2

I'm aware of the fact that there exists a natural isomorphism between a vector space $V$ and its double dual $V''$; what I'm asking is the following: let's define an application $F:V\to V'$ as $F(v)=v^T$ (the image of a vector is the application of the dual which corresponds to multiplying to the transpose of said vector); now for every $v\in V$ we can find $F(v)\in V'$ and vice versa, therefore $F$ is an isomorphism. Now I believe that could be considered a natural isomorphism since whatever are the basis for $V$ and $V'$, the coordinates of some vector with respect to said basis are unique and therefore to one vector in $V$ with respect to some basis $\mathscr{B}_V$ corresponds one and only application in $V'$ with respect to some basis $\mathscr{B}_{V'}$, even if $\mathscr{B}_{V'}$ is not the dual basis of $\mathscr{B}_V$.

Why isn't this a natural isomorphism? It seems to me I don't need to chose a basis for either $V$ or its dual in order to make this work...

bocceri
  • 212
  • 4
    The transpose is not defined in an arbitrary vector space. – Mikhail Katz Jan 11 '24 at 10:40
  • As a complement to @MikhailKatz's answer, what you describe works for instance in finite-dimensional spaces where indeed the space and its dual are isomorphic, and in Hilbert spaces where this shows that the space and a subspace of its dual (the continuous linear functionals only) are isomorphic. – Beleth Jan 11 '24 at 10:45
  • @Beleth, finite-dimensional spaces are isomorphic to their duals as you say, but they are not naturally isomorphic to their duals. – Mikhail Katz Jan 11 '24 at 10:51
  • @MikhailKatz yeah, right; didn't consider that, thanks! – bocceri Jan 11 '24 at 10:56
  • @MikhailKatz you are right, I understood the use of "natural" as "that is rather simple and intuitive", and not as the mathematical definition of a natural isomorphism. Thanks for the precision. – Beleth Jan 11 '24 at 11:00
  • 2
    Ah, you must be French :-) "Precision" in English does not mean précision. English translation of the latter is "clarification". @Beleth – Mikhail Katz Jan 11 '24 at 11:03
  • no, there isn't. –  Jan 11 '24 at 11:35
  • @MikhailKatz You got me Sherlock, thank you for the clarification :) Best wishes – Beleth Jan 11 '24 at 11:53

1 Answers1

3

If you're stil not convinced by the non-naturality of transposition, you'll surely agree on something more geometric, that there's no natural inner product.

Or even simpler, there's no natural length (norm) defined on all vectors in all spaces - doubling any length (norm) makes it a just as valid norm.

But any identification between $U$ and its dual $U^*$, $\eta:U\to U^*$ defines a scalar product on $U$ $$\langle u,v\rangle = \eta(u)(v).$$

If there was a natural isomorphism between $U$ and $U^*$, that would automatically define a scalar product and hence a norm (length) by $\|u\|=\sqrt{\langle u,u\rangle}$, i.e. turn all vector spaces into inner product spaces, which they initially are not :)

Al.G.
  • 1,802
  • The answers in the "Related" tab give some more rigorous treatment on this: https://math.stackexchange.com/questions/579739/why-it-is-important-for-isomorphism-between-vector-space-and-its-double-dual-spa?rq=1 https://math.stackexchange.com/questions/622589/in-categorical-terms-why-is-there-no-canonical-isomorphism-from-a-finite-dimens?rq=1 – Al.G. Jan 11 '24 at 13:20
  • that's very clear, thank you! – bocceri Jan 11 '24 at 13:39