Graduate Differential forms and vector calculus

[spaghetti3451]

Let

$0$-form $f =$ function $f$

$1$-form $\alpha^{1} =$ covariant expression for a vector $\bf{A}$

Then consider the following dictionary of symbolic identifications of expressions expressed in the language of differential forms on a manifold and expressions expressed in the language of vector calculus in $\mathbb{R}^{n}$.

$i_{\bf{v}}(\alpha^{1}) \iff \bf{v}\cdot{\bf{A}}$

$df \iff \text{grad}f$
Now, with $\alpha^{1} = a_{i}dx^{i}$,

why is $\textbf{d}i_{\bf{v}}(a_{i}dx^{i})\iff\text{grad}({\bf{v\cdot{A}}})\cdot{d\bf{x}}$

and not $\textbf{d}i_{\bf{v}}(a_{i}dx^{i})\iff\text{grad}({\bf{v\cdot{A}}})$?

 

[lavinia]

Let

$0$-form $f =$ function $f$

$1$-form $\alpha^{1} =$ covariant expression for a vector $\bf{A}$

Then consider the following dictionary of symbolic identifications of expressions expressed in the language of differential forms on a manifold and expressions expressed in the language of vector calculus in $\mathbb{R}^{n}$.

$i_{\bf{v}}(\alpha^{1}) \iff \bf{v}\cdot{\bf{A}}$

$df \iff \text{grad}f$
Now, with $\alpha^{1} = a_{i}dx^{i}$,

why is $\textbf{d}i_{\bf{v}}(a_{i}dx^{i})\iff\text{grad}({\bf{v\cdot{A}}})\cdot{d\bf{x}}$

and not $\textbf{d}i_{\bf{v}}(a_{i}dx^{i})\iff\text{grad}({\bf{v\cdot{A}}})$?

An exterior derivative is always a differential form not a vector.

 

[WWGD]

An exterior derivative is always a differential form not a vector.

Doesn't the vector space duality give you equality (up to non-natural isomorphism)?

 

[fresh_42]

Doesn't the vector space duality give you equality (up to non-natural isomorphism)?

The exterior derivation is an operator that shifts k-forms to (k+1)-forms. But this non-natural between 1-forms and vectors drives my nuts. To be or not to be?

 

[lavinia]

Doesn't the vector space duality give you equality (up to non-natural isomorphism)?

Yes but that is not the point.

If the isomorphism were natural then it might be reasonable not to distinguish them. The different isomorphisms define all of the possible metrics - I think. One could also say that a vector is the same as the operator that takes inner products with that vector. But again this operator is not a vector. It is a dual vector.

 

[fresh_42]

If the isomorphism were natural then it might be reasonable not to distinguish them.

I don't think that this is a legal implication. At a short glimpse into my homology book, duality seems to be a natural functor.

 

[lavinia]

I don't think that this is a legal implication. At a short glimpse into my homology book, duality seems to be a natural functor.

Can you explain this further? It seems that if there is no natural isomorphism then identifying the spaces is not natural. The isomorphism between a vector space and its dual seems to require a metric. But the double does not.

BTW: What is the title of your homology book?

 

[fresh_42]

Can you explain this further? It seems that if there is no natural isomorphism then identifying the spaces is not natural.

BTW: What is the title of your homology book?

I can, but it's in the wrong language and from the 70's (G. Brunner: Homological Algebra; even amazon apparently doesn't have it). But as I said, I didn't really searched for an exact definition of "natural" in a sense as "universal" is defined. I simply stumbled upon the definition (and a commutative diagram) of a functorial morphism, which the author said is also called morphism of functors or natural transformation, resp. natural equivalence in case of isomorphisms. To me it made sense, since duality transforms objects and morphisms between two categories in a - which I find - natural way and I couldn't see, why this diagram shouldn't commute in the case of the $*$-functor.

 

[zinq]

"Doesn't the vector space duality give you equality (up to non-natural isomorphism)?"

This amounts to nothing more than saying that the two real vector spaces have the same dimension.

"... duality seems to be a natural functor."

There is such a thing as a "functor", and there is such a thing as a "natural transformation". But I do not know the definition of a "natural functor".

For why an isomorphism between a vector space and its (algebraic) dual is not natural, see Wikipedia: https://en.wikipedia.org/wiki/Natur...le:_dual_of_a_finite-dimensional_vector_space.

 

[fresh_42]

"... duality seems to be a natural functor."

Yes, this is nonsense, see the correction in #9 to "natural transformation". But, as I said, I took only a glimpse on it, without further examination. Therefore thank you for clarifying this. I wouldn't have expected it to be a counterexample. Something learnt. (Does this count for tuesday or wednesday? )

 

[lavinia]

If there were a canonical isomorphism from the tangent bundle of a smooth manifold to the bundle of differential 1-forms then every manifold would have a canonical metric.