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# Orientation

## Orientation of curves

Question: What happens to the displacement when the motion changes direction?

Answer: It changes its sign.

In terms of integrals:

$$\int_a^b V(x) dx = - \int_b^a V(x) dx,$$ where $V$ is the velocity.

This is where it comes from: $$\int_{-[a,b]} f(x) dx = - \int_{[a,b]} f(x) dx = \int_{[b,a]} f(x) dx.$$

Note: There is a difference when you treat the domain of integration as

- an interval: $[ a, b ] = [ b, a ]$;

or as

- an
*oriented*interval: $[ a, b ] = -[ b, a ]$.

More generally, $$\int_{-A} f(x) dx = - \int_A f(x) dx.$$

This may be the flow integral, for example.

The orientation of a curve comes from its parametrization (see Parametric curves). Given a continuous function $$p: [ a, b ] \rightarrow {\bf R}^n,$$ the set $C = p( [ a, b ] )$ is a curve.

Consider a curve given by $p:[0,1] \rightarrow R$. Then $q:[0,1] \rightarrow R$ given by
$$q(s) = p(1 - s)$$
is another parametrization that produces the *opposite orientation* to the one from $p$.

For the curves in the picture there are two possible orientations (and infinite many parametrizations).

## Orientation of surfaces

Suppose we want to compute the flow of liquid through a region in space. For that we need to understand the direction of the flow with respect to the *orientation* of the surface it passes through. There are two ways to do specify the orientation:

- orient the curve that bounds the region and then use the "screw rule" to find the corresponding normal vector to the surface; and
- look at the tangent plane to the surface and deal with its orientation which is simply a choice of the normal vector.

We can approach the orientation of surfaces in a similar manner to curves. Indeed, the orientation of a surface comes from its parametrization (at least locally).

Let's understand orientation of the square, $Q$, in space first. We can still look at it as corresponding to the direction of the parametric curve (a closed curve!) that bounds the square. This is a more direct way.

Parametrize $Q$ as a map from the unit square in the plane to the $3$-dimensional space: $$R : {\bf R}^2 \rightarrow {\bf R}^3,$$ specifically:

- $R_1(s,t) = s$,
- $R_2(s,t) = t$,
- $R_3(s,t) = 0$.

What is its orientation? It is a unit vector perpendicular to both $R(e_1)$ and $R(e_2)$, where $e_1, e_2$ are the (fixed) basis vectors of ${\bf R}^2$. So, if $L$ is a line perpendicular to the square (or a surface), then the orientation is a basis for $L$ with unit vectors. And there are exactly two choices.

Note: the alternative parametrization of $Q$ gives the opposite orientation:

- $R_1(s,t) = 1 - s$,
- $R_2(s,t) = t$,
- $R_3(s,t) = 0$.

The orientation of a surface is determined by the orientation of its tangent plane (hyperplane) $L$. The set of all vectors perpendicular to $L$ is a vector subspace of dimension $1$. There are two possible unit vectors in it. Choosing one of them amount to *fixing the orientation*.

In the special case of $n = 3$, how do we find these vectors? With the cross product.

Given a basis of $L$, say $v_1$ and $v_2$, we take as unit vector $u$, defined as $$u = \frac{v_1 \times v_2}{|| v_1 \times v_2 ||}.$$

Where does the basis of $L$ come from? From the parametrization. Given $$p( 0, 0 ) = A,$$ find $$v_1 = ?$$ $$v_2 = ?$$

To summarize, the basis $\{ v_1, v_2 \}$ comes from that of $R^2$ by means of $p$: $$v_1 = p'( 0, 0 ) e_1,$$ $$v_2 = p'( 0, 0 ) e_2.$$

It should become clear now that what we are talking about here is equivalence classes of parametrizations (or, more generally, atlases) given by:

## Orientation of simplices

On the other hand, there are two equivalence classes of orderings of any set, in particular on the set of vertices of a simplex, as follows:

(Recall, even permutation is the composition of an even number of transpositions.)

For example, $ABC$

- isn't equivalent to $ACB$ (one flip), but
- is equivalent to $CAB$ (two flips).