Chapter 3: Forms
Section 2: 1-Forms
Once again:
My notes are in black.
My comments and homework solutions are in blue.[/color]
My questions are in red.[/color].
I'll pause 24 hours for discussion, questions, and corrections. If none are forthcoming, then I will post the next section of my notes tomorrow night at about the same time.
1-Forms
A
1-form \alpha is a linear function that maps vectors into real numbers. Since it is called "linear", we require it to satisfy:
\alpha (\mathbf{v}+\mathbf{w})=\alpha (\mathbf{v}) + \alpha (\mathbf {w})<br />
\alpha (k \mathbf{v})=k\alpha (\mathbf{v})<br />
Quick question:
Are "1-form" and "linear functional" synonymous?
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The geometric interpretation of \omega is that of a
plane whose graph passes through the origin in the dx-dy coordinate system. Fixing our attention on 1-forms on T_p\mathbb {R}^2, we see that our general 1-form is \omega = a dx +b dy. This is the equation of a plane in T_p\mathbb{R}^2 \times \mathbb{R}.
Just a note of clarification for students: "\times" denotes a
Cartesian product, which makes n-tuples out of elements of sets. For instance \mathbb{R} \times \mathbb{R} is the set of all ordered pairs of real numbers. And in our case, T_p\mathbb{R}^2 \times \mathbb{R} indicates that we are forming n-tuples from ordered pairs in T_p\mathbb{R}^2 (the coordinates for dx and dy) and a member of \mathbb{R} (the value of \omega).
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Illustrative Example
For \omega (<dx,dy>)=2dx+3dy, evaluate \omega (<-1,2>).
This is easily done by plugging in the components of <-1,2> into the right places in \omega:
\omega (<-1,2>)=(2)(-1)+(3)(2)=4
And we are to take note that \omega (<-1,2>) is just the dot product <-1.2> \cdot <2,3>
Note that we can make a vector out of the coefficients in \omega. We can call it < \omega >=<a,b>. This notation is not introduced until Section 2.3, but I think it would be nice to have it now for shorthand.
So a recipe for evaluating a 1-form on a given vector is:
<br />
\omega (V) = <\omega> \cdot V<br />
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This brings us to the main point of the section: the geometric interpretation of 1-forms.
David Bachman said:
Evaluating a 1-form on a vector is the same as projecting onto some line and then multiplying by some constant.
This of course has the huge advantage of being
independent of coordinates. Anyone who has studied relativity can see the value of this!
So now we know how to use a given 1-form to determine the projection of a vector onto a line, and we can then determine the scaling factor. What if we want to do things the other way around? What if I am
given a line L, a scaling factor k, and a vector V? Recall from vector calculus that the dot product is related to the projection of a vector onto a line:
proj_{\mathbf {u}}\mathbf {v}=\frac {\mathbf {u} \cdot \mathbf {v}}{|\mathbf {v}|}
So say I want to write down a differential form that projects vectors onto a line L: dy=c dx and scales them by a factor of k (this will be asked of us in the Exercises). Since the slope of L is c=\frac {c}{1}, it is readily seen that a vector that is parallel to L is W=<1,c>. Since we are looking for the projection of V onto a line parallel to W, we look at:
<br />
proj_WV=\frac {W \cdot V}{|W|}<br />
<br />
proj_WV=\frac {<1,c> \cdot V}{\sqrt {1+c^2}}<br />
Upon comparing this with our expression for \omega above, it should be clear that our vector W is nothing other than <\omega>. Furthermore, I can scale the projection by a factor of k by multiplying both sides of the above projection by that factor.
<br />
k{}proj_WV=k \frac{<1,c> \cdot V}{\sqrt {1+c^2}}<br />
So we can now find the differential form \omega that projects V onto dy=cdx and scales by a factor of k, because we have just derived a function that does that very thing. Recognizing that:
<br />
<\omega>=<\frac{k}{\sqrt {1+c^2}},\frac{ck}{\sqrt {1+c^2}}><br />
we have:
<br />
\omega=\frac{k}{\sqrt {1+c^2}}dx+\frac{ck}{\sqrt {1+c^2}}dy<br />
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1-Forms in \mathbb{R}^n
All of this straightforwardly generalizes to n dimensions. There is no need for elaboration.
