    – DonAntonio
    Nov 24, 2013 at 22:01
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    $\begingroup$ I think really it just means that if you have a canonical basis for $V$, then the natural isomorphism induces a canonical basis on $V^{**}$ (the one which is the image of the canonical basis of $V$ under the natural isomorphism). As you mentioned though, there is a precise meaning that is made clear in terms of category theory. $\endgroup$
    – Dan Rust
    Nov 24, 2013 at 22:21
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    $\begingroup$ "совершенно равноправные" means "totally equal in rights", not "totally equal". Generally, not even natural isomorphisms allow one to identify two objects, but naturally goes a long way to making this possible in particular cases. $\endgroup$ Nov 25, 2013 at 2:17

1 Answer 1


3) Of course you can define natural isomorphisms without using the language of category theory, but category theory was (in part) invented in order to efficiently express this notion, so there is little reason to (apart from trying to improve understanding).

2a) Philosophically, what makes an isomorphism between two objects natural is that constructing the isomorphism does not require more information than constructing the objects.

For example, in order to construct $V^*$ or $V^{**}$, it is enough to know that $V$ is a vector space. What I mean is that in order to build the two sets $V^*=\{f\colon V\to\mathbb F\mid f\text{ is linear}\}$, $V^{**}=\{f\colon V^*\to\mathbb F\mid f\text{ is linear}\}$, and to give them the structure of vector spaces, you need know nothing more than what a vector space is (hence what "linear" is), and that $V$ is a vector space.

Most likely in your book, to construct an injective linear map $\phi\colon V\to V^*$, you used additional information about the vector space $V$: probably a choice of basis. For $\phi$ to also be surjective, and hence an isomorphism, you need even more information: that the basis was finite*.

On the other hand, in order to construct the linear map $\psi_V\colon V\to V^{**}$, however, you don't need any extra information other than that $V$ is a vector space. You simply define $\psi_V(v)\in V^{**}$ by specifying how the $\psi(v)$ acts on elements $f\in V^*$: you declare $\psi_V(v)(f)=f(v)$. (To know that it is an isomorphism, you still need the extra information that $V$ is finite-dimensional, otherwise $\psi_V$ is just injective).

2b) Mathematically, the fact that an isomorphism is natural, i.e. does not depend on structural information beyond that contained in the objects, is captured by defining an isomorphism to be natural if its construction is preserved by structure-preserving (e.g. linear) maps. This is most easily expressed using the language of category theory (since the language of category theory was invented to express this).

Concretely, you can see that the isomorphism between $V$ and $V^*$ is not natural since the isomorphism based on $\{e_1,\dots,e_n\}$ likely defines the dual basis $e_i^*(\sum a_je_j)=a_i$. But all this is doing is defining an inner-product (non-degenerate bilinear form if the base field $\mathbb F\neq\mathbb R$) $\left<\cdot,\cdot\right>$ on $V$ such that $\left<e_i,e_j\right>=\delta_{ij}=\begin{cases}0&i\neq j\\1&i=j\end{cases}$. The basis (when $\mathbb F=\mathbb R$) allows us to identify $V$ as our usual $n$-dimensional space with coordinates, the inner product becomes the dot product, and the isomorphism to $V^*$ sends a vector $\vec v$ to the functional that projects orthogonally onto $\vec v$. Then the only linear transformations of $V$ to itself that preserve the isomorphism to $V^*$ are the orthonormal transformations, ones that preserve the length and (unsigned) angle between vectors. All other transformations of $V$ to itself break the isomorphism, and hence we can conclude the isomorphism is not natural.

1) The use of natural transformations is that they ensure that the maps you are constructing reflect genuine properties of the mathematical object, rather than being an artifact of the specific way in which you present the object, and hence a consequence of properties that the object "in-and-of-itself" does not have.

*More generally, the extra information for $\phi$ that you need is that of a bilinear form, which the basis allows you to define, and for $\phi^{-1}$ you need the bilinear form to be non-degenerate, which is what finiteness of the basis allows you to build. However, there are other ways to build non-degenerate bilinear forms, e.g. $L^2$ inner products.

EDIT: in response to comments, note that it makes no sense to talk about natural isomorphisms between "unnatural constructions". I said that philosophically, a natural construction is one that depends on no extra information. Mathematically, it is a functor: in addition to telling us how to construct a new object $F(V)$ out of an old object $V$, a natural construction also tells us how the construction behaves if we are given a structure-preserving map $f\colon V\to W$, by giving a structure-preserving map $F(f)$ between $F(V)$ and $F(W)$.

There are two types of natural constructions: covariant and contravariant, depending on whether $F(f\circ g)=F(f)\circ F(g)$ or $F(f\circ g)=F(g)\circ F(f)$. In particular, one shows the construction of $V^*$ is (contravariantly) natural by defining/constructing, for any $f\colon V\to W$, an $f^*\colon W^*\to V^*$ given by $f^*\colon g\mapsto g\circ f$ for any $g\colon W\to\mathbb F$, and showing that $(f_1\circ f_2)^*(g)=g\circ f_1\circ f_2=(f_2^*\circ f_1^*)(g)$.

Now, a natural transformation between $I$ (the identity construction $I(V)=V$ and $I(f)=f$ are isomorphisms $\phi_V\colon I(V)\to F(V)$ such that: $ \require{AMScd} \begin{CD} I(V) @>{\phi_V}>> F(V)\\ @VfVV @VVF(f)V \\ I(W) @>{\phi_W}>> I(W) \end{CD}$ for covariant $F$ or $\begin{CD} I(V) @>{\phi_V}>> F(V)\\ @VfVV @AAF(f)A \\ I(W) @>{\phi_W}>> I(W) \end{CD}$ for contravariant $F$. Now it is easy to see that there is no natural isomorphism between $V$ and $V^*$. For suppose that there were, so that we have $\begin{CD} V @>{\phi_V}>> V^*\\ @VfVV @AAf^*A \\ W @>{\phi_W}>> W^* \end{CD}$ for every $f\colon V\to W$. Then clearly, $\phi_V^{-1}\circ f^*\circ\phi_W\circ f$ would be the identity, so $f^{-1}\colon W\to V$ would be given by $\phi_V^{-1}\circ f^*\circ\phi_W$. But not every linear map is an isomorphism (in particular, the $f(v)=0$ is a linear map that is not an isomorphism for any choice of $V$ and $W$), so contradiction.

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    $\begingroup$ This is a great answer. Thank you. $\endgroup$
    – Dan Rust
    Nov 24, 2013 at 23:38
  • $\begingroup$ Thank you very much! It's seems much, much more reasonable now! Can I ask another couple of questions, to make it clear? First, do I understand your example right: say, if there is an isomorphism $V \mapsto V^*$, then we can always find such transformation $g: V \to V$, that for some $x \in V: x \mapsto y$, but $g(x) \not\to g(y)$? While for natural isomorphism the correspondence between elements of $V$ and $V^{**}$ will be preserved whatever transformation is applied to $V, V^{**}$? $\endgroup$
    – lithuak
    Nov 25, 2013 at 0:59
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    $\begingroup$ Second, we can prove that any isomorphism built with basis selection is non-natural. But does this prove that there is no other way at all to build natural isomorphism between $V$ and $V^{*}$? $\endgroup$
    – lithuak
    Nov 25, 2013 at 1:02
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    $\begingroup$ @izhak , you have understood correctly. I have answered your second question as an edit. $\endgroup$ Nov 25, 2013 at 2:05

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