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In this question, I am considering only regular manifolds.

A Trivial Bundle

The circle $S^1$ is known to have a trivial tangent bundle.

As a subset of $\mathbb{R}^4$, the tangent bundle of $S^1$ is given by

$$\{(x,y\, ; u,v) \in \mathbb{R}^4:x^2+y^2=1 \ \text{and} \ ux+vy=0\}$$

A Non-Trivial Bundle

The sphere $S^2$ is known to have a non-trivial tangent bundle.

As a subset of $\mathbb{R}^6$, the tangent bundle of $S^2$ is given by

$$\{(x,y,z\, ;u,v,w) \in \mathbb{R}^6:x^2+y^2+z^2=1 \ \text{and} \ ux+vy+wz=0\}$$

Comments

The tangent bundle $TS^1$ is trivial and so can be expressed as a Cartesian product. In this case, we have $TS^1 \simeq S^1 \times \mathbb{R}$. It is impossible to express $TS^2$ as a Cartesian product.

My Question

Is it possible to know if a manifold is, topologically, a Cartesian product (like $TS^1$) by looking at the defining equations? Similarly, Is it possible to know if a manifold is not, topologically, a Cartesian product (like $TS^2$) by looking at the defining equations?

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  • $\begingroup$ The answers so far seem to treat manifolds that arise as tangent bundles. Is that what you intended, or are you referring to more general manifolds (that may not be $TM$ for some $M$) and the question of when they can be written as a Cartesian product? (E.g., choosing arbitrary examples, $\mathbb{R}^3 \backslash \{0\}$ is the product of the simpler manifolds $S^2$ and $\mathbb{R}$, whereas $S^3$ is not a nontrivial product.) $\endgroup$ – Phillip Andreae Jul 22 '14 at 13:14
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If a smooth $n$-manifold $M \subset \mathbb{R}^m$ is globally defined as the preimage of a regular value of a smooth function $f:\mathbb{R}^n \to \mathbb{R}^{m-n}$, then we have the desired setup $$TM=\{(x,v) \in \mathbb{R}^{m} \times \mathbb{R}^{m} \mid f(x)=0 \text{ and } df_{x}(v)=0\}.$$

If you read his answer more carefully, @jef808 was saying that we could try to use the defining equations to find linearly independent global sections. For example, one obvious solution to the equation $ux+vy=0$ is $(u,v)=(-y,x)$. Then the map $S^1\to TS^1$ given by $(x,y) \mapsto ((x,y),(-y,x))$ defines a nonvanishing vector field. We can generalize this construction to $S^{2n-1}$ and its tangent bundle \begin{align*} TS^{2n-1} &= \bigg \{((x_1,y_1,\ldots,x_n,y_n),(u_1,v_1,\ldots,u_n,v_n)) \in \mathbb{R}^{2n} \times \mathbb{R}^{2n} \mid \\ & \qquad \qquad \sum_{k=1}^n x_k^2 +\sum_{k=1}^n y_k^2=1 \text{ and } \sum_{k=1}^n u_k x_k + \sum_{k=1}^n v_k y_k =0 \bigg \} \end{align*} by defining a map $$(x_1,y_1,\ldots,x_n,y_n)\mapsto ((x_1,y_1,\ldots,x_n,y_n),(-y_1,x_1,\ldots,-y_n,x_n)).$$ However, for $n >1$, we need several linearly independent nonvanishing sections. As we found out circa 1960, this is only possible for $S^1$, $S^3$, and $S^7$; try googling "parallelizability of spheres". So even when the defining equations are very simple, it may be practically impossible to see whether or not you can construct enough linearly independent global sections of the tangent bundle. Similarly, if there was another way to use the defining equations to demonstrate triviality of a bundle, then Milnor, Bott, et cetera probably would have used it in the case of spheres.

One can ask if there are special cases where the defining equations shed light on the issue. I don't know if there are interesting families of special cases, but there are certainly ad hoc methods on a case-by-case basis:

