In a lot of resources that I have read it is mentioned that the isomorphism between $V$ and $V^*$ is non-canonical, but I was never sure that I properly understood precisely what this means. I haven't studied category theory so I couldn't really understand some other answers on math.stack based on category theory arguments.

I will try to explain how I understand things and maybe someone could comment on that.


Let $V$ be a finite-dimensional vector space. Its dual $V^*$ is the space of linear maps from $V$ to $\mathbb{F}$: $V^* = L(V,\mathbb{F})$. Now let $v$ be an arbitrary vector from $V$. If I fix a basis $F = (f_1,\ldots,f_n)$ for $V$ then I can write $v = \sum_{i=1}^n [v]_F^i f_i$, for some unique coefficient vector $[v]_F\in\mathbb{F}^n$. Then as far as I understand the "non-canonical isomorphism" that depends on this basis is $v\mapsto\alpha = \sum_{i=1}^n [\alpha]^{F^*}_if^i = \sum_{i=1}^n [v]^i_F f^i$ where $F^* = (f^1,\ldots,f^n)$ is the dual basis w.r.t. $F$ such that $f^i(f_j) = \delta^i_j$. Now let me choose a different basis $G = (g_1,\ldots,g_n)$ for $V$, then the dual basis w.r.t. it is $G^*=(g^1,\ldots,g^n)$ such that $g^i(g_j) = \delta^i_j$. Now I can write the vector $v$ in terms of the basis $G$: $v = \sum_{i=1}^n [v]_G^i g_i$, and I can form a functional $\beta = \sum_{i=1}^n [\beta]^{G^*}_ig^i = \sum_{i=1}^n [v]^i_G g^i$.

For $\alpha$ and $\beta$ to be well-defined (i.e. so that the values $\alpha(u)$ and $\beta(u)$ are fixed for every $u\in V$ irrespective of the coordinate representation), their coordinates must transform in an appropriate manner under a change of basis. Let $P$ be the change of basis matrix such that $g_i = \sum_{j=1}^n f_j P^j_i$ then in order to have $f^i(f_j) =\delta^i_k \implies g^i(g_j) = \delta^i_j$ I need that $g^i = \sum_{j=1}^n (P^{-1})^i_j f^j$ since: $$g^i(g_j) = \sum_{k=1}^n (P^{-1})^i_k f^k \left(\sum_{l=1}^n P^l_j f_l\right) = \sum_{k=1}^{n}\sum_{l=1}^n(P^{-1})^i_kP^l_jf^k(f_l) = \sum_{k=1}^n(P^{-1})^i_kP^k_j = \delta^i_j.$$ If I make it so that I transform the coordinates as follows: $[\alpha]^{G^*} = [\alpha]^{F^*}P$ and $[v]_G = P^{-1}[v]_F$, then I should get the same result for $\alpha(v)$ regardless of the chosen basis. Of course now $[\alpha]^{G^*}\ne[v]_G^T=[\beta]^{G^*}$ in general showing that $\alpha\ne \beta$.

The Question

Is the "non-canonical isomorphism" referring to the fact that $\alpha\ne \beta$ in the general case? That is, the construction of $\alpha$ was a function of my choice of $f_1,\ldots,f_n$, while the construction of $\beta$ was a function of my choice of $g_1,\ldots,g_n$. I assume that $\alpha$ could miraculously be equal to $\beta$ in some cases (e.g. if $f_1=g_1$ and $[v]^i_F=0=[v]^i_G$ for $i>1$), but in general it is not, so the non-canonicity is referring to the dependence on the choice of basis? If I had a canonical basis, for example $(e_1,\ldots,e_n)$ in $\mathbb{R}^n$, would this then be considered a canonical choice instead?

On the other hand, if I have a non-degenerate bilinear form $B:V\times V\to\mathbb{F}$, I can construct an isomorphism $\gamma_v = B(v,\_)$, that is independent of the basis. So is the word canonic referring to independence of the choice of basis in the initial definition of the functional?


