New Bundles from Old Bundles

We can construct new bundles from the given bundles. Let denote the given bundle $\xi =\left(E,\pi ,B\right)$.

Induced Bundles(Pullback Bundle)

Let $f:B^{\prime }\to B$. Then we define the Pullback bundle $f^{\ast }\xi =\left(f^{\ast }(E),{\pi }^{\prime },B^{\prime }\right)$ where $f^{\ast }(E)=B^{\prime }{\times }_{B}E$.

Restriction Bundle

For the bundle $\xi =\left(E,\pi ,B\right)$, restircted bundle of $\xi$ is $\xi {|}_{\bar{B}}=\left({\pi }^{-1}\bar{B},\pi |\bar{B},\bar{B}\right)$. In Pullback language, it is a pullback of $\bar{B}\hookrightarrow B$.

Product Bundle

Let ${\xi }_1$ and ${\xi }_2$ be bundles over $B({\xi }_1)$ and $B({\xi }_2)$. The Product bundle ${\xi }_1\times {\xi }_2$ is

For example, $T(M_1\times M_2)=T{M}_1\times T{M}_2$.

Whitney Sum

Let ${\xi }_1$ and $\xi$ be vector bundles over $B$. The Whitney Sum ${\xi }_1\oplus {\xi }_2$ is defined as

$${\xi }_1\oplus {\xi }_2:-{\Delta }^{\ast }({\xi }_1\times {\xi }_2)$$

where $\Delta$ is a diagonal map $\Delta :B\to B\times B$. As a diagram,

Example

${\Sigma }_{S^2}^3=T{S}^2\oplus \nu (S^2\hookrightarrow {\mathbb{R}}^3)$.

Subbundle

A subbundle $\eta$ of $\xi$ is $E(\eta )\subset E(\xi )$ such that ${\pi }^{-1}{|}_{E(\eta )}$ is a vector bundle.

Orthogonal Complements

From the natural question, if we have a sub-bundle $\eta \subset \xi$, does there exist a complementary sub-bundle so that $\eta$ splits as a Whitney sum? If $\xi$ is a bundle with a metric, we can construct it as follows.

Let $F_b({\eta }^{\perp })$ denote the subspace of $F_b(\xi )$ which is

$$F_b({\eta }^{\perp })=\left\{v\in F_b(\xi ):v\cdot w=0,\forall w\in F_b(\eta )\right\}.$$

Let $E({\eta }^{\perp })\subset E(\xi )$ denote the union of the $F_b({\eta }^{\perp })$.

Then we call ${\eta }^{\perp }$ as the orthogonal complement of $\eta$ in $\xi$.

Quotient Bundle

Given a Vector bundle $\eta \subset \xi$, define the quotient bundle $\xi /\eta$. If $\eta$ has a Euclidean metric, then ${\eta }^{\perp }\cong \xi /\eta$. |Proof

Normal Bundle

Normal Bundle of submanifold $M$ of $N$ is defined as

$$\nu (M\hookrightarrow N)=TN{|}_M/TM.$$

Embedded case

If $N\subset {\mathbb{R}}^k$ or $N$ has a Riemannian metric, $(TM{)}^{\perp }\subset TN{|}_M$, then

So that $TN{|}_M=TM\oplus T{M}^{\perp }$.

Example

For example, let’s consider ${\mathbb{S}}^2\subset {\mathbb{R}}^3$.

$$T{\mathbb{R}}^3{|}_{{\mathbb{S}}^2}=T{S}^2\oplus \nu ({\mathbb{S}}^2\hookrightarrow {\mathbb{R}}^3).$$

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If $B$ is paracompact(e.g. CW complex), then bundles of $B$ form an exact category.

We can give $\beta$ a metric, such that

For example,

$$\begin{array}{rl}TN{|}_M& =TM\oplus T{M}^{\perp }\\ & \cong TM\oplus \nu (M\hookrightarrow N)\end{array}$$

Immersion

For any immersion, we have a normal bundle of the immersion $f$ ${\nu }_f$.

If N has a Riemannian metric, ${\nu }_f=f^{\ast }TN/TM$, for any immersion $f:M\to N$. So that here is a Whitney sum decomposition

$$TM=TN|M\oplus f^{\ast }TN/TM.$$

Other Bundles

	Vector Space	Vector Bundle $\xi$	Comment
Hom	$Hom(V,W)$	$Hom(\xi ,\eta )$
Dual	$V^{\ast }$	${\xi }^{\ast }=Hom(\xi ,{\Sigma }_B^1)$	${\Sigma }_B^1$ is the Trivial Bundle
Product	$V\otimes W$	$xi\otimes \eta$
Tensor Algebra	${\bigwedge }^{k}V$	${\bigwedge }^k\xi$
Exterior Algebra	${\bigwedge }^{\ast }V$	${\bigwedge }^{\ast }\xi$
etc