2

In this Wikipedia article here what is "the line bundle $\Omega^n(M)$"?

It seems to me that there can be many different line bundles on a smooth manifold $M$ so it's not clear to me what speaking of $the$ line bundle here means.

Also, it's not clear to me how a volume form can be a section of a line bundle. A section of a line bundle takes a point $m$ on $M$ and returns a line in the tangent space at $m$.

But a volume form $\omega$ is, at each point $m$, an $n$-linear map $\omega_m$ into $\mathbb R$.

As far as I understand, normally in differential geometry, one really means the kernel of a differential form if one says things like "the hyperplane field defined by $\omega$.

But the kernel of $\omega_m$ will have dimension $n-1$ which is not a line.

What am I missing? How is a volume form a section of a line bundle?

self-learner
  • 1,177
  • 8
  • 25
  • 2
    That notation refers to the top exterior power of the cotangent bundle, which is indeed a line bundle. It's uniquely determined by the smooth structure so it makes sense to use "the." Its sections are differential forms of top degree, and in particular they include volume forms. Your definition of a section of a line bundle is incorrect. – Qiaochu Yuan Apr 25 '16 at 05:45
  • @QiaochuYuan I guess my problem is that I don't see how $\Omega^n(M)$ is a line bundle. – self-learner Apr 25 '16 at 06:46
  • 3
    The "line" of the line bundle is the space of all alternating $n$-linear maps on the tangent bundle at a point, not a single such map. The set of all these maps forms a 1-dimensional vector space, hence a "line". – Eric Wofsey Apr 25 '16 at 06:49
  • @QiaochuYuan I just realised I don't see where my definition of section of a line bundle is wrong: a section is supposed to take a point on the manifold and return a point in the fibre (here a line), no? – self-learner Apr 25 '16 at 06:56
  • @EricWofsey Oh, this was not clear to me at all. I was thinking of tangent lines of the manifold the entire time! – self-learner Apr 25 '16 at 06:57
  • @self-learner: a line bundle need not have anything to do with the tangent bundle. What you wrote down is the definition of a section of a $1$-dimensional subbundle of the tangent bundle. – Qiaochu Yuan Apr 25 '16 at 20:35
  • 1
    Incidentally, there are three distinct but closely-related concepts: 1. The (canonical) line bundle $K_{M} = \bigwedge^{n}(T^{}M)$; 2. The sheaf of germs of smooth sections $\Omega^{n}(M)$; 3. The space of smooth sections $A^{n}(M)$, whose elements are smooth $n$-forms on $M$. (These notations are common, but admittedly not universal.) The point is, it's a bit sloppy, and potentially confusing, to write $\Omega^{n}(M) = \bigwedge^{n}(T^{}M)$, as wikipedia does. – Andrew D. Hwang Apr 26 '16 at 14:31

1 Answers1

1

It said "the line bundle $\Omega^n(M) = \wedge_n(T^*M)$". The line bundle $\Omega^n(M)$ is a particular line bundle, the determinant line bundle, not some arbitrary line bundle that they are deciding to call $\Omega^n(M)$. A line bundle is just a 1-dimensional vector bundle. In general, a vector bundle doesn't need to have anything to do with the tangent bundle.

A section of $\Omega^n(M)$ would be a smooth map taking an element $m \in M$ to an antisymmetric $n$-form, which is to say an antisymmetric $n$-linear map into $\mathbb{R}$. This sounds awfully similar to your definition of a volume form. Further, the collection of $n$-forms in an $n$-dimensional space is a 1-dimensional vector space, so $\Omega^n(M)$ is a line bundle.

Derek Elkins left SE
  • 24,967
  • I'm sorry I don't understand your 2nd paragraph: According to Wikipedia a section is a map from the base space into the bundle. So how can a section of a line bundle return an antisymmetric $n$-linear map into $\mathbb R$? It has to return a line, I believe. – self-learner Apr 25 '16 at 06:24