If I take a basis for the space of polynomials of at most degree $1$ then I could choose the monomial basis $(f_1,f_2) = (1,x)$ or the Lagrange basis $(g_1,g_2) = \left(\frac{x-x_0}{x_1-x_0}, \frac{x-x_1}{x_0-x_1}\right)$. The dual bases are given as $(f^1,f^2) = \left(\delta_0, \delta_0 \circ \frac{d}{dx}\right)$ and $(g^1,g^2) = \left(\delta_{x_1},\delta_{x_0}\right)$, where $\delta_p(f) = f(p)$. Now if I have the vector $v = a\frac{x-x_0}{x_1-x_0} + b\frac{x-x_1}{x_0-x_1}$ then $v = \frac{bx_1-ax_0}{x_1-x_0}+\frac{a-b}{x_1-x_0}x = p\cdot 1 + qx $. This induces the two functionals $\alpha = p\delta_0 + q\delta_0 \circ \frac{d}{dx}$ and $\beta = a\delta_{x_1}+b\delta_{x_0}$. Now $\beta(v) = a^2+b^2$ while $\alpha(v) = p^2+q^2$, and in general those are unequal, thus $\alpha\ne\beta$.

However if I were to take for example the inner product $\langle v, w\rangle = \int_{0}^{1} v(t) w(t)\,dt$ then I could define the "canonical isomorphism" $\alpha_v = \langle v, \_\rangle$. Now a subsequent question is - didn't I just swap the choice of the initial basis for a choice of an inner product? Because I could of course take a number of other inner products that would result in different functionals. In what sense is the choice of inner product more canonical than the choice of basis? Or did I misunderstand this, and a choice of inner product doesn't result in a canonical isomorphism either. But if I go with that and I presuppose that the choice is canonical only when a canonical inner product exists, wouldn't this be the same argument as presupposing that a canonical basis exists? Basically I have trouble making sense of what "canonical" is referring to in the first place.

Canonical Isomorphism to the Double Dual $V^{**}$

The isomorphism $v\mapsto v^{**}(\phi) = \phi(v)$ is considered "canonical", I assume because there is no choice whatsoever in the above definition - I neither choose a basis nor an inner product. So is the above the only thing that "canonical" can refer to? Or is a vector space with a bilinear form isomorphism to $V^*$ also considered "canonical" if I assume that there is some distinguished canonical bilinear form? And similarly a choice $v\mapsto \alpha = \sum_{i=1}[\alpha]^{E^*}_ie^i = \sum_{i=1}^n[v]^i_E e_i$ would be considered "canonical" if there is some distinguished "canonical" basis $E$? Essentially, what do I need for something to be considered canonic?


Attempt at a solution of the exercise from coiso

For $(V,\mathbb{F}_2)$ with $\dim = 1$, with $V=\{0,x_1\}$ and $V^*=\{0^*,y^1\}$ I can construct the table: \begin{equation} \begin{array}{c|c} & 0^* & y^1 \\ \hline 0 & 0 & 0 \\ \hline x_1 & 0 & 1 \end{array} \end{equation} If I assign $0\ne x_1\mapsto 0^*$ and $0\mapsto y^1$ then $0^* = (x_1)^* = (x_1+0)^* = 0^* + y^1 = y^1$, and the assignment is not injective hence not an isomorphism. So there is only one valid choice and then $V\cong V^*$.

For $(V,\mathbb{F}_2)$ with $\dim = 2$, with $V=\{0,x_1,x_2,x_{12}=x_1+x_2\}$ and $V^* = \{0^*, y^1, y^2, y^{12}\}$ I can construct the table: \begin{equation} \begin{array}{c|c|c|c|c} & 0^* & y^1 & y^2 & y^{12} \\ \hline 0 & 0 & 0 & 0 & 0 \\ \hline x_1 & 0 & 1 & 0 & 1 \\ \hline x_2 & 0 & 0 & 1 & 1 \\ \hline x_{12} & 0 & 1 & 1 & 0 \end{array} \end{equation} I cannot have any other columns in the table as otherwise the functionals won't be linear. One possible assignment, when taking $x_1,x_2$ as basis vectors for $V$ is $0\mapsto 0^*$ and $x_J\mapsto y^J$. I could of course take $x_1,x_{12}$ instead as a basis, then the dual basis must be $x_1\mapsto y^{12}$ and $x_{12}\mapsto y^2$. Then the only option left is $x_2\mapsto y^1$. I don't get why any of the two assignments should be more canonical, unless I am missing something and the second assignment breaks linearity, which doesn't seem obvious from the table.

