3
$\begingroup$

I have read similar questions with regards to what the first fundamental form is. I couldn't find my answers due to them assuming extra knowledge and/or using a different book which presents the topics differently. I am studying differential geometry from Pressley (2nd ed.). Here is an excerpt from the book:

enter image description here

I understand the first fundamental form is a dot product. It takes two vectors from the tangent space and outputs a number. Now, what does this have to do with $du$ and $dv$? And why would they name it so it coincides with the notation from integration? According to the definition provided, $du$ is a function that outputs the first component of a vector in your tangent plane, and $dv$ outputs the second component. Then he takes the dot product of the two vectors and now the dot product is written in terms of the basis vectors $\pmb\sigma_u$ and $\pmb\sigma_v$.

(1) He says the first fundamental form is $Edu^2+2Fdudv+Gdv^2$, and says how the coefficients and $du$, $dv$ depend on the choice of surface patch, but the first fundamental form doesn't. How can that be? If the fundamental form is $Edu^2+2Fdudv+Gdv^2$ then surely it inherits the dependence of its components.

Now we have the following:

enter image description here

(2) What was the purpose of inventing all this machinery? So you can calculate the arc length using only the vectors that span the tangent plane?

(3) Why is this property intrinsic? Someone living on the surface would calculate the length of the curve by looking at the curve and measuring the unit tangent vectors of the curve. But now it seems more complicated, that he somehow needs access to the surface patch parametrization ($\pmb\sigma$) in order to get $\pmb\sigma_u$ and $\pmb\sigma_v$

(4) Lastly, what is the connection between the $du$'s and integration? What is the relation between these two functions and the integrand we are used to seeing all the time. They are obviously functions, not infinitesimals.

$\endgroup$
2
  • 1
    $\begingroup$ Just to gauge the level at which this should be answered, do you understand dual vector spaces and tensor products? $\endgroup$ Mar 8 '18 at 16:16
  • $\begingroup$ @ArturAraujo Thank you for asking. No, I do not. This course is offered after only having learned multivariable calculus, ordinary differential equations, and basic linear algebra. So pretty much the lowest level possible. $\endgroup$
    – user462561
    Mar 8 '18 at 16:29
3
+50
$\begingroup$

I'll try to answer your first three questions.

(1) As you observed, the first fundamental form is nothing but the inner product in $\mathbb{R}^3$, restricted to the two dimensional tangent space. However, when we make the choice of a surface patch, we are then working in the space $\mathbb{R}^2$. In general, the tangent vectors in this space won't correspond in any simple manner to the tangent vectors of the original surface, rather the correspondence will depend upon at what point of the surface we're looking at the tangent vectors. That means in terms of the vectors in $\mathbb{R}^2$, the first fundamental form won't be a dot product, but something more complicated: it will be a symmetric bilinear form, and that's what exactly the formula describes. If you make a different choice of surface patch, you'd still like to compute the same fundamental form, but since you made a different choice of a surface patch, you'll have to adjust the values of $E$, $F$ and $G$ as to ensure you're computing the same function. I'd invite you to work out a simple example, say the unit sphere with two different surface patches.

(2) Essentially, yes. It's often easier in differential geometry to do all your calculations using coordinate patches (especially if you find clever coordinate patches).

(3) In fact, for someone living on the surface, surface patches are all they have: this definition ensures no matter what surface patch they pick, they measure the same length.

$\endgroup$
2
+100
$\begingroup$

4) The forms $du$ and $dv$ and integration:

A linear form is a linear real function, i.e., a linear map whose range is $\mathbb{R}$.

Example: $\mathbb{R}\ni(x,y)\mapsto x\in \mathbb{R}$.

Important examples of linear forms are the differentials of real functions at a given point. The differential of a real function $f$ at a point $x$ is the linear form $d_xf$ that for every vector $v$ generates the derivative of $f$ at $x$ in the direction $v$. In your multivariable calculus course, you have probably encountered the likes of $$ d_x f(v) = \sum_j \frac{\partial f}{\partial x_j}v_j $$ where the $v_j$ are the coordinates of $v$ is some basis. Now it can be useful to write $d_x f$ without referring to its action on $v$. This is where the linear forms $dx_j$ become handy. Letting $dx_j: v\mapsto v_j$, you get $$ d_x f = \sum_j \frac{\partial f}{\partial x_j}d x_j. $$ This is something that physicists in particular appreciate a lot. From there you can invent your own forms without referring to any function $f$. Simply pick real functions $w_j(x)$ and let $$ w(x) = \sum_j w_j(x)d x_j, \quad w(x)(v) = \sum_j w_j(x)v_j. $$ Keep in mind that both equations for $w(x)$ and $d_xf$ are equalities relating linear forms and have a priori nothing to do with infinitesimals (whatever that means). Now we can construct the integral of a form $w$ along a path $C$. Let $[a,b]\ni t\mapsto x(t)$ be a parametrization of $C$, then $$ \int_C w = \int_a^b w(x(t))(\dot x(t))dt = \int_a^b\sum_j w_j(x(t))\dot x_j(t) d t. $$ In particular, $$ \int_C f(x_j)dx_j = \int_a^b f(x_j)\dot x_j(t) d t, $$ which, using the change of variable $t\to x_j$ is something you would write, well, $\int f(x_j)d x_j$. I hope this shows that the notation makes sense and naturally extends the one used in integrating a function of a single variable.

Finally, you can now look differently at the $dt$: it is also a linear form on the vector space $\mathbb{R}$. In fact any function $f(t)$ defines a form $f(t)dt$ and can be integrated according to $\int f = \int f(t)dt$.

1,3) The notation $dx_j$ is ambiguous: what $dx_j$ is ultimately depends on what basis we choose to determine the coordinate $v_j$ of $v$. When you change basis, you change all the forms $dx_j$ AND all the functions $w_j$ that appear in the expression of $w$. In fact, the functions $w_j$ change exactly such that $w(x)(v)$ yields the same result whether you calculate it using the old basis or the new one. The same thing happens with the first fundamental form. Coordinates and basis change, the form does not.

2) What for? Given two tangent vectors $v$ and $w$, you can of course calculate $\langle v,w \rangle$ directly with $v\cdot w = \sum_j v_jw_j$ without using $E$, $F$ and $G$. Same for lengths of arcs. But I can think of at least two reasons why $E$, $F$ and $G$ are useful. The first is to connect the lengths of arcs drawn on a surface to the notion of area and later curvature; why these should be connected is not a trivial matter I believe. The second reason is that they allow to calculate dot products and lengths over flat maps without referring to 3D coordinates. For instance, pick up any map of the globe, draw two vectors emanating from the same point and try to calculate their dot product. You will find that reconstructing the 3D vectors is not convenient and using the functions $E$, $F$ and $G$ is much more straightforward.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy