That's incorrect if V is infinite-dimensional. A (0,1)-tensor is just supposed to be an element of V but with your definition you get an element of the bidual of V. Which is not isomorphic to V when dim V is infinite. And even when dim V is finite, you need to choose a basis of V to find an isomorphism with the bidual. From a math point of view, that's just no good.
No, the isomorphism between V and V** (for finite-dimensional V) is canonical. The canonical isomorphism T:V->V** is easy to construct: map a vector v in V to the element of V** which takes an element w from V* and applies it to v: T(v)(w) = w(v).
GP is giving you an element of V**. You want to turn it into a vector. To do that, please make the inverse isomorphism explicit without using a basis. I'll wait...
Why? All I need to prove is that I have a canonical linear map going in one direction, and that this map is a bijection. (Since I already constructed such a bijection, I could just answer "for any z in V**, take the unique element v in V such that for all w in V*, w(v) = z(w)".) Do you disagree that I provided such a map? You are correct that V** is not isomorphic to V when V is infinite-dimensional, but your statement that "when dim V is finite, you need to choose a basis of V to find an isomorphism with the bidual" is incorrect. This is elementary textbook stuff (e.g. first chapter of Wald's GR book), so I won't argue further.
I responded to a sibling comment with an explanation: https://news.ycombinator.com/item?id=41234913 But good on you for the condescension even though you didn't understand my point ;)
You said even when dim V is finite you need a basis to find an isomorphism with V**. But that’s not true. You’re right if you mean V*, but not the bidual.
Are you even trying to understand my point? Yes, in that direction, explicitly constructing an element of the bidual of V from an element of V is easy. To explicitly find an element of V from the element of the bidual, you need to choose a basis. Just try it, come on! Write down the inverse isomorphism. Let \alpha be an element of V**. Then v \in V such that f(v) = \alpha (where f is the isomorphism you wrote down) is given by...?
I will help you: if (e_i) is a basis of V and (e_i^*) is its dual basis, then v = \sum_i \alpha(e_i^*) e_i. Can you find such a formula without mentioning the word "basis"?
There's both directions of the isomorphism explicitly defined in a programming language. No choice of basis needed to define the maps, only to prove that the constructor for Bidual really gives you all linear functionals on the dual.
You are right about infinite dimensions, wrong about finite dimensions. V and V* are naturally isomorphic for finite dimensions.
In finite dimensions, V and V* are isomorphic, but not naturally so. The isomorphism requires additional information. You can specify a basis to get the isomorphism, but many bases will give the same isomorphism. The exact amount of information that you need is a metric. If you have a metric, then every orthonormal basis in that metric will give the same isomorphism.
You need to correct the 2nd sentence to say that V and V** are naturally isomorphic. V and V* are only unnaturally isomorphic. All of this holds only in finite dimensions, of course.
You have a typo in your first line, and I answered a sibling comment about that. Metrics are irrelevant to the discussion (and I presume you wanted to write "norm" instead of "metric").
As for why I said metric, see https://en.wikipedia.org/wiki/Metric_tensor. Which is technically a concept from differential geometry rather than linear algebra. But then again, tensors are literally the topic that started this. And it is only in differential geometry that I've ever cared about mapping from V to V*.
This comment is in a discussion about an article titled, Tensors, the geometric tool that solved Einstein's relativity problem. Therefore, "tensors are literally the topic that started this discussion."
Hopefully that's a hint that you should attempt to figure out what someone might be talking about before going to schoolyard insults.
Schoolyard insult? Uh? Can you quote the part of my comment that would be the "schoolyard insult"?
> This comment is in a discussion about an article titled, Tensors, the geometric tool that solved Einstein's relativity problem. Therefore, "tensors are literally the topic that started this discussion."
Again, are manifold involved in any way in the definition of tensors and their properties? No? Then why are you even mentioning "metric tensors"? (Which aren't even tensors, but tensor fields...)
Can you provide some examples of important tensors in physics for which the underlying vector space is infinite dimensional? I’m most familiar with the setting of tensor fields on manifolds, in which case the vector bundle consists of finite dimensional vector spaces. Nevertheless, I suppose in the absence of a pseudo-Riemannian metric one lacks a natural isomorphism between vectors/dual vectors. Does this “bidual” distinction arise in that case as well?
