1

Let $V$ be a finite dimensional vector space.

The issue of there being no canonical isomorphism between $V$ and its dual $V^*$ is commonly explained by stating that such an isomorphism would require additional structure involving arbitrary choice such as the choice of a basis, or equivalently, an inner product.

Clearly $V$ becomes canonically isomorphic to $V^*$ when endowed with an inner product, but for the existence of a canonical isomorphism, this observation gives a sufficient condition and not a necessary one.

Even resorting to category theory, and allowing for the generality of dinatural transformations, the best we get is that there is no collection of simultaneously structure preserving maps between the identity functor and $(-)^*$ other than the collection of zero maps, as discussed here.

How do we prove that augmentation of additional structure is a necessary condition for a canonical isomorphism $V \to V^*$ to exist?

Pang
  • 407
  • 5
  • 9
user
  • 649
  • Why does the result you mention in your fourth paragraph not answer your question? – Eric Wofsey Mar 11 '19 at 18:36
  • @EricWofsey How does that paragraph necessarily mean that out of all possible things that could be necessary to fix this, we need additional structure involving arbitrary choices? – user Mar 11 '19 at 18:38
  • Well, you haven't given any definition of what "we need additional structure involving arbitrary choices" means. It's not at all obvious how this should be defined rigorously. One perfectly reasonable definition would be the non-existence of a dinatural isomorphism. It is fine if you prefer a different definition, but you need to explain what your definition is. – Eric Wofsey Mar 11 '19 at 18:40
  • @EricWofsey In an effort to make it a bit more concrete, how does the non-existence of such a dinatural transformation concretely show us that we need to pick an equivalent of an inner product? It's not me jumping to vaguely defined conclusions out of the blue; this is commonplace in literature and I'm just trying to get a proper understanding of what's going on. – user Mar 11 '19 at 18:47
  • I don't think it's true that you need to pick an equivalent of an inner product. You just need to pick something that is not determined by just the vector space structure (since if you had a definition that used only the vector space structure, it would be dinatural, at least with respect to isomorphisms). – Eric Wofsey Mar 11 '19 at 18:48
  • Maybe your actual question is about my last parenthetical: why is "dinatural with respect to isomorphisms" a reasonable way to make "using only the vector space structure" precise? – Eric Wofsey Mar 11 '19 at 18:53
  • So very strictly speaking, the mainstream statement "There is no canonical isomorphism between $V \to V^*$ because this requires the choice of (some equivalent of) an inner product" is incorrect? And the best we can say is that $V$, just by itself, won't support such an isomorphism to it's dual? Logically, this is quite different! – user Mar 11 '19 at 18:55
  • 2
    Well, it's close to correct. Note that an isomorphism $V\to V^*$ is easily seen to be equivalent to a nondegenerate bilinear form on $V$. So, you don't necessarily have the symmetry or positivity that is usually assumed of an inner product, but it's a similar concept. – Eric Wofsey Mar 11 '19 at 19:00
  • @EricWofsey if all that we can infer is that $V$ by its structure alone can't support an isomorphism $V \to V^*$, how does looking at this mapping more generally as a nondegenerate bilinear form help? – user Mar 11 '19 at 19:13
  • Let us continue this discussion in chat. – user Mar 11 '19 at 19:15

0 Answers0