Edit 2:

Other Isomorphisms to the Double Dual

Thinking about this I realized that I could define the map $I_{\lambda} : V\to V^{**}$, $v\mapsto I_{\lambda}(V) = \lambda v^{**}$ such that $I_{\lambda}(v)(\phi) = \lambda v^{**}(\phi) = \lambda \phi(v)$ for $\lambda \ne 0$. This map looks like it is bijective, and it is linear since $I_{\lambda}(au + bv)(\phi) = \lambda \phi(au+bv) = a\lambda\phi(u)+b\lambda\phi(v) = (aI(u)+bI(v))(\phi)$, so it must be an isomorphism. Yet I suppose this would not be termed canonical, or will it? To my understanding it is the map $I_{1}(v)(\phi) = \phi(v)$ that is the canonical isomorphism, i.e., where $\lambda=1$. Clearly the above family of isomorphisms $\{I_{\lambda}\,:\,\lambda \ne 0\}$ do not depend on a choice of basis or inner product, and yet I do not think that they would be termed canonical. So I am even more confused now, as clearly "canonical" is not referring to a choice of basis or inner product in this case. Or is it that all $I_{\lambda}$ are considered canonical isomorphisms?

I also found this video, I don't know how relevant it is to the above setting though. From the video it becomes clear that canonical isomorphism is a kind of equality that is true for a certain subset of statements, but not all. So the question is what is canonical referring to in the above cases. Are all $I_{\lambda}$ canonical isomorphisms or is $I_1$ the only canonical isomorphism? Is $v\mapsto B(v, \_)$ canonical if there is only a single non-degenerate bilinear form $B$ given, i.e. $(V,B)\stackrel{canon}{\cong} (V^*, B^*) = (V^*, B(B^{-1}, B^{-1})$ but $V\stackrel{non-canon}{\cong} V^*$, and also $(V,(B_1, B_2)) \stackrel{non-canon}{\cong} (V^*, (B^*_1, B^*_2))$. Is the isomorphism canonical if a single basis is specified but no more, i.e. $(V, E) \stackrel{canon}{\cong} (V^*, E^*)$ but $(V, (E_1, E_2)) \stackrel{non-canon}{\cong} (V, (E_1^*, E_2^*))$? Also in the answers Mikhail Katz mentioned that in differential geometry the map may be considered non-canonical if it is not continuous, but I assume this is slightly different, as I don't need a notion of continuity to talk about the algebraic dual and double dual as far as I know.

Edit 4:

What I Gathered so Far

There are many isomorphisms $V\to V^{**}$ but the canonical one is given by the evaluation functional $\Lambda:V\to V^{**}$: $$\Lambda(v)(\phi) := \phi(v), \quad \forall \phi\in V^*.$$

According to coiso also the isomorphisms $I_{\lambda}(v)(\phi) = \lambda\phi(v)$ for $\lambda\ne 0$ can be termed canonical, but they are not the canonical isomorphism, which is instead given by the evaluation functional $\Lambda(v)$. As far as I understood the motivation for calling those canonical is because they do not require making a choice of basis, and we care greatly about results being coordinate invariant in linear algebra, differential geometry, physics, etc.

The isomorphism $J : (V,B) \to (V^*,B^*)$ where $J(v) = B(v,\_)$ is also canonical if $B$ is a non-degenerate bilinear form (conjugate-linear in one of the slots in the complex-valued case). If $B$ is degenerate then the latter is not bijective so not an isomorphism, but I guess that restricted to the subspace where it is non-degenerate it will be a bijection and thus a canonical isomorphism on that subspace.

Arturo Magidin made it clear that the isomorphism $J$ is canonical because it is between the vector space along with its inner product/bilinear form $(V,B)$ and its counterpart $(V^*,B^*)$. As far as I understood a map $J_B : V\to V^*$ where $J_B(v) = B(v,\_)$ would not provide a canonical isomorphism between $V$ and $V^*$ by themselves, since one could choose a different bilinear form, which would result in a different identification.