Example (Torus). Even though we know that the torus has a trivial tangent bundle, it's not so apparent from a particular embedding $M \subset \mathbb{R}^3$, say $$M=\{(x,y,z) \in \mathbb{R}^3 \mid f(x,y,z)=(x^2 +y^2+z^2+3)^2-16(x^2+y^2)=0\}.$$ We know that points $((x,y,z),(u,v,w))$ in the tangent space satisfy \begin{align*} df_{(x,y,z)}(u,v,w)&=\begin{bmatrix} -32x +4x(3+x^2+y^2+z^2) \\ -32y +4y(3+x^2+y^2+z^2) \\ 4z(3+x^2+y^2+z^2) \end{bmatrix} \begin{bmatrix} u \\ v \\ w \end{bmatrix} \\ &=-16(2x u+2yv)+2(3+x^2+y^2+z^2)(2xu+2yv+2zw) \\ &=0.\end{align*} Similar to the $S^1$ case, we have an obvious solution of $(u,v,w)=(-y,x,0)$. Call this first vector field $\nu$. To find a second (linearly independent) vector field $\eta$, we could start by assuming $\eta=(u,v,w)$ is perpendicular to $\nu=(-y,x,0)$, i.e. $\eta=c_1(x,y,0)+c_2(0,0,1)$ for some $c_1,c_2$ (that are functions of $x,y,z$). Plugging $u=c_1 x$, $v=c_1 y$, $w=c_2$ into the above equation, we must have $$-16(2x^2 c_1+2y^2 c_1)+2(3+x^2+y^2+z^2)(2x^2 c_1+2y^2c_1+2z c_2) =0,$$ equivalently written as \begin{align*}2x^2 c_1+2y^2c_1+2z c_2 &=\frac{16(2x^2 c_1+2y^2 c_1)}{2(3+x^2+y^2+z^2)} \\ c_2 &=\frac{c_1}{z} \left(\frac{8(x^2 +y^2 )}{3+x^2+y^2+z^2} -x^2 +y^2\right). \end{align*} If we set $c_1=z$, this gives us a reasonably simple expression for $c_2$. One can easily check that $\nu(x,y,z)=(-y,x,0)$ and $\eta(x,y,z)=(x z, y z, c_2(x,y,z))$ define two linearly independent nonvanishing vector fields on $M$. Thus $M$ has a trivial tangent bundle.

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Since a rank $k$ vector bundle $E \to M$ is trivial iff it admits $k$ global sections which are everywhere linearly independent (in the fibers), one way to do this would be to try and write those sections explicitly from the defining equations. Of course this is a lot easier if you already know that your bundle is trivial. For example with your description of $TS^1$ the global section $\theta \mapsto (\cos \theta, \sin \theta, -\sin \theta, \cos \theta)$ shows that $TS^1$ is trivial.

In principle, having those equations for the presentation of the bundle should help finding trivializing sections. On the other hand, showing that some bundle is not trivial from this crude presentation looks rather hard since we'd lack geometric intuition...

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  • $\begingroup$ Thanks, but this doesn't really help. Being able to parametrize the manifold itself would prove that the tangent bundle is trivial. Not being able to find a parametrization for a global section does not prove that the tangent bundle is non-trivial. $\endgroup$ – Fly by Night Jul 21 '14 at 23:35
  • $\begingroup$ I think I'm missing your point, what do you mean by parametrizing the manifold? Also, not being able to find a parametrization for a global (non-zero) section is the definition of a rank 1 vector bundle not being trivial (as a vector bundle), the whole thing with linearly independent sections is the analog in higher rank. For example, we could take a parametrization of the mobius band in $\mathbb{R}^3$ and show that it's non-trivial as a vector bundle by directly showing it doesn't admit a global non-zero section. $\endgroup$ – jef808 Jul 22 '14 at 3:31
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    $\begingroup$ What about the parametrization $(\theta, \phi) \mapsto (\cos \theta \sin \phi, \sin \theta \sin \phi, \cos \phi)$ of the two sphere $S^2$ by $[0,2\pi) \times [0,\pi]$? Is this not an example of what you call a parametrization? $\endgroup$ – jef808 Jul 22 '14 at 16:29
  • $\begingroup$ No, it fails to be regular when $\phi = 0$. All of $[0,2\pi) \times \{0\}$ gets sent to the single point $(0,0,1)$. The sphere is well known as an example of a non-parametrizable manifold; there is no global, regular parametrization. This shows up in the so-called hairy ball theorem. The points where the parametrization is not regular correspond to the tufts in the combing. $\endgroup$ – Fly by Night Jul 22 '14 at 18:45
  • $\begingroup$ I believe you mean "parallelizable", not "parametrizable". $\endgroup$ – Kyle Jul 22 '14 at 18:47
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I agree with your comment:

Being able to parametrize the manifold itself would prove that the tangent bundle is trivial. Not being able to find a parametrization for a global section does not prove that the tangent bundle is non-trivial.

And I think this is essentially "how valuable" a parameterization can be for this problem (at least most of the time). A lot of great technology has been developed (e.g. Stiefel-Whitney classes) in order to answer this question, so I doubt there are direct ways of approaching it with a parametrization. I suggest Milnor-Stasheff: Characteristic Classes as a great reference which develops techniques for answering the question of whether a bundle is trivial early.

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