# Why orientation on $S^1$ is the one that counter-clockwise-pointing vectors are positive.

I am totally disoriented by orientation..

So in the text, it says

The closed unit vall $B^2$ in $\mathbb{R}^2$ inherits the standard orientation from $\mathbb{R}^2$. The induced orientation on $S^1$ is the one for which the counter-clockwise-pointing vectors are positive.

So according to my understanding, $S^1$ is oriented by $\{n_x, v_1\}$, where $n_x$ is the outward unit normal, and $v_1$ is a basis vector for $S^1$. Is this right? What is $v_1$? just $(1)$ as the standard basis?

Then, how can I get the conclusion that the orientation on $S^1$ is the one for which the counter-clockwise-pointing vectors are positive? We suppose to exam whether the determinant of linear transformation between standard basis $\{(1,0),(0,1)\}$ and $\{n_x, v_1\}$ is positive right? But how can I compute it, since I don't really understand the numerical representation of $\{n_x, v_1\}$ - therefore I can't carry out the computation.

Very much confused, sorry if the statement does not make much sense...

According to Guillemin and Pollack's Differential Topology

Definition: Orientation of $V$, a finite-dimensional real vector space: Let $\beta, \beta^\prime$ be ordered basis of $V$, then there is a unique linear isomorphism $A: V \to V$ such that $\beta = A \beta^\prime$. The sign given an ordered basis $\beta$ is called its orientation.

Definition: Orientation of $X$, a manifold with boundary: A smooth choice of orientations for all the tangent space $T_x(X).$

• What do you mean by "$v_1$ is a basis vector for $S^1$"? Since $S^1$ is not a vector space, it doesn't have a basis in the standard sense. Perhaps it would help to reproduce the definition of orientation with which you're working-there are several. – Kevin Carlson Jul 19 '13 at 1:50
• Thanks @KevinCarlson, I am updating. – 1LiterTears Jul 19 '13 at 1:51

It should help to write the general element of $S^1$ as $x = (\cos\theta, \sin\theta)$, and view this as a unit normal $n_x$ at $x$. The counterclockwise tangent vector at $x$ (a unit basis for the tangent line $T_x S^1$) is $v_1 = (-\sin\theta, \cos\theta)$....

• Oh that's very helpful, let me continue work on that, thanks a lot Andrew! – 1LiterTears Jul 19 '13 at 2:08
• Nice to see you here, Andy :) We need more geometers :) – Ted Shifrin Jul 19 '13 at 2:08
• Ah, I got it! So I want to show that the determinant of linear transformation between $\{(1,0),(0,1)\}$ and $\{\cos \theta, \sin \theta), (-\sin \theta, \cos \theta)\}$ is positive. – 1LiterTears Jul 19 '13 at 2:20
• Thanks so much ! :-D May I ask about Holy Cross? It seems a wonderful school. :) – 1LiterTears Jul 19 '13 at 2:34

@AndrewD.Hwang answered my question. To confirm my understanding is correct, here's the complete solution following Andy's answer.

To write the general element of $S^1$ as $x = (\cos\theta, \sin\theta)$, and view this as a unit normal $n_x$ at $x$. The counterclockwise tangent vector at $x$ (a unit basis for the tangent line $T_x S^1$) is $v_1 = (-\sin\theta, \cos\theta)$, hence we want to verify that the determinant of linear transformation between $\{(1,0),(0,1)\}$ and $\{(\cos \theta, \sin \theta), (-\sin \theta, \cos \theta)\}$ is positive.

Remember the basis is ordered, to solve $$A\begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} \cos \theta \\ \sin \theta \end{pmatrix}$$ We obtain $$A = \begin{pmatrix} \cos \theta & * \\ \sin \theta & * \end{pmatrix},$$ * means anything, which we will be determined by second condition that $$A\begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} - \sin \theta \\ \cos \theta \end{pmatrix}.$$ We obtain $$A = \begin{pmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{pmatrix}.$$

Clearly, $\det A = 1$.