The isomorphism $K_F:V\to V^*$: $$K_F(v) = \sum_{i=1}^n [v]^i_F f^i \in V^*,$$ where $F = (f_1,\ldots,f_n)$ is a basis for $V$ and $F^*=(f^1,\ldots,f^n)$ is the dual basis for $V^*$ such that $f^i(f_j) = \delta^i_j$, is not a canonical isomorphism between $V$ and $V^*$ in general. This is because it depend on a choice of basis $F$, and as illustrated by my above examples a different basis can result in a different result.

A special case when there is a canonical isomorphism is when $\dim V\leq 2$ and the underlying field is $\mathbb{F}_2$ as given in the solution by coiso. I believe that also a one-dimensional space is canonically isomorphic to its dual (even if it is not over $\mathbb{F}_2$) since the map $K_F$ and $K_G$ actually produce the same functionals irrespective of the choice of $F$ and $G$ (the scalar difference gets canceled out by the change of basis).

Finally, supposedly also $K : (V,F) \to (V^*,F^*)$ where $K(v) = K_F(v)$ should also be a canonical isomorphism since I am now identifying a vector space along with a basis. The latter is more restrictive than providing a non-degenerate bilinear form however.

  • $\begingroup$ It is "non-canonical" in the sense that a particular vector $v\in V$ is mapped to different functions $\mathsf{f}$ depending on your choice of basis. By contrast, the isomorphism $V\to V^{**}$ always maps $v$ to the same element of $V^{**}$. $\endgroup$ Commented May 21 at 1:34
  • $\begingroup$ The "Setting" section seems unnecessarily convoluted to me. I can only assume the "transforms as" discussion can be translated into the language of natural transformations between functors in category theory. | Also, fun fact: there is a canonical iso $V\cong V^\ast$ if $V$ is $\dim\le 2$ over the finite field $\Bbb F_2$. I leave it as an exercise to construct it when $\dim =2$. $\endgroup$
    – coiso
    Commented May 21 at 2:14
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    $\begingroup$ The canonical map maps $v$ to the same element regardless of any choice of basis: "evaluate at $v$." $\endgroup$ Commented May 21 at 16:21
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    $\begingroup$ There are always multiple isomorphisms because there are multiple automorphisms(except for the $0$-dimensional space and the $1$-dimensional space over the field of two elements). The assertion is not "any map you define is 'canonical'". The assertion is that "there is a canonical choice of isomorphism", which is the one I described, not any ones you may decide to confuse yourself with. $\endgroup$ Commented May 21 at 16:39
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    $\begingroup$ No, it's not what we do.No, you are not getting contradictory answers. Yes, you are confused. Changing the inner product changes the space you are considering. because an inner product space is both the vector space and the inner product on it. again, "canonical" does not mean "there is no other way of doing it". $\endgroup$ Commented May 21 at 19:06

4 Answers 4


Yes, something is "canonical" if it does not depend on any choices.

This will depend on what kind of mathematical object you're talking about. For example, for (finite dim) vector spaces $V$ there is no canonical isomorphism $V\cong V^\ast$, but for (finite dim) inner product spaces $V$ there is a canonical isomorphism. An inner product is a positive-definite symmetric bilinear form, but allowing arbitrary nondegenerate bilinear form will do just as well via currying / musical isomorphisms.

I like to think of "canonicity" in game-theoretic terms. Consider a game where you and a teammate are put in separate rooms and given the same mathematical object, and your goal is to both come up with the same construction. For example, you may be both given the same vector space and want to construct an isomorphism $V\cong V^\ast$. You and your teammate are allowed to strategize before going into the rooms, but you don't have any information about the object you will be given other than what kind of mathematical object it will be. Can you and your teammate devise a strategy that guarantees you both come up with the same construction? If your method of construction involves arbitrary choices which the end result depends on, then that will not be such a strategy. If the object you're given is known ahead of time to be an inner product space, then there is such a strategy.

Because I could of course take a number of other inner products that would result in different functionals. In what sense is the choice of inner product more canonical than the choice of basis?

