1 Basic denitions 1.1 1-forms A dierential 1-form on R3 is an expression of the

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1 Basic denitions
1.1 1-forms
A dierential 1-form on R3 is an expression of the form
f dx + g dy + h dz.
Here f,g, and h are scalar functions in three variables x,y, and z.
This expression looks strange. Just take it as an abstract expression for now.
1.2 Addition of 1-forms
We can add two 1-forms by the formula
(f dx + g dy + h dz) + (f0 dx + g0 dy + h0 dz) = (f + f0) dx + (g + g0) dy + (h + h0) dz.
For example
y2 dx + (x + z) dy + 7 dz + xz dx + y dy + sin(z) dz = (y2 + xz) dx + (x + y + z) dy + (7 + sin(t)) dz.
1.3 Multiplication of 1-forms
We dene the wedge product of two 1-forms. We do this such that, for example, dxdy = dydx. This means that this product isn’t “commutative”. When you switch the order, then you have to put a minus sign in front. In that way it behaves a bit like the cross product of vectors. Problem 1. Prove that for u a variable, dudu = 0. We do one example of this. Let = y2 dx + (x + z) dy + 7 dz, = xz dx + y dy + sin(z) dz, then = (y2 dx + (x + z)dy + 7 dz)(xz dx + y dy + sin(z)dz) = y3 dxdy + y2 sin(z)dxdz + (x + z)xz dydx + (x + z)sin(z)dydz + 7xz dzdx + 7y dzdy = (x + z)sin(z)7y)dydz + (7xzy2 sin(z))dzdx + (y3 (x + z)xz)dxdy.
The order of the wedges in the nal answer is important. Note that this is not a 1-form, but a 2-form as we will dene in a bit.
Problem 2. Let = xdx+z dy, = sin(xz)dx+z dy+y dz. Find .
1.4 0-forms, 2-forms, and 3-forms
A 0-form is just a scalar function f.
A 2-form is an expression of the form f dydz + g dzdx + hdxdy.
Again f,g, and h are scalar functions.
A 3-form is an expression of the form f dxdydz. where f is a scalar function.
From the example above, we see that the product of two 1-forms is a 2-form.
1.5 Addition and multiplication of 0-forms, 2-forms, and 3forms
You can consider the dy dz, dz dx, and dxdy is a type of basis for the space of 2-forms. Addition is done “componentwise”. One can’t add “dierent” forms. But we can multiply dierent forms. The product of a 2-form and a 1-form is a 3-form. For example consider the following problem. Problem 3. Let = dx+dz and = dydz +xdzdx+y dxdy. Find . Note, this will be a 3-form. Part of this problem is guring out how this is done.
2 Dierential operator
We now dene the derivative operator on the dierential forms. For a 0-form (scalar function) f we dene
df =
f x
dx +
f y
dy +
f z
dz.
We note that the derivative operator of a 0-form is a 1-form. We can also apply the operator to 1-forms. We use the formula d(f dx + g dy + hdz) = d(f)dx + d(g)dy + d(h)dz.
As an example, consider f(x,y,z) = x2yz + z. Then
df =
f x
dx +
f y
dy +
f z
dz
= (2xyz)dx + (x2z)dy + (x2y + 1)dz.
Problem 4a. Find
d(x2 dx + zxdy + dx).
Problem 4b. Let
= 3(dxdy) + 5(dxdz) + 7(dydz))
Find d.
Problem 4c. For f and g 0-forms, prove that
d(fg) = fd(g) + gd(f)
Problem 4d. Prove that for and 1-forms
d() = (d) d. How does this work for 2-forms?
Problem 4e. Let
= (xy)dydz + (yz)dzdx + (xy + z)dxdy.
Find d.
2.1 Connection to what we already know from Calculus 3
We now make the following identications • A 0-form is a scalar function (the is basically true by denition) • A 1-form f dx + g dy + hdz is the vector eld (f,g,h). • A 2-form f dydz + g dzdx + hdxdy is the vector eld (f,g,h). • A 3-form f dxdydz is the scalar function f. With these identications do the following problem.
Problem 5. In the following we consider forms dened on R3. Show that 1. If f is a 0 form, then df is the gradient of f.
2. If is a 1-form, then d is the curl of .
3. If is a 2-form, then d is the divergence of
4. If is a 3-form, then d = 0.
With this in mind, we have the following theorem
Theorem. For a dierential form , = d for another dierential form if and only if d = 0.
Problem 6. Prove this theorem. That is, for example, prove that if is a 1-form and that = d for a 0-form , then d = 0.
3 Uses of dierential forms
We recall the dierential operator “of” 0-forms:
df =
f x
dx +
f y
dy +
f z
dz.
For example, if f(x) = x2+1, then df = 2xdx. We see the similarity between writing this and writing that df dx = 2x.
3.1 Integrals
Recall Stoke’s Theorem from class. Under several assumptions, we have that ZC ~ F ·d~r =ZZ S curlF ·dS.
Here C is the boundary of S. We often write things like S = C.
For integration, we can in general integrate a n-form over a n-dimensional object. So 1-forms can be integrated over curves (recall intervals are curves). 2-forms can be integrated over 2-dimensional objects.
Stokes theorem now looks like this ZS d =ZS .
3.2 Integrals of 0-forms
We integrate 0-forms f using the denition ZC f = f(b)f(a) where C is a curve from a to b.
3.3 Integrals of 1-forms
Integrating 1-forms is equivalent to our usual Riemann integral. Consider the example where f is a function of one variable x. Say f is dened on the interval/curve C = [a,b]. Note that C = [b,a] (the reverse orientation). Then we say that C = {a,b}. Then df = f0(x)dx is a 1-form. We then have ZC df =ZC f0(x)dx = f(b)f(a). Note that we only have integrals of 1-forms that are dierentials (of 0-forms). This is our usual Fundamental Theorem of Calculus.
Let’s now consider a line-integral, or an integral of a 1-form.
Example: Consider the 1-form y dx + xdy. Let C be the curve C(t) = (cos(t),sin(t)), 0 t 2. Then we can form and nd the integral ZC y dx + xdy =Z[0,2]sin(t)d(cost) + cos(t)d(sint) =Z[0,2]sin(t)(sin(t))dt + cos(t)cos(t)dt =Z[0,2] sin2(t) + cos2(t) dt = 2.
Integrating 1-forms is the same as line integrals.
Problem 7. Let C be the vertical line segment from (3,4,5) to (3,4,0). Find ZC y dx + z dy + xdz. You rst need to nd a parametrization of the line segment.
3.4 Integrals of 2-forms
Integrating 2-forms is the same as nding ux integrals over a vector eld. Problem 8. Let S be part of the paraboloid z = 1x2 y2 where z 0. This is a surface. Dene the orientation to be outward. Now compute the integral ZZ S y dydz + z dzdx + xdxdy.
This integral is basically the same as an integral over the surface S of a vector eld. Make sense of this.

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