Chap 10,12. There are vectors. Then there are vectors.

DimBulb · 05 Aug 2009, 18:22

I think I should preface any post by saying I might not know what I am talking about.

A major difficulty with my understanding of chapters 10 and 12 stemmed from a notion of vectors, or vector space, that is different than my previous understanding of these terms. I have figured things out, I think. Some things I figured out are not mentioned in these chapters, at least not explicitly. I am going to try to share my understanding here. It may help others, or others may help my understanding if it is not correct.

The vectors I learned about in high school and college are what I guess would be called Euclidean vectors (http://en.wikipedia.org/wiki/Euclidean_vector). They are like pointed sticks. They have a magnitude and a direction. They usually are described as having a standard basis, which in two dimensions would be $\boldsymbol\i$ , $\boldsymbol\j$ , or $\hat{x}$ , $\hat{y}$ , that is, unit length vectors pointing in the x and y directions of the Cartesian coordinate system. A basis means you can make any vector in the vector space by adding multiples of those basis. Examples,

$\boldsymbol a = a_x \boldsymbol \i + a_y \boldsymbol \j$
$\boldsymbol b = b_x \boldsymbol \i + b_y \boldsymbol \j$

For these types of vectors you have a thing called a scalar product or dot product (http://en.wikipedia.org/wiki/Dot_product) which is

$\boldsymbol a \bullet \boldsymbol b = | \boldsymbol a | | \boldsymbol b | \cos \theta_{ab}$

where $\theta_{ab}$ in the angle between the two vectors. In the basis described above

$\boldsymbol \i \bullet \boldsymbol \i = 1$
$\boldsymbol \j \bullet \boldsymbol \j = 1$
$\boldsymbol \i \bullet \boldsymbol \j = 0$
$\boldsymbol \j \bullet \boldsymbol \i = 0$

and because the scalar product is distributive and associative with scalars

$\boldsymbol a \bullet \boldsymbol b = a_x b_x \boldsymbol \i \bullet \boldsymbol \i + a_x b_y \boldsymbol \i \bullet \boldsymbol \j + a_y b_x \boldsymbol \j \bullet \boldsymbol \i + a_y b_y \boldsymbol \j \bullet \boldsymbol \j = a_x b_x + a_y b_y$

RTR introduces a new concept of vector space. And looking across the Internet and into other books, this concept seems to be boiled down to, if it obeys vector algebra, then it is a vector space. This means that, yes Euclidean vectors form a vector space. But so do things like

$\boldsymbol a = a_x \frac{\partial}{\partial x} + a_y \frac{\partial}{\partial y}$

The basis here are the $\frac{\partial}{\partial x}$ and $\frac{\partial}{\partial y}$ . But you can not do a scalar product for two of these things. This kind of vector will only form a scalar product with another kind of vector, with a different basis, called a covector, or 1-form:

$\boldsymbol b = b_x dx + b_y dy$

Here the basis are

and

and the scalar product between the vector and covector then makes sense:

$\frac{\partial}{\partial x} \bullet dx = 1$
$\frac{\partial}{\partial y} \bullet dy = 1$
$\frac{\partial}{\partial y} \bullet dx = 0$
$\frac{\partial}{\partial x} \bullet dy = 0$

I look at something like

$\frac{\partial}{\partial x} \bullet dx = 1$

as saying "how much does x change as x changes", which of course is 1. While

$\frac{\partial}{\partial x} \bullet dy = 0$

is saying "how much does y change as x changes" and of course is zero, since the direction of x is that direction in which y does not change.

Penrose says we should not think of things like

as being infinitesimals. I do not get that. In my mind they still work as infinitesimals.

So, just to be clear here, a covector or 1-form, is a kind of vector. It is a dual of the other kind of vector because these kind of vectors do not form scalar products with their same kind, they have to mate with their opposite kind.

The other thing that confused the heck out of me was Penrose saying a thing like

, a 1-form or covector, should be thought of as specifying all the directions that were not the direction the thing is pointing, the directions that the vector (that opposite kind of thing) must point so that the scalar product is zero. In our two dimensional case, the directions

specifies are the +/- y directions, because

$\frac{\partial}{\partial y} \bullet dx = 0$

Penrose says the 1-form is a kind of contour line. If the vector (the opposite kind of thing) is parallel to this contour line then the scalar product is zero. If the vector was perpendicular to this contour line, the scalar product would be of maximum magnitude, because it points either fully "uphill" or "downhill". This was confusing because with Euclidean vectors if one vector is parallel to another, their scalar product would be maximum in magnitude, and if perpendicular their scalar product would be zero. It seemed opposite.

Look,

, is a type of vector, it is pointing in the

direction. But when you specify one direction, you also are specifying all the directions that are not that direction. So, a 1-form in n-dimensional space does specify the (n-1) directions that are not in the direction that the covector is pointing. These other directions are normal or perpendicular to the direction the covector is pointing. So in three dimensions a covector does specify a two dimensional plane normal to the direction of the covector. In two dimensions a covector specifies a one dimensional line normal to the direction of the covector. These infinitesimally thin (there's that word again) n-1 dimensional slices of n-space are what are used in doing integrals, but when it comes to taking the scalar product, I am sticking to the image of two pointed sticks forming some angle $\theta$ .

To further clarify or confuse, I would comment that since p-forms and p-vectors obey the algebraic vector rules, they are also vector spaces. So are matrices, and real numbers, and polynomial, and tensors.