# Vector bundle example in Hatcher

In Hatchers notes on K-Theory he gives the following example of a vector bundle that I don't understand. Define $$E:=\{(x,v) \in S^n \times \mathbb{R}^{n+1} \ | \ x \perp v\}$$, that is, we regard the inner product $$\langle \cdot , \cdot \rangle: \mathbb{R}^{n+1} \times \mathbb{R}^{n+1} \to \mathbb{R}$$ restricted to $$S^n \times \mathbb{R}^{n+1}$$. He then says that one should think of $$v$$ as a tangent vector $$S^n$$ but I don't know if that is truly necessary. He defines $$p:E \to S^n, \ (x,v) \mapsto x$$ and constructs local trivializations the following way. Choose an $$x \in S^n$$ and let $$U_x \subseteq S^n$$ be the open hemisphere containing $$x$$ and bounded by the hyperplane through the origin orthogonal to $$x$$. Define $$h_x:p^{-1}(U_x) \to U_x \times p^{-1}(x)$$ by $$h_x(y,v)=(y, \pi_x(v))$$ where $$\pi_x$$ is the orthogonal projection onto the hyperplane $$p^{-1}(x)$$. Then $$h_x$$ is a local trivialization since $$\pi_x$$ restricts to an isomorphism of $$p^{-1}(y)$$ onto $$p^{-1}(x)$$ for each $$y \in U_x$$.

$$(1)$$ By open hemisphere he means the subset of $$S^n$$ where the last coordinate is greater than $$0$$ right? Why is this necessary and why do we not regard the entire hemisphere?

$$(2)$$ If $$n=2$$ and $$x=(0,0,1)$$, then the hyperplane is the $$xy$$-plane, so nothing changes when restricting the open hemisphere and $$U_x$$ is simply the open hemisphere, right? So when moving $$x$$ on $$S^2$$ it just tilts the hyperplane by the same angle and thus "cuts off" a bit of the open hemisphere, doesn't it?

$$(3)$$ When defining $$h_x$$ we have $$p^{-1}(U_x)=U_x \times \mathbb{R}^{n+1}$$ and one regards $$\pi_x: p^{-1}(x) \bigoplus p^{-1}(x)^{\perp} \to p^{-1}(x),$$ where $$p^{-1}(x) \bigoplus p^{-1}(x)^{\perp}=S^n \times \mathbb{R}^{n+1}.$$ Why is $$p^{-1}(x)$$ a hyperplane?

$$(4)$$ Why does $$\pi_x$$ restrict to an isomorphism?

• For (1), because you need to work with open sets. And your interpretation isn't quite right: the open hemisphere will change as $x$ changes, so it need not be just $x_{n+1}=0$. Commented Jan 10, 2023 at 14:28
• @Randall Ah of course! So when he says that its bounded by the hyperplane, it should also mean that the points on the plane itself are not part of $U_x$ but only up to that plane, because otherwise its not guaranteed to be open, right? Commented Jan 10, 2023 at 14:31
• The inner product is a bilinear map $\mathbb R^{n+1} \times \mathbb R^{n+1} \to \mathbb R$. Commented Jan 10, 2023 at 14:31
• @PaulFrost But we regard a subset of $S^n \times \mathbb{R}^{n+1}$, so we have to regard the inner product on the added dimension, don't we? Commented Jan 10, 2023 at 14:32
• No, we have to take the standard inner product on $\mathbb R^{n+1}$. Of course we have $\mathbb R^{n+1} \times \mathbb R^{n+1} = \mathbb R^{2n+2}$, but this is irrelevant for the purpose of the question. Commented Jan 10, 2023 at 14:36

(1) No. Given $$x \in S^n$$, he considers the hyperplane through the origin orthogonal to $$x$$. This is the $$n$$-dimensional linear subspace $$E_x = \{ v \in \mathbb R^{n+1} \mid \langle v, x \rangle = 0 \} \subset \mathbb R^{n+1} . \tag{1}$$ Note that $$E= \bigcup_{x \in S^n} \{x\} \times E_x . \tag{2}$$ $$E_x$$ separates $$S^n$$ into the two disjoint open hemispheres $$H_x^\pm = \{ v \in S^n \mid \pm \langle v, x \rangle > 0 \} .\tag{3}$$ We have $$x \in H_x^+$$. Thus the open hemisphere containing $$x$$ and bounded by the hyperplane $$E_x$$ is the set $$U_x = H_x^+$$.

(2) First question: $$U_x$$ is the upper open hemisphere $$\{(x_1,x_2,x_3) \in S^2 \mid x_3 > 0\}$$. Second question: No. If we move $$x$$, then $$U_x$$ is dragged along with $$x$$. For example, if $$x = (1,0,0)$$, then $$U_x = \{(x_1,x_2,x_3) \in S^2 \mid x_1 > 0\}$$.

(3) By $$(2)$$ and the definiton of $$p : E \to S^n$$ we have for $$x \in S^n$$ $$p^{-1}(x) = \{x\} \times E_x . \tag{4}$$ Hatcher identifies $$\{x\} \times E_x$$ with $$E_x$$. When he speaks about the orthogonal projection onto the hyperplane $$p^{-1}(x)$$ he means the linear map $$\pi_x : \mathbb R^{n+1} \to E_x, \pi_x(v) = v - \langle v, x \rangle x.$$ Note that

1. $$\pi_x(v) \in E_x$$ because $$\langle \pi_x(v),x \rangle = \langle v, x \rangle - \langle v, x \rangle \langle x, x \rangle = 0$$ (since $$\langle x, x \rangle = \lVert x \rVert^2 = 1$$ for $$x \in S^n$$).

