# Dual basis and differential forms

I was reading about dual spaces and dual bases in the book Linear Algebra by Friedberg, Spence and Insel (FSI) and they give an example of a linear functional, f_i (x) = a_i where [x]_β = [a_1 a_2 ... a_n] denotes the matrix representation of x in terms of the basis β = {x_1, x_2, ..., x_n} of V. Now they go on to prove that {f_1, f_2, ..., f_n} is in fact the dual basis of β for V* by actually never using the fact that f_i (x) = a_i, but rather that f_i (x_j) = δ_ij, where δ_ij denotes the Kronecker delta function. I also happened to have read other references by Halmos and Lang and they did not go about actually finding the linear functionals, but rather use the fact that a linear functional that satisfies φ_i (x_j) = δ_ij exists. Now I suppose my question is, doesn't it so happen that the dual basis of β ALWAYS equals {f_1, f_2, ..., f_n} as defined by FSI? Because since φ and f agree on the basis elements in turns out that φ_i = f_i for all x.

Now here is where differential forms come in. I was reading do Carmo's book on Differential forms and he says that the basis of R³ is {dx_i; i = 1,2,3} where x_i represents the i-th coordinate function. So I guess its true that dx_i = f_i(x) from above then right? I, to be honest, find this a bit unexpected as x_i is precisely f_i when we're using the canonical basis, yet differentiating doesn't mess it up. So I actually went on to check it on some values that dx_i really does equal f_i(x). And so it seems to also verify what I was asking in the first paragraph.

Is this all right? I'd appreciate someone clarifying some of this stuff to me.

Thanks!

EDIT: I guess another way of putting my question is, is the dual basis unique, in the sense that the linear functionals that form the basis are always identically equal to f_i(x) as given by FSI?

Last edited:

Hurkyl
Staff Emeritus
Gold Member
Another way of putting your question is:

If you have two lists f and g of linear functionals satisfying:
fi(xj) = δ_ij
gi(xj) = δ_ij​

Can you prove that, for each i,
fi = gi
?

I guess I could since they agree on basis elements they should be equal - that's a result from linear algebra. I guess that answers my question, but the differential form example sorta throws me off...

I suppose the purpose of 'rewriting' the f_i (x) as dx_i is probably just more convenient as it is "differential" forms.

...purpose of 'rewriting' the f_i (x) as dx_i is probably just more convenient as it is "differential" forms.
The purpose is that do Carmo's is interested in differential forms defined on a manifold (or in an open set of it), not just on a single point. Therefore, given a manifold M and $p \in M$ and R3 is its tangent space at p, then the dual basis to the canonical one is indeed dx_i(p) and is exactly the same as the dual basis as defined in linear algebra.

This notation is useful when, for example, you have a function f defined on an open set U of M and you need df, which is then written (usually supressing p):

$$df(p) = \sum_{i = 1}^{n}a_{i}\left(p\right)dx_{i}\left(p\right)$$

Where the a_i's are (usually) smooth functions on U.

The purpose is that do Carmo's is interested in differential forms defined on a manifold (or in an open set of it), not just on a single point. Therefore, given a manifold M and $p \in M$ and R3 is its tangent space at p, then the dual basis to the canonical one is indeed dx_i(p) and is exactly the same as the dual basis as defined in linear algebra.

This notation is useful when, for example, you have a function f defined on an open set U of M and you need df, which is then written (usually supressing p):

$$df(p) = \sum_{i = 1}^{n}a_{i}\left(p\right)dx_{i}\left(p\right)$$

Where the a_i's are (usually) smooth functions on U.

Ahh that makes sense to me now. Thanks a lot JSuarez! Also thanks to Hurkyl for helping me out too! 