There is no intrinsic sense in which one is more canonical than the other. One could just as well consider "vector space equipped with a basis" as its own kind of mathematical object, and then there is a canonical choice of isomorphism to the dual. However, this is arguably a less "natural" kind of mathematical object. In "nature" (Euclidean space) there is no built-in choice of coordinates but there is a built-in inner product (the dot product which can be defined solely from angles and distances).

Note, by the way, that a basis determines an inner product but not vice-versa. So it is a more "cost-effective" choice to yield a duality (i.e. isomorphism $V\cong V^\ast$). In fact, there is an equivalence between dualities and nondegenerate bilinear forms (via musical isomorphisms).

Responding to edits:

Or is it that all $I_\lambda$ are considered canonical isomorphisms?

They are all canonical. The fact that $I_1$ is called "the" canonical isomorphism adds into the mix the usage of the word "canonical" for "standard," which is a social thing.

Imagine you and your teammate playing the game are told you will be given a certain kind of mathematical object, and in the strategy session your teammate tells you "hey there's lots of different strategies we have, in fact one strategy for each $\lambda\ne0$, but I've heard when people play this game they tend to use the strategy with $\lambda=1$ so let's do that too."

I don't get why any of the two assignments should be more canonical

Remember in this version of the game, you're not given a basis. So I wouldn't call the vectors $x_1,x_2,x_{12}$, I'd call them $x_1,x_2,x_3$, and the dual vectors $x_1^\ast,x_2^\ast,x_3^\ast$.

But notice something. In your labelling, you made $x_1^\ast,x_2^\ast,x_3^\ast$ have respective kernels $\{0,x_2\},\{0,x_1\},\{0,x_3\}$. That means $x_3^\ast$ has kernel associated to $x_3$, but $x_1^\ast,x_2^\ast$ have kernels associated to $x_2,x_1$ respectively. This automatically distinguishes $x_3,x_3^\ast$ as "special."

What's to prevent your teammate from making $x_2,x_2^\ast$ the "special" ones, or $x_1,x_1^\ast$? After all, your labelling of the vectors $x_1,x_2,x_3$ as first, second, third is an arbitrary choice that you made, and there is no guarantee your teammate makes the same choice as you. If they make a different choice of 1/2/3, they get a different isomorphsim.

There are six isomorphism between $V$ and $V^\ast$ (when $V\cong\Bbb F_2^2$), but only one of them is canonical. One way to guarantee your isomorphism is canonical is to describe what $x_i^\ast$ is in a way that doesn't depend on how the other two $x_j,x_k$ elements are labelled...

  • $\begingroup$ I agree that inner products or bilinear forms carry less information than a basis, and are a less restrictive requirement (e.g. every orthogonal transformation of a basis results in the same Gramian). What I gathered from your answer is that $v\mapsto v^{**}(\phi) := \phi(v)$ is always canonical. And $v\mapsto B(v,\_)$ is canonical only if I have an inner product space that comes with one distinguished canonical bilinear form $B$. Similarly $v\mapsto \alpha = \sum_{i=1}^n [\alpha]^{E^*}_i e^i = \sum_{i=1}^n [v]_E^i e^i$ is canonical only if I have a canonical basis $E$. $\endgroup$
    – lightxbulb
    Commented May 21 at 12:21
  • $\begingroup$ From your analogy it sounds like for the bilinear form canonicity you and your teammate need to be given only a single canonical bilinear form. Similarly for being provided a single basis. I guess that if I am given a single basis $f_1,\ldots,f_n\in V$ that is considered distinguished, then the map $f_i \mapsto e_i\in\mathbb{R}^n$ is "canonical" now, and I could just work in $\mathbb{R}^n$ using the standard basis there, while $B$ distinguishes bases only up to isometries. So a basis $E$ would define $[B]_E$ while just a bilinear form would define $[B]_{UE}$ for all $U\,:\,B(Uv,Uw) = B(v,w)$. $\endgroup$
    – lightxbulb
    Commented May 21 at 12:34
  • $\begingroup$ I could have named the elements $0,\triangleright, \Delta, \nabla\in V$ and $0^*,\circ, \bullet, \odot\in V^*$ table corresponding to that order $\circ(*) = (0,1,0,1)$, $\bullet(*)=(0,1,1,0)$, $\odot(*) = (0,0,1,1)$, and how you order the table seems to be subjective (you are allowed to swap columns and rows), so I do not get how one of the assignments would be more canonical. In your example you say $x_3, x_3^*$ is special, but if I enumerate things as $(x_1,x_2,x_3) \to (x_3', x_1', x_2')$ then now $x_1', x_1^*$ is special. The teammate would not be able to tell without a table convention. $\endgroup$
    – lightxbulb
    Commented May 21 at 15:52
  • $\begingroup$ @lightxbulb Here is the solution. You and your teammate agree ahead of time that for each $v\in V$ the dual vector $v^\ast\in V^\ast$ will have kernel $\{0,v\}$. Then both you and your teammate will get the same isomorphism. $\endgroup$
    – coiso
    Commented May 21 at 15:59
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    $\begingroup$ I said what $v^\ast$ will be and how to agree with your teammate what it will be before the game starts. You ask what the dual of $\Delta$ is, for example: the dual of $\Delta$ we decide ahead of time will be the unique functional with kernel $\{0,\Delta\}$. $\endgroup$
    – coiso
    Commented May 21 at 16:19

