# Coefficients in the linear combination of a k-form

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## Main Question or Discussion Point

Hello,
We defined a k-form on a smooth manifold M as a transfromation

Where the right space is the one of the alternating k-linear forms over the tangent space in p.
If we suppose we know, that we get a basis of this space by using the wedge-product and a basis of the dual space, then we might get for a given form:

Can someone explain to me what the coefficient in this linear combination is? I know that they are functions and in some cases scalars. But I don't know the cases. According to the definition at the beginning I suppose, that they are funtions for a given p and become scalars if we have argument values of the tangent space. Is this correct?
Greetings
Maxi

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## Answers and Replies

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fresh_42
Mentor
You have $\omega \in \Lambda^k TM$ which is a vector space with basis vectors $dx^{i_1}\wedge \ldots \wedge dx^{i_k}$. Thus the $\omega_{i_1 \ldots i_k}$ are scalars, the coefficients in the linear combination which represents $\omega$ according to this basis.

The evaluation $p \longmapsto \omega_p$ at a certain point affects only the differential forms, resp. the function those forms apply to, not the scalar coefficients.

E.g. if $k=1$ and $\omega = 2dx$ we get $p \longmapsto \omega_p = (f \longmapsto 2df_p)$ or $f(p+v)=f(p)+df_p\cdot v + o(v)=f(p)+\frac{1}{2}\omega_p(f)\cdot v+o(v)$. The coefficient is not affected.

mathwonk
Homework Helper
What fresh says is of course exactly right, but in one special case, you can say more about these coefficients. if you have a set of k one- forms, each written in terms of the n basic one forms dx1,...,dxn, then each of your one - forms needs n coefficients. If you wedge together those k one - forms, you get a k-form expressible in terms of the standard coordinate k-forms, as you have written, with some coefficients you have asked about. Those cofficients in this case are exactly all k by k sub determinants of the n by k matrix of coefficients you began with, the ones expressing the k arbitrary one - forms in terms of the n standard coordinate one-forms. Note there are exactly "n choose k" such k by k subdeterminants.

Indeed if you prove independently that the space of n-forms (at one point) is one dimensional, then one way to define the determinant of a matrix is as the coefficient needed to write the wedge product of its rows in terms of the standard n-form. More abstractly, a linear endomorphism of an n dimensional vector space induces a linear endomorphism of its space of n-linear alternating functions, which must be multiplication by a scalar, which scalar is then defined as the determinant of the original endomorphism.

Hello,
@ fresh_42
What is f in your notation? Is it a curve?
I understand what you said about the elements of the basis, thus to say that they are just scalars. But I found in a Wikipedia article (https://en.wikipedia.org/wiki/Differential_form#Differential_calculus) that we might find a linear combination where these factors are smooth functions from $$U \subset \mathbb{R}^n \rightarrow \mathbb{R}$$. Because of this I'm going to add more detail the context in which my question arose.
We defined the outer derivative on $$\mathbb{R}^n$$ as follows: Let $$w$$ be a k-form on a open set U

Then we define the exteriour derivative by

where

is the (normal) differential of the function $$w_{i_1 \cdots i_k}$$ (Please note that we used here the Einstein notation, so we have a sum over j.).
As you said I expected to be the coefficients scalars, but in this case they are just smooth functions, what puzzled me. Can you explain that to me?
@mathwonk
We had the following:

Where $$V^*$$ is the dual space and j counts the lines and i the columns. What do you mean by subdeterminants?
Best wishes
Maxi

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fresh_42
Mentor
What is f in your notation? Is it a curve?
A smooth function the differential form $\omega$ applies to, so in general no curve, simply a variable.

The moment you write $d\omega$ you do no longer consider a certain and thus fixed vector in $\Lambda^k$ but investigate what happens by infinitesimal changes of coordinates. You switch from $\vec{v}=v_0\vec{e}_0$ to $\vec{v}(t)=v_0(t)\vec{e}_0$. It is literally the transition from algebra to calculus. Of course now you will have to say in which way these coordinates totter, i.e. what is $t \mapsto v_0(t)$ or in general and local coordinates again, what is $d\omega_{i_1\ldots i_k} = \sum_{j}\dfrac{\partial \omega_{i_1\ldots i_k}}{\partial x^j}dx^j\,.$

If you want to know, why this might confusing you, have a look at the 10 items list at the end of section 1 herein:
https://www.physicsforums.com/insights/journey-manifold-su2mathbbc-part/

I'm not sure whether it will confuses you further or actually be of help, but at least it might serve as a quick reference guide (5 parts):
https://www.physicsforums.com/insights/the-pantheon-of-derivatives-i/

Last edited:
Hello,
thank you for your answer, I'm going to read through these pages. I'll give a sign in case of further questions.

Hello,
I have another question in this context. Let us say we want to talk about the function as object, thus to say the function itself as element of $$\Lambda ^k V.$$ We derived the coefficients in our proofs by using the smooth k-form itself. We set the coefficients as $$w_{i_1...i_k}=w(e_{i_1},...,e_{i_k})$$. From this and the fact that we made other proofs in $$\Lambda^n V$$ where V was n-dimensional and $$w$$ had the form $$w= f dx_1\wedge...\wedge dx_n$$ with f a smooth function from $$U \subset \mathbb{R}^n \rightarrow \mathbb{R}.$$ I'm still a little bit puzzled, but think this might be caused by vieweing the function as object. Is that right?