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Okay first of all here's what I understand of the two notions —

A natural Isomorphism between two sets is an isomorphism that can be constructed without any choices involved, like choosing a basis or an inner product or something of that nature to define the isomorphism.

Equality of 2 sets implies that the 2 sets contain the exact same elements.

My questions are particularly in regard to the natural isomorphism between a vector space ($V$) & its double dual ($(V^*)^*$) for finite dimensional vector spaces.

Question 1: We can prove that a vector space ($V$) is naturally isomorphic to its double dual double dual ($(V^*)^*$) but we cannot prove that they are equal, correct? Saying $V = (V^*)^*$ is an additional assertion that has nothing to do with the mathematics, right?

Question 2: If so what's the motivation for making this assertion? I understand that this allows us to think of vectors as (1,0) tensors but is there a deeper reason for doing this?

Question 3: Consider a real inner product space $(V, <.,.>)$ equipped with a special distinguished inner product, then a map $$ \phi : V \longrightarrow V^*$$ $$ \ \ : v \longrightarrow <., v>$$ then $\phi$ ought to be a natural isomorphism too, right? Does this mean that we can equate $V = V^*$ such that $v = \phi (v)$ & draw no distinction between vectors & covectors?

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    $\begingroup$ It is not generally "okay", as you will quickly get tangled up in contradictions if you do this too often (e.g., what happens if you have two inner products on the same space $V$ and use both to identify $V$ with $V^\ast$ ?). But it is possible in "restricted environments". Think of it like the use of pronouns: They shorten your text, but it's your responsibility to ensure that you are not creating ambiguity, e.g. by having two masculine nouns in one sentence and then referring to "him" in the next. Whatever argument you make using these identifactions should be ... $\endgroup$ – darij grinberg Dec 17 '16 at 2:44
  • $\begingroup$ ... translateable into one where you don't use them, by "filling in" the omitted isomorphisms. $\endgroup$ – darij grinberg Dec 17 '16 at 2:46
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    $\begingroup$ It's maybe worth adding that this concept is most relevant when you have many objects to consider, not just one. A natural construction works for all of them at once (this is why naturality is really about functorial constructions on entire categories of objects). A non-canonical operation requires choice(s) for each object separately. $\endgroup$ – Jake Levinson Dec 17 '16 at 4:16
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    $\begingroup$ I'm the \langle \rangle fairy, here to let you know that $\langle, \rangle$ plays nicer with TeX than $<, >$ does :) $\endgroup$ – Patrick Stevens Dec 17 '16 at 9:41
  • $\begingroup$ @Jake ... and not just on the objects but the arrows too! $\endgroup$ – user14972 Dec 19 '16 at 1:29
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A natural isomorphism is a special case of a natural transformation. More precisely, it is a natural transformation with an inverse. This is the correct technical definition. Since natural transformations are morphisms between functors, then a natural isomorphism is really between functors or "constructions" rather than between specific vector spaces or specific sets.

For instance, when $V$ is finite-dimensional the natural isomorphism $V \cong V^{**}$ is really between the identity functor and the double dual functor on the category of finite-dimensional vector spaces.

Saying that $V = V^{**}$ is a convenient "abuse of notation." If we are working in set-theoretic foundations it is not literally true. Justifying this common and useful abuse of notation is one of the benefits of the univalence axiom in homotopy type theory, which is a new alternative foundation for mathematics.

With regards to the canonical isomorphism $V \cong V^*$ when $V$ is an inner product space, this isomorphism cannot be natural in the usual sense since the identity functor is covariant whereas the duality functor is contravariant.

See this mathoverflow thread for a discussion of naturality which you might find helpful.

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  • $\begingroup$ Hi sorry. See I come from a physics background so I'm really not familiar with homotopy type theory. But I think physics does operate on set theoretic foundations so can you please tell me then why it makes sense to take tensor product of vectors in physics? Aren't tensor products defined between 2 tensors (or tensor spaces)? If a vector space cannot always be thought of as its double dual, then it doesn't make sense to think of vectors as tensors so it doesn't really make sense to talk about tensor product of vectors. Yet we seem to do it everywhere, especially in quantum mechanics. $\endgroup$ – Nameless Paladin Dec 19 '16 at 2:43
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    $\begingroup$ Well, thinking of a vector space as being "equal" to its double dual is just omitting the natural isomorphism from the notation, to keep it clean and tidy. There's nothing wrong with doing that. I'm a bit confused about the tensor product part of your question: the tensor product is just a bifunctor on the category of vector spaces, and duality doesn't really matter when defining it. $\endgroup$ – ಠ_ಠ Dec 19 '16 at 4:29
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    $\begingroup$ Also, physics doesn't really operate on any sort of mathematical foundation; physicists were doing physics long before logicians and mathematicians even began to consider foundations. Currently, set theoretic foundations are the "default" assumption in mathematics, but most working mathematicians and physicists don't think rigorously about foundations much, if at all, though they do influence the language we use. $\endgroup$ – ಠ_ಠ Dec 19 '16 at 12:52
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A natural Isomorphism between 2 sets is an isomorphism that can be constructed without any choices involved, like choosing a basis or an inner product or something of that nature to define the isomorphism.

I have no idea what this means. If one is talking about isomorphism between two sets, then it means bijection. For sets, there is no vector space structure and hence "choosing a basis" is nonsense.

We can prove that a vector space $V$ is naturally isomorphic to its double dual $V^{**}$

This is not always true. Consider the infinite dimensional (topological) vector spaces. (Also, "algebraic dual spaces" and "continuous dual spaces" are two different notions.)

but we cannot prove that they are equal, correct?

Yes, if "equal" is in the sense of sets.

If so what's the motivation for making this assertion?

You mean isomorphism between $V$ and $V^{**}$? Answers should be in

Why do we care about dual spaces?

What is the motivation/application of dual spaces and transposes?

Consider a real inner product space $(V, \langle.,.\rangle)$ equipped with a special distinguished inner product, ...

What is the "special distinguished inner product"? A related topic to your third question is called Riesz representation theorem.

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