Two additional remarks:

  1. It comes up a lot in geometry that we are trying to identify two different vector spaces that just don't have a canonical isomorphism between them, and this matters a lot. For a simple example, imagine you are talking about wind velocity on earth. You can say "10 km/hr East" and this works fine at Buenos Aires and Stockholm but is ambiguous at the north and south poles. So even though there is a "locally consistent" set of isomorphisms between the "wind-speed" vector spaces at various points on earth, there isn't a global one. Providing canonical isomorphisms is a way to provide global ones (when they exist -- they don't in the case of wind speed on a sphere!). All this rant is just to say why you should care about canonical isomorphisms beyond pedantry -- canonical isomorphisms are the ones that tend to "glue together."

  2. I understand that you haven't studied category theory, but the subject was historically invented precisely to make our intuition about what is canonical precise, and so it is worth learning a definition or two in there. "The vector spaces $V$ and $V^{**}$ are canonically isomorphic" means "there is a natural transformation from the identity functor (on the category of f.d. vector spaces over a field) to the double-dual functor, which is an isomorphism of functors." This is false for the single-dual functor (well, it's hard to state, because the single-dual functor reverses arrows, but any precise version you can state of it will be false).

(A possibly apocryphal quotation from Saunders MacLane: "I didn't invent categories to study functors. I invented them to study natural transformations.")

  • $\begingroup$ I saw the other posts explaining this in terms of category theory, but considering I have no idea what a "natural transformation", "functor", or "category" is, I could not understand those. My current understanding is that "non-canonic" is more or less equivalent to having multiple possible options that one considers equivalent. Specifying a single distinguished bilinear form mitigates this by reducing the choices to one, as does specifying a distinguished basis (which in some sense defines a canonical isomorphism to $\mathbb{F}^n$). So "non-canonical" choice $\approx$ non-unique choice? $\endgroup$
    – lightxbulb
    Commented May 21 at 13:00
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    $\begingroup$ I recommend reading just enough of the Wikipedia to get those three definitions. $\endgroup$
    – hunter
    Commented May 21 at 13:17

The fact that vector spaces of the same dimension are not necessarily isomorphic in a natural way can be illustrated geometrically as follows. Consider the family of tangent planes $V_p, p\in S^2$ to the unit sphere $S^2$. Each of the planes $V_p$ as $p$ runs over $S^2$ is isomorphic to the $(x,y)$-plane $\mathbb R^2$, but I claim that one couldn't choose such isomorphisms in such a way that they would vary continuously in $p$ (if they don't vary continuously, obviously it's not a natural or canonical choice).

Suppose one could choose such isomorphisms $\phi_p : \mathbb R^2 \to V_p$ varying continuously in $p$. Now choose your favorite vector, say $\frac{\partial}{\partial x}$ in the plane. Then apply ${\phi_p}_*$ at all points. We then obtain a continuous nonvanishing vector field ${\phi_p}_*(\frac{\partial}{\partial x})$ on the 2-sphere.