2. $$\pi_x(v) = v$$ for $$v \in E_x$$ because in that case $$\langle v, x \rangle = 0$$.

3. $$\ker \pi_x = Span(x)$$ = one-dimensional subspace of $$\mathbb R^{n+1}$$ spanned by $$x$$. In fact, $$v \in \ker \pi_x$$ means that $$v = \langle v, x \rangle x \in Span(x)$$ and $$v = \lambda x$$ with $$\lambda \in \mathbb R$$ gives $$\pi_x(v) = \lambda x - \langle \lambda x, x \rangle x = \lambda x - \lambda \lVert x \rVert^2 x = 0$$.

The proper definition of $$h_x$$ is then $$h_x : p^{-1}(U_x) \to U_x \times E_x, h_x(y,v) = (y,\pi_x(v)) .$$

(4) The claim is that for each $$y \in U_x$$ we get an isomorphism $$h_{x,y} : p^{-1}(y) = \{y\} \times E_y \stackrel{h_x}{\to} \{y\} \times E_x .$$ This means that the restriction $$\pi_{x,y} : E_y \stackrel{\pi_x}{\to} E_x$$ is an isomorphism of vector spaces. Since $$E_y, E_x$$ are both $$n$$-dimensional, it suffices to show that it is surjective. Consider the equation $$w = v - \langle v, x \rangle x .\tag{5}$$ We have to find a solution $$v \in E_y$$ (which means $$\langle v, y \rangle = 0$$). Applying the scalar product with $$y$$ to $$(5)$$, we get $$\langle w, y \rangle = -\langle v, x \rangle \langle x, y \rangle$$ and therefore $$\langle v, x \rangle = - \frac{\langle w, y \rangle}{\langle x, y \rangle} .$$ This gives $$v = w - \frac{\langle w, y \rangle}{\langle x, y \rangle}x . \tag{6}$$ Defining $$\rho_{x,y} : E_x \to E_y, \rho(w) = w - \frac{\langle w, y \rangle}{\langle x, y \rangle}x , \tag{7}$$ we see that $$\pi_{x,y} \circ \rho_{x,y} = id .\tag{8}$$ This can of course be verified directly without solving $$(5)$$. If you want, you can also verify that $$\rho_{x,y} \circ \pi_{x,y} = id$$, but it is unnecessary.

By the way, we can now easily see that $$h_x$$ is a homeomorphism. Its inverse is given by $$h^{-1}_x(y,w) = (y,\rho_{x,y}(w)) = (y, w - \frac{\langle w, y \rangle}{\langle x, y \rangle}x)$$ which is obviously continuous.

Remark.

To clear doubts about domain and range of $$h_x$$, let us summarize what we did above.

1. $$p^{-1}(U_x)$$ is a subpace of $$U_x \times \mathbb R^{n+1}$$.

2. The map $$H_x : U_x \times \mathbb R^{n+1} \to U_x \times E_x, H_x(y,v) = (y, v - \langle v, x \rangle x) = (y, \pi_x(v))$$, is well-defined and continuous.

3. $$h_x = H_x \mid_{p^{-1}(U_x)} : p^{-1}(U_x) \to U_x \times E_x$$, is well-defined and continuous.

4. The map $$\bar \rho_x : U_x \times E_x \to U_x \times \mathbb R^{n+1}, \bar \rho_x(y,w) = (y, w - \frac{\langle w, y \rangle}{\langle x, y \rangle}x)$$, is well-defined and continuous. Since $$\langle w - \frac{\langle w, y \rangle}{\langle x, y \rangle}x, y \rangle =0$$, we have $$w - \frac{\langle w, y \rangle}{\langle x, y \rangle}x \in E_y$$. Therefore $$\bar \rho_x ( U_x \times E_x) \subset p^{-1}(U_x)$$.

5. $$\rho_x : U_x \times E_x \stackrel{\bar \rho_x}{\to} p^{-1}(U_x)$$, is well-defined and continuous.

6. $$\rho_x \circ h_x = id$$ and $$h_x \circ \rho_x = id$$. Hence $$h_x$$ is a homeomorphism.

• There are a couple of points I don't understand in your answer. $(1)$ Could you give some insights how you came up with these sets? I appear to lack knowledge here and appear to have no real idea why this is what he meant. $(2)$ So basically my mistake here was that nothing is cut off but rather the whole $U_x$ is just tilted by same angle that $x$ is rotated by, right? $(3)$ I dont understand what your $p:D \to S^n$ is here. Commented Jan 10, 2023 at 17:28
• (1) Which sets do you mean? (2) Yes. Draw a picture for $n = 1$ to see what is going on. (3) $D$ was a typo. Commented Jan 10, 2023 at 17:32
• $(1)$ I mean $E_x$ and the $H_x$. $(3)$ Ah I also made a mistake in my question and wrote $B$ instead of $S^n$... Now I realize another mistake, I am making a lot of silly mistakes I am sorry. I thought that $p^{-1}(x)$ is $\{x\} \times \mathbb{R}^{n+1}$ but obviously that is not the case, since $E$ is defined to be a specific subset. Commented Jan 10, 2023 at 17:36
• $E_x$ is the definition of hyperplane through the origin orthogonal to $x$. The $H_x^\pm$ are the hemispheres "above" and "below" this hyperplane. Take $n=1$ to visualize this. Commented Jan 10, 2023 at 18:27
• You shoud consult a book on linear algebra dealing especially with inner products. But I cannot give a recommendation, for almost half a century I never looked into such a book. Commented Jan 10, 2023 at 18:42