However, such a vector field does not exist by the hairy ball theorem. The contradition shows that the isomorphisms in question could not be chosen canonically.

Another point is that if we had a canonical isomorphism $\phi : V \to V^*$, we could use it to define a canonical metric on many manifold, by using the pairing $\langle v, w \rangle_p = (\phi_p v) (w)$. But we know that a general differentiable manifold does not possess any canonical metric.

For a similar reason, even a 1-dimensional space is not canonically isomorphic to its dual.

  • $\begingroup$ "if they don't vary continuously, obviously it's not a natural or canonical choice" - so in this case I guess "non-canonical" and "not natural" refers to the non-existence of a continuous map (in terms of $p$) from $TS^2$ to $\mathbb{R}^2$, and you are left with infinitely many maps that are not continuous, so there is no canonical choice from those. What if I had some other manifold than $S^2$ where I have several continuous maps? I guess the choice would still not be canonical, since I have to pick from several. Also this canonicity seems to be different than the one for $V\cong V^*$, is it? $\endgroup$
    – lightxbulb
    Commented May 21 at 12:42
  • $\begingroup$ @lightxbulb, I added a comment about $V$ and $V^*$. Certainly there do exist parallelizable manifolds where one can choose such isomorphisms continuously, but $S^2$ is perhaps the simplest one where one can't. $\endgroup$ Commented May 21 at 12:45
  • $\begingroup$ So for parallelizable manifolds with many continuous maps, this is still not canonic because of the many choices, right? Regarding the canonical metric: Let $f_1,\ldots,f_n$ in $V$ (not necessarily an arbitrary differentiable manifold) and specify this as a canonical basis. Now setting $\langle v, w\rangle := \sum_{i=1}^n f^i(v)f^i\sum_{j=1}^n f^j(w)f_j = \sum_{i=1}^n f^i(v)f_i(w)$. Does this make my above metric canonical in some sense? Basically, is non-canonicity referring to having to choose from multiple equivalent options, and e.g. having a basis fixed reduces the choices to a single one $\endgroup$
    – lightxbulb
    Commented May 21 at 12:55
  • $\begingroup$ Parallelization can be canonical. E.g. with a copy of $S^3$ as a Riemannian manifold, one can pick an arbitrary basepoint to call $1$, which makes it a copy of $S^3$ in the quaternions, then one map from $T_pS^3$ to $T_qS^3$ is left-multiplication by $q\overline{p}$. (One could also pick right-multiplication by $\overline{p}q$.) This involves an arbitrary choice of basepoint, but (exercise, if you know quaternions) the end result does not actually depend on this basepoint. Indeed, you can describe all that with left/right isoclinic rotations without a basepoint or mentioning quaternions. $\endgroup$
    – coiso
    Commented May 21 at 14:59

Questions like this are often subtle. There are definitions for a canonical map or a natural transformation, but often the words "canonical" or "natural" are used in a not-entirely-technical sense, and judgements that something is or is not "canonical" can change depending on the context (as I mention below).

One way I think about things is that when we look around in the world, we don't find vectors lying around. I don't walk down the street and see arrows everywhere, there's no museum where they keep the One True Basis. So, when we use vector spaces to model the world, there's always a question of "which basis do you pick"? This has practical consequences, because sometimes one basis is easier to work with than another, but if our model is going to be any good, then the answer we get at the end of the day needs to be independent of the particular basis we choose, because the actual world we're modelling is definitely independent of the basis we chose. Keeping this perspective in mind, I'll try to answer your questions.

$V$ is not canonically isomorphic to $V^\ast$: I think your intuition for why people say a finite-dimensional vector space is not "canonically isomorphic" to its dual is pretty spot-on. When you choose the isomorphism that takes a given basis to the dual basis, that isomorphism doesn't "play well" with other linear maps. You gave a great example that changing the basis for $V$ does not induce the same change of basis of $V^\ast$. This is exactly the sort of behavior we don't want if we're using vector spaces to model something.

Inner products: Yes, endowing a vector space with an inner product (or any non-degenerate bilinear form) does cause it to have a canonical isomorphism with its dual. But, as you seem to have intuited, inner products themselves are generally speaking not "canonical".

The way I think about this is in computer science terms: you have a data type called "vector space", and a different data type called "inner product space". An inner product space has all the data of a vector space $V$ plus the additional data of the inner product $\langle \cdot, \cdot \rangle$. If you start off with the data of an inner product space, then that gives you a canonical choice of isomorphism between $V$ and $V^\ast$ via $v \mapsto \langle v, \cdot \rangle$. But, if you only start with the data of a vector space, there are many incompatible inner products that could be imposed on it, e.g. I could simply declare $\begin{bmatrix}1\\0\end{bmatrix}, \begin{bmatrix}1\\1\end{bmatrix}$ to be an orthonormal basis for $\mathbb{R}^2$. You can't stop me.

(Sometimes, we're in an even larger context than inner product spaces, and that larger context makes some choice of inner product natural. If we're modelling the world, we choose the inner product so orthogonal vectors literally meet at right angles.)

The canonical isomorphism with the double dual: The canonical isomorphism from $V$ to $V^{\ast \ast}$ is canonical in all of the senses I've mentioned. It doesn't depend on our choice of basis, it plays nicely with other linear maps, it doesn't require the auxiliary data of an inner product.

You raise an interesting point about having other isomorphisms between $V$ and $V^{\ast \ast}$. I would say two things about these other isomorphisms:

  1. For a f.d. vector space over an arbitrary field, you aren't guaranteed to have any isomorphism $V \to V^{\ast \ast}$ other than $v \mapsto (\phi \mapsto \phi(v))$, because a field isn't guaranteed to have any nonzero elements aside from 1. Indeed, there's a perfectly good field whose only elements are 0 and 1. This is canonicity of a somewhat different kind than what we've talked about so far: the canonical isomorphism should be independent of what field you're talking about.
  2. In the case where your field does have other elements, all these different isomorphisms are very similar to one another! I don't want to work out the formalism right now, but I would bet there's a strong sense in which these isomorphisms are all "canonically equivalent" to one another.
  • $\begingroup$ I saw the definition of canonical map on wikipedia yes, but it is not formal (it's discussed in the video). I think you meant to write $V^{**}$ for the double dual section. I agree that in $\mathbb{F}_2$ I have only $I_1$, however that doesn't resolve my issue with fields of characteristic that is not two. In 1D the choice of $\lambda$ is equivalent to a choice of basis. On the other hand a choice of basis in 1D seems to be immaterial for the functionals since $\alpha = \beta$, i.e. the isomorphism $V\cong V^*$ is supposedly canonical in 1D. $\endgroup$
    – lightxbulb
    Commented May 21 at 17:12
  • $\begingroup$ Good catch, I've edited all the incorrect $V^\ast$s to be $V^{\ast \ast}$s. I agree that the definition of "canonical map" is not formal, if you're looking for a formal definition I think you're out of luck. As far as I know, at present there isn't a formal definition of when something is or is not "canonical", we just know it when we see it. $\endgroup$
    – Ivan Aidun
    Commented May 22 at 15:15
  • $\begingroup$ Re: the choice of $\lambda$, as I said in my post, you should regard $\lambda=1$ as the most "natural" choice, because it's the choice you can make independent of what field the vector space is defined over. This choice is both canonical in the sense that it doesn't require a particular choice of basis, but also in the sense that it doesn't require a particular choice of field. $\endgroup$
    – Ivan Aidun
    Commented May 22 at 15:35
  • $\begingroup$ As far as your example with a 1D vector space, in this case the dual basis seems to transform correctly because coincidentally every $1 \times 1$ matrix is symmetric. However, there are other things that I could want a canonical isomorphism to respect, such as maps to other vector spaces. If I have a linear map from $V \to W$ for some other vector space $W$, I would expect a canonical isomorphism to give a map $V^\ast \to W^\ast$, but this is not what you get when you consider the dual basis. $\endgroup$
    – Ivan Aidun
    Commented May 22 at 15:38
  • $\begingroup$ Can you elaborate on the $V\to W$ and $V^*\to W^*$ with some simple example in 1D? $\endgroup$
    – lightxbulb
    Commented May 22 at 20:50

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